Mechanics

Fullerton, Californiawww.lightandmatter.com

copyright 1998-2011 Benjamin Crowell

rev. August 6, 2011

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Brief Contents

0 Introduction and review 111 Scaling and estimation 37

Motion in one dimension2 Velocity and relative motion 593 Acceleration and free fall 974 Force and motion 1175 Analysis of forces 145

Motion in three dimensions6 Newton’s laws in three dimensions 1777 Vectors 1898 Vectors and motion 2039 Circular motion 223

10 Gravity 241

Conservation laws11 Conservation of energy 27112 Simplifying the energy zoo 29713 Work: the transfer of mechanical energy 31514 Conservation of momentum 34715 Conservation of angular momentum 381

Vibrations16 Vibrations 44917 Resonance 467

Contents

0 Introduction and review0.1 The scientific method . . . . . . 110.2 What is physics? . . . . . . . . 14

Isolated systems and reductionism, 16.

0.3 How to learn physics. . . . . . . 170.4 Self-evaluation . . . . . . . . . 190.5 Basics of the metric system . . . . 20

The metric system, 20.—The second,21.—The meter, 22.—The kilogram,22.—Combinations of metric units,22.—Checking units, 23.

0.6 The newton, the metric unit of force 240.7 Less common metric prefixes . . . 250.8 Scientific notation . . . . . . . . 250.9 Conversions . . . . . . . . . . 26

Should that exponent be positive, ornegative?, 27.

0.10 Significant figures . . . . . . . 28Summary . . . . . . . . . . . . . 31Problems . . . . . . . . . . . . . 33Exercise 0: Models and idealization . . 35

1 Scaling and estimation1.1 Introduction . . . . . . . . . . 37

Area and volume, 37.

1.2 Scaling of area and volume . . . . 39Galileo on the behavior of nature on largeand small scales, 40.—Scaling of area andvolume for irregularly shaped objects, 43.

1.3 Order-of-magnitude estimates . . . 47Summary . . . . . . . . . . . . . 50Problems . . . . . . . . . . . . . 51

Exercise 1: Scaling applied to leaves. . 55

Motion in one dimension

2 Velocity and relative motion2.1 Types of motion . . . . . . . . . 59

Rigid-body motion distinguished from mo-tion that changes an object’s shape,59.—Center-of-mass motion as opposed torotation, 59.—Center-of-mass motion inone dimension, 63.

2.2 Describing distance and time . . . 63A point in time as opposed to duration,64.—Position as opposed to change inposition, 65.—Frames of reference, 66.

2.3 Graphs of motion; velocity . . . . 67Motion with constant velocity,67.—Motion with changing velocity, 68.

2.4 The principle of inertia . . . . . . 69Physical effects relate only to a change invelocity, 69.—Motion is relative, 70.

2.5 Addition of velocities . . . . . . . 72Addition of velocities to describe relativemotion, 72.—Negative velocities in relative

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motion, 72.

2.6 Graphs of velocity versus time. . . 74

2.7 ? Relativity . . . . . . . . . . . 75

Time is not absolute, 75.—Distortionof space and time, 78.—Proof that theLorentz transformation preserves area, 89.

Summary . . . . . . . . . . . . . 90

Problems . . . . . . . . . . . . . 92

3 Acceleration and free fall

3.1 The motion of falling objects . . . 97

How the speed of a falling object increaseswith time, 99.—A contradiction in Aris-totle’s reasoning, 100.—What is gravity?,100.

3.2 Acceleration . . . . . . . . . . 101

Definition of acceleration for linear v − tgraphs, 101.—The acceleration of gravityis different in different locations., 102.

3.3 Positive and negative acceleration . 104

3.4 Varying acceleration . . . . . . . 108

3.5 Algebraic results for constantacceleration . . . . . . . . . . . . 108

Summary . . . . . . . . . . . . . 110

Problems . . . . . . . . . . . . . 111

4 Force and motion4.1 Force . . . . . . . . . . . . . 118

We need only explain changes in motion,not motion itself., 118.—Motion changesdue to an interaction between two objects.,119.—Forces can all be measured on thesame numerical scale., 119.—More thanone force on an object, 120.—Objects canexert forces on each other at a distance.,120.—Weight, 121.—Positive and negativesigns of force, 121.

4.2 Newton’s first law . . . . . . . . 121

More general combinations of forces, 123.

4.3 Newton’s second law . . . . . . 125

Some applications of calculus, 126.—Ageneralization, 127.—The relationship be-tween mass and weight, 128.

4.4 What force is not . . . . . . . . 131

Force is not a property of one object.,131.—Force is not a measure of an ob-ject’s motion., 131.—Force is not energy.,131.—Force is not stored or used up.,132.—Forces need not be exerted by livingthings or machines., 132.—A force is thedirect cause of a change in motion., 132.

4.5 Inertial and noninertial frames ofreference . . . . . . . . . . . . . 133

4.6 Numerical techniques . . . . . . 136

Summary . . . . . . . . . . . . . 140

Problems . . . . . . . . . . . . . 141

Exercise 4: Force and motion . . . . . 144

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5 Analysis of forces5.1 Newton’s third law . . . . . . . . 145

A mnemonic for using newton’s third lawcorrectly, 148.

5.2 Classification and behavior of forces 149Normal forces, 152.—Gravitationalforces, 153.—Static and kinetic friction,153.—Fluid friction, 157.

5.3 Analysis of forces . . . . . . . . 1585.4 Transmission of forces by low-massobjects . . . . . . . . . . . . . . 1615.5 Objects under strain . . . . . . . 1635.6 Simple machines: the pulley . . . 164Summary . . . . . . . . . . . . . 166Problems . . . . . . . . . . . . . 168

Motion in three dimensions

6 Newton’s laws in threedimensions6.1 Forces have no perpendicular effects 177

Relationship to relative motion, 179.

6.2 Coordinates and components . . . 180Projectiles move along parabolas., 183.

6.3 Newton’s laws in three dimensions . 183Summary . . . . . . . . . . . . . 185Problems . . . . . . . . . . . . . 186

7 Vectors7.1 Vector notation . . . . . . . . . 189

Drawing vectors as arrows, 191.

7.2 Calculations with magnitude anddirection. . . . . . . . . . . . . . 1927.3 Techniques for adding vectors . . . 194

Addition of vectors given their

components, 194.—Addition of vectorsgiven their magnitudes and directions,194.—Graphical addition of vectors, 194.

7.4 ? Unit vector notation . . . . . . 1967.5 ? Rotational invariance . . . . . . 196Summary . . . . . . . . . . . . . 198Problems . . . . . . . . . . . . . 200

8 Vectors and motion8.1 The velocity vector . . . . . . . 2048.2 The acceleration vector . . . . . 2068.3 The force vector and simple machines2098.4 More about calculus with vectors . 210Summary . . . . . . . . . . . . . 213Problems . . . . . . . . . . . . . 214Exercise 8: Vectors and motion . . . . 220

9 Circular motion9.1 Conceptual framework . . . . . . 223

Circular motion does not produce an out-ward force, 223.—Circular motion does notpersist without a force, 224.—Uniform andnonuniform circular motion, 225.—Only aninward force is required for uniform circu-lar motion., 226.—In uniform circular mo-tion, the acceleration vector is inward, 227.

9.2 Uniform circular motion. . . . . . 2299.3 Nonuniform circular motion . . . . 232Summary . . . . . . . . . . . . . 234Problems . . . . . . . . . . . . . 235

10 Gravity10.1 Kepler’s laws . . . . . . . . . 24210.2 Newton’s law of gravity . . . . . 244

The sun’s force on the planets obeys aninverse square law., 244.—The forces be-tween heavenly bodies are the same type offorce as terrestrial gravity., 245.—Newton’slaw of gravity, 246.

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10.3 Apparent weightlessness . . . . 25010.4 Vector addition of gravitationalforces . . . . . . . . . . . . . . . 25010.5 Weighing the earth . . . . . . . 25310.6 ? Dark energy . . . . . . . . . 257Summary . . . . . . . . . . . . . 259Problems . . . . . . . . . . . . . 261Exercise 10: The shell theorem . . . . 267

Conservation laws

11 Conservation of energy11.1 The search for a perpetual motionmachine. . . . . . . . . . . . . . 27111.2 Energy . . . . . . . . . . . . 27211.3 A numerical scale of energy . . . 276

How new forms of energy are discovered,279.

11.4 Kinetic energy . . . . . . . . . 281Energy and relative motion, 282.

11.5 Power . . . . . . . . . . . . 28311.6 ? Massless particles . . . . . . 285

Failure of newton’s laws, 285.—Ultrarelativisticmotion, 286.

Summary . . . . . . . . . . . . . 290Problems . . . . . . . . . . . . . 292

12 Simplifying the energy zoo12.1 Heat is kinetic energy . . . . . . 29812.2 Potential energy: energy of distanceor closeness . . . . . . . . . . . . 300

An equation for gravitational potentialenergy, 301.

12.3 All energy is potential or kinetic . . 30412.4 ? E=mc2 . . . . . . . . . . . 306Summary . . . . . . . . . . . . . 310Problems . . . . . . . . . . . . . 311

13 Work: the transfer of me-chanical energy13.1 Work: the transfer of mechanicalenergy . . . . . . . . . . . . . . 315

The concept of work, 315.—Calculatingwork as force multiplied by distance,316.—machines can increase force, but notwork., 318.—No work is done withoutmotion., 318.—Positive and negative work,319.

13.2 Work in three dimensions . . . . 321A force perpendicular to the motion doesno work., 321.—Forces at other angles, 322.

13.3 The dot product . . . . . . . . 32313.4 Varying force . . . . . . . . . 32513.5 Work and potential energy . . . . 32613.6 ? When does work equal force timesdistance? . . . . . . . . . . . . . 32913.7 ? Uniqueness of the dot product . 33213.8 ? A dot product for relativity? . . . 332

Clock time, 333.—Four-vectors, 334.


14 Conservation of momentum14.1 Momentum . . . . . . . . . . 348

A conserved quantity of motion,348.—Momentum, 349.—Generalization ofthe momentum concept, 351.—Momentumcompared to kinetic energy, 352.

14.2 Collisions in one dimension . . . 354The discovery of the neutron, 357.

14.3 ? Relationship of momentum to thecenter of mass . . . . . . . . . . . 359

Momentum in different frames of reference,360.—The center of mass frame ofreference, 361.

14.4 Momentum transfer. . . . . . . 362The rate of change of momentum,362.—The area under the force-time graph,

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364.

14.5 Momentum in three dimensions . 365The center of mass, 366.—Counting equa-tions and unknowns, 367.—Calculationswith the momentum vector, 368.

14.6 Applications of calculus . . . . . 36914.7 ? Relativistic momentum . . . . 370Summary . . . . . . . . . . . . . 374Problems . . . . . . . . . . . . . 376

15 Conservation of angularmomentum15.1 Conservation of angular momentum383

Restriction to rotation in a plane, 387.

15.2 Angular momentum in planetarymotion . . . . . . . . . . . . . . 38715.3 Two theorems about angularmomentum . . . . . . . . . . . . 38915.4 Torque: the rate of transfer of angu-lar momentum . . . . . . . . . . . 391

Torque distinguished from force,391.—Relationship between force andtorque, 392.—The torque due to gravity,395.

15.5 Statics . . . . . . . . . . . . 399Equilibrium, 399.—Stable and unstableequilibria, 402.

15.6 Simple machines: the lever . . . 40315.7 Rigid-body rotation . . . . . . . 405

Kinematics, 405.—Relations between an-gular quantities and motion of a point,406.—Dynamics, 408.—Iterated integrals,410.—Finding moments of inertia byintegration, 413.

15.8 Angular momentum in threedimensions . . . . . . . . . . . . 418

Rigid-body kinematics in threedimensions, 418.—Angular momentum inthree dimensions, 420.—Rigid-body dy-namics in three dimensions, 425.

15.9 ? Proof of Kepler’s elliptical orbit law42715.10 Some theorems and proofs . . . 430

Uniqueness of the cross product,430.—Choice of axis theorem, 431.—Spintheorem, 431.—Parallel axis theorem, 432.


Exercise 15: Torque . . . . . . . . . 445

Vibrations and resonance

16 Vibrations16.1 Period, frequency, and amplitude . 45016.2 Simple harmonic motion. . . . . 453

Why are sine-wave vibrations so common?,453.—Period is approximately indepen-dent of amplitude, if the amplitude issmall., 458.

Summary . . . . . . . . . . . . . 460Problems . . . . . . . . . . . . . 461Exercise 16: Vibrations . . . . . . . 466

17 Resonance17.1 Energy in vibrations . . . . . . 46817.2 Energy lost from vibrations. . . . 470

Numerical treatment, 470.—Analytictreatment, 471.

17.3 Putting energy into vibrations . . 47317.4 ? Proofs . . . . . . . . . . . 482

Statement 2: maximum amplitude atresonance, 483.—Statement 3: ampli-tude at resonance proportional to Q,483.—Statement 4: FWHM related to Q,483.

Summary . . . . . . . . . . . . . 485Problems . . . . . . . . . . . . . 487Exercise 17: Resonance . . . . . . . 491

A Appendix 1: Programming withPython 492Hints . . . . . . . . . . . . . . . 494Solutions . . . . . . . . . . . . . 495Photo credits . . . . . . . . . . . . 515

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The Mars Climate Orbiter is pre-pared for its mission. The lawsof physics are the same every-where, even on Mars, so theprobe could be designed basedon the laws of physics as discov-ered on earth. There is unfor-tunately another reason why thisspacecraft is relevant to the top-ics of this chapter: it was de-stroyed attempting to enter Mars’atmosphere because engineersat Lockheed Martin forgot to con-vert data on engine thrusts frompounds into the metric unit offorce (newtons) before giving theinformation to NASA. Conver-sions are important!

Chapter 0

Introduction and review

If you drop your shoe and a coin side by side, they hit the ground atthe same time. Why doesn’t the shoe get there first, since gravity ispulling harder on it? How does the lens of your eye work, and whydo your eye’s muscles need to squash its lens into different shapes inorder to focus on objects nearby or far away? These are the kindsof questions that physics tries to answer about the behavior of lightand matter, the two things that the universe is made of.

0.1 The scientific methodUntil very recently in history, no progress was made in answeringquestions like these. Worse than that, the wrong answers writtenby thinkers like the ancient Greek physicist Aristotle were acceptedwithout question for thousands of years. Why is it that scientificknowledge has progressed more since the Renaissance than it hadin all the preceding millennia since the beginning of recorded his-tory? Undoubtedly the industrial revolution is part of the answer.Building its centerpiece, the steam engine, required improved tech-

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a / Science is a cycle of the-ory and experiment.

b / A satirical drawing of analchemist’s laboratory. H. Cock,after a drawing by Peter Brueghelthe Elder (16th century).

niques for precise construction and measurement. (Early on, it wasconsidered a major advance when English machine shops learned tobuild pistons and cylinders that fit together with a gap narrowerthan the thickness of a penny.) But even before the industrial rev-olution, the pace of discovery had picked up, mainly because of theintroduction of the modern scientific method. Although it evolvedover time, most scientists today would agree on something like thefollowing list of the basic principles of the scientific method:

(1) Science is a cycle of theory and experiment. Scientific the-ories 1 are created to explain the results of experiments that werecreated under certain conditions. A successful theory will also makenew predictions about new experiments under new conditions. Even-tually, though, it always seems to happen that a new experimentcomes along, showing that under certain conditions the theory isnot a good approximation or is not valid at all. The ball is thenback in the theorists’ court. If an experiment disagrees with thecurrent theory, the theory has to be changed, not the experiment.

(2) Theories should both predict and explain. The requirement ofpredictive power means that a theory is only meaningful if it predictssomething that can be checked against experimental measurementsthat the theorist did not already have at hand. That is, a theoryshould be testable. Explanatory value means that many phenomenashould be accounted for with few basic principles. If you answerevery “why” question with “because that’s the way it is,” then yourtheory has no explanatory value. Collecting lots of data withoutbeing able to find any basic underlying principles is not science.

(3) Experiments should be reproducible. An experiment shouldbe treated with suspicion if it only works for one person, or onlyin one part of the world. Anyone with the necessary skills andequipment should be able to get the same results from the sameexperiment. This implies that science transcends national and eth-nic boundaries; you can be sure that nobody is doing actual sciencewho claims that their work is “Aryan, not Jewish,” “Marxist, notbourgeois,” or “Christian, not atheistic.” An experiment cannot bereproduced if it is secret, so science is necessarily a public enterprise.

As an example of the cycle of theory and experiment, a vital steptoward modern chemistry was the experimental observation that thechemical elements could not be transformed into each other, e.g.,lead could not be turned into gold. This led to the theory thatchemical reactions consisted of rearrangements of the elements in

1The term “theory” in science does not just mean “what someone thinks,” oreven “what a lot of scientists think.” It means an interrelated set of statementsthat have predictive value, and that have survived a broad set of empiricaltests. Thus, both Newton’s law of gravity and Darwinian evolution are scientifictheories. A “hypothesis,” in contrast to a theory, is any statement of interestthat can be empirically tested. That the moon is made of cheese is a hypothesis,which was empirically tested, for example, by the Apollo astronauts.

12 Chapter 0 Introduction and review

different combinations, without any change in the identities of theelements themselves. The theory worked for hundreds of years, andwas confirmed experimentally over a wide range of pressures andtemperatures and with many combinations of elements. Only inthe twentieth century did we learn that one element could be trans-formed into one another under the conditions of extremely high pres-sure and temperature existing in a nuclear bomb or inside a star.That observation didn’t completely invalidate the original theory ofthe immutability of the elements, but it showed that it was only anapproximation, valid at ordinary temperatures and pressures.

self-check AA psychic conducts seances in which the spirits of the dead speak tothe participants. He says he has special psychic powers not possessedby other people, which allow him to “channel” the communications withthe spirits. What part of the scientific method is being violated here?. Answer, p. 508

The scientific method as described here is an idealization, andshould not be understood as a set procedure for doing science. Sci-entists have as many weaknesses and character flaws as any othergroup, and it is very common for scientists to try to discredit otherpeople’s experiments when the results run contrary to their own fa-vored point of view. Successful science also has more to do withluck, intuition, and creativity than most people realize, and therestrictions of the scientific method do not stifle individuality andself-expression any more than the fugue and sonata forms stifledBach and Haydn. There is a recent tendency among social scien-tists to go even further and to deny that the scientific method evenexists, claiming that science is no more than an arbitrary social sys-tem that determines what ideas to accept based on an in-group’scriteria. I think that’s going too far. If science is an arbitrary socialritual, it would seem difficult to explain its effectiveness in buildingsuch useful items as airplanes, CD players, and sewers. If alchemyand astrology were no less scientific in their methods than chem-istry and astronomy, what was it that kept them from producinganything useful?

Discussion questionsConsider whether or not the scientific method is being applied in the fol-lowing examples. If the scientific method is not being applied, are thepeople whose actions are being described performing a useful humanactivity, albeit an unscientific one?

A Acupuncture is a traditional medical technique of Asian origin inwhich small needles are inserted in the patient’s body to relieve pain.Many doctors trained in the west consider acupuncture unworthy of ex-perimental study because if it had therapeutic effects, such effects couldnot be explained by their theories of the nervous system. Who is beingmore scientific, the western or eastern practitioners?

Section 0.1 The scientific method 13

B Goethe, a German poet, is less well known for his theory of color.He published a book on the subject, in which he argued that scientificapparatus for measuring and quantifying color, such as prisms, lensesand colored filters, could not give us full insight into the ultimate meaningof color, for instance the cold feeling evoked by blue and green or theheroic sentiments inspired by red. Was his work scientific?

C A child asks why things fall down, and an adult answers “because ofgravity.” The ancient Greek philosopher Aristotle explained that rocks fellbecause it was their nature to seek out their natural place, in contact withthe earth. Are these explanations scientific?

D Buddhism is partly a psychological explanation of human suffering,and psychology is of course a science. The Buddha could be said tohave engaged in a cycle of theory and experiment, since he worked bytrial and error, and even late in his life he asked his followers to challengehis ideas. Buddhism could also be considered reproducible, since theBuddha told his followers they could find enlightenment for themselvesif they followed a certain course of study and discipline. Is Buddhism ascientific pursuit?

0.2 What is physics?Given for one instant an intelligence which could comprehendall the forces by which nature is animated and the respectivepositions of the things which compose it...nothing would beuncertain, and the future as the past would be laid out beforeits eyes.

Pierre Simon de Laplace

Physics is the use of the scientific method to find out the basicprinciples governing light and matter, and to discover the implica-tions of those laws. Part of what distinguishes the modern outlookfrom the ancient mind-set is the assumption that there are rules bywhich the universe functions, and that those laws can be at least par-tially understood by humans. From the Age of Reason through thenineteenth century, many scientists began to be convinced that thelaws of nature not only could be known but, as claimed by Laplace,those laws could in principle be used to predict everything aboutthe universe’s future if complete information was available aboutthe present state of all light and matter. In subsequent sections,I’ll describe two general types of limitations on prediction using thelaws of physics, which were only recognized in the twentieth century.

Matter can be defined as anything that is affected by gravity,i.e., that has weight or would have weight if it was near the Earthor another star or planet massive enough to produce measurablegravity. Light can be defined as anything that can travel from oneplace to another through empty space and can influence matter, buthas no weight. For example, sunlight can influence your body byheating it or by damaging your DNA and giving you skin cancer.The physicist’s definition of light includes a variety of phenomena


c / This telescope picture showstwo images of the same distantobject, an exotic, very luminousobject called a quasar. This isinterpreted as evidence that amassive, dark object, possiblya black hole, happens to bebetween us and it. Light rays thatwould otherwise have missed theearth on either side have beenbent by the dark object’s gravityso that they reach us. The actualdirection to the quasar is presum-ably in the center of the image,but the light along that central linedoesn’t get to us because it isabsorbed by the dark object. Thequasar is known by its catalognumber, MG1131+0456, or moreinformally as Einstein’s Ring.

that are not visible to the eye, including radio waves, microwaves,x-rays, and gamma rays. These are the “colors” of light that do nothappen to fall within the narrow violet-to-red range of the rainbowthat we can see.

self-check BAt the turn of the 20th century, a strange new phenomenon was discov-ered in vacuum tubes: mysterious rays of unknown origin and nature.These rays are the same as the ones that shoot from the back of yourTV’s picture tube and hit the front to make the picture. Physicists in1895 didn’t have the faintest idea what the rays were, so they simplynamed them “cathode rays,” after the name for the electrical contactfrom which they sprang. A fierce debate raged, complete with national-istic overtones, over whether the rays were a form of light or of matter.What would they have had to do in order to settle the issue? .

Answer, p. 508

Many physical phenomena are not themselves light or matter,but are properties of light or matter or interactions between lightand matter. For instance, motion is a property of all light and somematter, but it is not itself light or matter. The pressure that keepsa bicycle tire blown up is an interaction between the air and thetire. Pressure is not a form of matter in and of itself. It is asmuch a property of the tire as of the air. Analogously, sisterhoodand employment are relationships among people but are not peoplethemselves.

Some things that appear weightless actually do have weight, andso qualify as matter. Air has weight, and is thus a form of mattereven though a cubic inch of air weighs less than a grain of sand. Ahelium balloon has weight, but is kept from falling by the force of thesurrounding more dense air, which pushes up on it. Astronauts inorbit around the Earth have weight, and are falling along a curvedarc, but they are moving so fast that the curved arc of their fallis broad enough to carry them all the way around the Earth in acircle. They perceive themselves as being weightless because theirspace capsule is falling along with them, and the floor therefore doesnot push up on their feet.

Optional Topic: Modern Changes in the Definition of Light andMatterEinstein predicted as a consequence of his theory of relativity that lightwould after all be affected by gravity, although the effect would be ex-tremely weak under normal conditions. His prediction was borne outby observations of the bending of light rays from stars as they passedclose to the sun on their way to the Earth. Einstein’s theory also impliedthe existence of black holes, stars so massive and compact that theirintense gravity would not even allow light to escape. (These days thereis strong evidence that black holes exist.)

Einstein’s interpretation was that light doesn’t really have mass, butthat energy is affected by gravity just like mass is. The energy in a light

Section 0.2 What is physics? 15

d / Reductionism.

beam is equivalent to a certain amount of mass, given by the famousequation E = mc2, where c is the speed of light. Because the speedof light is such a big number, a large amount of energy is equivalent toonly a very small amount of mass, so the gravitational force on a lightray can be ignored for most practical purposes.

There is however a more satisfactory and fundamental distinctionbetween light and matter, which should be understandable to you if youhave had a chemistry course. In chemistry, one learns that electronsobey the Pauli exclusion principle, which forbids more than one electronfrom occupying the same orbital if they have the same spin. The Pauliexclusion principle is obeyed by the subatomic particles of which matteris composed, but disobeyed by the particles, called photons, of which abeam of light is made.

Einstein’s theory of relativity is discussed more fully in book 6 of thisseries.

The boundary between physics and the other sciences is notalways clear. For instance, chemists study atoms and molecules,which are what matter is built from, and there are some scientistswho would be equally willing to call themselves physical chemistsor chemical physicists. It might seem that the distinction betweenphysics and biology would be clearer, since physics seems to dealwith inanimate objects. In fact, almost all physicists would agreethat the basic laws of physics that apply to molecules in a test tubework equally well for the combination of molecules that constitutesa bacterium. (Some might believe that something more happens inthe minds of humans, or even those of cats and dogs.) What differ-entiates physics from biology is that many of the scientific theoriesthat describe living things, while ultimately resulting from the fun-damental laws of physics, cannot be rigorously derived from physicalprinciples.

Isolated systems and reductionism

To avoid having to study everything at once, scientists isolate thethings they are trying to study. For instance, a physicist who wantsto study the motion of a rotating gyroscope would probably preferthat it be isolated from vibrations and air currents. Even in biology,where field work is indispensable for understanding how living thingsrelate to their entire environment, it is interesting to note the vitalhistorical role played by Darwin’s study of the Galapagos Islands,which were conveniently isolated from the rest of the world. Anypart of the universe that is considered apart from the rest can becalled a “system.”

Physics has had some of its greatest successes by carrying thisprocess of isolation to extremes, subdividing the universe into smallerand smaller parts. Matter can be divided into atoms, and the be-havior of individual atoms can be studied. Atoms can be split apart


into their constituent neutrons, protons and electrons. Protons andneutrons appear to be made out of even smaller particles calledquarks, and there have even been some claims of experimental ev-idence that quarks have smaller parts inside them. This methodof splitting things into smaller and smaller parts and studying howthose parts influence each other is called reductionism. The hope isthat the seemingly complex rules governing the larger units can bebetter understood in terms of simpler rules governing the smallerunits. To appreciate what reductionism has done for science, it isonly necessary to examine a 19th-century chemistry textbook. Atthat time, the existence of atoms was still doubted by some, elec-trons were not even suspected to exist, and almost nothing wasunderstood of what basic rules governed the way atoms interactedwith each other in chemical reactions. Students had to memorizelong lists of chemicals and their reactions, and there was no way tounderstand any of it systematically. Today, the student only needsto remember a small set of rules about how atoms interact, for in-stance that atoms of one element cannot be converted into anothervia chemical reactions, or that atoms from the right side of the pe-riodic table tend to form strong bonds with atoms from the leftside.

Discussion questions

A I’ve suggested replacing the ordinary dictionary definition of lightwith a more technical, more precise one that involves weightlessness. It’sstill possible, though, that the stuff a lightbulb makes, ordinarily called“light,” does have some small amount of weight. Suggest an experimentto attempt to measure whether it does.

B Heat is weightless (i.e., an object becomes no heavier when heated),and can travel across an empty room from the fireplace to your skin,where it influences you by heating you. Should heat therefore be con-sidered a form of light by our definition? Why or why not?

C Similarly, should sound be considered a form of light?

0.3 How to learn physicsFor as knowledges are now delivered, there is a kind of con-tract of error between the deliverer and the receiver; for hethat delivereth knowledge desireth to deliver it in such a formas may be best believed, and not as may be best examined;and he that receiveth knowledge desireth rather present sat-isfaction than expectant inquiry.

Francis Bacon

Many students approach a science course with the idea that theycan succeed by memorizing the formulas, so that when a problem

Section 0.3 How to learn physics 17

is assigned on the homework or an exam, they will be able to plugnumbers in to the formula and get a numerical result on their cal-culator. Wrong! That’s not what learning science is about! Thereis a big difference between memorizing formulas and understandingconcepts. To start with, different formulas may apply in differentsituations. One equation might represent a definition, which is al-ways true. Another might be a very specific equation for the speedof an object sliding down an inclined plane, which would not be trueif the object was a rock drifting down to the bottom of the ocean.If you don’t work to understand physics on a conceptual level, youwon’t know which formulas can be used when.

Most students taking college science courses for the first timealso have very little experience with interpreting the meaning of anequation. Consider the equation w = A/h relating the width of arectangle to its height and area. A student who has not developedskill at interpretation might view this as yet another equation tomemorize and plug in to when needed. A slightly more savvy stu-dent might realize that it is simply the familiar formula A = whin a different form. When asked whether a rectangle would havea greater or smaller width than another with the same area buta smaller height, the unsophisticated student might be at a loss,not having any numbers to plug in on a calculator. The more ex-perienced student would know how to reason about an equationinvolving division — if h is smaller, and A stays the same, then wmust be bigger. Often, students fail to recognize a sequence of equa-tions as a derivation leading to a final result, so they think all theintermediate steps are equally important formulas that they shouldmemorize.

When learning any subject at all, it is important to become asactively involved as possible, rather than trying to read throughall the information quickly without thinking about it. It is a goodidea to read and think about the questions posed at the end of eachsection of these notes as you encounter them, so that you know youhave understood what you were reading.

Many students’ difficulties in physics boil down mainly to diffi-culties with math. Suppose you feel confident that you have enoughmathematical preparation to succeed in this course, but you arehaving trouble with a few specific things. In some areas, the briefreview given in this chapter may be sufficient, but in other areasit probably will not. Once you identify the areas of math in whichyou are having problems, get help in those areas. Don’t limp alongthrough the whole course with a vague feeling of dread about some-thing like scientific notation. The problem will not go away if youignore it. The same applies to essential mathematical skills that youare learning in this course for the first time, such as vector addition.

Sometimes students tell me they keep trying to understand a


certain topic in the book, and it just doesn’t make sense. The worstthing you can possibly do in that situation is to keep on staringat the same page. Every textbook explains certain things badly —even mine! — so the best thing to do in this situation is to lookat a different book. Instead of college textbooks aimed at the samemathematical level as the course you’re taking, you may in somecases find that high school books or books at a lower math levelgive clearer explanations.

Finally, when reviewing for an exam, don’t simply read backover the text and your lecture notes. Instead, try to use an activemethod of reviewing, for instance by discussing some of the discus-sion questions with another student, or doing homework problemsyou hadn’t done the first time.

0.4 Self-evaluationThe introductory part of a book like this is hard to write, becauseevery student arrives at this starting point with a different prepara-tion. One student may have grown up outside the U.S. and so maybe completely comfortable with the metric system, but may havehad an algebra course in which the instructor passed too quicklyover scientific notation. Another student may have already takencalculus, but may have never learned the metric system. The fol-lowing self-evaluation is a checklist to help you figure out what youneed to study to be prepared for the rest of the course.

If you disagree with this state-ment. . .

you should study this section:

I am familiar with the basic metricunits of meters, kilograms, and sec-onds, and the most common metricprefixes: milli- (m), kilo- (k), andcenti- (c).

section 0.5 Basic of the Metric Sys-tem

I know about the newton, a unit offorce

section 0.6 The newton, the MetricUnit of Force

I am familiar with these less com-mon metric prefixes: mega- (M),micro- (µ), and nano- (n).

section 0.7 Less Common MetricPrefixes

I am comfortable with scientific no-tation.

section 0.8 Scientific Notation

I can confidently do metric conver-sions.

section 0.9 Conversions

I understand the purpose and use ofsignificant figures.

section 0.10 Significant Figures

It wouldn’t hurt you to skim the sections you think you alreadyknow about, and to do the self-checks in those sections.

Section 0.4 Self-evaluation 19

0.5 Basics of the metric systemThe metric system

Units were not standardized until fairly recently in history, sowhen the physicist Isaac Newton gave the result of an experimentwith a pendulum, he had to specify not just that the string was 377/8 inches long but that it was “37 7/8 London inches long.” Theinch as defined in Yorkshire would have been different. Even afterthe British Empire standardized its units, it was still very inconve-nient to do calculations involving money, volume, distance, time, orweight, because of all the odd conversion factors, like 16 ounces ina pound, and 5280 feet in a mile. Through the nineteenth century,schoolchildren squandered most of their mathematical education inpreparing to do calculations such as making change when a customerin a shop offered a one-crown note for a book costing two pounds,thirteen shillings and tuppence. The dollar has always been decimal,and British money went decimal decades ago, but the United Statesis still saddled with the antiquated system of feet, inches, pounds,ounces and so on.

Every country in the world besides the U.S. has adopted a sys-tem of units known in English as the “metric system.” This systemis entirely decimal, thanks to the same eminently logical people whobrought about the French Revolution. In deference to France, thesystem’s official name is the Systeme International, or SI, meaningInternational System. (The phrase “SI system” is therefore redun-dant.)

The wonderful thing about the SI is that people who live incountries more modern than ours do not need to memorize howmany ounces there are in a pound, how many cups in a pint, howmany feet in a mile, etc. The whole system works with a single,consistent set of prefixes (derived from Greek) that modify the basicunits. Each prefix stands for a power of ten, and has an abbreviationthat can be combined with the symbol for the unit. For instance,the meter is a unit of distance. The prefix kilo- stands for 103, so akilometer, 1 km, is a thousand meters.

The basic units of the metric system are the meter for distance,the second for time, and the gram for mass.

The following are the most common metric prefixes. You shouldmemorize them.

prefix meaning examplekilo- k 103 60 kg = a person’s masscenti- c 10−2 28 cm = height of a piece of papermilli- m 10−3 1 ms = time for one vibration of a guitar

string playing the note D

The prefix centi-, meaning 10−2, is only used in the centimeter;


a hundredth of a gram would not be written as 1 cg but as 10 mg.The centi- prefix can be easily remembered because a cent is 10−2

dollars. The official SI abbreviation for seconds is “s” (not “sec”)and grams are “g” (not “gm”).

The second

The sun stood still and the moon halted until the nation hadtaken vengeance on its enemies. . .

Joshua 10:12-14

Absolute, true, and mathematical time, of itself, and from itsown nature, flows equably without relation to anything exter-nal. . .

Isaac Newton

When I stated briefly above that the second was a unit of time,it may not have occurred to you that this was not really much ofa definition. The two quotes above are meant to demonstrate howmuch room for confusion exists among people who seem to mean thesame thing by a word such as “time.” The first quote has been inter-preted by some biblical scholars as indicating an ancient belief thatthe motion of the sun across the sky was not just something thatoccurred with the passage of time but that the sun actually causedtime to pass by its motion, so that freezing it in the sky would havesome kind of a supernatural decelerating effect on everyone exceptthe Hebrew soldiers. Many ancient cultures also conceived of timeas cyclical, rather than proceeding along a straight line as in 1998,1999, 2000, 2001,... The second quote, from a relatively modernphysicist, may sound a lot more scientific, but most physicists to-day would consider it useless as a definition of time. Today, thephysical sciences are based on operational definitions, which meansdefinitions that spell out the actual steps (operations) required tomeasure something numerically.

Now in an era when our toasters, pens, and coffee pots tell us thetime, it is far from obvious to most people what is the fundamentaloperational definition of time. Until recently, the hour, minute, andsecond were defined operationally in terms of the time required forthe earth to rotate about its axis. Unfortunately, the Earth’s ro-tation is slowing down slightly, and by 1967 this was becoming anissue in scientific experiments requiring precise time measurements.The second was therefore redefined as the time required for a cer-tain number of vibrations of the light waves emitted by a cesiumatoms in a lamp constructed like a familiar neon sign but with theneon replaced by cesium. The new definition not only promises tostay constant indefinitely, but for scientists is a more convenientway of calibrating a clock than having to carry out astronomicalmeasurements.

Section 0.5 Basics of the metric system 21

e / The original definition ofthe meter.

f / A duplicate of the Pariskilogram, maintained at the Dan-ish National Metrology Institute.

self-check CWhat is a possible operational definition of how strong a person is? .

Answer, p. 508

The meter

The French originally defined the meter as 10−7 times the dis-tance from the equator to the north pole, as measured through Paris(of course). Even if the definition was operational, the operation oftraveling to the north pole and laying a surveying chain behind youwas not one that most working scientists wanted to carry out. Fairlysoon, a standard was created in the form of a metal bar with twoscratches on it. This was replaced by an atomic standard in 1960,and finally in 1983 by the current definition, which is that the me-ter is the distance traveled by light in a vacuum over a period of(1/299792458) seconds.

The kilogram

The third base unit of the SI is the kilogram, a unit of mass.Mass is intended to be a measure of the amount of a substance,but that is not an operational definition. Bathroom scales work bymeasuring our planet’s gravitational attraction for the object beingweighed, but using that type of scale to define mass operationallywould be undesirable because gravity varies in strength from placeto place on the earth.

There’s a surprising amount of disagreement among physics text-books about how mass should be defined, but here’s how it’s actuallyhandled by the few working physicists who specialize in ultra-high-precision measurements. They maintain a physical object in Paris,which is the standard kilogram, a cylinder made of platinum-iridiumalloy. Duplicates are checked against this mother of all kilogramsby putting the original and the copy on the two opposite pans of abalance. Although this method of comparison depends on gravity,the problems associated with differences in gravity in different geo-graphical locations are bypassed, because the two objects are beingcompared in the same place. The duplicates can then be removedfrom the Parisian kilogram shrine and transported elsewhere in theworld. It would be desirable to replace this at some point with auniversally accessible atomic standard rather than one based on aspecific artifact, but as of 2010 the technology for automated count-ing of large numbers of atoms has not gotten good enough to makethat work with the desired precision.

Combinations of metric units

Just about anything you want to measure can be measured withsome combination of meters, kilograms, and seconds. Speed can bemeasured in m/s, volume in m3, and density in kg/m3. Part of whatmakes the SI great is this basic simplicity. No more funny units like


a cord of wood, a bolt of cloth, or a jigger of whiskey. No moreliquid and dry measure. Just a simple, consistent set of units. TheSI measures put together from meters, kilograms, and seconds makeup the mks system. For example, the mks unit of speed is m/s, notkm/hr.

Checking units

A useful technique for finding mistakes in one’s algebra is toanalyze the units associated with the variables.

Checking units example 1. Jae starts from the formula V = 1

3Ah for the volume of a cone,where A is the area of its base, and h is its height. He wants tofind an equation that will tell him how tall a conical tent has to bein order to have a certain volume, given its radius. His algebragoes like this:

V =13

Ah[1]

A = πr2[2]

V =13πr2h[3]

h =πr2

3V[4]

Is his algebra correct? If not, find the mistake.

. Line 4 is supposed to be an equation for the height, so the unitsof the expression on the right-hand side had better equal meters.The pi and the 3 are unitless, so we can ignore them. In terms ofunits, line 4 becomes

m =m2

m3 =1m

.

This is false, so there must be a mistake in the algebra. The unitsof lines 1, 2, and 3 check out, so the mistake must be in the stepfrom line 3 to line 4. In fact the result should have been

h =3Vπr2 .

Now the units check: m = m3/m2.

Discussion question

A Isaac Newton wrote, “. . . the natural days are truly unequal, thoughthey are commonly considered as equal, and used for a measure oftime. . . It may be that there is no such thing as an equable motion, wherebytime may be accurately measured. All motions may be accelerated or re-tarded. . . ” Newton was right. Even the modern definition of the secondin terms of light emitted by cesium atoms is subject to variation. For in-stance, magnetic fields could cause the cesium atoms to emit light witha slightly different rate of vibration. What makes us think, though, that a

Section 0.5 Basics of the metric system 23

pendulum clock is more accurate than a sundial, or that a cesium atomis a more accurate timekeeper than a pendulum clock? That is, how canone test experimentally how the accuracies of different time standardscompare?

0.6 The newton, the metric unit of forceA force is a push or a pull, or more generally anything that canchange an object’s speed or direction of motion. A force is requiredto start a car moving, to slow down a baseball player sliding in tohome base, or to make an airplane turn. (Forces may fail to changean object’s motion if they are canceled by other forces, e.g., theforce of gravity pulling you down right now is being canceled by theforce of the chair pushing up on you.) The metric unit of force isthe Newton, defined as the force which, if applied for one second,will cause a 1-kilogram object starting from rest to reach a speed of1 m/s. Later chapters will discuss the force concept in more detail.In fact, this entire book is about the relationship between force andmotion.

In section 0.5, I gave a gravitational definition of mass, but bydefining a numerical scale of force, we can also turn around and de-fine a scale of mass without reference to gravity. For instance, if aforce of two Newtons is required to accelerate a certain object fromrest to 1 m/s in 1 s, then that object must have a mass of 2 kg.From this point of view, mass characterizes an object’s resistanceto a change in its motion, which we call inertia or inertial mass.Although there is no fundamental reason why an object’s resistanceto a change in its motion must be related to how strongly gravityaffects it, careful and precise experiments have shown that the in-ertial definition and the gravitational definition of mass are highlyconsistent for a variety of objects. It therefore doesn’t really matterfor any practical purpose which definition one adopts.

Discussion question

A Spending a long time in weightlessness is unhealthy. One of themost important negative effects experienced by astronauts is a loss ofmuscle and bone mass. Since an ordinary scale won’t work for an astro-naut in orbit, what is a possible way of monitoring this change in mass?(Measuring the astronaut’s waist or biceps with a measuring tape is notgood enough, because it doesn’t tell anything about bone mass, or aboutthe replacement of muscle with fat.)


g / This is a mnemonic tohelp you remember the most im-portant metric prefixes. The word“little” is to remind you that thelist starts with the prefixes usedfor small quantities and buildsupward. The exponent changesby 3, except that of course thatwe do not need a special prefixfor 100, which equals one.

0.7 Less common metric prefixes

The following are three metric prefixes which, while less commonthan the ones discussed previously, are well worth memorizing.

prefix meaning examplemega- M 106 6.4 Mm = radius of the earthmicro- µ 10−6 10 µm = size of a white blood cellnano- n 10−9 0.154 nm = distance between carbon

nuclei in an ethane molecule

Note that the abbreviation for micro is the Greek letter mu, µ— a common mistake is to confuse it with m (milli) or M (mega).

There are other prefixes even less common, used for extremelylarge and small quantities. For instance, 1 femtometer = 10−15 m isa convenient unit of distance in nuclear physics, and 1 gigabyte =109 bytes is used for computers’ hard disks. The international com-mittee that makes decisions about the SI has recently even addedsome new prefixes that sound like jokes, e.g., 1 yoctogram = 10−24 gis about half the mass of a proton. In the immediate future, how-ever, you’re unlikely to see prefixes like “yocto-” and “zepto-” usedexcept perhaps in trivia contests at science-fiction conventions orother geekfests.

self-check DSuppose you could slow down time so that according to your perception,a beam of light would move across a room at the speed of a slow walk.If you perceived a nanosecond as if it was a second, how would youperceive a microsecond? . Answer, p. 508

0.8 Scientific notationMost of the interesting phenomena in our universe are not on thehuman scale. It would take about 1,000,000,000,000,000,000,000bacteria to equal the mass of a human body. When the physicistThomas Young discovered that light was a wave, it was back in thebad old days before scientific notation, and he was obliged to writethat the time required for one vibration of the wave was 1/500 ofa millionth of a millionth of a second. Scientific notation is a lessawkward way to write very large and very small numbers such asthese. Here’s a quick review.

Scientific notation means writing a number in terms of a productof something from 1 to 10 and something else that is a power of ten.For instance,

32 = 3.2× 101

320 = 3.2× 102

3200 = 3.2× 103 . . .

Each number is ten times bigger than the previous one.

Section 0.7 Less common metric prefixes 25

Since 101 is ten times smaller than 102 , it makes sense to usethe notation 100 to stand for one, the number that is in turn tentimes smaller than 101 . Continuing on, we can write 10−1 to standfor 0.1, the number ten times smaller than 100 . Negative exponentsare used for small numbers:

3.2 = 3.2× 100

0.32 = 3.2× 10−1

0.032 = 3.2× 10−2 . . .

A common source of confusion is the notation used on the dis-plays of many calculators. Examples:

3.2× 106 (written notation)3.2E+6 (notation on some calculators)3.26 (notation on some other calculators)

The last example is particularly unfortunate, because 3.26 reallystands for the number 3.2 × 3.2 × 3.2 × 3.2 × 3.2 × 3.2 = 1074, atotally different number from 3.2 × 106 = 3200000. The calculatornotation should never be used in writing. It’s just a way for themanufacturer to save money by making a simpler display.

self-check EA student learns that 104 bacteria, standing in line to register for classesat Paramecium Community College, would form a queue of this size:

The student concludes that 102 bacteria would form a line of this length:

Why is the student incorrect? . Answer, p. 508

0.9 ConversionsI suggest you avoid memorizing lots of conversion factors betweenSI units and U.S. units, but two that do come in handy are:

1 inch = 2.54 cm

An object with a weight on Earth of 2.2 pounds-force has amass of 1 kg.

The first one is the present definition of the inch, so it’s exact. Thesecond one is not exact, but is good enough for most purposes. (U.S.units of force and mass are confusing, so it’s a good thing they’re notused in science. In U.S. units, the unit of force is the pound-force,


and the best unit to use for mass is the slug, which is about 14.6kg.)

More important than memorizing conversion factors is under-standing the right method for doing conversions. Even within theSI, you may need to convert, say, from grams to kilograms. Differ-ent people have different ways of thinking about conversions, butthe method I’ll describe here is systematic and easy to understand.The idea is that if 1 kg and 1000 g represent the same mass, thenwe can consider a fraction like

103 g

1 kg

to be a way of expressing the number one. This may bother you. Forinstance, if you type 1000/1 into your calculator, you will get 1000,not one. Again, different people have different ways of thinkingabout it, but the justification is that it helps us to do conversions,and it works! Now if we want to convert 0.7 kg to units of grams,we can multiply kg by the number one:

0.7 kg× 103 g

1 kg

If you’re willing to treat symbols such as “kg” as if they were vari-ables as used in algebra (which they’re really not), you can thencancel the kg on top with the kg on the bottom, resulting in

0.7kg× 103 g

1kg= 700 g .

To convert grams to kilograms, you would simply flip the fractionupside down.

One advantage of this method is that it can easily be applied toa series of conversions. For instance, to convert one year to units ofseconds,

1year× 365days

1year× 24hours

1day× 60min

1hour× 60 s

1min=

= 3.15× 107 s .

Should that exponent be positive, or negative?

A common mistake is to write the conversion fraction incorrectly.For instance the fraction

103 kg

1 g(incorrect)

does not equal one, because 103 kg is the mass of a car, and 1 g isthe mass of a raisin. One correct way of setting up the conversionfactor would be

10−3 kg

1 g(correct) .

Section 0.9 Conversions 27

You can usually detect such a mistake if you take the time to checkyour answer and see if it is reasonable.

If common sense doesn’t rule out either a positive or a negativeexponent, here’s another way to make sure you get it right. Thereare big prefixes and small prefixes:

big prefixes: k Msmall prefixes: m µ n

(It’s not hard to keep straight which are which, since “mega” and“micro” are evocative, and it’s easy to remember that a kilometeris bigger than a meter and a millimeter is smaller.) In the exampleabove, we want the top of the fraction to be the same as the bottom.Since k is a big prefix, we need to compensate by putting a smallnumber like 10−3 in front of it, not a big number like 103.

. Solved problem: a simple conversion page 33, problem 1

. Solved problem: the geometric mean page 33, problem 7

Discussion question

A Each of the following conversions contains an error. In each case,explain what the error is.

(a) 1000 kg× 1 kg1000 g = 1 g

(b) 50 m× 1 cm100 m = 0.5 cm

(c) “Nano” is 10−9, so there are 10−9 nm in a meter.

(d) “Micro” is 10−6, so 1 kg is 106 µg.

0.10 Significant figuresAn engineer is designing a car engine, and has been told that thediameter of the pistons (which are being designed by someone else)is 5 cm. He knows that 0.02 cm of clearance is required for a pistonof this size, so he designs the cylinder to have an inside diameter of5.04 cm. Luckily, his supervisor catches his mistake before the cargoes into production. She explains his error to him, and mentallyputs him in the “do not promote” category.

What was his mistake? The person who told him the pistonswere 5 cm in diameter was wise to the ways of significant figures,as was his boss, who explained to him that he needed to go backand get a more accurate number for the diameter of the pistons.That person said “5 cm” rather than “5.00 cm” specifically to avoidcreating the impression that the number was extremely accurate. Inreality, the pistons’ diameter was 5.13 cm. They would never havefit in the 5.04-cm cylinders.

The number of digits of accuracy in a number is referred to asthe number of significant figures, or “sig figs” for short. As in the


example above, sig figs provide a way of showing the accuracy of anumber. In most cases, the result of a calculation involving severalpieces of data can be no more accurate than the least accurate pieceof data. In other words, “garbage in, garbage out.” Since the 5cm diameter of the pistons was not very accurate, the result of theengineer’s calculation, 5.04 cm, was really not as accurate as hethought. In general, your result should not have more than thenumber of sig figs in the least accurate piece of data you startedwith. The calculation above should have been done as follows:

5 cm (1 sig fig)

+0.04 cm (1 sig fig)

=5 cm (rounded off to 1 sig fig)

The fact that the final result only has one significant figure thenalerts you to the fact that the result is not very accurate, and wouldnot be appropriate for use in designing the engine.

Note that the leading zeroes in the number 0.04 do not countas significant figures, because they are only placeholders. On theother hand, a number such as 50 cm is ambiguous — the zero couldbe intended as a significant figure, or it might just be there as aplaceholder. The ambiguity involving trailing zeroes can be avoidedby using scientific notation, in which 5 × 101 cm would imply onesig fig of accuracy, while 5.0× 101 cm would imply two sig figs.

self-check FThe following quote is taken from an editorial by Norimitsu Onishi in theNew York Times, August 18, 2002.

Consider Nigeria. Everyone agrees it is Africa’s most populousnation. But what is its population? The United Nations says114 million; the State Department, 120 million. The World Banksays 126.9 million, while the Central Intelligence Agency puts itat 126,635,626.

What should bother you about this? . Answer, p. 508

Dealing correctly with significant figures can save you time! Of-ten, students copy down numbers from their calculators with eightsignificant figures of precision, then type them back in for a latercalculation. That’s a waste of time, unless your original data hadthat kind of incredible precision.

The rules about significant figures are only rules of thumb, andare not a substitute for careful thinking. For instance, $20.00 +$0.05 is $20.05. It need not and should not be rounded off to $20.In general, the sig fig rules work best for multiplication and division,and we also apply them when doing a complicated calculation thatinvolves many types of operations. For simple addition and subtrac-tion, it makes more sense to maintain a fixed number of digits afterthe decimal point.

Section 0.10 Significant figures 29

When in doubt, don’t use the sig fig rules at all. Instead, in-tentionally change one piece of your initial data by the maximumamount by which you think it could have been off, and recalculatethe final result. The digits on the end that are completely reshuffledare the ones that are meaningless, and should be omitted.

self-check GHow many significant figures are there in each of the following mea-surements?

(1) 9.937 m

(2) 4.0 s

(3) 0.0000000000000037 kg . Answer, p. 508


SummarySelected vocabularymatter . . . . . . Anything that is affected by gravity.light . . . . . . . . Anything that can travel from one place to an-

other through empty space and can influencematter, but is not affected by gravity.

operational defi-nition . . . . . . .

A definition that states what operationsshould be carried out to measure the thing be-ing defined.

Systeme Interna-tional . . . . . . .

A fancy name for the metric system.

mks system . . . The use of metric units based on the meter,kilogram, and second. Example: meters persecond is the mks unit of speed, not cm/s orkm/hr.

mass . . . . . . . A numerical measure of how difficult it is tochange an object’s motion.

significant figures Digits that contribute to the accuracy of ameasurement.

Notationm . . . . . . . . . meter, the metric distance unitkg . . . . . . . . . kilogram, the metric unit of masss . . . . . . . . . . second, the metric unit of timeM- . . . . . . . . . the metric prefix mega-, 106

k- . . . . . . . . . the metric prefix kilo-, 103

m- . . . . . . . . . the metric prefix milli-, 10−3

µ- . . . . . . . . . the metric prefix micro-, 10−6

n- . . . . . . . . . the metric prefix nano-, 10−9

Summary

Physics is the use of the scientific method to study the behaviorof light and matter. The scientific method requires a cycle of the-ory and experiment, theories with both predictive and explanatoryvalue, and reproducible experiments.

The metric system is a simple, consistent framework for measure-ment built out of the meter, the kilogram, and the second plus a setof prefixes denoting powers of ten. The most systematic method fordoing conversions is shown in the following example:

370 ms× 10−3 s

1 ms= 0.37 s

Mass is a measure of the amount of a substance. Mass can bedefined gravitationally, by comparing an object to a standard masson a double-pan balance, or in terms of inertia, by comparing theeffect of a force on an object to the effect of the same force on astandard mass. The two definitions are found experimentally tobe proportional to each other to a high degree of precision, so we

Summary 31

usually refer simply to “mass,” without bothering to specify whichtype.

A force is that which can change the motion of an object. Themetric unit of force is the Newton, defined as the force required toaccelerate a standard 1-kg mass from rest to a speed of 1 m/s in 1s.

Scientific notation means, for example, writing 3.2× 105 ratherthan 320000.

Writing numbers with the correct number of significant figurescorrectly communicates how accurate they are. As a rule of thumb,the final result of a calculation is no more accurate than, and shouldhave no more significant figures than, the least accurate piece ofdata.


ProblemsKey√

A computerized answer check is available online.∫A problem that requires calculus.

? A difficult problem.

1 Convert 134 mg to units of kg, writing your answer in scientificnotation. . Solution, p. 495

2 The speed of light is 3.0 × 108 m/s. Convert this to furlongsper fortnight. A furlong is 220 yards, and a fortnight is 14 days. Aninch is 2.54 cm.

√

3 Express each of the following quantities in micrograms:(a) 10 mg, (b) 104 g, (c) 10 kg, (d) 100× 103 g, (e) 1000 ng.

√

4 In the last century, the average age of the onset of puberty forgirls has decreased by several years. Urban folklore has it that thisis because of hormones fed to beef cattle, but it is more likely to bebecause modern girls have more body fat on the average and pos-sibly because of estrogen-mimicking chemicals in the environmentfrom the breakdown of pesticides. A hamburger from a hormone-implanted steer has about 0.2 ng of estrogen (about double theamount of natural beef). A serving of peas contains about 300ng of estrogen. An adult woman produces about 0.5 mg of estrogenper day (note the different unit!). (a) How many hamburgers woulda girl have to eat in one day to consume as much estrogen as anadult woman’s daily production? (b) How many servings of peas?√

5 Compute the following things. If they don’t make sense be-cause of units, say so.(a) 3 cm + 5 cm(b) 1.11 m + 22 cm(c) 120 miles + 2.0 hours(d) 120 miles / 2.0 hours

6 Your backyard has brick walls on both ends. You measure adistance of 23.4 m from the inside of one wall to the inside of theother. Each wall is 29.4 cm thick. How far is it from the outsideof one wall to the outside of the other? Pay attention to significantfigures.

7 The usual definition of the mean (average) of two numbers aand b is (a+b)/2. This is called the arithmetic mean. The geometricmean, however, is defined as (ab)1/2 (i.e., the square root of ab). Forthe sake of definiteness, let’s say both numbers have units of mass.

Problems 33

Problem 10.

Problem 11.

(a) Compute the arithmetic mean of two numbers that have unitsof grams. Then convert the numbers to units of kilograms andrecompute their mean. Is the answer consistent? (b) Do the samefor the geometric mean. (c) If a and b both have units of grams,what should we call the units of ab? Does your answer make sensewhen you take the square root? (d) Suppose someone proposes toyou a third kind of mean, called the superduper mean, defined as(ab)1/3. Is this reasonable? . Solution, p. 495

8 The distance to the horizon is given by the expression√

2rh,where r is the radius of the Earth, and h is the observer’s heightabove the Earth’s surface. (This can be proved using the Pythagoreantheorem.) Show that the units of this expression make sense. (Seeexample 1 on p. 23 for an example of how to do this.) Don’t try toprove the result, just check its units.

9 In an article on the SARS epidemic, the May 7, 2003 NewYork Times discusses conflicting estimates of the disease’s incuba-tion period (the average time that elapses from infection to the firstsymptoms). “The study estimated it to be 6.4 days. But other sta-tistical calculations ... showed that the incubation period could beas long as 14.22 days.” What’s wrong here?

10 (a) Based on the definitions of the sine, cosine, and tangent,what units must they have? (b) A cute formula from trigonometrylets you find any angle of a triangle if you know the lengths ofits sides. Using the notation shown in the figure, and letting s =(a+ b+ c)/2 be half the perimeter, we have

tanA/2 =

√(s− b)(s− c)s(s− a)

.

Show that the units of this equation make sense. In other words,check that the units of the right-hand side are the same as youranswer to part a of the question. . Solution, p. 495

11 The photo shows the corner of a bag of pretzels. What’swrong here?


Exercise 0: Models and idealizationEquipment:

coffee filters

ramps (one per group)

balls of various sizes

sticky tape

vacuum pump and “guinea and feather” apparatus (one)

The motion of falling objects has been recognized since ancient times as an important piece ofphysics, but the motion is inconveniently fast, so in our everyday experience it can be hard totell exactly what objects are doing when they fall. In this exercise you will use several techniquesto get around this problem and study the motion. Your goal is to construct a scientific model offalling. A model means an explanation that makes testable predictions. Often models containsimplifications or idealizations that make them easier to work with, even though they are notstrictly realistic.

1. One method of making falling easier to observe is to use objects like feathers that we knowfrom everyday experience will not fall as fast. You will use coffee filters, in stacks of varioussizes, to test the following two hypotheses and see which one is true, or whether neither is true:

Hypothesis 1A: When an object is dropped, it rapidly speeds up to a certain natural fallingspeed, and then continues to fall at that speed. The falling speed is proportional to the object’sweight. (A proportionality is not just a statement that if one thing gets bigger, the other doestoo. It says that if one becomes three times bigger, the other also gets three times bigger, etc.)

Hypothesis 1B: Different objects fall the same way, regardless of weight.

Test these hypotheses and discuss your results with your instructor.

2. A second way to slow down the action is to let a ball roll down a ramp. The steeper theramp, the closer to free fall. Based on your experience in part 1, write a hypothesis about whatwill happen when you race a heavier ball against a lighter ball down the same ramp, startingthem both from rest.

Hypothesis:

Show your hypothesis to your instructor, and then test it.

You have probably found that falling was more complicated than you thought! Is there morethan one factor that affects the motion of a falling object? Can you imagine certain idealizedsituations that are simpler? Try to agree verbally with your group on an informal model offalling that can make predictions about the experiments described in parts 3 and 4.

3. You have three balls: a standard “comparison ball” of medium weight, a light ball, and aheavy ball. Suppose you stand on a chair and (a) drop the light ball side by side with thecomparison ball, then (b) drop the heavy ball side by side with the comparison ball, then (c)join the light and heavy balls together with sticky tape and drop them side by side with thecomparison ball.

Use your model to make a prediction:

Test your prediction.

Exercise 0: Models and idealization 35

4. Your instructor will pump nearly all the air out of a chamber containing a feather and aheavier object, then let them fall side by side in the chamber.

Use your model to make a prediction:


a / Amoebas this size areseldom encountered.

Life would be very different if youwere the size of an insect.

Chapter 1

Scaling and estimation

1.1 Introduction

Why can’t an insect be the size of a dog? Some skinny stretched-out cells in your spinal cord are a meter tall — why does naturedisplay no single cells that are not just a meter tall, but a meterwide, and a meter thick as well? Believe it or not, these are ques-tions that can be answered fairly easily without knowing much moreabout physics than you already do. The only mathematical tech-nique you really need is the humble conversion, applied to area andvolume.

Area and volume

Area can be defined by saying that we can copy the shape ofinterest onto graph paper with 1 cm × 1 cm squares and count thenumber of squares inside. Fractions of squares can be estimated byeye. We then say the area equals the number of squares, in units ofsquare cm. Although this might seem less “pure” than computingareas using formulae like A = πr2 for a circle or A = wh/2 for atriangle, those formulae are not useful as definitions of area becausethey cannot be applied to irregularly shaped areas.

Units of square cm are more commonly written as cm2 in science.

37

Of course, the unit of measurement symbolized by “cm” is not analgebra symbol standing for a number that can be literally multipliedby itself. But it is advantageous to write the units of area that wayand treat the units as if they were algebra symbols. For instance,if you have a rectangle with an area of 6m2 and a width of 2 m,then calculating its length as (6 m2)/(2 m) = 3 m gives a resultthat makes sense both numerically and in terms of units. Thisalgebra-style treatment of the units also ensures that our methodsof converting units work out correctly. For instance, if we acceptthe fraction

100 cm

1 mas a valid way of writing the number one, then one times one equalsone, so we should also say that one can be represented by

100 cm

1 m× 100 cm

1 m,

which is the same as10000 cm2

1 m2.

That means the conversion factor from square meters to square cen-timeters is a factor of 104, i.e., a square meter has 104 square cen-timeters in it.

All of the above can be easily applied to volume as well, usingone-cubic-centimeter blocks instead of squares on graph paper.

To many people, it seems hard to believe that a square meterequals 10000 square centimeters, or that a cubic meter equals amillion cubic centimeters — they think it would make more sense ifthere were 100 cm2 in 1 m2, and 100 cm3 in 1 m3, but that would beincorrect. The examples shown in figure b aim to make the correctanswer more believable, using the traditional U.S. units of feet andyards. (One foot is 12 inches, and one yard is three feet.)

b / Visualizing conversions ofarea and volume using traditionalU.S. units.

self-check ABased on figure b, convince yourself that there are 9 ft2 in a square yard,and 27 ft3 in a cubic yard, then demonstrate the same thing symbolically(i.e., with the method using fractions that equal one). . Answer, p.508

38 Chapter 1 Scaling and estimation

. Solved problem: converting mm2 to cm2 page 51, problem 2

. Solved problem: scaling a liter page 51, problem 1

Discussion question

A How many square centimeters are there in a square inch? (1 inch =2.54 cm) First find an approximate answer by making a drawing, then de-rive the conversion factor more accurately using the symbolic method.

c / Galileo Galilei (1564-1642) was a Renaissance Italian who brought thescientific method to bear on physics, creating the modern version of thescience. Coming from a noble but very poor family, Galileo had to dropout of medical school at the University of Pisa when he ran out of money.Eventually becoming a lecturer in mathematics at the same school, hebegan a career as a notorious troublemaker by writing a burlesque ridi-culing the university’s regulations — he was forced to resign, but found anew teaching position at Padua. He invented the pendulum clock, inves-tigated the motion of falling bodies, and discovered the moons of Jupiter.The thrust of his life’s work was to discredit Aristotle’s physics by con-fronting it with contradictory experiments, a program that paved the wayfor Newton’s discovery of the relationship between force and motion. Inchapter 3 we’ll come to the story of Galileo’s ultimate fate at the hands ofthe Church.

1.2 Scaling of area and volumeGreat fleas have lesser fleasUpon their backs to bite ’em.And lesser fleas have lesser still,And so ad infinitum.

Jonathan Swift

Now how do these conversions of area and volume relate to thequestions I posed about sizes of living things? Well, imagine thatyou are shrunk like Alice in Wonderland to the size of an insect.One way of thinking about the change of scale is that what usedto look like a centimeter now looks like perhaps a meter to you,because you’re so much smaller. If area and volume scaled accordingto most people’s intuitive, incorrect expectations, with 1 m2 beingthe same as 100 cm2, then there would be no particular reasonwhy nature should behave any differently on your new, reducedscale. But nature does behave differently now that you’re small.For instance, you will find that you can walk on water, and jumpto many times your own height. The physicist Galileo Galilei hadthe basic insight that the scaling of area and volume determineshow natural phenomena behave differently on different scales. Hefirst reasoned about mechanical structures, but later extended hisinsights to living things, taking the then-radical point of view that atthe fundamental level, a living organism should follow the same laws

Section 1.2 Scaling of area and volume 39

d / The small boat holds upjust fine.

e / A larger boat built withthe same proportions as thesmall one will collapse under itsown weight.

f / A boat this large needs tohave timbers that are thickercompared to its size.

of nature as a machine. We will follow his lead by first discussingmachines and then living things.

Galileo on the behavior of nature on large and small scales

One of the world’s most famous pieces of scientific writing isGalileo’s Dialogues Concerning the Two New Sciences. Galileo wasan entertaining writer who wanted to explain things clearly to laypeo-ple, and he livened up his work by casting it in the form of a dialogueamong three people. Salviati is really Galileo’s alter ego. Simpliciois the stupid character, and one of the reasons Galileo got in troublewith the Church was that there were rumors that Simplicio repre-sented the Pope. Sagredo is the earnest and intelligent student, withwhom the reader is supposed to identify. (The following excerptsare from the 1914 translation by Crew and de Salvio.)

SAGREDO: Yes, that is what I mean; and I refer especially tohis last assertion which I have always regarded as false. . . ;namely, that in speaking of these and other similar machinesone cannot argue from the small to the large, because manydevices which succeed on a small scale do not work on alarge scale. Now, since mechanics has its foundations in ge-ometry, where mere size [ is unimportant], I do not see thatthe properties of circles, triangles, cylinders, cones and othersolid figures will change with their size. If, therefore, a largemachine be constructed in such a way that its parts bear toone another the same ratio as in a smaller one, and if thesmaller is sufficiently strong for the purpose for which it isdesigned, I do not see why the larger should not be able towithstand any severe and destructive tests to which it may besubjected.

Salviati contradicts Sagredo:

SALVIATI: . . . Please observe, gentlemen, how facts whichat first seem improbable will, even on scant explanation, dropthe cloak which has hidden them and stand forth in naked andsimple beauty. Who does not know that a horse falling from aheight of three or four cubits will break his bones, while a dogfalling from the same height or a cat from a height of eightor ten cubits will suffer no injury? Equally harmless would bethe fall of a grasshopper from a tower or the fall of an ant fromthe distance of the moon.

The point Galileo is making here is that small things are sturdierin proportion to their size. There are a lot of objections that could beraised, however. After all, what does it really mean for something tobe “strong”, to be “strong in proportion to its size,” or to be strong“out of proportion to its size?” Galileo hasn’t given operationaldefinitions of things like “strength,” i.e., definitions that spell outhow to measure them numerically.


g / Galileo discusses planksmade of wood, but the conceptmay be easier to imagine withclay. All three clay rods in thefigure were originally the sameshape. The medium-sized onewas twice the height, twice thelength, and twice the width ofthe small one, and similarly thelarge one was twice as big asthe medium one in all its lineardimensions. The big one hasfour times the linear dimensionsof the small one, 16 times thecross-sectional area when cutperpendicular to the page, and64 times the volume. That meansthat the big one has 64 times theweight to support, but only 16times the strength compared tothe smallest one.

Also, a cat is shaped differently from a horse — an enlargedphotograph of a cat would not be mistaken for a horse, even if thephoto-doctoring experts at the National Inquirer made it look like aperson was riding on its back. A grasshopper is not even a mammal,and it has an exoskeleton instead of an internal skeleton. The wholeargument would be a lot more convincing if we could do some iso-lation of variables, a scientific term that means to change only onething at a time, isolating it from the other variables that might havean effect. If size is the variable whose effect we’re interested in see-ing, then we don’t really want to compare things that are differentin size but also different in other ways.

SALVIATI: . . . we asked the reason why [shipbuilders] em-ployed stocks, scaffolding, and bracing of larger dimensionsfor launching a big vessel than they do for a small one; and[an old man] answered that they did this in order to avoid thedanger of the ship parting under its own heavy weight, a dan-ger to which small boats are not subject?

After this entertaining but not scientifically rigorous beginning,Galileo starts to do something worthwhile by modern standards.He simplifies everything by considering the strength of a woodenplank. The variables involved can then be narrowed down to thetype of wood, the width, the thickness, and the length. He alsogives an operational definition of what it means for the plank tohave a certain strength “in proportion to its size,” by introducingthe concept of a plank that is the longest one that would not snapunder its own weight if supported at one end. If you increasedits length by the slightest amount, without increasing its width orthickness, it would break. He says that if one plank is the sameshape as another but a different size, appearing like a reduced orenlarged photograph of the other, then the planks would be strong“in proportion to their sizes” if both were just barely able to supporttheir own weight.

h / 1. This plank is as long as itcan be without collapsing underits own weight. If it was a hun-dredth of an inch longer, it wouldcollapse. 2. This plank is madeout of the same kind of wood. It istwice as thick, twice as long, andtwice as wide. It will collapse un-der its own weight.


Also, Galileo is doing something that would be frowned on inmodern science: he is mixing experiments whose results he has ac-tually observed (building boats of different sizes), with experimentsthat he could not possibly have done (dropping an ant from theheight of the moon). He now relates how he has done actual ex-periments with such planks, and found that, according to this op-erational definition, they are not strong in proportion to their sizes.The larger one breaks. He makes sure to tell the reader how impor-tant the result is, via Sagredo’s astonished response:

SAGREDO: My brain already reels. My mind, like a cloudmomentarily illuminated by a lightning flash, is for an instantfilled with an unusual light, which now beckons to me andwhich now suddenly mingles and obscures strange, crudeideas. From what you have said it appears to me impossibleto build two similar structures of the same material, but ofdifferent sizes and have them proportionately strong.

In other words, this specific experiment, using things like woodenplanks that have no intrinsic scientific interest, has very wide impli-cations because it points out a general principle, that nature actsdifferently on different scales.

To finish the discussion, Galileo gives an explanation. He saysthat the strength of a plank (defined as, say, the weight of the heav-iest boulder you could put on the end without breaking it) is pro-portional to its cross-sectional area, that is, the surface area of thefresh wood that would be exposed if you sawed through it in themiddle. Its weight, however, is proportional to its volume.1

How do the volume and cross-sectional area of the longer plankcompare with those of the shorter plank? We have already seen,while discussing conversions of the units of area and volume, thatthese quantities don’t act the way most people naively expect. Youmight think that the volume and area of the longer plank would bothbe doubled compared to the shorter plank, so they would increasein proportion to each other, and the longer plank would be equallyable to support its weight. You would be wrong, but Galileo knowsthat this is a common misconception, so he has Salviati address thepoint specifically:

SALVIATI: . . . Take, for example, a cube two inches on aside so that each face has an area of four square inchesand the total area, i.e., the sum of the six faces, amountsto twenty-four square inches; now imagine this cube to besawed through three times [with cuts in three perpendicularplanes] so as to divide it into eight smaller cubes, each oneinch on the side, each face one inch square, and the total

1Galileo makes a slightly more complicated argument, taking into accountthe effect of leverage (torque). The result I’m referring to comes out the sameregardless of this effect.


i / The area of a shape isproportional to the square of itslinear dimensions, even if theshape is irregular.

surface of each cube six square inches instead of twenty-four in the case of the larger cube. It is evident therefore,that the surface of the little cube is only one-fourth that ofthe larger, namely, the ratio of six to twenty-four; but the vol-ume of the solid cube itself is only one-eighth; the volume,and hence also the weight, diminishes therefore much morerapidly than the surface. . . You see, therefore, Simplicio, thatI was not mistaken when . . . I said that the surface of a smallsolid is comparatively greater than that of a large one.

The same reasoning applies to the planks. Even though theyare not cubes, the large one could be sawed into eight small ones,each with half the length, half the thickness, and half the width.The small plank, therefore, has more surface area in proportion toits weight, and is therefore able to support its own weight while thelarge one breaks.

Scaling of area and volume for irregularly shaped objects

You probably are not going to believe Galileo’s claim that thishas deep implications for all of nature unless you can be convincedthat the same is true for any shape. Every drawing you’ve seen sofar has been of squares, rectangles, and rectangular solids. Clearlythe reasoning about sawing things up into smaller pieces would notprove anything about, say, an egg, which cannot be cut up into eightsmaller egg-shaped objects with half the length.

Is it always true that something half the size has one quarterthe surface area and one eighth the volume, even if it has an irreg-ular shape? Take the example of a child’s violin. Violins are madefor small children in smaller size to accomodate their small bodies.Figure i shows a full-size violin, along with two violins made withhalf and 3/4 of the normal length.2 Let’s study the surface area ofthe front panels of the three violins.

Consider the square in the interior of the panel of the full-sizeviolin. In the 3/4-size violin, its height and width are both smallerby a factor of 3/4, so the area of the corresponding, smaller squarebecomes 3/4×3/4 = 9/16 of the original area, not 3/4 of the originalarea. Similarly, the corresponding square on the smallest violin hashalf the height and half the width of the original one, so its area is1/4 the original area, not half.

The same reasoning works for parts of the panel near the edge,such as the part that only partially fills in the other square. Theentire square scales down the same as a square in the interior, andin each violin the same fraction (about 70%) of the square is full, sothe contribution of this part to the total area scales down just thesame.

2The customary terms “half-size” and “3/4-size” actually don’t describe thesizes in any accurate way. They’re really just standard, arbitrary marketinglabels.


j / The muffin comes out ofthe oven too hot to eat. Breakingit up into four pieces increasesits surface area while keepingthe total volume the same. Itcools faster because of thegreater surface-to-volume ratio.In general, smaller things havegreater surface-to-volume ratios,but in this example there is noeasy way to compute the effectexactly, because the small piecesaren’t the same shape as theoriginal muffin.

Since any small square region or any small region covering partof a square scales down like a square object, the entire surface areaof an irregularly shaped object changes in the same manner as thesurface area of a square: scaling it down by 3/4 reduces the area bya factor of 9/16, and so on.

In general, we can see that any time there are two objects withthe same shape, but different linear dimensions (i.e., one looks like areduced photo of the other), the ratio of their areas equals the ratioof the squares of their linear dimensions:

A1

A2=

(L1

L2

)2

.

Note that it doesn’t matter where we choose to measure the linearsize, L, of an object. In the case of the violins, for instance, it couldhave been measured vertically, horizontally, diagonally, or even fromthe bottom of the left f-hole to the middle of the right f-hole. Wejust have to measure it in a consistent way on each violin. Since allthe parts are assumed to shrink or expand in the same manner, theratio L1/L2 is independent of the choice of measurement.

It is also important to realize that it is completely unnecessaryto have a formula for the area of a violin. It is only possible toderive simple formulas for the areas of certain shapes like circles,rectangles, triangles and so on, but that is no impediment to thetype of reasoning we are using.

Sometimes it is inconvenient to write all the equations in termsof ratios, especially when more than two objects are being compared.A more compact way of rewriting the previous equation is

A ∝ L2 .

The symbol “∝” means “is proportional to.” Scientists and engi-neers often speak about such relationships verbally using the phrases“scales like” or “goes like,” for instance “area goes like length squared.”

All of the above reasoning works just as well in the case of vol-ume. Volume goes like length cubed:

V ∝ L3 .

If different objects are made of the same material with the samedensity, ρ = m/V , then their masses, m = ρV , are proportionalto L3, and so are their weights. (The symbol for density is ρ, thelower-case Greek letter “rho.”)

An important point is that all of the above reasoning aboutscaling only applies to objects that are the same shape. For instance,a piece of paper is larger than a pencil, but has a much greatersurface-to-volume ratio.


k / Example 1. The big trian-gle has four times more area thanthe little one.

l / A tricky way of solving ex-ample 1, explained in solution #2.

One of the first things I learned as a teacher was that studentswere not very original about their mistakes. Every group of studentstends to come up with the same goofs as the previous class. Thefollowing are some examples of correct and incorrect reasoning aboutproportionality.

Scaling of the area of a triangle example 1. In figure k, the larger triangle has sides twice as long. Howmany times greater is its area?

Correct solution #1: Area scales in proportion to the square of thelinear dimensions, so the larger triangle has four times more area(22 = 4).

Correct solution #2: You could cut the larger triangle into four ofthe smaller size, as shown in fig. (b), so its area is four timesgreater. (This solution is correct, but it would not work for a shapelike a circle, which can’t be cut up into smaller circles.)

Correct solution #3: The area of a triangle is given by

A = bh/2, where b is the base and h is the height. The areas ofthe triangles are

A1 = b1h1/2A2 = b2h2/2

= (2b1)(2h1)/2= 2b1h1

A2/A1 = (2b1h1)/(b1h1/2)= 4

(Although this solution is correct, it is a lot more work than solution#1, and it can only be used in this case because a triangle is asimple geometric shape, and we happen to know a formula for itsarea.)

Correct solution #4: The area of a triangle is A = bh/2. Thecomparison of the areas will come out the same as long as theratios of the linear sizes of the triangles is as specified, so let’sjust say b1 = 1.00 m and b2 = 2.00 m. The heights are then alsoh1 = 1.00 m and h2 = 2.00 m, giving areas A1 = 0.50 m2 andA2 = 2.00 m2, so A2/A1 = 4.00.

(The solution is correct, but it wouldn’t work with a shape forwhose area we don’t have a formula. Also, the numerical cal-culation might make the answer of 4.00 appear inexact, whereassolution #1 makes it clear that it is exactly 4.)

Incorrect solution: The area of a triangle is A = bh/2, and if youplug in b = 2.00 m and h = 2.00 m, you get A = 2.00 m2, sothe bigger triangle has 2.00 times more area. (This solution isincorrect because no comparison has been made with the smallertriangle.)


m / Example 2. The big spherehas 125 times more volume thanthe little one.

n / Example 3. The 48-point“S” has 1.78 times more areathan the 36-point “S.”

Scaling of the volume of a sphere example 2. In figure m, the larger sphere has a radius that is five timesgreater. How many times greater is its volume?

Correct solution #1: Volume scales like the third power of thelinear size, so the larger sphere has a volume that is 125 timesgreater (53 = 125).

Correct solution #2: The volume of a sphere is V = (4/3)πr3, so

V1 =43πr3

1

V2 =43πr3

2

=43π(5r1)3

=500

3πr3

1

V2/V1 =(

5003πr3

1

)/

(43πr3

1

)= 125

Incorrect solution: The volume of a sphere is V = (4/3)πr3, so

V1 =43πr3

1

V2 =43πr3

2

=43π · 5r3

1

=203πr3

1

V2/V1 =(

203πr3

1

)/

(43πr3

1

)= 5

(The solution is incorrect because (5r1)3 is not the same as 5r31 .)

Scaling of a more complex shape example 3. The first letter “S” in figure n is in a 36-point font, the second in48-point. How many times more ink is required to make the larger“S”? (Points are a unit of length used in typography.)

Correct solution: The amount of ink depends on the area to becovered with ink, and area is proportional to the square of thelinear dimensions, so the amount of ink required for the second“S” is greater by a factor of (48/36)2 = 1.78.

Incorrect solution: The length of the curve of the second “S” islonger by a factor of 48/36 = 1.33, so 1.33 times more ink isrequired.

(The solution is wrong because it assumes incorrectly that thewidth of the curve is the same in both cases. Actually both the


width and the length of the curve are greater by a factor of 48/36,so the area is greater by a factor of (48/36)2 = 1.78.)

. Solved problem: a telescope gathers light page 51, problem 3

. Solved problem: distance from an earthquake page 51, problem 8


A A toy fire engine is 1/30 the size of the real one, but is constructedfrom the same metal with the same proportions. How many times smalleris its weight? How many times less red paint would be needed to paintit?

B Galileo spends a lot of time in his dialog discussing what reallyhappens when things break. He discusses everything in terms of Aristo-tle’s now-discredited explanation that things are hard to break, becauseif something breaks, there has to be a gap between the two halves withnothing in between, at least initially. Nature, according to Aristotle, “ab-hors a vacuum,” i.e., nature doesn’t “like” empty space to exist. Of course,air will rush into the gap immediately, but at the very moment of breaking,Aristotle imagined a vacuum in the gap. Is Aristotle’s explanation of whyit is hard to break things an experimentally testable statement? If so, howcould it be tested experimentally?

1.3 Order-of-magnitude estimatesIt is the mark of an instructed mind to rest satisfied with thedegree of precision that the nature of the subject permits andnot to seek an exactness where only an approximation of thetruth is possible.

Aristotle

It is a common misconception that science must be exact. Forinstance, in the Star Trek TV series, it would often happen thatCaptain Kirk would ask Mr. Spock, “Spock, we’re in a pretty badsituation. What do you think are our chances of getting out ofhere?” The scientific Mr. Spock would answer with something like,“Captain, I estimate the odds as 237.345 to one.” In reality, hecould not have estimated the odds with six significant figures ofaccuracy, but nevertheless one of the hallmarks of a person with agood education in science is the ability to make estimates that arelikely to be at least somewhere in the right ballpark. In many suchsituations, it is often only necessary to get an answer that is off by nomore than a factor of ten in either direction. Since things that differby a factor of ten are said to differ by one order of magnitude, suchan estimate is called an order-of-magnitude estimate. The tilde,∼, is used to indicate that things are only of the same order ofmagnitude, but not exactly equal, as in

odds of survival ∼ 100 to one .

The tilde can also be used in front of an individual number to em-phasize that the number is only of the right order of magnitude.

Section 1.3 Order-of-magnitude estimates 47

Although making order-of-magnitude estimates seems simple andnatural to experienced scientists, it’s a mode of reasoning that iscompletely unfamiliar to most college students. Some of the typicalmental steps can be illustrated in the following example.

Cost of transporting tomatoes example 4. Roughly what percentage of the price of a tomato comes fromthe cost of transporting it in a truck?

. The following incorrect solution illustrates one of the main waysyou can go wrong in order-of-magnitude estimates.

Incorrect solution: Let’s say the trucker needs to make a $400profit on the trip. Taking into account her benefits, the cost of gas,and maintenance and payments on the truck, let’s say the totalcost is more like $2000. I’d guess about 5000 tomatoes would fitin the back of the truck, so the extra cost per tomato is 40 cents.That means the cost of transporting one tomato is comparable tothe cost of the tomato itself. Transportation really adds a lot to thecost of produce, I guess.

The problem is that the human brain is not very good at esti-mating area or volume, so it turns out the estimate of 5000 tomatoesfitting in the truck is way off. That’s why people have a hard timeat those contests where you are supposed to estimate the number ofjellybeans in a big jar. Another example is that most people thinktheir families use about 10 gallons of water per day, but in realitythe average is about 300 gallons per day. When estimating areaor volume, you are much better off estimating linear dimensions,and computing volume from the linear dimensions. Here’s a bettersolution:

Better solution: As in the previous solution, say the cost of thetrip is $2000. The dimensions of the bin are probably 4 m × 2m × 1 m, for a volume of 8 m3. Since the whole thing is justan order-of-magnitude estimate, let’s round that off to the nearestpower of ten, 10 m3. The shape of a tomato is complicated, and Idon’t know any formula for the volume of a tomato shape, but sincethis is just an estimate, let’s pretend that a tomato is a cube, 0.05 m× 0.05 m × 0.05 m, for a volume of 1.25×10−4 m3. Since this is justa rough estimate, let’s round that to 10−4m3. We can find the totalnumber of tomatoes by dividing the volume of the bin by the volumeof one tomato: 10 m3/10−4 m3 = 105 tomatoes. The transportationcost per tomato is $2000/105 tomatoes=$0.02/tomato. That meansthat transportation really doesn’t contribute very much to the costof a tomato.

Approximating the shape of a tomato as a cube is an example ofanother general strategy for making order-of-magnitude estimates.A similar situation would occur if you were trying to estimate howmany m2 of leather could be produced from a herd of ten thousand


o / Consider a spherical cow.

cattle. There is no point in trying to take into account the shape ofthe cows’ bodies. A reasonable plan of attack might be to considera spherical cow. Probably a cow has roughly the same surface areaas a sphere with a radius of about 1 m, which would be 4π(1 m)2.Using the well-known facts that pi equals three, and four times threeequals about ten, we can guess that a cow has a surface area of about10 m2, so the herd as a whole might yield 105 m2 of leather.

The following list summarizes the strategies for getting a goodorder-of-magnitude estimate.

1. Don’t even attempt more than one significant figure of preci-sion.

2. Don’t guess area, volume, or mass directly. Guess linear di-mensions and get area, volume, or mass from them.

3. When dealing with areas or volumes of objects with complexshapes, idealize them as if they were some simpler shape, acube or a sphere, for example.

4. Check your final answer to see if it is reasonable. If you esti-mate that a herd of ten thousand cattle would yield 0.01 m2

of leather, then you have probably made a mistake with con-version factors somewhere.

Section 1.3 Order-of-magnitude estimates 49

SummaryNotation∝ . . . . . . . . . is proportional to∼ . . . . . . . . . on the order of, is on the order of

Summary

Nature behaves differently on large and small scales. Galileoshowed that this results fundamentally from the way area and vol-ume scale. Area scales as the second power of length, A ∝ L2, whilevolume scales as length to the third power, V ∝ L3.

An order of magnitude estimate is one in which we do not at-tempt or expect an exact answer. The main reason why the unini-tiated have trouble with order-of-magnitude estimates is that thehuman brain does not intuitively make accurate estimates of areaand volume. Estimates of area and volume should be approachedby first estimating linear dimensions, which one’s brain has a feelfor.


Problem 1.

ProblemsKey√



1 The one-liter cube in the photo has been marked off intosmaller cubes, with linear dimensions one tenth those of the bigone. What is the volume of each of the small cubes?

. Solution, p. 495

2 How many cm2 is 1 mm2? . Solution, p. 495

3 Compare the light-gathering powers of a 3-cm-diameter tele-scope and a 30-cm telescope. . Solution, p. 495

4 The traditional Martini glass is shaped like a cone with thepoint at the bottom. Suppose you make a Martini by pouring ver-mouth into the glass to a depth of 3 cm, and then adding gin to bringthe depth to 6 cm. What are the proportions of gin and vermouth?

. Solution, p. 495

5 How many cubic inches are there in a cubic foot? The answeris not 12.

√

6 Assume a dog’s brain is twice as great in diameter as a cat’s,but each animal’s brain cells are the same size and their brains arethe same shape. In addition to being a far better companion andmuch nicer to come home to, how many times more brain cells doesa dog have than a cat? The answer is not 2.

7 The population density of Los Angeles is about 4000 people/km2.That of San Francisco is about 6000 people/km2. How many timesfarther away is the average person’s nearest neighbor in LA than inSan Francisco? The answer is not 1.5.

√

8 One step on the Richter scale corresponds to a factor of 100in terms of the energy absorbed by something on the surface of theEarth, e.g., a house. For instance, a 9.3-magnitude quake wouldrelease 100 times more energy than an 8.3. The energy spreads outfrom the epicenter as a wave, and for the sake of this problem we’llassume we’re dealing with seismic waves that spread out in threedimensions, so that we can visualize them as hemispheres spreadingout under the surface of the earth. If a certain 7.6-magnitude earth-quake and a certain 5.6-magnitude earthquake produce the sameamount of vibration where I live, compare the distances from myhouse to the two epicenters. . Solution, p. 496

Problems 51

Problem 12.

9 The central portion of a CD is taken up by the hole and somesurrounding clear plastic, and this area is unavailable for storingdata. The radius of the central circle is about 35% of the outerradius of the data-storing area. What percentage of the CD’s areais therefore lost?

√

10 A taxon (plural taxa) is a group of living things. For ex-ample, Homo sapiens and Homo neanderthalensis are both taxa —specifically, they are two different species within the genus Homo.Surveys by botanists show that the number of plant taxa nativeto a given contiguous land area A is usually approximately propor-tional to A1/3. (a) There are 70 different species of lupine nativeto Southern California, which has an area of about 200, 000 km2.The San Gabriel Mountains cover about 1, 600 km2. Suppose thatyou wanted to learn to identify all the species of lupine in the SanGabriels. Approximately how many species would you have to fa-miliarize yourself with? . Answer, p. 513

√

(b) What is the interpretation of the fact that the exponent, 1/3, isless than one?

11 X-ray images aren’t only used with human subjects but also,for example, on insects and flowers. In 2003, a team of researchersat Argonne National Laboratory used x-ray imagery to find for thefirst time that insects, although they do not have lungs, do notnecessarily breathe completely passively, as had been believed pre-viously; many insects rapidly compress and expand their trachea,head, and thorax in order to force air in and out of their bodies.One difference between x-raying a human and an insect is that if amedical x-ray machine was used on an insect, virtually 100% of thex-rays would pass through its body, and there would be no contrastin the image produced. Less penetrating x-rays of lower energieshave to be used. For comparison, a typical human body mass isabout 70 kg, whereas a typical ant is about 10 mg. Estimate theratio of the thicknesses of tissue that must be penetrated by x-raysin one case compared to the other.

√

12 Radio was first commercialized around 1920, and ever sincethen, radio signals from our planet have been spreading out acrossour galaxy. It is possible that alien civilizations could detect thesesignals and learn that there is life on earth. In the 90 years that thesignals have been spreading at the speed of light, they have createda sphere with a radius of 90 light-years. To show an idea of thesize of this sphere, I’ve indicated it in the figure as a tiny whitecircle on an image of a spiral galaxy seen edge on. (We don’t havesimilar photos of our own Milky Way galaxy, because we can’t seeit from the outside.) So far we haven’t received answering signalsfrom aliens within this sphere, but as time goes on, the sphere willexpand as suggested by the dashed outline, reaching more and morestars that might harbor extraterrestrial life. Approximately whatyear will it be when the sphere has expanded to fill a volume 100times greater than the volume it fills today in 2010?

√


Problem 19.

13 The Earth’s surface is about 70% water. Mars’s diameter isabout half the Earth’s, but it has no surface water. Compare theland areas of the two planets.

√

14 In Europe, a piece of paper of the standard size, called A4,is a little narrower and taller than its American counterpart. Theratio of the height to the width is the square root of 2, and this hassome useful properties. For instance, if you cut an A4 sheet from leftto right, you get two smaller sheets that have the same proportions.You can even buy sheets of this smaller size, and they’re called A5.There is a whole series of sizes related in this way, all with the sameproportions. (a) Compare an A5 sheet to an A4 in terms of area andlinear size. (b) The series of paper sizes starts from an A0 sheet,which has an area of one square meter. Suppose we had a seriesof boxes defined in a similar way: the B0 box has a volume of onecubic meter, two B1 boxes fit exactly inside an B0 box, and so on.What would be the dimensions of a B0 box?

√

15 Estimate the volume of a human body, in cm3.

16 Estimate the number of blades of grass on a football field.

17 In a computer memory chip, each bit of information (a 0 ora 1) is stored in a single tiny circuit etched onto the surface of asilicon chip. The circuits cover the surface of the chip like lots in ahousing development. A typical chip stores 64 Mb (megabytes) ofdata, where a byte is 8 bits. Estimate (a) the area of each circuit,and (b) its linear size.

18 Suppose someone built a gigantic apartment building, mea-suring 10 km × 10 km at the base. Estimate how tall the buildingwould have to be to have space in it for the entire world’s populationto live.

19 (a) Using the microscope photo in the figure, estimate themass of a one cell of the E. coli bacterium, which is one of themost common ones in the human intestine. Note the scale at thelower right corner, which is 1 µm. Each of the tubular objects inthe column is one cell. (b) The feces in the human intestine aremostly bacteria (some dead, some alive), of which E. coli is a largeand typical component. Estimate the number of bacteria in yourintestines, and compare with the number of human cells in yourbody, which is believed to be roughly on the order of 1013. (c)Interpreting your result from part b, what does this tell you aboutthe size of a typical human cell compared to the size of a typicalbacterial cell?

Problems 53

Albert Einstein, and his mous-tache, problem 21.

20 A hamburger chain advertises that it has sold 10 billionBongo Burgers. Estimate the total mass of feed required to raisethe cows used to make the burgers.

21 Estimate the mass of one of the hairs in Albert Einstein’smoustache, in units of kg.

22 Estimate the number of man-hours required for building theGreat Wall of China. . Solution, p. 496

23 According to folklore, every time you take a breath, you areinhaling some of the atoms exhaled in Caesar’s last words. Is thistrue? If so, how many?


Exercise 1: Scaling applied to leavesEquipment:

leaves of three sizes, having roughly similar proportions of length, width, and thickness

balance

Each group will have one leaf, and should measure its surface area and volume, and determineits surface-to-volume ratio. For consistency, every group should use units of cm2 and cm3, andshould only find the area of one side of the leaf. The area can be found by tracing the area ofthe leaf on graph paper and counting squares. The volume can be found by weighing the leafand assuming that its density is 1 g/cm3 (the density of water). What implications do yourresults have for the plants’ abilities to survive in different environments?

Exercise 1: Scaling applied to leaves 55

Motion in one dimension

56

58

a / Rotation.

b / Simultaneous rotation andmotion through space.

c / One person might say that thetipping chair was only rotating ina circle about its point of contactwith the floor, but another coulddescribe it as having both rotationand motion through space.

Chapter 2

Velocity and relativemotion

2.1 Types of motionIf you had to think consciously in order to move your body, youwould be severely disabled. Even walking, which we consider tobe no great feat, requires an intricate series of motions that yourcerebrum would be utterly incapable of coordinating. The task ofputting one foot in front of the other is controlled by the more prim-itive parts of your brain, the ones that have not changed much sincethe mammals and reptiles went their separate evolutionary ways.The thinking part of your brain limits itself to general directivessuch as “walk faster,” or “don’t step on her toes,” rather than mi-cromanaging every contraction and relaxation of the hundred or somuscles of your hips, legs, and feet.

Physics is all about the conscious understanding of motion, butwe’re obviously not immediately prepared to understand the mostcomplicated types of motion. Instead, we’ll use the divide-and-conquer technique. We’ll first classify the various types of motion,and then begin our campaign with an attack on the simplest cases.To make it clear what we are and are not ready to consider, we needto examine and define carefully what types of motion can exist.

Rigid-body motion distinguished from motion that changesan object’s shape

Nobody, with the possible exception of Fred Astaire, can simplyglide forward without bending their joints. Walking is thus an ex-ample in which there is both a general motion of the whole objectand a change in the shape of the object. Another example is themotion of a jiggling water balloon as it flies through the air. We arenot presently attempting a mathematical description of the way inwhich the shape of an object changes. Motion without a change inshape is called rigid-body motion. (The word “body” is often usedin physics as a synonym for “object.”)

Center-of-mass motion as opposed to rotation

A ballerina leaps into the air and spins around once before land-ing. We feel intuitively that her rigid-body motion while her feetare off the ground consists of two kinds of motion going on simul-

59

e / No matter what point youhang the pear from, the stringlines up with the pear’s centerof mass. The center of masscan therefore be defined as theintersection of all the lines madeby hanging the pear in this way.Note that the X in the figureshould not be interpreted asimplying that the center of massis on the surface — it is actuallyinside the pear.

f / The circus performers hangwith the ropes passing throughtheir centers of mass.

taneously: a rotation and a motion of her body as a whole throughspace, along an arc. It is not immediately obvious, however, whatis the most useful way to define the distinction between rotationand motion through space. Imagine that you attempt to balance achair and it falls over. One person might say that the only motionwas a rotation about the chair’s point of contact with the floor, butanother might say that there was both rotation and motion downand to the side.

d / The leaping dancer’s motion is complicated, but the motion ofher center of mass is simple.

It turns out that there is one particularly natural and useful wayto make a clear definition, but it requires a brief digression. Everyobject has a balance point, referred to in physics as the center ofmass. For a two-dimensional object such as a cardboard cutout, thecenter of mass is the point at which you could hang the object froma string and make it balance. In the case of the ballerina (who islikely to be three-dimensional unless her diet is particularly severe),it might be a point either inside or outside her body, dependingon how she holds her arms. Even if it is not practical to attach astring to the balance point itself, the center of mass can be definedas shown in figure e.

Why is the center of mass concept relevant to the question ofclassifying rotational motion as opposed to motion through space?As illustrated in figures d and g, it turns out that the motion of anobject’s center of mass is nearly always far simpler than the motionof any other part of the object. The ballerina’s body is a large objectwith a complex shape. We might expect that her motion would bemuch more complicated than the motion of a small, simply-shaped

60 Chapter 2 Velocity and relative motion

h / An improperly balancedwheel has a center of mass thatis not at its geometric center.When you get a new tire, themechanic clamps little weights tothe rim to balance the wheel.

i / This toy was intentionallydesigned so that the mushroom-shaped piece of metal on topwould throw off the center ofmass. When you wind it up, themushroom spins, but the centerof mass doesn’t want to move,so the rest of the toy tends tocounter the mushroom’s motion,causing the whole thing to jumparound.

object, say a marble, thrown up at the same angle as the angle atwhich she leapt. But it turns out that the motion of the ballerina’scenter of mass is exactly the same as the motion of the marble. Thatis, the motion of the center of mass is the same as the motion theballerina would have if all her mass was concentrated at a point. Byrestricting our attention to the motion of the center of mass, we cantherefore simplify things greatly.

g / The same leaping dancer, viewed from above. Her center ofmass traces a straight line, but a point away from her center of mass,such as her elbow, traces the much more complicated path shown by thedots.

We can now replace the ambiguous idea of “motion as a wholethrough space” with the more useful and better defined conceptof “center-of-mass motion.” The motion of any rigid body can becleanly split into rotation and center-of-mass motion. By this defini-tion, the tipping chair does have both rotational and center-of-massmotion. Concentrating on the center of mass motion allows us tomake a simplified model of the motion, as if a complicated objectlike a human body was just a marble or a point-like particle. Sciencereally never deals with reality; it deals with models of reality.

Note that the word “center” in “center of mass” is not meantto imply that the center of mass must lie at the geometrical centerof an object. A car wheel that has not been balanced properly hasa center of mass that does not coincide with its geometrical center.An object such as the human body does not even have an obviousgeometrical center.

It can be helpful to think of the center of mass as the averagelocation of all the mass in the object. With this interpretation,we can see for example that raising your arms above your headraises your center of mass, since the higher position of the arms’mass raises the average. We won’t be concerned right now withcalculating centers of mass mathematically; the relevant equationsare in chapter 4 of Conservation Laws.

Ballerinas and professional basketball players can create an illu-sion of flying horizontally through the air because our brains intu-itively expect them to have rigid-body motion, but the body doesnot stay rigid while executing a grand jete or a slam dunk. The legsare low at the beginning and end of the jump, but come up higher at

Section 2.1 Types of motion 61

j / A fixed point on the dancer’s body follows a trajectory that is flat-ter than what we expect, creating an illusion of flight.

the middle. Regardless of what the limbs do, the center of mass willfollow the same arc, but the low position of the legs at the beginningand end means that the torso is higher compared to the center ofmass, while in the middle of the jump it is lower compared to thecenter of mass. Our eye follows the motion of the torso and triesto interpret it as the center-of-mass motion of a rigid body. Butsince the torso follows a path that is flatter than we expect, thisattempted interpretation fails, and we experience an illusion thatthe person is flying horizontally.

k / Example 1.

The center of mass as an average example 1. Explain how we know that the center of mass of each object isat the location shown in figure k.

. The center of mass is a sort of average, so the height of thecenters of mass in 1 and 2 has to be midway between the twosquares, because that height is the average of the heights of thetwo squares. Example 3 is a combination of examples 1 and2, so we can find its center of mass by averaging the horizontalpositions of their centers of mass. In example 4, each squarehas been skewed a little, but just as much mass has been movedup as down, so the average vertical position of the mass hasn’tchanged. Example 5 is clearly not all that different from example4, the main difference being a slight clockwise rotation, so just as


l / The high-jumper’s bodypasses over the bar, but hiscenter of mass passes under it.

m / Self-check B.

in example 4, the center of mass must be hanging in empty space,where there isn’t actually any mass. Horizontally, the center ofmass must be between the heels and toes, or else it wouldn’t bepossible to stand without tipping over.

Another interesting example from the sports world is the highjump, in which the jumper’s curved body passes over the bar, butthe center of mass passes under the bar! Here the jumper lowers hislegs and upper body at the peak of the jump in order to bring hiswaist higher compared to the center of mass.

Later in this course, we’ll find that there are more fundamentalreasons (based on Newton’s laws of motion) why the center of massbehaves in such a simple way compared to the other parts of anobject. We’re also postponing any discussion of numerical methodsfor finding an object’s center of mass. Until later in the course, wewill only deal with the motion of objects’ centers of mass.

Center-of-mass motion in one dimension

In addition to restricting our study of motion to center-of-massmotion, we will begin by considering only cases in which the centerof mass moves along a straight line. This will include cases suchas objects falling straight down, or a car that speeds up and slowsdown but does not turn.

Note that even though we are not explicitly studying the morecomplex aspects of motion, we can still analyze the center-of-massmotion while ignoring other types of motion that might be occurringsimultaneously . For instance, if a cat is falling out of a tree andis initially upside-down, it goes through a series of contortions thatbring its feet under it. This is definitely not an example of rigid-bodymotion, but we can still analyze the motion of the cat’s center ofmass just as we would for a dropping rock.

self-check AConsider a person running, a person pedaling on a bicycle, a personcoasting on a bicycle, and a person coasting on ice skates. In whichcases is the center-of-mass motion one-dimensional? Which cases areexamples of rigid-body motion? . Answer, p. 509

self-check BThe figure shows a gymnast holding onto the inside of a big wheel.From inside the wheel, how could he make it roll one way or the other?. Answer, p. 509

2.2 Describing distance and timeCenter-of-mass motion in one dimension is particularly easy to dealwith because all the information about it can be encapsulated in twovariables: x, the position of the center of mass relative to the origin,and t, which measures a point in time. For instance, if someone

Section 2.2 Describing distance and time 63

supplied you with a sufficiently detailed table of x and t values, youwould know pretty much all there was to know about the motion ofthe object’s center of mass.

A point in time as opposed to duration

In ordinary speech, we use the word “time” in two differentsenses, which are to be distinguished in physics. It can be used,as in “a short time” or “our time here on earth,” to mean a lengthor duration of time, or it can be used to indicate a clock reading, asin “I didn’t know what time it was,” or “now’s the time.” In sym-bols, t is ordinarily used to mean a point in time, while ∆t signifiesan interval or duration in time. The capital Greek letter delta, ∆,means “the change in...,” i.e. a duration in time is the change ordifference between one clock reading and another. The notation ∆tdoes not signify the product of two numbers, ∆ and t, but ratherone single number, ∆t. If a matinee begins at a point in time t = 1o’clock and ends at t = 3 o’clock, the duration of the movie was thechange in t,

∆t = 3 hours− 1 hour = 2 hours .

To avoid the use of negative numbers for ∆t, we write the clockreading “after” to the left of the minus sign, and the clock reading“before” to the right of the minus sign. A more specific definitionof the delta notation is therefore that delta stands for “after minusbefore.”

Even though our definition of the delta notation guarantees that∆t is positive, there is no reason why t can’t be negative. If tcould not be negative, what would have happened one second beforet = 0? That doesn’t mean that time “goes backward” in the sensethat adults can shrink into infants and retreat into the womb. Itjust means that we have to pick a reference point and call it t = 0,and then times before that are represented by negative values of t.An example is that a year like 2007 A.D. can be thought of as apositive t value, while one like 370 B.C. is negative. Similarly, whenyou hear a countdown for a rocket launch, the phrase “t minus tenseconds” is a way of saying t = −10 s, where t = 0 is the time ofblastoff, and t > 0 refers to times after launch.

Although a point in time can be thought of as a clock reading, itis usually a good idea to avoid doing computations with expressionssuch as “2:35” that are combinations of hours and minutes. Timescan instead be expressed entirely in terms of a single unit, such ashours. Fractions of an hour can be represented by decimals ratherthan minutes, and similarly if a problem is being worked in termsof minutes, decimals can be used instead of seconds.

self-check COf the following phrases, which refer to points in time, which refer totime intervals, and which refer to time in the abstract rather than as ameasurable number?


(1) “The time has come.”

(2) “Time waits for no man.”

(3) “The whole time, he had spit on his chin.” . Answer, p. 509

The Leibniz notation and infinitesimals

∆ is the Greek version of “D,” suggesting that there is a rela-tionship between ∆t and the notation dt from calculus. The “d”notation was invented by Leibniz around 1675 to suggest the word“difference.” The idea was that a dt would be like a ∆t that wasextremely small — smaller than any real number, and yet greaterthan zero. These infinitestimal numbers were the way the world’sgreatest mathematicians thought about calculus for the next twohundred years. For example, dy/dx meant the number you got whenyou divided dy by dx. The use of infinitesimal numbers was seenas a natural part of the process of generalization that had alreadyseen the invention of fractions and irrational numbers by the ancientGreeks, zero and negative numbers in India, and complex numbersin Renaissance Italy. By the end of the 19th century, mathemati-cians had begun making formal mathematical descriptions of num-ber systems, and they had succeeded in making nice tidy schemesout of all of these categories except for infinitesimals. Having runinto a brick wall, they decided to rebuild calculus using the notionof a limit. Depending on when and where you got your education incalculus, you may have been warned severely that dy and dx werenot numbers, and that dy/dx didn’t mean dividing one by another.

But in the 1960’s, the logician Abraham Robinson at Yale provedthat infinitesimals could be tamed and domesticated; they were nomore self-contradictory than negative numbers or fractions. Thereis a handy rule for making sure that you don’t come to incorrectconclusions by using infinitesimals. The rule is that you can applyany axiom of the real number system to infinitesimals, and the resultwill be correct, provided that the statement can be put in the formlike “for any number . . . ,” but not “for any set of numbers . . . ” Wecarry over the axiom, reinterpreting “number” to mean any memberof the enriched number system that includes both the real numbersand the infinitesimals.

Logic and infinitesimals example 2There is an axiom of the real number system that for any numbert , t + 0 = t . This applies to infinitesimals as well, so that dt + 0 =dt .

Position as opposed to change in position

As with time, a distinction should be made between a pointin space, symbolized as a coordinate x, and a change in position,symbolized as ∆x.

As with t,x can be negative. If a train is moving down the

Section 2.2 Describing distance and time 65

tracks, not only do you have the freedom to choose any point alongthe tracks and call it x = 0, but it’s also up to you to decide whichside of the x = 0 point is positive x and which side is negative x.

Since we’ve defined the delta notation to mean “after minusbefore,” it is possible that ∆x will be negative, unlike ∆t which isguaranteed to be positive. Suppose we are describing the motionof a train on tracks linking Tucson and Chicago. As shown in thefigure, it is entirely up to you to decide which way is positive.

n / Two equally valid ways of de-scribing the motion of a train fromTucson to Chicago. In example 1,the train has a positive ∆x as itgoes from Enid to Joplin. In 2,the same train going forward inthe same direction has a negative∆x .

Note that in addition to x and ∆x, there is a third quantity wecould define, which would be like an odometer reading, or actualdistance traveled. If you drive 10 miles, make a U-turn, and driveback 10 miles, then your ∆x is zero, but your car’s odometer readinghas increased by 20 miles. However important the odometer readingis to car owners and used car dealers, it is not very important inphysics, and there is not even a standard name or notation for it.The change in position, ∆x, is more useful because it is so mucheasier to calculate: to compute ∆x, we only need to know the be-ginning and ending positions of the object, not all the informationabout how it got from one position to the other.

self-check DA ball falls vertically, hits the floor, bounces to a height of one meter,falls, and hits the floor again. Is the ∆x between the two impacts equalto zero, one, or two meters? . Answer, p. 509

Frames of reference

The example above shows that there are two arbitrary choicesyou have to make in order to define a position variable, x. You haveto decide where to put x = 0, and also which direction will be posi-tive. This is referred to as choosing a coordinate system or choosinga frame of reference. (The two terms are nearly synonymous, butthe first focuses more on the actual x variable, while the second ismore of a general way of referring to one’s point of view.) As long asyou are consistent, any frame is equally valid. You just don’t wantto change coordinate systems in the middle of a calculation.


o / Motion with constant ve-locity.

p / Motion that decreases xis represented with negativevalues of ∆x and v .

q / Motion with changing ve-locity.

Have you ever been sitting in a train in a station when suddenlyyou notice that the station is moving backward? Most people woulddescribe the situation by saying that you just failed to notice thatthe train was moving — it only seemed like the station was moving.But this shows that there is yet a third arbitrary choice that goesinto choosing a coordinate system: valid frames of reference candiffer from each other by moving relative to one another. It mightseem strange that anyone would bother with a coordinate systemthat was moving relative to the earth, but for instance the frame ofreference moving along with a train might be far more convenientfor describing things happening inside the train.

2.3 Graphs of motion; velocity

Motion with constant velocity

In example o, an object is moving at constant speed in one di-rection. We can tell this because every two seconds, its positionchanges by five meters.

In algebra notation, we’d say that the graph of x vs. t showsthe same change in position, ∆x = 5.0 m, over each interval of∆t = 2.0 s. The object’s velocity or speed is obtained by calculatingv = ∆x/∆t = (5.0 m)/(2.0 s) = 2.5 m/s. In graphical terms, thevelocity can be interpreted as the slope of the line. Since the graphis a straight line, it wouldn’t have mattered if we’d taken a longertime interval and calculated v = ∆x/∆t = (10.0 m)/(4.0 s). Theanswer would still have been the same, 2.5 m/s.

Note that when we divide a number that has units of meters byanother number that has units of seconds, we get units of metersper second, which can be written m/s. This is another case wherewe treat units as if they were algebra symbols, even though they’renot.

In example p, the object is moving in the opposite direction: astime progresses, its x coordinate decreases. Recalling the definitionof the ∆ notation as “after minus before,” we find that ∆t is stillpositive, but ∆x must be negative. The slope of the line is thereforenegative, and we say that the object has a negative velocity, v =∆x/∆t = (−5.0 m)/(2.0 s) = −2.5 m/s. We’ve already seen thatthe plus and minus signs of ∆x values have the interpretation oftelling us which direction the object moved. Since ∆t is alwayspositive, dividing by ∆t doesn’t change the plus or minus sign, andthe plus and minus signs of velocities are to be interpreted in thesame way. In graphical terms, a positive slope characterizes a linethat goes up as we go to the right, and a negative slope tells us thatthe line went down as we went to the right.

Section 2.3 Graphs of motion; velocity 67

Motion with changing velocity

Now what about a graph like figure q? This might be a graphof a car’s motion as the driver cruises down the freeway, then slowsdown to look at a car crash by the side of the road, and then speedsup again, disappointed that there is nothing dramatic going on suchas flames or babies trapped in their car seats. (Note that we arestill talking about one-dimensional motion. Just because the graphis curvy doesn’t mean that the car’s path is curvy. The graph is notlike a map, and the horizontal direction of the graph represents thepassing of time, not distance.)

If we apply the equation v = ∆x/∆t to this example, we willget the wrong answer, because the ∆x/∆t gives a single number,but the velocity is clearly changing. This is an example of a goodgeneral rule that tells you when you need to use your differentialcalculus. Any time a rate of change is measured by an expression ofthe form ∆ . . . /∆ . . ., the result will only be right when the rate ofchange is constant. When the rate of change is varying, we need togeneralize the expression by making it into a derivative. Just as aninfinitesimally small1 ∆t is notated dt, an infinitesimally small ∆xis a dx. The velocity is then the derivative dx/dt.

Units of velocity example 3. Verify that the units of v = dx/dt make sense.

. We expect the velocity to have units of meters per second, andit does come out to have those units, since dx has units of me-ters and dt seconds. This ability to check the units of derivativesis one of the main reason that Leibniz designed his notation forderivatives the way he did.

An insect pest example 4. An insect pest from the United States is inadvertently releasedin a village in rural China. The pests spread outward at a rateof s kilometers per year, forming a widening circle of contagion.Find the number of square kilometers per year that become newlyinfested. Check that the units of the result make sense. Interpretthe result.

. Let t be the time, in years, since the pest was introduced. Theradius of the circle is r = st , and its area is a = πr2 = π(st)2. Thederivative is

dadt

= (2πs2)t

The units of s are km/year, so squaring it gives km2/year2. The 2and the π are unitless, and multiplying by t gives units of km2/year,which is what we expect for da/dt , since it represents the numberof square kilometers per year that become infested.

1see p. 65


Interpreting the result, we notice a couple of things. First, the rateof infestation isn’t constant; it’s proportional to t , so people mightnot pay so much attention at first, but later on the effort requiredto combat the problem will grow more and more quickly. Second,we notice that the result is proportional to s2. This suggests thatanything that could be done to reduce s would be very helpful.For instance, a measure that cut s in half would reduce da/dt bya factor of four.

2.4 The principle of inertiaPhysical effects relate only to a change in velocity

Consider two statements of a kind that was at one time madewith the utmost seriousness:

People like Galileo and Copernicus who say the earth is ro-tating must be crazy. We know the earth can’t be moving.Why, if the earth was really turning once every day, then ourwhole city would have to be moving hundreds of leagues inan hour. That’s impossible! Buildings would shake on theirfoundations. Gale-force winds would knock us over. Treeswould fall down. The Mediterranean would come sweepingacross the east coasts of Spain and Italy. And furthermore,what force would be making the world turn?

All this talk of passenger trains moving at forty miles an houris sheer hogwash! At that speed, the air in a passenger com-partment would all be forced against the back wall. People inthe front of the car would suffocate, and people at the backwould die because in such concentrated air, they wouldn’t beable to expel a breath.

Some of the effects predicted in the first quote are clearly justbased on a lack of experience with rapid motion that is smooth andfree of vibration. But there is a deeper principle involved. In eachcase, the speaker is assuming that the mere fact of motion musthave dramatic physical effects. More subtly, they also believe that aforce is needed to keep an object in motion: the first person thinksa force would be needed to maintain the earth’s rotation, and thesecond apparently thinks of the rear wall as pushing on the air tokeep it moving.

Common modern knowledge and experience tell us that thesepeople’s predictions must have somehow been based on incorrectreasoning, but it is not immediately obvious where the fundamentalflaw lies. It’s one of those things a four-year-old could infuriate youby demanding a clear explanation of. One way of getting at the fun-damental principle involved is to consider how the modern conceptof the universe differs from the popular conception at the time ofthe Italian Renaissance. To us, the word “earth” implies a planet,

Section 2.4 The principle of inertia 69

s / Why does Aristotle lookso sad? Has he realized thathis entire system of physics iswrong?

t / The earth spins. Peoplein Shanghai say they’re at restand people in Los Angeles aremoving. Angelenos say the sameabout the Shanghainese.

u / The jets are at rest. TheEmpire State Building is moving.

one of the nine planets of our solar system, a small ball of rock anddirt that is of no significance to anyone in the universe except formembers of our species, who happen to live on it. To Galileo’s con-temporaries, however, the earth was the biggest, most solid, mostimportant thing in all of creation, not to be compared with the wan-dering lights in the sky known as planets. To us, the earth is justanother object, and when we talk loosely about “how fast” an ob-ject such as a car “is going,” we really mean the car-object’s velocityrelative to the earth-object.

r / This Air Force doctor volunteered to ride a rocket sled as a medicalexperiment. The obvious effects on his head and face are not because ofthe sled’s speed but because of its rapid changes in speed: increasingin 2 and 3, and decreasing in 5 and 6. In 4 his speed is greatest, butbecause his speed is not increasing or decreasing very much at thismoment, there is little effect on him.

Motion is relative

According to our modern world-view, it really isn’t that reason-able to expect that a special force should be required to make theair in the train have a certain velocity relative to our planet. Afterall, the “moving” air in the “moving” train might just happen tohave zero velocity relative to some other planet we don’t even knowabout. Aristotle claimed that things “naturally” wanted to be atrest, lying on the surface of the earth. But experiment after exper-iment has shown that there is really nothing so special about beingat rest relative to the earth. For instance, if a mattress falls out of


Discussion question A.

Discussion question B.

Discussion question D.

the back of a truck on the freeway, the reason it rapidly comes torest with respect to the planet is simply because of friction forcesexerted by the asphalt, which happens to be attached to the planet.

Galileo’s insights are summarized as follows:

The principle of inertiaNo force is required to maintain motion with constant velocity ina straight line, and absolute motion does not cause any observablephysical effects.

There are many examples of situations that seem to disprove theprinciple of inertia, but these all result from forgetting that frictionis a force. For instance, it seems that a force is needed to keep asailboat in motion. If the wind stops, the sailboat stops too. Butthe wind’s force is not the only force on the boat; there is also africtional force from the water. If the sailboat is cruising and thewind suddenly disappears, the backward frictional force still exists,and since it is no longer being counteracted by the wind’s forwardforce, the boat stops. To disprove the principle of inertia, we wouldhave to find an example where a moving object slowed down eventhough no forces whatsoever were acting on it.

self-check EWhat is incorrect about the following supposed counterexamples to theprinciple of inertia?

(1) When astronauts blast off in a rocket, their huge velocity does causea physical effect on their bodies — they get pressed back into theirseats, the flesh on their faces gets distorted, and they have a hard timelifting their arms.

(2) When you’re driving in a convertible with the top down, the wind inyour face is an observable physical effect of your absolute motion. .

Answer, p. 509

. Solved problem: a bug on a wheel page 93, problem 13


A A passenger on a cruise ship finds, while the ship is docked, thathe can leap off of the upper deck and just barely make it into the poolon the lower deck. If the ship leaves dock and is cruising rapidly, will thisadrenaline junkie still be able to make it?

B You are a passenger in the open basket hanging under a heliumballoon. The balloon is being carried along by the wind at a constantvelocity. If you are holding a flag in your hand, will the flag wave? If so,which way? [Based on a question from PSSC Physics.]

C Aristotle stated that all objects naturally wanted to come to rest, withthe unspoken implication that “rest” would be interpreted relative to thesurface of the earth. Suppose we go back in time and transport Aristotleto the moon. Aristotle knew, as we do, that the moon circles the earth; he

Section 2.4 The principle of inertia 71

said it didn’t fall down because, like everything else in the heavens, it wasmade out of some special substance whose “natural” behavior was to goin circles around the earth. We land, put him in a space suit, and kickhim out the door. What would he expect his fate to be in this situation? Ifintelligent creatures inhabited the moon, and one of them independentlycame up with the equivalent of Aristotelian physics, what would they thinkabout objects coming to rest?

D The glass is sitting on a level table in a train’s dining car, but thesurface of the water is tilted. What can you infer about the motion of thetrain?

2.5 Addition of velocitiesAddition of velocities to describe relative motion

Since absolute motion cannot be unambiguously measured, theonly way to describe motion unambiguously is to describe the motionof one object relative to another. Symbolically, we can write vPQfor the velocity of object P relative to object Q.

Velocities measured with respect to different reference points canbe compared by addition. In the figure below, the ball’s velocityrelative to the couch equals the ball’s velocity relative to the truckplus the truck’s velocity relative to the couch:

vBC = vBT + vTC

= 5 cm/s + 10 cm/s

= 15 cm/s

The same equation can be used for any combination of threeobjects, just by substituting the relevant subscripts for B, T, andC. Just remember to write the equation so that the velocities beingadded have the same subscript twice in a row. In this example, ifyou read off the subscripts going from left to right, you get BC . . . =. . .BTTC. The fact that the two “inside” subscripts on the right arethe same means that the equation has been set up correctly. Noticehow subscripts on the left look just like the subscripts on the right,but with the two T’s eliminated.

Negative velocities in relative motion

My discussion of how to interpret positive and negative signs ofvelocity may have left you wondering why we should bother. Whynot just make velocity positive by definition? The original reasonwhy negative numbers were invented was that bookkeepers decidedit would be convenient to use the negative number concept for pay-ments to distinguish them from receipts. It was just plain easier thanwriting receipts in black and payments in red ink. After adding upyour month’s positive receipts and negative payments, you either got


v / These two highly competent physicists disagree on absolute ve-locities, but they would agree on relative velocities. Purple Dinoconsiders the couch to be at rest, while Green Dino thinks of the truck asbeing at rest. They agree, however, that the truck’s velocity relative to thecouch is vTC = 10 cm/s, the ball’s velocity relative to the truck is vBT = 5cm/s, and the ball’s velocity relative to the couch is vBC = vBT + vTC = 15cm/s.

a positive number, indicating profit, or a negative number, showinga loss. You could then show that total with a high-tech “+” or “−”sign, instead of looking around for the appropriate bottle of ink.

Nowadays we use positive and negative numbers for all kindsof things, but in every case the point is that it makes sense toadd and subtract those things according to the rules you learnedin grade school, such as “minus a minus makes a plus, why this istrue we need not discuss.” Adding velocities has the significanceof comparing relative motion, and with this interpretation negativeand positive velocities can be used within a consistent framework.For example, the truck’s velocity relative to the couch equals thetruck’s velocity relative to the ball plus the ball’s velocity relativeto the couch:

vTC = vTB + vBC

= −5 cm/s + 15 cm/s

= 10 cm/s

If we didn’t have the technology of negative numbers, we would havehad to remember a complicated set of rules for adding velocities: (1)if the two objects are both moving forward, you add, (2) if one ismoving forward and one is moving backward, you subtract, but (3)if they’re both moving backward, you add. What a pain that would

Section 2.5 Addition of velocities 73

w / Graphs of x and v ver-sus t for a car accelerating awayfrom a traffic light, and thenstopping for another red light.

have been.

. Solved problem: two dimensions page 93, problem 10


A Interpret the general rule vAB = −vBA in words.

B Wa-Chuen slips away from her father at the mall and walks up thedown escalator, so that she stays in one place. Write this in terms ofsymbols.

2.6 Graphs of velocity versus timeSince changes in velocity play such a prominent role in physics, weneed a better way to look at changes in velocity than by laboriouslydrawing tangent lines on x-versus-t graphs. A good method is todraw a graph of velocity versus time. The examples on the left showthe x− t and v− t graphs that might be produced by a car startingfrom a traffic light, speeding up, cruising for a while at constantspeed, and finally slowing down for a stop sign. If you have an airfreshener hanging from your rear-view mirror, then you will see aneffect on the air freshener during the beginning and ending periodswhen the velocity is changing, but it will not be tilted during theperiod of constant velocity represented by the flat plateau in themiddle of the v − t graph.

Students often mix up the things being represented on these twotypes of graphs. For instance, many students looking at the topgraph say that the car is speeding up the whole time, since “thegraph is becoming greater.” What is getting greater throughout thegraph is x, not v.

Similarly, many students would look at the bottom graph andthink it showed the car backing up, because “it’s going backwardsat the end.” But what is decreasing at the end is v, not x. Havingboth the x − t and v − t graphs in front of you like this is oftenconvenient, because one graph may be easier to interpret than theother for a particular purpose. Stacking them like this means thatcorresponding points on the two graphs’ time axes are lined up witheach other vertically. However, one thing that is a little counter-intuitive about the arrangement is that in a situation like this oneinvolving a car, one is tempted to visualize the landscape stretchingalong the horizontal axis of one of the graphs. The horizontal axes,however, represent time, not position. The correct way to visualizethe landscape is by mentally rotating the horizon 90 degrees coun-terclockwise and imagining it stretching along the upright axis of thex-t graph, which is the only axis that represents different positionsin space.


x / This Global PositioningSystem (GPS) system, runningon a smartphone attached to abike’s handlebar, depends onEinstein’s theory of relativity.Time flows at a different ratesaboard a GPS satellite than itdoes on the bike, and the GPSsoftware has to take this intoaccount.

y / The atomic clock has itsown ticket and its own seat.

z / Newton’s laws do not dis-tinguish past from future. Thefootball could travel in eitherdirection while obeying Newton’slaws.

2.7 ? RelativityTime is not absolute

So far we’ve been discussing relativity according to Galileo andNewton, but there is also relativity according to Einstein. WhenEinstein first began to develop the theory of relativity, around 1905,the only real-world observations he could draw on were ambiguousand indirect. Today, the evidence is part of everyday life. For ex-ample, every time you use a GPS receiver, x, you’re using Einstein’stheory of relativity. Somewhere between 1905 and today, technologybecame good enough to allow conceptually simple experiments thatstudents in the early 20th century could only discuss in terms like“Imagine that we could. . . ” A good jumping-on point is 1971. Inthat year, J.C. Hafele and R.E. Keating of the U.S. Naval Observa-tory brought atomic clocks aboard commercial airliners, y, and wentaround the world, once from east to west and once from west to east.(The clocks had their own tickets, and occupied their own seats.)Hafele and Keating observed that there was a discrepancy betweenthe times measured by the traveling clocks and the times measuredby similar clocks that stayed at the lab in Washington. The east-going clock lost time, ending up off by −59± 10 nanoseconds, whilethe west-going one gained 273± 7 ns.

The correspondence principle

This establishes that time doesn’t work the way Newton believedit did when he wrote that “Absolute, true, and mathematical time,of itself, and from its own nature flows equably without regard toanything external. . . ” We are used to thinking of time as absoluteand universal, so it is disturbing to find that it can flow at a differentrate for observers in different frames of reference. Nevertheless, theeffects that Hafele and Keating observed were small. This makessense: Newton’s laws have already been thoroughly tested by ex-periments under a wide variety of conditions, so a new theory likerelativity must agree with Newton’s to a good approximation, withinthe Newtonian theory’s realm of applicability. This requirement ofbackward-compatibility is known as the correspondence principle.

Causality

It’s also reassuring that the effects on time were small comparedto the three-day lengths of the plane trips. There was thereforeno opportunity for paradoxical scenarios such as one in which theeast-going experimenter arrived back in Washington before he leftand then convinced himself not to take the trip. A theory thatmaintains this kind of orderly relationship between cause and effectis said to satisfy causality.

Causality is like a water-hungry front-yard lawn in Los Angeles:we know we want it, but it’s not easy to explain why. Even in plain

Section 2.7 ? Relativity 75

aa / All three clocks are movingto the east. Even though thewest-going plane is moving to thewest relative to the air, the airis moving to the east due to theearth’s rotation.

old Newtonian physics, there is no clear distinction between pastand future. In figure z, number 18 throws the football to number25, and the ball obeys Newton’s laws of motion. If we took a videoof the pass and played it backward, we would see the ball flying from25 to 18, and Newton’s laws would still be satisfied. Nevertheless,we have a strong psychological impression that there is a forwardarrow of time. I can remember what the stock market did last year,but I can’t remember what it will do next year. Joan of Arc’s mil-itary victories against England caused the English to burn her atthe stake; it’s hard to accept that Newton’s laws provide an equallygood description of a process in which her execution in 1431 causedher to win a battle in 1429. There is no consensus at this pointamong physicists on the origin and significance of time’s arrow, andfor our present purposes we don’t need to solve this mystery. In-stead, we merely note the empirical fact that, regardless of whatcausality really means and where it really comes from, its behavioris consistent. Specifically, experiments show that if an observer in acertain frame of reference observes that event A causes event B, thenobservers in other frames agree that A causes B, not the other wayaround. This is merely a generalization about a large body of ex-perimental results, not a logically necessary assumption. If Keatinghad gone around the world and arrived back in Washington beforehe left, it would have disproved this statement about causality.

Time distortion arising from motion and gravity

Hafele and Keating were testing specific quantitative predictionsof relativity, and they verified them to within their experiment’serror bars. Let’s work backward instead, and inspect the empiricalresults for clues as to how time works.

The two traveling clocks experienced effects in opposite direc-tions, and this suggests that the rate at which time flows dependson the motion of the observer. The east-going clock was moving inthe same direction as the earth’s rotation, so its velocity relative tothe earth’s center was greater than that of the clock that remainedin Washington, while the west-going clock’s velocity was correspond-ingly reduced. The fact that the east-going clock fell behind, andthe west-going one got ahead, shows that the effect of motion is tomake time go more slowly. This effect of motion on time was pre-dicted by Einstein in his original 1905 paper on relativity, writtenwhen he was 26.

If this had been the only effect in the Hafele-Keating experiment,then we would have expected to see effects on the two flying clocksthat were equal in size. Making up some simple numbers to keep thearithmetic transparent, suppose that the earth rotates from west toeast at 1000 km/hr, and that the planes fly at 300 km/hr. Thenthe speed of the clock on the ground is 1000 km/hr, the speed ofthe clock on the east-going plane is 1300 km/hr, and that of the


ab / The correspondence princi-ple requires that the relativisticdistortion of time become smallfor small velocities.

west-going clock 700 km/hr. Since the speeds of 700, 1000, and1300 km/hr have equal spacing on either side of 1000, we wouldexpect the discrepancies of the moving clocks relative to the one inthe lab to be equal in size but opposite in sign.

In fact, the two effects are unequal in size: −59 ns and 273 ns.This implies that there is a second effect involved, simply due to theplanes’ being up in the air. Relativity predicts that time’s rate offlow also changes with height in a gravitational field. Einstein didn’tfigure out how to incorporate gravity into relativity until 1915, aftermuch frustration and many false starts. The simpler version of thetheory without gravity is known as special relativity, the full versionas general relativity. We’ll restrict ourselves to special relativity inthis book, and that means that what we want to focus on right nowis the distortion of time due to motion, not gravity.

We can now see in more detail how to apply the correspondenceprinciple. The behavior of the three clocks in the Hafele-Keatingexperiment shows that the amount of time distortion increases asthe speed of the clock’s motion increases. Newton lived in an erawhen the fastest mode of transportation was a galloping horse, andthe best pendulum clocks would accumulate errors of perhaps aminute over the course of several days. A horse is much slowerthan a jet plane, so the distortion of time would have had a relativesize of only ∼ 10−15 — much smaller than the clocks were capableof detecting. At the speed of a passenger jet, the effect is about10−12, and state-of-the-art atomic clocks in 1971 were capable ofmeasuring that. A GPS satellite travels much faster than a jet air-plane, and the effect on the satellite turns out to be ∼ 10−10. Thegeneral idea here is that all physical laws are approximations, andapproximations aren’t simply right or wrong in different situations.Approximations are better or worse in different situations, and thequestion is whether a particular approximation is good enough in agiven situation to serve a particular purpose. The faster the motion,the worse the Newtonian approximation of absolute time. Whetherthe approximation is good enough depends on what you’re tryingto accomplish. The correspondence principle says that the approxi-mation must have been good enough to explain all the experimentsdone in the centuries before Einstein came up with relativity.

By the way, don’t get an inflated idea of the importance of theHafele-Keating experiment. Special relativity had already been con-firmed by a vast and varied body of experiments decades before 1971.The only reason I’m giving such a prominent role to this experiment,which was actually more important as a test of general relativity, isthat it is conceptually very direct.


ac / Two events are given aspoints on a graph of positionversus time. Joan of Arc helps torestore Charles VII to the throne.At a later time and a differentposition, Joan of Arc is sentencedto death.

ad / A change of units dis-torts an x-t graph. This graphdepicts exactly the same eventsas figure ac. The only change isthat the x and t coordinates aremeasured using different units,so the grid is compressed in tand expanded in x .

ae / A convention we’ll useto represent a distortion of timeand space.

Distortion of space and time

The Lorentz transformation

Relativity says that when two observers are in different frames ofreference, each observer considers the other one’s perception of timeto be distorted. We’ll also see that something similar happens totheir observations of distances, so both space and time are distorted.What exactly is this distortion? How do we even conceptualize it?

The idea isn’t really as radical as it might seem at first. Wecan visualize the structure of space and time using a graph withposition and time on its axes. These graphs are familiar by now,but we’re going to look at them in a slightly different way. Before, weused them to describe the motion of objects. The grid underlyingthe graph was merely the stage on which the actors played theirparts. Now the background comes to the foreground: it’s time andspace themselves that we’re studying. We don’t necessarily needto have a line or a curve drawn on top of the grid to represent aparticular object. We may, for example, just want to talk aboutevents, depicted as points on the graph as in figure ac. A distortionof the Cartesian grid underlying the graph can arise for perfectlyordinary reasons that Isaac Newton would have readily accepted.For example, we can simply change the units used to measure timeand position, as in figure ad.

We’re going to have quite a few examples of this type, so I’lladopt the convention shown in figure ae for depicting them. Figureae summarizes the relationship between figures ac and ad in a morecompact form. The gray rectangle represents the original coordinategrid of figure ac, while the grid of black lines represents the newversion from figure ad. Omitting the grid from the gray rectanglemakes the diagram easier to decode visually.

Our goal of unraveling the mysteries of special relativity amountsto nothing more than finding out how to draw a diagram like aein the case where the two different sets of coordinates representmeasurements of time and space made by two different observers,each in motion relative to the other. Galileo and Newton thoughtthey knew the answer to this question, but their answer turnedout to be only approximately right. To avoid repeating the samemistakes, we need to clearly spell out what we think are the basicproperties of time and space that will be a reliable foundation forour reasoning. I want to emphasize that there is no purely logicalway of deciding on this list of properties. The ones I’ll list are simplya summary of the patterns observed in the results from a large bodyof experiments. Furthermore, some of them are only approximate.For example, property 1 below is only a good approximation whenthe gravitational field is weak, so it is a property that applies tospecial relativity, not to general relativity.


af / A Galilean version of therelationship between two framesof reference. As in all suchgraphs in this chapter, the originalcoordinates, represented by thegray rectangle, have a time axisthat goes to the right, and aposition axis that goes straightup.

Experiments show that:

1. No point in time or space has properties that make it differentfrom any other point.

2. Likewise, all directions in space have the same properties.

3. Motion is relative, i.e., all inertial frames of reference areequally valid.

4. Causality holds, in the sense described on page 75.

5. Time depends on the state of motion of the observer.

Most of these are not very subversive. Properties 1 and 2 dateback to the time when Galileo and Newton started applying thesame universal laws of motion to the solar system and to the earth;this contradicted Aristotle, who believed that, for example, a rockwould naturally want to move in a certain special direction (down)in order to reach a certain special location (the earth’s surface).Property 3 is the reason that Einstein called his theory “relativity,”but Galileo and Newton believed exactly the same thing to be true,as dramatized by Galileo’s run-in with the Church over the questionof whether the earth could really be in motion around the sun.Property 4 would probably surprise most people only because itasserts in such a weak and specialized way something that they feeldeeply must be true. The only really strange item on the list is 5,but the Hafele-Keating experiment forces it upon us.

If it were not for property 5, we could imagine that figure afwould give the correct transformation between frames of referencein motion relative to one another. Let’s say that observer 1, whosegrid coincides with the gray rectangle, is a hitch-hiker standing bythe side of a road. Event A is a raindrop hitting his head, andevent B is another raindrop hitting his head. He says that A and Boccur at the same location in space. Observer 2 is a motorist whodrives by without stopping; to him, the passenger compartment ofhis car is at rest, while the asphalt slides by underneath. He saysthat A and B occur at different points in space, because during thetime between the first raindrop and the second, the hitch-hiker hasmoved backward. On the other hand, observer 2 says that events Aand C occur in the same place, while the hitch-hiker disagrees. Theslope of the grid-lines is simply the velocity of the relative motionof each observer relative to the other.

Figure af has familiar, comforting, and eminently sensible behav-ior, but it also happens to be wrong, because it violates property5. The distortion of the coordinate grid has only moved the verticallines up and down, so both observers agree that events like B andC are simultaneous. If this was really the way things worked, thenall observers could synchronize all their clocks with one another for


ag / A transformation that leadsto disagreements about whethertwo events occur at the sametime and place. This is not justa matter of opinion. Either thearrow hit the bull’s-eye or it didn’t.

ah / A nonlinear transforma-tion.

once and for all, and the clocks would never get out of sync. Thiscontradicts the results of the Hafele-Keating experiment, in whichall three clocks were initially synchronized in Washington, but laterwent out of sync because of their different states of motion.

It might seem as though we still had a huge amount of wiggleroom available for the correct form of the distortion. It turns out,however, that properties 1-5 are sufficient to prove that there is onlyone answer, which is the one found by Einstein in 1905. To see whythis is, let’s work by a process of elimination.

Figure ag shows a transformation that might seem at first glanceto be as good a candidate as any other, but it violates property 3,that motion is relative, for the following reason. In observer 2’sframe of reference, some of the grid lines cross one another. Thismeans that observers 1 and 2 disagree on whether or not certainevents are the same. For instance, suppose that event A marks thearrival of an arrow at the bull’s-eye of a target, and event B is thelocation and time when the bull’s-eye is punctured. Events A andB occur at the same location and at the same time. If one observersays that A and B coincide, but another says that they don’t, wehave a direct contradiction. Since the two frames of reference infigure ag give contradictory results, one of them is right and oneis wrong. This violates property 3, because all inertial frames ofreference are supposed to be equally valid. To avoid problems likethis, we clearly need to make sure that none of the grid lines evercross one another.

The next type of transformation we want to kill off is shown infigure ah, in which the grid lines curve, but never cross one another.The trouble with this one is that it violates property 1, the unifor-mity of time and space. The transformation is unusually “twisty”at A, whereas at B it’s much more smooth. This can’t be correct,because the transformation is only supposed to depend on the rela-tive state of motion of the two frames of reference, and that giveninformation doesn’t single out a special role for any particular pointin spacetime. If, for example, we had one frame of reference rotatingrelative to the other, then there would be something special aboutthe axis of rotation. But we’re only talking about inertial frames ofreference here, as specified in property 3, so we can’t have rotation;each frame of reference has to be moving in a straight line at con-stant speed. For frames related in this way, there is nothing thatcould single out an event like A for special treatment compared toB, so transformation ah violates property 1.

The examples in figures ag and ah show that the transformationwe’re looking for must be linear, meaning that it must transformlines into lines, and furthermore that it has to take parallel lines toparallel lines. Einstein wrote in his 1905 paper that “. . . on accountof the property of homogeneity [property 1] which we ascribe to time


aj / In the units that are mostconvenient for relativity, the trans-formation has symmetry about a45-degree diagonal line.

and space, the [transformation] must be linear.”2 Applying this toour diagrams, the original gray rectangle, which is a special typeof parallelogram containing right angles, must be transformed intoanother parallelogram. There are three types of transformations,figure ai, that have this property. Case I is the Galilean transfor-mation of figure af on page 79, which we’ve already ruled out.

ai / Three types of transformations that preserve parallelism. Theirdistinguishing feature is what they do to simultaneity, as shown by whathappens to the left edge of the original rectangle. In I, the left edgeremains vertical, so simultaneous events remain simultaneous. In II, theleft edge turns counterclockwise. In III, it turns clockwise.

Case II can also be discarded. Here every point on the grid ro-tates counterclockwise. What physical parameter would determinethe amount of rotation? The only thing that could be relevant wouldbe v, the relative velocity of the motion of the two frames of referencewith respect to one another. But if the angle of rotation was pro-portional to v, then for large enough velocities the grid would haveleft and right reversed, and this would violate property 4, causality:one observer would say that event A caused a later event B, butanother observer would say that B came first and caused A.

The only remaining possibility is case III, which I’ve redrawn infigure aj with a couple of changes. This is the one that Einsteinpredicted in 1905. The transformation is known as the Lorentztransformation, after Hendrik Lorentz (1853-1928), who partiallyanticipated Einstein’s work, without arriving at the correct inter-pretation. The distortion is a kind of smooshing and stretching, assuggested by the hands. Also, we’ve already seen in figures ac-ae onpage 78 that we’re free to stretch or compress everything as much aswe like in the horizontal and vertical directions, because this simplycorresponds to choosing different units of measurement for time anddistance. In figure aj I’ve chosen units that give the whole drawinga convenient symmetry about a 45-degree diagonal line. Ordinarilyit wouldn’t make sense to talk about a 45-degree angle on a graphwhose axes had different units. But in relativity, the symmetric ap-pearance of the transformation tells us that space and time ought

2A. Einstein, “On the Electrodynamics of Moving Bodies,” Annalen derPhysik 17 (1905), p. 891, tr. Saha and Bose, 1920


ak / Interpretation of the Lorentztransformation. The slope in-dicated in the figure gives therelative velocity of the two framesof reference. Events A and B thatwere simultaneous in frame 1are not simultaneous in frame 2,where event A occurs to the rightof the t = 0 line represented bythe left edge of the grid, but eventB occurs to its left.

to be treated on the same footing, and measured in the same units.

As in our discussion of the Galilean transformation, slopes areinterpreted as velocities, and the slope of the near-horizontal lines infigure ak is interpreted as the relative velocity of the two observers.The difference between the Galilean version and the relativistic oneis that now there is smooshing happening from the other side aswell. Lines that were vertical in the original grid, representing si-multaneous events, now slant over to the right. This tells us that, asrequired by property 5, different observers do not agree on whetherevents that occur in different places are simultaneous. The Hafele-Keating experiment tells us that this non-simultaneity effect is fairlysmall, even when the velocity is as big as that of a passenger jet,and this is what we would have anticipated by the correspondenceprinciple. The way that this is expressed in the graph is that if wepick the time unit to be the second, then the distance unit turns outto be hundreds of thousands of miles. In these units, the velocityof a passenger jet is an extremely small number, so the slope v infigure ak is extremely small, and the amount of distortion is tiny —it would be much too small to see on this scale.

The only thing left to determine about the Lorentz transforma-tion is the size of the transformed parallelogram relative to the sizeof the original one. Although the drawing of the hands in figure ajmay suggest that the grid deforms like a framework made of rigidcoat-hanger wire, that is not the case. If you look carefully at thefigure, you’ll see that the edges of the smooshed parallelogram areactually a little longer than the edges of the original rectangle. Infact what stays the same is not lengths but areas. The proof of thisfact is straightforward, but a little lengthy, so I’ve relegated it tosection 2.7.3. Oversimplifying a little, the basic idea of the proof isthat it wouldn’t make sense if the area was increased by the Lorentztransformation, because then area would have to be decreased bya Lorentz transformation corresponding to motion in the oppositedirection, and this would violate property 2 on page 79, which statesthat all directions in space have the same properties.

The G factor

With a little algebra and geometry (homework problem 18, page94), one can use the equal-area property to show that the factor G(Greek letter gamma) defined in figure al is given by the equation

G =1√

1− v2.

If you’ve had good training in physics, the first thing you probablythink when you look at this equation is that it must be nonsense,


al / The G factor.

because its units don’t make sense. How can we take somethingwith units of velocity squared, and subtract it from a unitless 1?But remember that this is expressed in our special relativistic units,in which the same units are used for distance and time. In thissystem, velocities are always unitless. This sort of thing happensfrequently in physics. For instance, before James Joule discoveredconservation of energy, nobody knew that heat and mechanical en-ergy were different forms of the same thing, so instead of measuringthem both in units of joules as we would do now, they measuredheat in one unit (such as calories) and mechanical energy in another(such as foot-pounds). In ordinary metric units, we just need anextra convension factor c, and the equation becomes

G =1√

1−(vc

)2 .

Here’s why we care about G. Figure al defines it as the ratio oftwo times: the time between two events as expressed in one coordi-nate system, and the time between the same two events as measuredin the other one. The interpretation is:

Time dilationA clock runs fastest in the frame of reference of an observerwho is at rest relative to the clock. An observer in motionrelative to the clock at speed v perceives the clock as runningmore slowly by a factor of G.

Since the Lorentz transformation treats time and distance entirelysymmetrically, we could just as well have defined G using the uprightx axis in figure al, and we therefore have a similar interpretation interms of space:

Length contractionA meter-stick appears longest to an observer who is at restrelative to it. An observer moving relative to the meter-stickat v observes the stick to be shortened by a factor of G.

self-check FWhat is G when v = 0? What does this mean? . Answer, p. 509


an / Muons accelerated tonearly c undergo radioactive de-cay much more slowly than theywould according to an observerat rest with respect to the muons.The first two data-points (unfilledcircles) were subject to largesystematic errors.

am / Apparatus used for the testof relativistic time dilation de-scribed in example 5. The promi-nent black and white blocks arelarge magnets surrounding a cir-cular pipe with a vacuum inside.(c) 1974 by CERN.

Large time dilation example 5The time dilation effect in the Hafele-Keating experiment was verysmall. If we want to see a large time dilation effect, we can’t doit with something the size of the atomic clocks they used; the ki-netic energy would be greater than the total megatonnage of allthe world’s nuclear arsenals. We can, however, accelerate sub-atomic particles to speeds at which G is large. For experimentalparticle physicists, relativity is something you do all day beforeheading home and stopping off at the store for milk. An early, low-precision experiment of this kind was performed by Rossi and Hallin 1941, using naturally occurring cosmic rays. Figure am showsa 1974 experiment3 of a similar type which verified the time di-lation predicted by relativity to a precision of about one part perthousand.

Particles called muons (named after the Greek letter µ, “myoo”)were produced by an accelerator at CERN, near Geneva. A muonis essentially a heavier version of the electron. Muons undergoradioactive decay, lasting an average of only 2.197 µs before theyevaporate into an electron and two neutrinos. The 1974 experi-ment was actually built in order to measure the magnetic proper-ties of muons, but it produced a high-precision test of time dilationas a byproduct. Because muons have the same electric chargeas electrons, they can be trapped using magnetic fields. Muonswere injected into the ring shown in figure am, circling aroundit until they underwent radioactive decay. At the speed at whichthese muons were traveling, they had G = 29.33, so on the av-erage they lasted 29.33 times longer than the normal lifetime. Inother words, they were like tiny alarm clocks that self-destructedat a randomly selected time. Figure an shows the number of ra-dioactive decays counted, as a function of the time elapsed af-ter a given stream of muons was injected into the storage ring.3Bailey at al., Nucl. Phys. B150(1979) 1


ap / In the garage’s frame ofreference, 1, the bus is moving,and can fit in the garage. In thebus’s frame of reference, thegarage is moving, and can’t holdthe bus.

The two dashed lines show the rates of decay predicted with andwithout relativity. The relativistic line is the one that agrees withexperiment.

An example of length contraction example 6Figure ao shows an artist’s rendering of the length contractionfor the collision of two gold nuclei at relativistic speeds in theRHIC accelerator in Long Island, New York, which went on linein 2000. The gold nuclei would appear nearly spherical (or justslightly lengthened like an American football) in frames movingalong with them, but in the laboratory’s frame, they both appeardrastically foreshortened as they approach the point of collision.The later pictures show the nuclei merging to form a hot soup, inwhich experimenters hope to observe a new form of matter.

ao / Colliding nuclei show rela-tivistic length contraction.

The garage paradox example 7One of the most famous of all the so-called relativity paradoxeshas to do with our incorrect feeling that simultaneity is well de-fined. The idea is that one could take a schoolbus and drive it atrelativistic speeds into a garage of ordinary size, in which it nor-mally would not fit. Because of the length contraction, the buswould supposedly fit in the garage. The paradox arises when weshut the door and then quickly slam on the brakes of the bus.An observer in the garage’s frame of reference will claim that thebus fit in the garage because of its contracted length. The driver,however, will perceive the garage as being contracted and thuseven less able to contain the bus. The paradox is resolved whenwe recognize that the concept of fitting the bus in the garage “allat once” contains a hidden assumption, the assumption that itmakes sense to ask whether the front and back of the bus cansimultaneously be in the garage. Observers in different framesof reference moving at high relative speeds do not necessarilyagree on whether things happen simultaneously. The person inthe garage’s frame can shut the door at an instant he perceives tobe simultaneous with the front bumper’s arrival at the back wall ofthe garage, but the driver would not agree about the simultaneityof these two events, and would perceive the door as having shutlong after she plowed through the back wall.


aq / A proof that causalityimposes a universal speed limit.In the original frame of reference,represented by the square, eventA happens a little before event B.In the new frame, shown by theparallelogram, A happens aftert = 0, but B happens before t = 0;that is, B happens before A. Thetime ordering of the two eventshas been reversed. This can onlyhappen because events A and Bare very close together in timeand fairly far apart in space. Theline segment connecting A andB has a slope greater than 1,meaning that if we wanted to bepresent at both events, we wouldhave to travel at a speed greaterthan c (which equals 1 in theunits used on this graph). You willfind that if you pick any two pointsfor which the slope of the linesegment connecting them is lessthan 1, you can never get them tostraddle the new t = 0 line in thisfunny, time-reversed way. Sincedifferent observers disagree onthe time order of events like Aand B, causality requires thatinformation never travel fromA to B or from B to A; if it did,then we would have time-travelparadoxes. The conclusion is thatc is the maximum speed of causeand effect in relativity.

The universal speed c

Let’s think a little more about the role of the 45-degree diagonalin the Lorentz transformation. Slopes on these graphs are inter-preted as velocities. This line has a slope of 1 in relativistic units,but that slope corresponds to c in ordinary metric units. We al-ready know that the relativistic distance unit must be extremelylarge compared to the relativistic time unit, so c must be extremelylarge. Now note what happens when we perform a Lorentz transfor-mation: this particular line gets stretched, but the new version ofthe line lies right on top of the old one, and its slope stays the same.In other words, if one observer says that something has a velocityequal to c, every other observer will agree on that velocity as well.(The same thing happens with −c.)

. Velocities don’t simply add and subtract.

This is surprising, since we expect, as in section 2.5.1, that avelocity c in one frame should become c + v in a frame movingat velocity v relative to the first one. But velocities are measuredby dividing a distance by a time, and both distance and time aredistorted by relativistic effects, so we actually shouldn’t expect theordinary arithmetic addition of velocities to hold in relativity; it’san approximation that’s valid at velocities that are small comparedto c. Problem 22 on p. 96 shows that relativistically, combiningvelocities u and v gives not u + v but (u + v)/(1 + uv) (in unitswhere c = 1).

. A universal speed limit

For example, suppose Janet takes a trip in a spaceship, andaccelerates until she is moving at 0.6c relative to the earth. Shethen launches a space probe in the forward direction at a speedrelative to her ship of 0.6c. We might think that the probe was thenmoving at a velocity of 1.2c, but in fact the answer is still less thanc (problem 21, page 95). This is an example of a more general factabout relativity, which is that c represents a universal speed limit.This is required by causality, as shown in figure aq.

. Light travels at c.

Now consider a beam of light. We’re used to talking casuallyabout the “speed of light,” but what does that really mean? Motionis relative, so normally if we want to talk about a velocity, we haveto specify what it’s measured relative to. A sound wave has a certainspeed relative to the air, and a water wave has its own speed relativeto the water. If we want to measure the speed of an ocean wave, forexample, we should make sure to measure it in a frame of referenceat rest relative to the water. But light isn’t a vibration of a physicalmedium; it can propagate through the near-perfect vacuum of outerspace, as when rays of sunlight travel to earth. This seems like aparadox: light is supposed to have a specific speed, but there is no


way to decide what frame of reference to measure it in. The wayout of the paradox is that light must travel at a velocity equal to c.Since all observers agree on a velocity of c, regardless of their frameof reference, everything is consistent.

. The Michelson-Morley experiment

The constancy of the speed of light had in fact already beenobserved when Einstein was an 8-year-old boy, but because nobodycould figure out how to interpret it, the result was largely ignored.In 1887 Michelson and Morley set up a clever apparatus to measureany difference in the speed of light beams traveling east-west andnorth-south. The motion of the earth around the sun at 110,000km/hour (about 0.01% of the speed of light) is to our west during theday. Michelson and Morley believed that light was a vibration of amysterious medium called the ether, so they expected that the speedof light would be a fixed value relative to the ether. As the earthmoved through the ether, they thought they would observe an effecton the velocity of light along an east-west line. For instance, if theyreleased a beam of light in a westward direction during the day, theyexpected that it would move away from them at less than the normalspeed because the earth was chasing it through the ether. They weresurprised when they found that the expected 0.01% change in thespeed of light did not occur.

ar / The Michelson-Morley experi-ment, shown in photographs, anddrawings from the original 1887paper. 1. A simplified draw-ing of the apparatus. A beam oflight from the source, s, is par-tially reflected and partially trans-mitted by the half-silvered mirrorh1. The two half-intensity parts ofthe beam are reflected by the mir-rors at a and b, reunited, and ob-served in the telescope, t. If theearth’s surface was supposed tobe moving through the ether, thenthe times taken by the two lightwaves to pass through the mov-ing ether would be unequal, andthe resulting time lag would bedetectable by observing the inter-ference between the waves whenthey were reunited. 2. In the realapparatus, the light beams werereflected multiple times. The ef-fective length of each arm wasincreased to 11 meters, whichgreatly improved its sensitivity tothe small expected difference inthe speed of light. 3. In anearlier version of the experiment,they had run into problems withits “extreme sensitiveness to vi-bration,” which was “so great thatit was impossible to see the in-terference fringes except at briefintervals . . . even at two o’clockin the morning.” They thereforemounted the whole thing on amassive stone floating in a pool ofmercury, which also made it pos-sible to rotate it easily. 4. A photoof the apparatus.


Discussion question B


A A person in a spaceship moving at 99.99999999% of the speedof light relative to Earth shines a flashlight forward through dusty air, sothe beam is visible. What does she see? What would it look like to anobserver on Earth?

B A question that students often struggle with is whether time andspace can really be distorted, or whether it just seems that way. Comparewith optical illusions or magic tricks. How could you verify, for instance,that the lines in the figure are actually parallel? Are relativistic effects thesame, or not?

C On a spaceship moving at relativistic speeds, would a lecture seemeven longer and more boring than normal?

D Mechanical clocks can be affected by motion. For example, it wasa significant technological achievement to build a clock that could sailaboard a ship and still keep accurate time, allowing longitude to be deter-mined. How is this similar to or different from relativistic time dilation?

E Figure ao from page 85, depicting the collision of two nuclei at theRHIC accelerator, is reproduced below. What would the shapes of the twonuclei look like to a microscopic observer riding on the left-hand nucleus?To an observer riding on the right-hand one? Can they agree on what ishappening? If not, why not — after all, shouldn’t they see the same thingif they both compare the two nuclei side-by-side at the same instant intime?

as / Discussion question E: colliding nuclei show relativistic lengthcontraction.

F If you stick a piece of foam rubber out the window of your car whiledriving down the freeway, the wind may compress it a little. Does it makesense to interpret the relativistic length contraction as a type of strainthat pushes an object’s atoms together like this? How does this relate todiscussion question E?


Proof that the Lorentz transformation preserves area

On page 82, I claimed that area must be preserved by Lorentztransformations. Here is the proof of that fact.

First, consider a process, P, in which we start with a square ofunit area, and perform two successive Lorentz transformations, onewith v and one with −v. Let the first transformation scale the areaof the square by a factor R, which we want to prove equals 1. Sincethe second transformation undoes the effect of the first one, its effectis to scale the area of the intermediate parallelogram back down bya factor of 1/R.

Now consider a slightly different process Q, involving the threesteps shown in the figure below: (1) perform a Lorentz transforma-tion with velocity v; (2) chop the resulting parallelogram into littlesquares; and (3) transform each square with a −v transformation.We already know how a +v transformation acts on the area of asquare: it makes it into a parallelogram with an area that is greaterby a factor of R. In step 3, we are also acting on a square, but witha −v transformation. Property 2 of spacetime on page 79 statesthat all directions in space have the same properties, and thereforea −v transformation acting on a square must have the same effecton area as a +v transformation. (By property 1, homogeneity, itdoesn’t matter that these squares are located in different places.)The result is that after completing process Q, we have scaled thearea by a factor of R2.

Comparing processes P and Q, we see that they must both re-store the original square to itself. (In step 2 of process Q, there willplaces near the edges of the parallelogram where we would only havea fraction of a square, but the total area of this ragged edge regioncan be made as small as desired by making the squares sufficientlysmall.) Since P scales the area by 1, and Q scales the area by R2, weconclude that R2 = 1. This means that either R = +1 or R = −1,and R must be independent of v. But for v = 0 we have R = +1,so R = +1 for all values of v.


SummarySelected vocabularycenter of mass . . the balance point of an objectvelocity . . . . . . the rate of change of position; the slope of the

tangent line on an x− t graph.

Notationx . . . . . . . . . . a point in spacet . . . . . . . . . . a point in time, a clock reading∆ . . . . . . . . . “change in;” the value of a variable afterwards

minus its value before∆x . . . . . . . . a distance, or more precisely a change in x,

which may be less than the distance traveled;its plus or minus sign indicates direction

∆t . . . . . . . . . a duration of timev . . . . . . . . . . velocityvAB . . . . . . . . the velocity of object A relative to object B

Other terminology and notationdisplacement . . a name for the symbol ∆xspeed . . . . . . . the absolute value of the velocity, i.e., the ve-

locity stripped of any information about itsdirection

Summary

An object’s center of mass is the point at which it can be bal-anced. For the time being, we are studying the mathematical de-scription only of the motion of an object’s center of mass in casesrestricted to one dimension. The motion of an object’s center ofmass is usually far simpler than the motion of any of its other parts.

It is important to distinguish location, x, from distance, ∆x,and clock reading, t, from time interval ∆t. When an object’s x− tgraph is linear, we define its velocity as the slope of the line, ∆x/∆t.When the graph is curved, we generalize the definition so that thevelocity is the derivative dx/dt.

Galileo’s principle of inertia states that no force is required tomaintain motion with constant velocity in a straight line, and abso-lute motion does not cause any observable physical effects. Thingstypically tend to reduce their velocity relative to the surface of ourplanet only because they are physically rubbing against the planet(or something attached to the planet), not because there is anythingspecial about being at rest with respect to the earth’s surface. Whenit seems, for instance, that a force is required to keep a book slidingacross a table, in fact the force is only serving to cancel the contraryforce of friction.

Absolute motion is not a well-defined concept, and if two ob-servers are not at rest relative to one another they will disagreeabout the absolute velocities of objects. They will, however, agree


about relative velocities. If object A is in motion relative to objectB, and B is in motion relative to C, then A’s velocity relative to Cis given by vAC = vAB + vBC . Positive and negative signs are usedto indicate the direction of an object’s motion.

Modern experiments show that space and time only approxi-mately have the properties claimed by Galileo and Newton. Timeand space as seen by one observer are distorted compared to anotherobserver’s perceptions if they are moving relative to each other. Thisdistortion is quantified by the factor

γ =1√

1− v2

c2

,

where v is the relative velocity of the two observers, and c is auniversal velocity that is the same in all frames of reference. Lighttravels at c. A clock appears to run fastest to an observer who isnot in motion relative to it, and appears to run too slowly by afactor of γ to an observer who has a velocity v relative to the clock.Similarly, a meter-stick appears longest to an observer who sees itat rest, and appears shorter to other observers. Time and space arerelative, not absolute. In particular, there is no well-defined conceptof simultaneity. Velocities don’t add according to u + v but rather(u+ v)/(1 + uv) (in units where c = 1).

All of these strange effects, however, are very small when the rel-ative velocities are small compared to c. This makes sense, becauseNewton’s laws have already been thoroughly tested by experimentsat such speeds, so a new theory like relativity must agree with theold one in their realm of common applicability. This requirement ofbackwards-compatibility is known as the correspondence principle.

Summary 91

Problem 1.

ProblemsKey√



1 The graph represents the motion of a ball that rolls up a hilland then back down. When does the ball return to the location ithad at t = 0? . Solution, p. 496

2 The graph represents the velocity of a bee along a straightline. At t = 0, the bee is at the hive. (a) When is the bee farthestfrom the hive? (b) How far is the bee at its farthest point from thehive? (c) At t = 13 s, how far is the bee from the hive?

√

3 (a) Let θ be the latitude of a point on the Earth’s surface.Derive an algebra equation for the distance, L, traveled by thatpoint during one rotation of the Earth about its axis, i.e., over oneday, expressed in terms of θ and R, the radius of the earth. Check:Your equation should give L = 0 for the North Pole.(b) At what speed is Fullerton, at latitude θ = 34 , moving withthe rotation of the Earth about its axis? Give your answer in unitsof mi/h. [See the table in the back of the book for the relevantdata.]

√

4 A honeybee’s position as a function of time is given by x =10t− t3, where t is in seconds and x in meters. What is its velocityat t = 3.0 s?

√

5 Freddi Fish(TM) has a position as a function of time given byx = a/(b + t2). (a) Infer the units of the constants a and b. (b)Find her maximum speed. (c) Check that your answer has the rightunits.

√

6 A metal square expands and contracts with temperature, thelengths of its sides varying according to the equation ` = (1+αT )`o.Infer the units of α. Find the rate of change of its surface area withrespect to temperature. That is, find dA/dT . Check that youranswer has the right units, as in example 4 on page 68.

√


7 Let t be the time that has elapsed since the Big Bang. Inthat time, one would imagine that light, traveling at speed c, hasbeen able to travel a maximum distance ct. (In fact the distance isseveral times more than this, because according to Einstein’s theoryof general relativity, space itself has been expanding while the ray oflight was in transit.) The portion of the universe that we can observewould then be a sphere of radius ct, with volume v = (4/3)πr3 =(4/3)π(ct)3. Compute the rate dv/dt at which the volume of theobservable universe is increasing, and check that your answer hasthe right units, as in example 4 on page 68.

√

8 (a) Express the chain rule in Leibniz notation, and show thatit always results in an answer whose units make sense.(b) An object has a position as a function of time given by x =A cos(bt), where A and b are constants. Infer the units of A and b,and interpret their physical meanings.(c) Find the velocity of this object, and check that the chain rulehas indeed given an answer with the right units.

. Solution, p. 496

9 (a) Translate the following information into symbols, usingthe notation with two subscripts introduced in section 2.5. Eowynis riding on her horse at a velocity of 11 m/s. She twists around inher saddle and fires an arrow backward. Her bow fires arrows at 25m/s. (b) Find the velocity of the arrow relative to the ground.

10 Our full discussion of two- and three-dimensional motion ispostponed until the second half of the book, but here is a chance touse a little mathematical creativity in anticipation of that general-ization. Suppose a ship is sailing east at a certain speed v, and apassenger is walking across the deck at the same speed v, so thathis track across the deck is perpendicular to the ship’s center-line.What is his speed relative to the water, and in what direction is hemoving relative to the water? . Solution, p. 496

11 You’re standing in a freight train, and have no way to see out.If you have to lean to stay on your feet, what, if anything, does thattell you about the train’s velocity? Explain. . Solution, p. 496

12 Driving along in your car, you take your foot off the gas,and your speedometer shows a reduction in speed. Describe a frameof reference in which your car was speeding up during that sameperiod of time. (The frame of reference should be defined by anobserver who, although perhaps in motion relative to the earth, isnot changing her own speed or direction of motion.)

13 The figure shows the motion of a point on the rim of a rollingwheel. (The shape is called a cycloid.) Suppose bug A is riding onthe rim of the wheel on a bicycle that is rolling, while bug B is onthe spinning wheel of a bike that is sitting upside down on the floor.Bug A is moving along a cycloid, while bug B is moving in a circle.Both wheels are doing the same number of revolutions per minute.

Problems 93

Problem 13.

Problem 18.

Which bug has a harder time holding on, or do they find it equallydifficult? . Solution, p. 497

14 Astronauts in three different spaceships are communicatingwith each other. Those aboard ships A and B agree on the rate atwhich time is passing, but they disagree with the ones on ship C.(a) Alice is aboard ship A. How does she describe the motion of herown ship, in its frame of reference?(b) Describe the motion of the other two ships according to Alice.(c) Give the description according to Betty, whose frame of referenceis ship B.(d) Do the same for Cathy, aboard ship C.

15 What happens in the equation for γ when you put in anegative number for v? Explain what this means physically, andwhy it makes sense.

16 The Voyager 1 space probe, launched in 1977, is movingfaster relative to the earth than any other human-made object, at17,000 meters per second.(a) Calculate the probe’s γ.(b) Over the course of one year on earth, slightly less than one yearpasses on the probe. How much less? (There are 31 million secondsin a year.)

√

17 The earth is orbiting the sun, and therefore is contractedrelativistically in the direction of its motion. Compute the amountby which its diameter shrinks in this direction.

18 In this homework problem, you’ll fill in the steps of the alge-bra required in order to find the equation for γ on page 82. To keepthe algebra simple, let the time t in figure al equal 1, as suggestedin the figure accompanying this homework problem. The originalsquare then has an area of 1, and the transformed parallelogrammust also have an area of 1. (a) Prove that point P is at x = vγ, sothat its (t,x) coordinates are (γ, vγ). (b) Find the (t,x) coordinatesof point Q. (c) Find the length of the short diagonal connecting Pand Q. (d) Average the coordinates of P and Q to find the coordi-nates of the midpoint C of the parallelogram, and then find distanceOC. (e) Find the area of the parallelogram by computing twice thearea of triangle PQO. [Hint: You can take PQ to be the base ofthe triangle.] (f) Set this area equal to 1 and solve for γ to proveγ = 1/

√1− v2.

√

19 (a) Show that for v = (3/5)c, γ comes out to be a simplefraction.(b) Find another value of v for which γ is a simple fraction.

20 In the equation for the relativistic addition of velocities uand v, consider the limit in which u approaches 1, but v simulta-neously approaches −1. Give both a physical and a mathematicalinterpretation.


21 The figure illustrates a Lorentz transformation using theconventions employed in section 2.7.2. For simplicity, the transfor-mation chosen is one that lengthens one diagonal by a factor of 2.Since Lorentz transformations preserve area, the other diagonal isshortened by a factor of 2. Let the original frame of reference, de-picted with the square, be A, and the new one B. (a) By measuringwith a ruler on the figure, show that the velocity of frame B rela-tive to frame A is 0.6c. (b) Print out a copy of the page. With aruler, draw a third parallelogram that represents a second succes-sive Lorentz transformation, one that lengthens the long diagonalby another factor of 2. Call this third frame C. Use measurementswith a ruler to determine frame C’s velocity relative to frame A.Does it equal double the velocity found in part a? Explain why itshould be expected to turn out the way it does. A general equationfor this type of calculation is derived in problem 22.

√

Problems 95

22 The purpose of this problem is to prove the general re-sult w = (u + v)/(1 + uv) (in units where c = 1) for the kind ofcombination of velocities found graphically in problem 21. Supposethat we perform two Lorentz transformations, with velocities u andv, one after the other. Representing these transformations as dis-tortions of parallelograms, we stretch the stretching diagonals byfactors S(u) and S(v) (and compress the compressing ones by theinverses of these factors), so that the combined result is a stretchingby S(u)S(v). We want to prove that S(w) = S(u)S(v) gives the ex-pression claimed above for w. One can easily show by fiddling withthe result of part c of problem 18 that S(x) =

√(1 + x)/(1− x).

(a) Use these facts to write down an equation relating u, v, and w.(b) Solve for w in terms of u and v. (c) Show that your answerto part b satisfies the correspondence principle. (d) Show that it isconsistent with the constancy of c.


Galileo’s contradiction of Aristotle had serious consequences. He wasinterrogated by the Church authorities and convicted of teaching that theearth went around the sun as a matter of fact and not, as he had promisedpreviously, as a mere mathematical hypothesis. He was placed under per-manent house arrest, and forbidden to write about or teach his theories.Immediately after being forced to recant his claim that the earth revolvedaround the sun, the old man is said to have muttered defiantly “and yetit does move.” The story is dramatic, but there are some omissions inthe commonly taught heroic version. There was a rumor that the Sim-plicio character represented the Pope. Also, some of the ideas Galileoadvocated had controversial religious overtones. He believed in the exis-tence of atoms, and atomism was thought by some people to contradictthe Church’s doctrine of transubstantiation, which said that in the Catholicmass, the blessing of the bread and wine literally transformed them intothe flesh and blood of Christ. His support for a cosmology in which theearth circled the sun was also disreputable because one of its support-ers, Giordano Bruno, had also proposed a bizarre synthesis of Christianitywith the ancient Egyptian religion.

Chapter 3

Acceleration and free fall

3.1 The motion of falling objectsThe motion of falling objects is the simplest and most commonexample of motion with changing velocity. The early pioneers of

97

a / Galileo dropped a cannonballand a musketball simultaneouslyfrom a tower, and observed thatthey hit the ground at nearly thesame time. This contradictedAristotle’s long-accepted ideathat heavier objects fell faster.

physics had a correct intuition that the way things drop was a mes-sage directly from Nature herself about how the universe worked.Other examples seem less likely to have deep significance. A walkingperson who speeds up is making a conscious choice. If one stretch ofa river flows more rapidly than another, it may be only because thechannel is narrower there, which is just an accident of the local ge-ography. But there is something impressively consistent, universal,and inexorable about the way things fall.

Stand up now and simultaneously drop a coin and a bit of paperside by side. The paper takes much longer to hit the ground. That’swhy Aristotle wrote that heavy objects fell more rapidly. Europeansbelieved him for two thousand years.

Now repeat the experiment, but make it into a race between thecoin and your shoe. My own shoe is about 50 times heavier thanthe nickel I had handy, but it looks to me like they hit the ground atexactly the same moment. So much for Aristotle! Galileo, who hada flair for the theatrical, did the experiment by dropping a bulletand a heavy cannonball from a tall tower. Aristotle’s observationshad been incomplete, his interpretation a vast oversimplification.

It is inconceivable that Galileo was the first person to observe adiscrepancy with Aristotle’s predictions. Galileo was the one whochanged the course of history because he was able to assemble theobservations into a coherent pattern, and also because he carriedout systematic quantitative (numerical) measurements rather thanjust describing things qualitatively.

Why is it that some objects, like the coin and the shoe, have sim-ilar motion, but others, like a feather or a bit of paper, are different?Galileo speculated that in addition to the force that always pulls ob-jects down, there was an upward force exerted by the air. Anyonecan speculate, but Galileo went beyond speculation and came upwith two clever experiments to probe the issue. First, he experi-mented with objects falling in water, which probed the same issuesbut made the motion slow enough that he could take time measure-ments with a primitive pendulum clock. With this technique, heestablished the following facts:

• All heavy, streamlined objects (for example a steel rod droppedpoint-down) reach the bottom of the tank in about the sameamount of time, only slightly longer than the time they wouldtake to fall the same distance in air.

• Objects that are lighter or less streamlined take a longer timeto reach the bottom.

This supported his hypothesis about two contrary forces. Heimagined an idealized situation in which the falling object did not

98 Chapter 3 Acceleration and free fall

c / The v − t graph of a fallingobject is a line.

have to push its way through any substance at all. Falling in airwould be more like this ideal case than falling in water, but evena thin, sparse medium like air would be sufficient to cause obviouseffects on feathers and other light objects that were not streamlined.Today, we have vacuum pumps that allow us to suck nearly all theair out of a chamber, and if we drop a feather and a rock side byside in a vacuum, the feather does not lag behind the rock at all.

How the speed of a falling object increases with time

Galileo’s second stroke of genius was to find a way to make quan-titative measurements of how the speed of a falling object increasedas it went along. Again it was problematic to make sufficiently accu-rate time measurements with primitive clocks, and again he found atricky way to slow things down while preserving the essential physi-cal phenomena: he let a ball roll down a slope instead of dropping itvertically. The steeper the incline, the more rapidly the ball wouldgain speed. Without a modern video camera, Galileo had inventeda way to make a slow-motion version of falling.

b / Velocity increases more gradually on the gentle slope, but themotion is otherwise the same as the motion of a falling object.

Although Galileo’s clocks were only good enough to do accurateexperiments at the smaller angles, he was confident after makinga systematic study at a variety of small angles that his basic con-clusions were generally valid. Stated in modern language, what hefound was that the velocity-versus-time graph was a line. In the lan-guage of algebra, we know that a line has an equation of the formy = ax+ b, but our variables are v and t, so it would be v = at+ b.(The constant b can be interpreted simply as the initial velocity ofthe object, i.e., its velocity at the time when we started our clock,which we conventionally write as vo.)

self-check AAn object is rolling down an incline. After it has been rolling for a shorttime, it is found to travel 13 cm during a certain one-second interval.During the second after that, if goes 16 cm. How many cm will it travelin the second after that? . Answer, p. 509

Section 3.1 The motion of falling objects 99

d / Galileo’s experiments showthat all falling objects have thesame motion if air resistance isnegligible.

e / 1. Aristotle said that heavierobjects fell faster than lighterones. 2. If two rocks are tiedtogether, that makes an extra-heavy rock, which should fallfaster. 3. But Aristotle’s theorywould also predict that the lightrock would hold back the heavyrock, resulting in a slower fall.

A contradiction in Aristotle’s reasoning

Galileo’s inclined-plane experiment disproved the long-acceptedclaim by Aristotle that a falling object had a definite “natural fallingspeed” proportional to its weight. Galileo had found that the speedjust kept on increasing, and weight was irrelevant as long as airfriction was negligible. Not only did Galileo prove experimentallythat Aristotle had been wrong, but he also pointed out a logicalcontradiction in Aristotle’s own reasoning. Simplicio, the stupidcharacter, mouths the accepted Aristotelian wisdom:

SIMPLICIO: There can be no doubt but that a particularbody . . . has a fixed velocity which is determined by nature. . .

SALVIATI: If then we take two bodies whose natural speedsare different, it is clear that, [according to Aristotle], on unit-ing the two, the more rapid one will be partly held back bythe slower, and the slower will be somewhat hastened by theswifter. Do you not agree with me in this opinion?

SIMPLICIO: You are unquestionably right.

SALVIATI: But if this is true, and if a large stone moves with aspeed of, say, eight [unspecified units] while a smaller moveswith a speed of four, then when they are united, the systemwill move with a speed less than eight; but the two stoneswhen tied together make a stone larger than that which beforemoved with a speed of eight. Hence the heavier body moveswith less speed than the lighter; an effect which is contrary toyour supposition. Thus you see how, from your assumptionthat the heavier body moves more rapidly than the lighter one,I infer that the heavier body moves more slowly.

What is gravity?

The physicist Richard Feynman liked to tell a story about howwhen he was a little kid, he asked his father, “Why do things fall?”As an adult, he praised his father for answering, “Nobody knows whythings fall. It’s a deep mystery, and the smartest people in the worlddon’t know the basic reason for it.” Contrast that with the averageperson’s off-the-cuff answer, “Oh, it’s because of gravity.” Feynmanliked his father’s answer, because his father realized that simplygiving a name to something didn’t mean that you understood it.The radical thing about Galileo’s and Newton’s approach to sciencewas that they concentrated first on describing mathematically whatreally did happen, rather than spending a lot of time on untestablespeculation such as Aristotle’s statement that “Things fall becausethey are trying to reach their natural place in contact with theearth.” That doesn’t mean that science can never answer the “why”questions. Over the next month or two as you delve deeper intophysics, you will learn that there are more fundamental reasons whyall falling objects have v − t graphs with the same slope, regardless


f / Example 1.

g / Example 2.

of their mass. Nevertheless, the methods of science always imposelimits on how deep our explanation can go.

3.2 AccelerationDefinition of acceleration for linear v − t graphs

Galileo’s experiment with dropping heavy and light objects froma tower showed that all falling objects have the same motion, and hisinclined-plane experiments showed that the motion was described byv = at+vo. The initial velocity vo depends on whether you drop theobject from rest or throw it down, but even if you throw it down,you cannot change the slope, a, of the v − t graph.

Since these experiments show that all falling objects have lin-ear v − t graphs with the same slope, the slope of such a graph isapparently an important and useful quantity. We use the word accel-eration, and the symbol a, for the slope of such a graph. In symbols,a = ∆v/∆t. The acceleration can be interpreted as the amount ofspeed gained in every second, and it has units of velocity divided bytime, i.e., “meters per second per second,” or m/s/s. Continuing totreat units as if they were algebra symbols, we simplify “m/s/s” toread “m/s2.” Acceleration can be a useful quantity for describingother types of motion besides falling, and the word and the symbol“a” can be used in a more general context. We reserve the morespecialized symbol “g” for the acceleration of falling objects, whichon the surface of our planet equals 9.8 m/s2. Often when doingapproximate calculations or merely illustrative numerical examplesit is good enough to use g = 10 m/s2, which is off by only 2%.

Finding final speed, given time example 1. A despondent physics student jumps off a bridge, and falls forthree seconds before hitting the water. How fast is he going whenhe hits the water?

. Approximating g as 10 m/s2, he will gain 10 m/s of speed eachsecond. After one second, his velocity is 10 m/s, after two sec-onds it is 20 m/s, and on impact, after falling for three seconds,he is moving at 30 m/s.

Extracting acceleration from a graph example 2. The x − t and v − t graphs show the motion of a car startingfrom a stop sign. What is the car’s acceleration?

. Acceleration is defined as the slope of the v-t graph. The graphrises by 3 m/s during a time interval of 3 s, so the acceleration is(3 m/s)/(3 s) = 1 m/s2.

Incorrect solution #1: The final velocity is 3 m/s, and accelerationis velocity divided by time, so the acceleration is (3 m/s)/(10 s) =0.3 m/s2.

Section 3.2 Acceleration 101

The solution is incorrect because you can’t find the slope of agraph from one point. This person was just using the point at theright end of the v-t graph to try to find the slope of the curve.

Incorrect solution #2: Velocity is distance divided by time so v =(4.5 m)/(3 s) = 1.5 m/s. Acceleration is velocity divided by time,so a = (1.5 m/s)/(3 s) = 0.5 m/s2.

The solution is incorrect because velocity is the slope of thetangent line. In a case like this where the velocity is changing,you can’t just pick two points on the x-t graph and use them tofind the velocity.

Converting g to different units example 3. What is g in units of cm/s2?

. The answer is going to be how many cm/s of speed a fallingobject gains in one second. If it gains 9.8 m/s in one second, thenit gains 980 cm/s in one second, so g = 980 cm/s2. Alternatively,we can use the method of fractions that equal one:

9.8ms2 × 100 cm

1m=

980 cms2

. What is g in units of miles/hour2?

.

9.8 ms2 × 1 mile

1600 m×(

3600 s1 hour

)2

= 7.9× 104 mile/hour2

This large number can be interpreted as the speed, in miles perhour, that you would gain by falling for one hour. Note that we hadto square the conversion factor of 3600 s/hour in order to cancelout the units of seconds squared in the denominator.

. What is g in units of miles/hour/s?

.

9.8 ms2 × 1 mile

1600 m× 3600 s

1 hour= 22 mile/hour/s

This is a figure that Americans will have an intuitive feel for. Ifyour car has a forward acceleration equal to the acceleration of afalling object, then you will gain 22 miles per hour of speed everysecond. However, using mixed time units of hours and secondslike this is usually inconvenient for problem-solving. It would belike using units of foot-inches for area instead of ft2 or in2.

The acceleration of gravity is different in different locations.

Everyone knows that gravity is weaker on the moon, but actu-ally it is not even the same everywhere on Earth, as shown by thesampling of numerical data in the following table.


location latitude elevation (m) g (m/s2)north pole 90 N 0 9.8322Reykjavik, Iceland 64 N 0 9.8225Fullerton, California 34 N 0 9.7957Guayaquil, Ecuador 2 S 0 9.7806Mt. Cotopaxi, Ecuador 1 S 5896 9.7624Mt. Everest 28 N 8848 9.7643

The main variables that relate to the value of g on Earth are latitudeand elevation. Although you have not yet learned how g wouldbe calculated based on any deeper theory of gravity, it is not toohard to guess why g depends on elevation. Gravity is an attractionbetween things that have mass, and the attraction gets weaker withincreasing distance. As you ascend from the seaport of Guayaquilto the nearby top of Mt. Cotopaxi, you are distancing yourself fromthe mass of the planet. The dependence on latitude occurs becausewe are measuring the acceleration of gravity relative to the earth’ssurface, but the earth’s rotation causes the earth’s surface to fallout from under you. (We will discuss both gravity and rotation inmore detail later in the course.)

h / This false-color map showsvariations in the strength of theearth’s gravity. Purple areas havethe strongest gravity, yellow theweakest. The overall trend towardweaker gravity at the equator andstronger gravity at the poles hasbeen artificially removed to al-low the weaker local variations toshow up. The map covers onlythe oceans because of the tech-nique used to make it: satelliteslook for bulges and depressionsin the surface of the ocean. Avery slight bulge will occur over anundersea mountain, for instance,because the mountain’s gravita-tional attraction pulls water to-ward it. The US government orig-inally began collecting data likethese for military use, to correctfor the deviations in the paths ofmissiles. The data have recentlybeen released for scientific andcommercial use (e.g., searchingfor sites for off-shore oil wells).

Much more spectacular differences in the strength of gravity canbe observed away from the Earth’s surface:

Section 3.2 Acceleration 103

location g (m/s2)asteroid Vesta (surface) 0.3Earth’s moon (surface) 1.6Mars (surface) 3.7Earth (surface) 9.8Jupiter (cloud-tops) 26Sun (visible surface) 270typical neutron star (surface) 1012

black hole (center) infinite according to some theo-ries, on the order of 1052 accord-ing to others

A typical neutron star is not so different in size from a large asteroid,but is orders of magnitude more massive, so the mass of a bodydefinitely correlates with the g it creates. On the other hand, aneutron star has about the same mass as our Sun, so why is its gbillions of times greater? If you had the misfortune of being on thesurface of a neutron star, you’d be within a few thousand miles of allits mass, whereas on the surface of the Sun, you’d still be millionsof miles from most of its mass.


A What is wrong with the following definitions of g?

(1) “g is gravity.”

(2) “g is the speed of a falling object.”

(3) “g is how hard gravity pulls on things.”

B When advertisers specify how much acceleration a car is capableof, they do not give an acceleration as defined in physics. Instead, theyusually specify how many seconds are required for the car to go from restto 60 miles/hour. Suppose we use the notation “a” for the acceleration asdefined in physics, and “acar ad” for the quantity used in advertisements forcars. In the US’s non-metric system of units, what would be the units ofa and acar ad? How would the use and interpretation of large and small,positive and negative values be different for a as opposed to acar ad?

C Two people stand on the edge of a cliff. As they lean over the edge,one person throws a rock down, while the other throws one straight upwith an exactly opposite initial velocity. Compare the speeds of the rockson impact at the bottom of the cliff.

3.3 Positive and negative accelerationGravity always pulls down, but that does not mean it always speedsthings up. If you throw a ball straight up, gravity will first slowit down to v = 0 and then begin increasing its speed. When Itook physics in high school, I got the impression that positive signsof acceleration indicated speeding up, while negative accelerationsrepresented slowing down, i.e., deceleration. Such a definition wouldbe inconvenient, however, because we would then have to say thatthe same downward tug of gravity could produce either a positive


i / The ball’s acceleration staysthe same — on the way up, at thetop, and on the way back down.It’s always negative.

or a negative acceleration. As we will see in the following example,such a definition also would not be the same as the slope of the v− tgraph

Let’s study the example of the rising and falling ball. In the ex-ample of the person falling from a bridge, I assumed positive velocityvalues without calling attention to it, which meant I was assuminga coordinate system whose x axis pointed down. In this example,where the ball is reversing direction, it is not possible to avoid neg-ative velocities by a tricky choice of axis, so let’s make the morenatural choice of an axis pointing up. The ball’s velocity will ini-tially be a positive number, because it is heading up, in the samedirection as the x axis, but on the way back down, it will be a neg-ative number. As shown in the figure, the v − t graph does not doanything special at the top of the ball’s flight, where v equals 0. Itsslope is always negative. In the left half of the graph, there is anegative slope because the positive velocity is getting closer to zero.On the right side, the negative slope is due to a negative velocitythat is getting farther from zero, so we say that the ball is speedingup, but its velocity is decreasing!

To summarize, what makes the most sense is to stick with theoriginal definition of acceleration as the slope of the v − t graph,∆v/∆t. By this definition, it just isn’t necessarily true that thingsspeeding up have positive acceleration while things slowing downhave negative acceleration. The word “deceleration” is not usedmuch by physicists, and the word “acceleration” is used unblush-ingly to refer to slowing down as well as speeding up: “There was ared light, and we accelerated to a stop.”

Numerical calculation of a negative acceleration example 4. In figure i, what happens if you calculate the acceleration be-tween t = 1.0 and 1.5 s?

. Reading from the graph, it looks like the velocity is about−1 m/sat t = 1.0 s, and around −6 m/s at t = 1.5 s. The acceleration,figured between these two points, is

a =∆v∆t

=(−6 m/s)− (−1 m/s)

(1.5 s)− (1.0 s)= −10 m/s2 .

Even though the ball is speeding up, it has a negative accelera-tion.

Another way of convincing you that this way of handling the plusand minus signs makes sense is to think of a device that measuresacceleration. After all, physics is supposed to use operational defini-tions, ones that relate to the results you get with actual measuringdevices. Consider an air freshener hanging from the rear-view mirrorof your car. When you speed up, the air freshener swings backward.Suppose we define this as a positive reading. When you slow down,the air freshener swings forward, so we’ll call this a negative reading

Section 3.3 Positive and negative acceleration 105

on our accelerometer. But what if you put the car in reverse andstart speeding up backwards? Even though you’re speeding up, theaccelerometer responds in the same way as it did when you weregoing forward and slowing down. There are four possible cases:

motion of car accelerome-ter swings

slope ofv-t graph

directionof forceacting oncar

forward, speeding up backward + forwardforward, slowing down forward − backwardbackward, speeding up forward − backwardbackward, slowing down backward + forward

Note the consistency of the three right-hand columns — nature istrying to tell us that this is the right system of classification, notthe left-hand column.

Because the positive and negative signs of acceleration dependon the choice of a coordinate system, the acceleration of an objectunder the influence of gravity can be either positive or negative.Rather than having to write things like “g = 9.8 m/s2 or −9.8 m/s2”every time we want to discuss g’s numerical value, we simply defineg as the absolute value of the acceleration of objects moving underthe influence of gravity. We consistently let g = 9.8 m/s2, but wemay have either a = g or a = −g, depending on our choice of acoordinate system.

Acceleration with a change in direction of motion example 5. A person kicks a ball, which rolls up a sloping street, comes toa halt, and rolls back down again. The ball has constant accel-eration. The ball is initially moving at a velocity of 4.0 m/s, andafter 10.0 s it has returned to where it started. At the end, it hassped back up to the same speed it had initially, but in the oppositedirection. What was its acceleration?

. By giving a positive number for the initial velocity, the statementof the question implies a coordinate axis that points up the slopeof the hill. The “same” speed in the opposite direction shouldtherefore be represented by a negative number, -4.0 m/s. Theacceleration is

a = ∆v/∆t= (vf − vo)/10.0 s= [(−4.0 m/s)− (4.0 m/s)]/10.0s

= −0.80 m/s2 .

The acceleration was no different during the upward part of theroll than on the downward part of the roll.

Incorrect solution: Acceleration is ∆v/∆t, and at the end it’s notmoving any faster or slower than when it started, so ∆v=0 and


Discussion question C.

a = 0.

The velocity does change, from a positive number to a negativenumber.



A A child repeatedly jumps up and down on a trampoline. Discuss thesign and magnitude of his acceleration, including both the time when he isin the air and the time when his feet are in contact with the trampoline.

B The figure shows a refugee from a Picasso painting blowing on arolling water bottle. In some cases the person’s blowing is speeding thebottle up, but in others it is slowing it down. The arrow inside the bottleshows which direction it is going, and a coordinate system is shown at thebottom of each figure. In each case, figure out the plus or minus signs ofthe velocity and acceleration. It may be helpful to draw a v − t graph ineach case.

C Sally is on an amusement park ride which begins with her chair beinghoisted straight up a tower at a constant speed of 60 miles/hour. Despitestern warnings from her father that he’ll take her home the next time shemisbehaves, she decides that as a scientific experiment she really needsto release her corndog over the side as she’s on the way up. She doesnot throw it. She simply sticks it out of the car, lets it go, and watches itagainst the background of the sky, with no trees or buildings as referencepoints. What does the corndog’s motion look like as observed by Sally?Does its speed ever appear to her to be zero? What acceleration doesshe observe it to have: is it ever positive? negative? zero? What wouldher enraged father answer if asked for a similar description of its motionas it appears to him, standing on the ground?

D Can an object maintain a constant acceleration, but meanwhilereverse the direction of its velocity?

E Can an object have a velocity that is positive and increasing at thesame time that its acceleration is decreasing?

Section 3.3 Positive and negative acceleration 107

3.4 Varying accelerationSo far we have only been discussing examples of motion for whichthe acceleration is constant. As always, an expression of the form∆ . . . /∆ . . . for a rate of change must be generalized to a derivativewhen the rate of change isn’t constant. We therefore define the ac-celeration as a = dv/dt, which is the same as the second derivative,which Leibniz notated as

a =d2x

dt2.

The seemingly inconsistent placement of the twos on the top andbottom confuses all beginning calculus students. The motivationfor this funny notation is that acceleration has units of m/s2, andthe notation correctly suggests that: the top looks like it has units ofmeters, the bottom seconds2. The notation is not meant, however,to suggest that t is really squared.

3.5 Algebraic results for constant accelerationWhen an object is accelerating, the variables x, v, and t are allchanging continuously. It is often of interest to eliminate one ofthese and relate the other two to each other.

Constant acceleration example 6. How high does a diving board have to be above the water if thediver is to have as much as 1.0 s in the air?

. The diver starts at rest, and has an acceleration of 9.8 m/s2.We need to find a connection between the distance she travelsand time it takes. In other words, we’re looking for informationabout the function x(t), given information about the acceleration.To go from acceleration to position, we need to integrate twice:

x =∫ ∫

a dt dt

=∫

(at + vo) dt [vo is a constant of integration.]

=∫

at dt [vo is zero because she’s dropping from rest.]

=12

at2 + xo [xo is a constant of integration.]

=12

at2 [xo can be zero if we define it that way.]

Note some of the good problem-solving habits demonstrated here.We solve the problem symbolically, and only plug in numbers atthe very end, once all the algebra and calculus are done. Oneshould also make a habit, after finding a symbolic result, of check-ing whether the dependence on the variables make sense. Agreater value of t in this expression would lead to a greater value


for x ; that makes sense, because if you want more time in theair, you’re going to have to jump from higher up. A greater ac-celeration also leads to a greater height; this also makes sense,because the stronger gravity is, the more height you’ll need in or-der to stay in the air for a given amount of time. Now we plug innumbers.

x =12

(9.8 m/s2

)(1.0 s)2

= 4.9 m

Note that when we put in the numbers, we check that the unitswork out correctly,

(m/s2) (s)2 = m. We should also check that

the result makes sense: 4.9 meters is pretty high, but not unrea-sonable.

Under conditions of constant acceleration, we can relate velocityand time,

a =∆v

∆t,

or, as in the example 6, position and time,

x =1

2at2 + vot+ xo .

It can also be handy to have a relation involving velocity and posi-tion, eliminating time. Straightforward algebra gives

v2f = v2

o + 2a∆x ,

where vf is the final velocity, vo the initial velocity, and ∆x thedistance traveled.

. Solved problem: Dropping a rock on Mars page 112, problem 13

. Solved problem: The Dodge Viper page 112, problem 11

Section 3.5 Algebraic results for constant acceleration 109

SummarySelected vocabularygravity . . . . . . A general term for the phenomenon of attrac-

tion between things having mass. The attrac-tion between our planet and a human-sized ob-ject causes the object to fall.

acceleration . . . The rate of change of velocity; the slope of thetangent line on a v − t graph.

Notationvo . . . . . . . . . initial velocityvf . . . . . . . . . final velocitya . . . . . . . . . . accelerationg . . . . . . . . . . the acceleration of objects in free fall; the

strength of the local gravitational field

Summary

Galileo showed that when air resistance is negligible all fallingbodies have the same motion regardless of mass. Moreover, theirv− t graphs are straight lines. We therefore define a quantity calledacceleration as the derivative dv/dt. This definition has the advan-tage that a force with a given sign, representing its direction, alwaysproduces an acceleration with the same sign. The acceleration of ob-jects in free fall varies slightly across the surface of the earth, andgreatly on other planets.

For motion with constant acceleration, the following three equa-tions hold:

∆x = vo∆t+1

2a∆t2

v2f = v2

o + 2a∆x

a =∆v

∆t

They are not valid if the acceleration is changing.


Problem 6.

Problem 7.

Problem 9.

ProblemsKey√



1 On New Year’s Eve, a stupid person fires a pistol straight up.The bullet leaves the gun at a speed of 100 m/s. How long does ittake before the bullet hits the ground? . Solution, p. 497

2 What is the acceleration of a car that moves at a steadyvelocity of 100 km/h for 100 seconds? Explain your answer. [Basedon a problem by Hewitt.]

3 You are looking into a deep well. It is dark, and you cannotsee the bottom. You want to find out how deep it is, so you dropa rock in, and you hear a splash 3.0 seconds later. How deep is thewell?

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4 A honeybee’s position as a function of time is given by x =10t− t3, where t is in seconds and x in meters. What is its acceler-ation at t = 3.0 s? . Solution, p. 497

5 Alice drops a rock off a cliff. Bubba shoots a gun straightdown from the edge of the same cliff. Compare the accelerations ofthe rock and the bullet while they are in the air on the way down.[Based on a problem by Serway and Faughn.]

6 The top part of the figure shows the position-versus-time graphfor an object moving in one dimension. On the bottom part of thefigure, sketch the corresponding v-versus-t graph.

. Solution, p. 497

7 (a) The ball is released at the top of the ramp shown in thefigure. Friction is negligible. Use physical reasoning to draw v − tand a− t graphs. Assume that the ball doesn’t bounce at the pointwhere the ramp changes slope. (b) Do the same for the case wherethe ball is rolled up the slope from the right side, but doesn’t quitehave enough speed to make it over the top. . Solution, p. 497

8 You throw a rubber ball up, and it falls and bounces sev-eral times. Draw graphs of position, velocity, and acceleration asfunctions of time. . Solution, p. 498

9 A ball rolls down the ramp shown in the figure, consisting of acurved knee, a straight slope, and a curved bottom. For each part ofthe ramp, tell whether the ball’s velocity is increasing, decreasing,or constant, and also whether the ball’s acceleration is increasing,decreasing, or constant. Explain your answers. Assume there is noair friction or rolling resistance.

Problems 111

Problem 12.

10 Consider the following passage from Alice in Wonderland, inwhich Alice has been falling for a long time down a rabbit hole:

Down, down, down. Would the fall never come to an end? “Iwonder how many miles I’ve fallen by this time?” she said aloud.“I must be getting somewhere near the center of the earth. Let mesee: that would be four thousand miles down, I think” (for, you see,Alice had learned several things of this sort in her lessons in theschoolroom, and though this was not a very good opportunity forshowing off her knowledge, as there was no one to listen to her, stillit was good practice to say it over)...

Alice doesn’t know much physics, but let’s try to calculate theamount of time it would take to fall four thousand miles, startingfrom rest with an acceleration of 10 m/s2. This is really only a lowerlimit; if there really was a hole that deep, the fall would actuallytake a longer time than the one you calculate, both because thereis air friction and because gravity gets weaker as you get deeper (atthe center of the earth, g is zero, because the earth is pulling youequally in every direction at once).

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11 In July 1999, Popular Mechanics carried out tests to findwhich car sold by a major auto maker could cover a quarter mile(402 meters) in the shortest time, starting from rest. Because thedistance is so short, this type of test is designed mainly to favor thecar with the greatest acceleration, not the greatest maximum speed(which is irrelevant to the average person). The winner was theDodge Viper, with a time of 12.08 s. The car’s top (and presumablyfinal) speed was 118.51 miles per hour (52.98 m/s). (a) If a car,starting from rest and moving with constant acceleration, coversa quarter mile in this time interval, what is its acceleration? (b)What would be the final speed of a car that covered a quarter milewith the constant acceleration you found in part a? (c) Based onthe discrepancy between your answer in part b and the actual finalspeed of the Viper, what do you conclude about how its accelerationchanged over time? . Solution, p. 498

12 The photo shows Apollo 16 astronaut John Young jumpingon the moon and saluting at the top of his jump. The video footageof the jump shows him staying aloft for 1.45 seconds. Gravity onthe moon is 1/6 as strong as on the earth. Compute the height ofthe jump.

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13 If the acceleration of gravity on Mars is 1/3 that on Earth,how many times longer does it take for a rock to drop the samedistance on Mars? Ignore air resistance. . Solution, p. 499

14 You climb half-way up a tree, and drop a rock. Then youclimb to the top, and drop another rock. How many times greateris the velocity of the second rock on impact? Explain. (The answeris not two times greater.)


Problem 21.

15 Starting from rest, a ball rolls down a ramp, traveling adistance L and picking up a final speed v. How much of the distancedid the ball have to cover before achieving a speed of v/2? [Basedon a problem by Arnold Arons.] . Solution, p. 499

16 A toy car is released on one side of a piece of track that isbent into an upright U shape. The car goes back and forth. Whenthe car reaches the limit of its motion on one side, its velocity iszero. Is its acceleration also zero? Explain using a v − t graph.[Based on a problem by Serway and Faughn.]

17 A physics homework question asks, “If you start from restand accelerate at 1.54 m/s2 for 3.29 s, how far do you travel by theend of that time?” A student answers as follows:

1.54× 3.29 = 5.07 m

His Aunt Wanda is good with numbers, but has never taken physics.She doesn’t know the formula for the distance traveled under con-stant acceleration over a given amount of time, but she tells hernephew his answer cannot be right. How does she know?

18 Find the error in the following calculation. A student wantsto find the distance traveled by a car that accelerates from rest for5.0 s with an acceleration of 2.0 m/s2. First he solves a = ∆v/∆t for∆v = 10 m/s. Then he multiplies to find (10 m/s)(5.0 s) = 50 m.Do not just recalculate the result by a different method; if that wasall you did, you’d have no way of knowing which calculation wascorrect, yours or his.

19 Acceleration could be defined either as ∆v/∆t or as the slopeof the tangent line on the v − t graph. Is either one superior as adefinition, or are they equivalent? If you say one is better, give anexample of a situation where it makes a difference which one youuse.

20 If an object starts accelerating from rest, we have v2 =2a∆x for its speed after it has traveled a distance ∆x. Explain inwords why it makes sense that the equation has velocity squared, butdistance only to the first power. Don’t recapitulate the derivationin the book, or give a justification based on units. The point isto explain what this feature of the equation tells us about the wayspeed increases as more distance is covered.

21 The graph shows the acceleration of a chipmunk in a TVcartoon. It consists of two circular arcs and two line segments.At t = 0.00 s, the chipmunk’s velocity is −3.10 m/s. What is itsvelocity at t = 10.00 s?

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Problems 113

22 You take a trip in your spaceship to another star. Setting off,you increase your speed at a constant acceleration. Once you gethalf-way there, you start decelerating, at the same rate, so that bythe time you get there, you have slowed down to zero speed. You seethe tourist attractions, and then head home by the same method.(a) Find a formula for the time, T , required for the round trip, interms of d, the distance from our sun to the star, and a, the magni-tude of the acceleration. Note that the acceleration is not constantover the whole trip, but the trip can be broken up into constant-acceleration parts.(b) The nearest star to the Earth (other than our own sun) is Prox-ima Centauri, at a distance of d = 4× 1016 m. Suppose you use anacceleration of a = 10 m/s2, just enough to compensate for the lackof true gravity and make you feel comfortable. How long does theround trip take, in years?(c) Using the same numbers for d and a, find your maximum speed.Compare this to the speed of light, which is 3.0× 108 m/s. (Laterin this course, you will learn that there are some new things goingon in physics when one gets close to the speed of light, and that itis impossible to exceed the speed of light. For now, though, just usethe simpler ideas you’ve learned so far.)

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Problem 23. This spectacular series of photos from a 2011 paper by Bur-rows and Sutton (“Biomechanics of jumping in the flea,” J. Exp. Biology214:836) shows the flea jumping at about a 45-degree angle, but for thesake of this estimate just consider the case of a flea jumping vertically.

23 Some fleas can jump as high as 30 cm. The flea only has ashort time to build up speed — the time during which its center ofmass is accelerating upward but its feet are still in contact with theground. Make an order-of-magnitude estimate of the accelerationthe flea needs to have while straightening its legs, and state youranswer in units of g, i.e., how many “g’s it pulls.” (For comparison,fighter pilots black out or die if they exceed about 5 or 10 g’s.)


24 The speed required for a low-earth orbit is 7.9×103 m/s (seech. 10). When a rocket is launched into orbit, it goes up a little atfirst to get above almost all of the atmosphere, but then tips overhorizontally to build up to orbital speed. Suppose the horizontalacceleration is limited to 3g to keep from damaging the cargo (orhurting the crew, for a crewed flight). (a) What is the minimumdistance the rocket must travel downrange before it reaches orbitalspeed? How much does it matter whether you take into account theinitial eastward velocity due to the rotation of the earth? (b) Ratherthan a rocket ship, it might be advantageous to use a railgun design,in which the craft would be accelerated to orbital speeds along arailroad track. This has the advantage that it isn’t necessary to lifta large mass of fuel, since the energy source is external. Based onyour answer to part a, comment on the feasibility of this design forcrewed launches from the earth’s surface.

25 A person is parachute jumping. During the time betweenwhen she leaps out of the plane and when she opens her chute, heraltitude is given by an equation of the form

y = b− c(t+ ke−t/k

),

where e is the base of natural logarithms, and b, c, and k are con-stants. Because of air resistance, her velocity does not increase at asteady rate as it would for an object falling in vacuum.(a) What units would b, c, and k have to have for the equation tomake sense?(b) Find the person’s velocity, v, as a function of time. [You willneed to use the chain rule, and the fact that d(ex)/dx = ex.]

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(c) Use your answer from part (b) to get an interpretation of theconstant c. [Hint: e−x approaches zero for large values of x.](d) Find the person’s acceleration, a, as a function of time.

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(e) Use your answer from part (b) to show that if she waits longenough to open her chute, her acceleration will become very small.

Problems 115

Problem 26.

26 The figure shows a practical, simple experiment for determin-ing g to high precision. Two steel balls are suspended from electro-magnets, and are released simultaneously when the electric currentis shut off. They fall through unequal heights ∆x1 and ∆x2. Acomputer records the sounds through a microphone as first one balland then the other strikes the floor. From this recording, we canaccurately determine the quantity T defined as T = ∆t2 −∆t1, i.e.,the time lag between the first and second impacts. Note that sincethe balls do not make any sound when they are released, we haveno way of measuring the individual times ∆t2 and ∆t1.(a) Find an equation for g in terms of the measured quantities T ,∆x1 and ∆x2.

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(b) Check the units of your equation.(c) Check that your equation gives the correct result in the casewhere ∆x1 is very close to zero. However, is this case realistic?(d) What happens when ∆x1 = ∆x2? Discuss this both mathemat-ically and physically.


Isaac Newton

Chapter 4

Force and motion

If I have seen farther than others, it is because I have stoodon the shoulders of giants.

Newton, referring to Galileo

Even as great and skeptical a genius as Galileo was unable tomake much progress on the causes of motion. It was not until a gen-eration later that Isaac Newton (1642-1727) was able to attack theproblem successfully. In many ways, Newton’s personality was theopposite of Galileo’s. Where Galileo agressively publicized his ideas,

117

a / Aristotle said motion hadto be caused by a force. Toexplain why an arrow kept flyingafter the bowstring was no longerpushing on it, he said the airrushed around behind the arrowand pushed it forward. We knowthis is wrong, because an arrowshot in a vacuum chamber doesnot instantly drop to the flooras it leaves the bow. Galileoand Newton realized that a forcewould only be needed to changethe arrow’s motion, not to makeits motion continue.

Newton had to be coaxed by his friends into publishing a book onhis physical discoveries. Where Galileo’s writing had been popularand dramatic, Newton originated the stilted, impersonal style thatmost people think is standard for scientific writing. (Scientific jour-nals today encourage a less ponderous style, and papers are oftenwritten in the first person.) Galileo’s talent for arousing animos-ity among the rich and powerful was matched by Newton’s skill atmaking himself a popular visitor at court. Galileo narrowly escapedbeing burned at the stake, while Newton had the good fortune of be-ing on the winning side of the revolution that replaced King JamesII with William and Mary of Orange, leading to a lucrative postrunning the English royal mint.

Newton discovered the relationship between force and motion,and revolutionized our view of the universe by showing that thesame physical laws applied to all matter, whether living or nonliv-ing, on or off of our planet’s surface. His book on force and motion,the Mathematical Principles of Natural Philosophy, was un-contradicted by experiment for 200 years, but his other main work,Optics, was on the wrong track, asserting that light was composedof particles rather than waves. Newton was also an avid alchemist,a fact that modern scientists would like to forget.

4.1 Force

We need only explain changes in motion, not motion itself.

So far you’ve studied the measurement of motion in some detail,but not the reasons why a certain object would move in a certainway. This chapter deals with the “why” questions. Aristotle’s ideasabout the causes of motion were completely wrong, just like all hisother ideas about physical science, but it will be instructive to startwith them, because they amount to a road map of modern students’incorrect preconceptions.

Aristotle thought he needed to explain both why motion occursand why motion might change. Newton inherited from Galileo theimportant counter-Aristotelian idea that motion needs no explana-tion, that it is only changes in motion that require a physical cause.Aristotle’s needlessly complex system gave three reasons for motion:

Natural motion, such as falling, came from the tendency ofobjects to go to their “natural” place, on the ground, andcome to rest.

Voluntary motion was the type of motion exhibited by ani-mals, which moved because they chose to.

Forced motion occurred when an object was acted on by someother object that made it move.

118 Chapter 4 Force and motion

b / “Our eyes receive bluelight reflected from this paintingbecause Monet wanted to repre-sent water with the color blue.”This is a valid statement at onelevel of explanation, but physicsworks at the physical level ofexplanation, in which blue lightgets to your eyes because it isreflected by blue pigments in thepaint.

Motion changes due to an interaction between two objects.

In the Aristotelian theory, natural motion and voluntary mo-tion are one-sided phenomena: the object causes its own motion.Forced motion is supposed to be a two-sided phenomenon, becauseone object imposes its “commands” on another. Where Aristotleconceived of some of the phenomena of motion as one-sided andothers as two-sided, Newton realized that a change in motion wasalways a two-sided relationship of a force acting between two phys-ical objects.

The one-sided “natural motion” description of falling makes acrucial omission. The acceleration of a falling object is not causedby its own “natural” tendencies but by an attractive force betweenit and the planet Earth. Moon rocks brought back to our planet donot “want” to fly back up to the moon because the moon is their“natural” place. They fall to the floor when you drop them, justlike our homegrown rocks. As we’ll discuss in more detail later inthis course, gravitational forces are simply an attraction that occursbetween any two physical objects. Minute gravitational forces caneven be measured between human-scale objects in the laboratory.

The idea of natural motion also explains incorrectly why thingscome to rest. A basketball rolling across a beach slows to a stopbecause it is interacting with the sand via a frictional force, notbecause of its own desire to be at rest. If it was on a frictionlesssurface, it would never slow down. Many of Aristotle’s mistakesstemmed from his failure to recognize friction as a force.

The concept of voluntary motion is equally flawed. You mayhave been a little uneasy about it from the start, because it assumesa clear distinction between living and nonliving things. Today, how-ever, we are used to having the human body likened to a complexmachine. In the modern world-view, the border between the livingand the inanimate is a fuzzy no-man’s land inhabited by viruses,prions, and silicon chips. Furthermore, Aristotle’s statement thatyou can take a step forward “because you choose to” inappropriatelymixes two levels of explanation. At the physical level of explana-tion, the reason your body steps forward is because of a frictionalforce acting between your foot and the floor. If the floor was coveredwith a puddle of oil, no amount of “choosing to” would enable youto take a graceful stride forward.

Forces can all be measured on the same numerical scale.

In the Aristotelian-scholastic tradition, the description of mo-tion as natural, voluntary, or forced was only the broadest level ofclassification, like splitting animals into birds, reptiles, mammals,and amphibians. There might be thousands of types of motion,each of which would follow its own rules. Newton’s realization thatall changes in motion were caused by two-sided interactions made

Section 4.1 Force 119

it seem that the phenomena might have more in common than hadbeen apparent. In the Newtonian description, there is only one causefor a change in motion, which we call force. Forces may be of differ-ent types, but they all produce changes in motion according to thesame rules. Any acceleration that can be produced by a magneticforce can equally well be produced by an appropriately controlledstream of water. We can speak of two forces as being equal if theyproduce the same change in motion when applied in the same situ-ation, which means that they pushed or pulled equally hard in thesame direction.

The idea of a numerical scale of force and the newton unit wereintroduced in chapter 0. To recapitulate briefly, a force is when apair of objects push or pull on each other, and one newton is theforce required to accelerate a 1-kg object from rest to a speed of 1m/s in 1 second.

More than one force on an object

As if we hadn’t kicked poor Aristotle around sufficiently, histheory has another important flaw, which is important to discussbecause it corresponds to an extremely common student misconcep-tion. Aristotle conceived of forced motion as a relationship in whichone object was the boss and the other “followed orders.” It there-fore would only make sense for an object to experience one force ata time, because an object couldn’t follow orders from two sources atonce. In the Newtonian theory, forces are numbers, not orders, andif more than one force acts on an object at once, the result is foundby adding up all the forces. It is unfortunate that the use of theEnglish word “force” has become standard, because to many peopleit suggests that you are “forcing” an object to do something. Theforce of the earth’s gravity cannot “force” a boat to sink, becausethere are other forces acting on the boat. Adding them up gives atotal of zero, so the boat accelerates neither up nor down.

Objects can exert forces on each other at a distance.

Aristotle declared that forces could only act between objects thatwere touching, probably because he wished to avoid the type of oc-cult speculation that attributed physical phenomena to the influenceof a distant and invisible pantheon of gods. He was wrong, however,as you can observe when a magnet leaps onto your refrigerator orwhen the planet earth exerts gravitational forces on objects that arein the air. Some types of forces, such as friction, only operate be-tween objects in contact, and are called contact forces. Magnetism,on the other hand, is an example of a noncontact force. Althoughthe magnetic force gets stronger when the magnet is closer to yourrefrigerator, touching is not required.


c / Forces are applied to asaxophone. In this example,positive signs have been usedconsistently for forces to theright, and negative signs forforces to the left. (The forcesare being applied to differentplaces on the saxophone, but thenumerical value of a force carriesno information about that.)

Weight

In physics, an object’s weight, FW , is defined as the earth’sgravitational force on it. The SI unit of weight is therefore theNewton. People commonly refer to the kilogram as a unit of weight,but the kilogram is a unit of mass, not weight. Note that an object’sweight is not a fixed property of that object. Objects weigh morein some places than in others, depending on the local strength ofgravity. It is their mass that always stays the same. A baseballpitcher who can throw a 90-mile-per-hour fastball on earth wouldnot be able to throw any faster on the moon, because the ball’sinertia would still be the same.

Positive and negative signs of force

We’ll start by considering only cases of one-dimensional center-of-mass motion in which all the forces are parallel to the direction ofmotion, i.e., either directly forward or backward. In one dimension,plus and minus signs can be used to indicate directions of forces, asshown in figure c. We can then refer generically to addition of forces,rather than having to speak sometimes of addition and sometimes ofsubtraction. We add the forces shown in the figure and get 11 N. Ingeneral, we should choose a one-dimensional coordinate system withits x axis parallel the direction of motion. Forces that point alongthe positive x axis are positive, and forces in the opposite directionare negative. Forces that are not directly along the x axis cannot beimmediately incorporated into this scheme, but that’s OK, becausewe’re avoiding those cases for now.


A In chapter 0, I defined 1 N as the force that would accelerate a1-kg mass from rest to 1 m/s in 1 s. Anticipating the following section, youmight guess that 2 N could be defined as the force that would acceleratethe same mass to twice the speed, or twice the mass to the same speed.Is there an easier way to define 2 N based on the definition of 1 N?

4.2 Newton’s first lawWe are now prepared to make a more powerful restatement of theprinciple of inertia.

Newton’s first lawIf the total force on an object is zero, its center of mass continuesin the same state of motion.

In other words, an object initially at rest is predicted to remainat rest if the total force on it is zero, and an object in motion remainsin motion with the same velocity in the same direction. The converseof Newton’s first law is also true: if we observe an object movingwith constant velocity along a straight line, then the total force on

Section 4.2 Newton’s first law 121

it must be zero.

In a future physics course or in another textbook, you may en-counter the term “net force,” which is simply a synonym for totalforce.

What happens if the total force on an object is not zero? Itaccelerates. Numerical prediction of the resulting acceleration is thetopic of Newton’s second law, which we’ll discuss in the followingsection.

This is the first of Newton’s three laws of motion. It is notimportant to memorize which of Newton’s three laws are numbersone, two, and three. If a future physics teacher asks you somethinglike, “Which of Newton’s laws are you thinking of?,” a perfectlyacceptable answer is “The one about constant velocity when there’szero total force.” The concepts are more important than any spe-cific formulation of them. Newton wrote in Latin, and I am notaware of any modern textbook that uses a verbatim translation ofhis statement of the laws of motion. Clear writing was not in voguein Newton’s day, and he formulated his three laws in terms of a con-cept now called momentum, only later relating it to the concept offorce. Nearly all modern texts, including this one, start with forceand do momentum later.

An elevator example 1. An elevator has a weight of 5000 N. Compare the forces that thecable must exert to raise it at constant velocity, lower it at constantvelocity, and just keep it hanging.

. In all three cases the cable must pull up with a force of exactly5000 N. Most people think you’d need at least a little more than5000 N to make it go up, and a little less than 5000 N to let it down,but that’s incorrect. Extra force from the cable is only necessaryfor speeding the car up when it starts going up or slowing it downwhen it finishes going down. Decreased force is needed to speedthe car up when it gets going down and to slow it down when itfinishes going up. But when the elevator is cruising at constantvelocity, Newton’s first law says that you just need to cancel theforce of the earth’s gravity.

To many students, the statement in the example that the cable’supward force “cancels” the earth’s downward gravitational force im-plies that there has been a contest, and the cable’s force has won,vanquishing the earth’s gravitational force and making it disappear.That is incorrect. Both forces continue to exist, but because theyadd up numerically to zero, the elevator has no center-of-mass ac-celeration. We know that both forces continue to exist because theyboth have side-effects other than their effects on the car’s center-of-mass motion. The force acting between the cable and the car con-tinues to produce tension in the cable and keep the cable taut. The


earth’s gravitational force continues to keep the passengers (whomwe are considering as part of the elevator-object) stuck to the floorand to produce internal stresses in the walls of the car, which musthold up the floor.

Terminal velocity for falling objects example 2. An object like a feather that is not dense or streamlined does notfall with constant acceleration, because air resistance is nonneg-ligible. In fact, its acceleration tapers off to nearly zero within afraction of a second, and the feather finishes dropping at constantspeed (known as its terminal velocity). Why does this happen?

. Newton’s first law tells us that the total force on the feather musthave been reduced to nearly zero after a short time. There aretwo forces acting on the feather: a downward gravitational forcefrom the planet earth, and an upward frictional force from the air.As the feather speeds up, the air friction becomes stronger andstronger, and eventually it cancels out the earth’s gravitationalforce, so the feather just continues with constant velocity withoutspeeding up any more.

The situation for a skydiver is exactly analogous. It’s just that theskydiver experiences perhaps a million times more gravitationalforce than the feather, and it is not until she is falling very fastthat the force of air friction becomes as strong as the gravita-tional force. It takes her several seconds to reach terminal veloc-ity, which is on the order of a hundred miles per hour.

More general combinations of forces

It is too constraining to restrict our attention to cases whereall the forces lie along the line of the center of mass’s motion. Forone thing, we can’t analyze any case of horizontal motion, sinceany object on earth will be subject to a vertical gravitational force!For instance, when you are driving your car down a straight road,there are both horizontal forces and vertical forces. However, thevertical forces have no effect on the center of mass motion, becausethe road’s upward force simply counteracts the earth’s downwardgravitational force and keeps the car from sinking into the ground.

Later in the book we’ll deal with the most general case of manyforces acting on an object at any angles, using the mathematicaltechnique of vector addition, but the following slight generalizationof Newton’s first law allows us to analyze a great many cases ofinterest:

Suppose that an object has two sets of forces acting on it, oneset along the line of the object’s initial motion and another set per-pendicular to the first set. If both sets of forces cancel, then theobject’s center of mass continues in the same state of motion.

Section 4.2 Newton’s first law 123

d / Example 4.

A passenger riding the subway example 3. Describe the forces acting on a person standing in a subwaytrain that is cruising at constant velocity.

. No force is necessary to keep the person moving relative tothe ground. He will not be swept to the back of the train if thefloor is slippery. There are two vertical forces on him, the earth’sdownward gravitational force and the floor’s upward force, whichcancel. There are no horizontal forces on him at all, so of coursethe total horizontal force is zero.

Forces on a sailboat example 4. If a sailboat is cruising at constant velocity with the wind comingfrom directly behind it, what must be true about the forces actingon it?

. The forces acting on the boat must be canceling each otherout. The boat is not sinking or leaping into the air, so evidentlythe vertical forces are canceling out. The vertical forces are thedownward gravitational force exerted by the planet earth and anupward force from the water.

The air is making a forward force on the sail, and if the boat isnot accelerating horizontally then the water’s backward frictionalforce must be canceling it out.

Contrary to Aristotle, more force is not needed in order to maintaina higher speed. Zero total force is always needed to maintainconstant velocity. Consider the following made-up numbers:

boat moving ata low, constantvelocity

boat moving ata high, constantvelocity

forward force ofthe wind on thesail . . .

10,000 N 20,000 N

backward force ofthe water on thehull . . .

−10, 000 N −20, 000 N

total force on theboat . . .

0 N 0 N

The faster boat still has zero total force on it. The forward forceon it is greater, and the backward force smaller (more negative),but that’s irrelevant because Newton’s first law has to do with thetotal force, not the individual forces.

This example is quite analogous to the one about terminal velocityof falling objects, since there is a frictional force that increaseswith speed. After casting off from the dock and raising the sail,the boat will accelerate briefly, and then reach its terminal velocity,at which the water’s frictional force has become as great as thewind’s force on the sail.




A car crash example 5. If you drive your car into a brick wall, what is the mysteriousforce that slams your face into the steering wheel?

. Your surgeon has taken physics, so she is not going to believeyour claim that a mysterious force is to blame. She knows thatyour face was just following Newton’s first law. Immediately afteryour car hit the wall, the only forces acting on your head werethe same canceling-out forces that had existed previously: theearth’s downward gravitational force and the upward force fromyour neck. There were no forward or backward forces on yourhead, but the car did experience a backward force from the wall,so the car slowed down and your face caught up.


A Newton said that objects continue moving if no forces are actingon them, but his predecessor Aristotle said that a force was necessary tokeep an object moving. Why does Aristotle’s theory seem more plausible,even though we now believe it to be wrong? What insight was Aristotlemissing about the reason why things seem to slow down naturally? Givean example.

B In the figure what would have to be true about the saxophone’s initialmotion if the forces shown were to result in continued one-dimensionalmotion of its center of mass?

C This figure requires an ever further generalization of the precedingdiscussion. After studying the forces, what does your physical intuition tellyou will happen? Can you state in words how to generalize the conditionsfor one-dimensional motion to include situations like this one?

4.3 Newton’s second lawWhat about cases where the total force on an object is not zero,so that Newton’s first law doesn’t apply? The object will have anacceleration. The way we’ve defined positive and negative signsof force and acceleration guarantees that positive forces producepositive accelerations, and likewise for negative values. How muchacceleration will it have? It will clearly depend on both the object’smass and on the amount of force.

Experiments with any particular object show that its acceler-ation is directly proportional to the total force applied to it. Thismay seem wrong, since we know of many cases where small amountsof force fail to move an object at all, and larger forces get it going.This apparent failure of proportionality actually results from for-getting that there is a frictional force in addition to the force weapply to move the object. The object’s acceleration is exactly pro-portional to the total force on it, not to any individual force on it.In the absence of friction, even a very tiny force can slowly changethe velocity of a very massive object.

Section 4.3 Newton’s second law 125

Experiments also show that the acceleration is inversely propor-tional to the object’s mass, and combining these two proportional-ities gives the following way of predicting the acceleration of anyobject:

Newton’s second law

a = Ftotal/m ,

where

m is an object’s mass

Ftotal is the sum of the forces acting on it, and

a is the acceleration of the object’s center of mass.

We are presently restricted to the case where the forces of interestare parallel to the direction of motion.

An accelerating bus example 6. A VW bus with a mass of 2000 kg accelerates from 0 to 25 m/s(freeway speed) in 34 s. Assuming the acceleration is constant,what is the total force on the bus?

. We solve Newton’s second law for Ftotal = ma, and substitute∆v/∆t for a, giving

Ftotal = m∆v/∆t= (2000 kg)(25 m/s− 0 m/s)/(34 s)= 1.5 kN .

Some applications of calculus

Newton doesn’t care what frame of reference you use his lawsin, and this makes him different from Aristotle, who says there issomething special about the frame of reference attached firmly tothe dirt underfoot. Suppose that an object obeys Newton’s secondlaw in the dirt’s frame. It has some velocity that is a function oftime, and differentiating this function gives dv/dt = F/m. Supposewe change to the frame of reference of a train that is in motionrelative to the dirt at constant velocity c. Looking out the windowof the train, we see the object moving with velocity v − c. But thederivative of a constant is zero, so when we differentiate v − c, theconstant goes away, and we get exactly the same result. Newton isstill happy, although Aristotle feels a great disturbance in the force.


Often we know the forces acting on an object, and we want tofind its motion, i.e., its position as a function of time, x(t). SinceNewton’s second law predicts the acceleration d2x/dt2, we need tointegrate twice to find x. The first integration gives the velocity,and the constant of integration is also a velocity, which can be fixedby giving the object’s velocity at some initial time. In the secondintegration we pick up a second constant of integration, this onerelated to an initial position.

A force that tapers off to zero example 7. An object of mass m starts at rest at t = to. A force varying asF = bt−2, where b is a constant, begins acting on it. Find thegreatest speed it will ever have.

.

F = mdvdt

dv =Fm

dt∫dv =

∫Fm

dt

v = − bm

t−1 + vo ,

where vo is a constant of integration. The given initial condition isthat v = 0 at t = to, so we find that vo = b/mto. The negative termgets closer to zero with increasing time, so the maximum velocityis achieved by letting t approach infinity. That is, the object willnever stop speeding up, but it will also never surpass a certainspeed. In the limit t → ∞, we identify vo as the velocity that theobject will approach asymptotically.

A generalization

As with the first law, the second law can be easily generalizedto include a much larger class of interesting situations:

Suppose an object is being acted on by two sets of forces, oneset lying parallel to the object’s initial direction of motion andanother set acting along a perpendicular line. If the forcesperpendicular to the initial direction of motion cancel out,then the object accelerates along its original line of motionaccording to a = F‖/m, where F‖ is the sum of the forcesparallel to the line.


e / A coin slides across a ta-ble. Even for motion in onedimension, some of the forcesmay not lie along the line of themotion.

f / A simple double-pan bal-ance works by comparing theweight forces exerted by theearth on the contents of the twopans. Since the two pans areat almost the same location onthe earth’s surface, the valueof g is essentially the same foreach one, and equality of weighttherefore also implies equality ofmass.

g / Example 9.

A coin sliding across a table example 8Suppose a coin is sliding to the right across a table, e, and let’schoose a positive x axis that points to the right. The coin’s velocityis positive, and we expect based on experience that it will slowdown, i.e., its acceleration should be negative.

Although the coin’s motion is purely horizontal, it feels both ver-tical and horizontal forces. The Earth exerts a downward gravi-tational force F2 on it, and the table makes an upward force F3that prevents the coin from sinking into the wood. In fact, withoutthese vertical forces the horizontal frictional force wouldn’t exist:surfaces don’t exert friction against one another unless they arebeing pressed together.

Although F2 and F3 contribute to the physics, they do so onlyindirectly. The only thing that directly relates to the accelerationalong the horizontal direction is the horizontal force: a = F1/m.

The relationship between mass and weight

Mass is different from weight, but they’re related. An apple’smass tells us how hard it is to change its motion. Its weight measuresthe strength of the gravitational attraction between the apple andthe planet earth. The apple’s weight is less on the moon, but itsmass is the same. Astronauts assembling the International SpaceStation in zero gravity cannot just pitch massive modules back andforth with their bare hands; the modules are weightless, but notmassless.

We have already seen the experimental evidence that when weight(the force of the earth’s gravity) is the only force acting on an ob-ject, its acceleration equals the constant g, and g depends on whereyou are on the surface of the earth, but not on the mass of the ob-ject. Applying Newton’s second law then allows us to calculate themagnitude of the gravitational force on any object in terms of itsmass:

|FW | = mg .

(The equation only gives the magnitude, i.e. the absolute value, ofFW , because we’re defining g as a positive number, so it equals theabsolute value of a falling object’s acceleration.)

. Solved problem: Decelerating a car page 141, problem 1

Weight and mass example 9. Figure g shows masses of one and two kilograms hung from aspring scale, which measures force in units of newtons. Explainthe readings.

. Let’s start with the single kilogram. It’s not accelerating, soevidently the total force on it is zero: the spring scale’s upwardforce on it is canceling out the earth’s downward gravitationalforce. The spring scale tells us how much force it is being obliged


to supply, but since the two forces are equal in strength, thespring scale’s reading can also be interpreted as measuring thestrength of the gravitational force, i.e., the weight of the one-kilogram mass. The weight of a one-kilogram mass should be

FW = mg

= (1.0 kg)(9.8 m/s2) = 9.8 N ,

and that’s indeed the reading on the spring scale.

Similarly for the two-kilogram mass, we have

FW = mg

= (2.0 kg)(9.8 m/s2) = 19.6 N .

Calculating terminal velocity example 10. Experiments show that the force of air friction on a falling objectsuch as a skydiver or a feather can be approximated fairly wellwith the equation |Fair | = cρAv2, where c is a constant, ρ is thedensity of the air, A is the cross-sectional area of the object asseen from below, and v is the object’s velocity. Predict the object’sterminal velocity, i.e., the final velocity it reaches after a long time.

. As the object accelerates, its greater v causes the upward forceof the air to increase until finally the gravitational force and theforce of air friction cancel out, after which the object continuesat constant velocity. We choose a coordinate system in whichpositive is up, so that the gravitational force is negative and theforce of air friction is positive. We want to find the velocity at which

Fair + FW = 0 , i .e.,

cρAv2 −mg = 0 .

Solving for v gives

vterminal =√

mgcρA

self-check AIt is important to get into the habit of interpreting equations. This may bedifficult at first, but eventually you will get used to this kind of reasoning.

(1) Interpret the equation vterminal =√

mg/cρA in the case of ρ=0.

(2) How would the terminal velocity of a 4-cm steel ball compare to thatof a 1-cm ball?

(3) In addition to teasing out the mathematical meaning of an equation,we also have to be able to place it in its physical context. How generallyimportant is this equation? . Answer, p. 509


x (m) t (s)10 1.8420 2.8630 3.8040 4.6750 5.5360 6.3870 7.2380 8.1090 8.96100 9.83

h / Discussion question D.


A Show that the Newton can be reexpressed in terms of the threebasic mks units as the combination kg·m/s2.

B What is wrong with the following statements?

(1) “g is the force of gravity.”

(2) “Mass is a measure of how much space something takes up.”

C Criticize the following incorrect statement:

“If an object is at rest and the total force on it is zero, it stays at rest.There can also be cases where an object is moving and keeps on movingwithout having any total force on it, but that can only happen when there’sno friction, like in outer space.”

D Table h gives laser timing data for Ben Johnson’s 100 m dash at the1987 World Championship in Rome. (His world record was later revokedbecause he tested positive for steroids.) How does the total force on himchange over the duration of the race?


4.4 What force is notViolin teachers have to endure their beginning students’ screeching.A frown appears on the woodwind teacher’s face as she watches herstudent take a breath with an expansion of his ribcage but nonein his belly. What makes physics teachers cringe is their students’verbal statements about forces. Below I have listed several dictaabout what force is not.

Force is not a property of one object.

A great many of students’ incorrect descriptions of forces couldbe cured by keeping in mind that a force is an interaction of twoobjects, not a property of one object.

Incorrect statement: “That magnet has a lot of force.”

If the magnet is one millimeter away from a steel ball bearing, theymay exert a very strong attraction on each other, but if they were ameter apart, the force would be virtually undetectable. The magnet’sstrength can be rated using certain electrical units (ampere−meters2),but not in units of force.

Force is not a measure of an object’s motion.

If force is not a property of a single object, then it cannot beused as a measure of the object’s motion.

Incorrect statement: “The freight train rumbled down the tracks withawesome force.”

Force is not a measure of motion. If the freight train collides with astalled cement truck, then some awesome forces will occur, but if it hitsa fly the force will be small.

Force is not energy.

There are two main approaches to understanding the motion ofobjects, one based on force and one on a different concept, called en-ergy. The SI unit of energy is the Joule, but you are probably morefamiliar with the calorie, used for measuring food’s energy, and thekilowatt-hour, the unit the electric company uses for billing you.Physics students’ previous familiarity with calories and kilowatt-hours is matched by their universal unfamiliarity with measuringforces in units of Newtons, but the precise operational definitions ofthe energy concepts are more complex than those of the force con-cepts, and textbooks, including this one, almost universally place theforce description of physics before the energy description. Duringthe long period after the introduction of force and before the carefuldefinition of energy, students are therefore vulnerable to situationsin which, without realizing it, they are imputing the properties ofenergy to phenomena of force.

Incorrect statement: “How can my chair be making an upward force onmy rear end? It has no power!”

Power is a concept related to energy, e.g., a 100-watt lightbulb uses

Section 4.4 What force is not 131

up 100 joules per second of energy. When you sit in a chair, no energyis used up, so forces can exist between you and the chair without anyneed for a source of power.

Force is not stored or used up.

Because energy can be stored and used up, people think forcealso can be stored or used up.

Incorrect statement: “If you don’t fill up your tank with gas, you’ll runout of force.”

Energy is what you’ll run out of, not force.

Forces need not be exerted by living things or machines.

Transforming energy from one form into another usually requiressome kind of living or mechanical mechanism. The concept is notapplicable to forces, which are an interaction between objects, nota thing to be transferred or transformed.

Incorrect statement: “How can a wooden bench be making an upwardforce on my rear end? It doesn’t have any springs or anything inside it.”

No springs or other internal mechanisms are required. If the benchdidn’t make any force on you, you would obey Newton’s second law andfall through it. Evidently it does make a force on you!

A force is the direct cause of a change in motion.

I can click a remote control to make my garage door change frombeing at rest to being in motion. My finger’s force on the button,however, was not the force that acted on the door. When we speakof a force on an object in physics, we are talking about a force thatacts directly. Similarly, when you pull a reluctant dog along by itsleash, the leash and the dog are making forces on each other, notyour hand and the dog. The dog is not even touching your hand.

self-check BWhich of the following things can be correctly described in terms offorce?

(1) A nuclear submarine is charging ahead at full steam.

(2) A nuclear submarine’s propellers spin in the water.

(3) A nuclear submarine needs to refuel its reactor periodically. .

Answer, p. 509


A Criticize the following incorrect statement: “If you shove a bookacross a table, friction takes away more and more of its force, until finallyit stops.”

B You hit a tennis ball against a wall. Explain any and all incorrectideas in the following description of the physics involved: “The ball getssome force from you when you hit it, and when it hits the wall, it loses partof that force, so it doesn’t bounce back as fast. The muscles in your armare the only things that a force can come from.”


4.5 Inertial and noninertial frames of referenceOne day, you’re driving down the street in your pickup truck, onyour way to deliver a bowling ball. The ball is in the back of thetruck, enjoying its little jaunt and taking in the fresh air and sun-shine. Then you have to slow down because a stop sign is comingup. As you brake, you glance in your rearview mirror, and see yourtrusty companion accelerating toward you. Did some mysteriousforce push it forward? No, it only seems that way because you andthe car are slowing down. The ball is faithfully obeying Newton’sfirst law, and as it continues at constant velocity it gets ahead rela-tive to the slowing truck. No forces are acting on it (other than thesame canceling-out vertical forces that were always acting on it).1

The ball only appeared to violate Newton’s first law because therewas something wrong with your frame of reference, which was basedon the truck.

i / 1. In a frame of reference thatmoves with the truck, the bowl-ing ball appears to violate New-ton’s first law by accelerating de-spite having no horizontal forceson it. 2. In an inertial frame of ref-erence, which the surface of theearth approximately is, the bowl-ing ball obeys Newton’s first law.It moves equal distances in equaltime intervals, i.e., maintains con-stant velocity. In this frame ofreference, it is the truck that ap-pears to have a change in veloc-ity, which makes sense, since theroad is making a horizontal forceon it.

How, then, are we to tell in which frames of reference Newton’slaws are valid? It’s no good to say that we should avoid movingframes of reference, because there is no such thing as absolute restor absolute motion. All frames can be considered as being either atrest or in motion. According to an observer in India, the strip mallthat constituted the frame of reference in panel (b) of the figurewas moving along with the earth’s rotation at hundreds of miles perhour.

The reason why Newton’s laws fail in the truck’s frame of refer-

1Let’s assume for simplicity that there is no friction.

Section 4.5 Inertial and noninertial frames of reference 133

ence is not because the truck is moving but because it is accelerating.(Recall that physicists use the word to refer either to speeding up orslowing down.) Newton’s laws were working just fine in the movingtruck’s frame of reference as long as the truck was moving at con-stant velocity. It was only when its speed changed that there wasa problem. How, then, are we to tell which frames are acceleratingand which are not? What if you claim that your truck is not ac-celerating, and the sidewalk, the asphalt, and the Burger King areaccelerating? The way to settle such a dispute is to examine themotion of some object, such as the bowling ball, which we knowhas zero total force on it. Any frame of reference in which the ballappears to obey Newton’s first law is then a valid frame of reference,and to an observer in that frame, Mr. Newton assures us that allthe other objects in the universe will obey his laws of motion, notjust the ball.

Valid frames of reference, in which Newton’s laws are obeyed,are called inertial frames of reference. Frames of reference that arenot inertial are called noninertial frames. In those frames, objectsviolate the principle of inertia and Newton’s first law. While thetruck was moving at constant velocity, both it and the sidewalkwere valid inertial frames. The truck became an invalid frame ofreference when it began changing its velocity.

You usually assume the ground under your feet is a perfectlyinertial frame of reference, and we made that assumption above. Itisn’t perfectly inertial, however. Its motion through space is quitecomplicated, being composed of a part due to the earth’s daily rota-tion around its own axis, the monthly wobble of the planet causedby the moon’s gravity, and the rotation of the earth around the sun.Since the accelerations involved are numerically small, the earth isapproximately a valid inertial frame.

Noninertial frames are avoided whenever possible, and we willseldom, if ever, have occasion to use them in this course. Sometimes,however, a noninertial frame can be convenient. Naval gunners, forinstance, get all their data from radars, human eyeballs, and otherdetection systems that are moving along with the earth’s surface.Since their guns have ranges of many miles, the small discrepan-cies between their shells’ actual accelerations and the accelerationspredicted by Newton’s second law can have effects that accumulateand become significant. In order to kill the people they want to kill,they have to add small corrections onto the equation a = Ftotal/m.Doing their calculations in an inertial frame would allow them touse the usual form of Newton’s second law, but they would haveto convert all their data into a different frame of reference, whichwould require cumbersome calculations.

Discussion question

A If an object has a linear x − t graph in a certain inertial frame,


what is the effect on the graph if we change to a coordinate system witha different origin? What is the effect if we keep the same origin but re-verse the positive direction of the x axis? How about an inertial framemoving alongside the object? What if we describe the object’s motion ina noninertial frame?

Section 4.5 Inertial and noninertial frames of reference 135

4.6 Numerical techniquesEngineering majors are a majority of the students in the kind ofphysics course for which this book is designed, so most likely youfall into that category. Although you surely recognize that physicsis an important part of your training, if you’ve had any exposureto how engineers really work, you’re probably skeptical about theflavor of problem-solving taught in most science courses. You real-ize that not very many practical engineering calculations fall intothe narrow range of problems for which an exact solution can becalculated with a piece of paper and a sharp pencil. Real-life prob-lems are usually complicated, and typically they need to be solvedby number-crunching on a computer, although we can often gaininsight by working simple approximations that have algebraic solu-tions. Not only is numerical problem-solving more useful in real life,it’s also educational; as a beginning physics student, I really onlyfelt like I understood projectile motion after I had worked it bothways, using algebra and then a computer program.

In this section, we’ll start by seeing how to apply numericaltechniques to some simple problems for which we know the answer in“closed form,” i.e., a single algebraic expression without any calculusor infinite sums. After that, we’ll solve a problem that would havemade you world-famous if you could have done it in the seventeenthcentury using paper and a quill pen! Before you continue, you shouldread Appendix 1 on page 492 that introduces you to the Pythonprogramming language.

First let’s solve the trivial problem of finding the distance trav-eled by an object moving at speed v to in time t. This closed-formanswer is, of course, vt, but the point is to introduce the techniqueswe can use to solve other problems of this type. The basic idea isto divide the time up into n equal parts, and add up the distancestraveled in all the parts. The following Python function does thejob. Note that you shouldn’t type in the line numbers on the left,and you don’t need to type in the comments, either.

1 import math

2 def dist(n):

3 t = 1.0 # seconds

4 v = 1.0 # m/s

5 x=0 # Initialize the position.

6 dt = t/n # Divide t into n equal parts.

7 for i in range(n):

8 dx = v*dt # tiny distance traveled in dt

9 x = x+dx # Change x.

10 return x

Of course line 5 shows how silly this example is — if we knew dx =vdt, then presumably we knew x = vt, which was the answer to


the whole problem — but the point is to illustrate the techniquewith the simplest possible example. How far do we move in 1 s at aconstant speed of 1 m/s? If we do this,

>>> print(dist(10))

1.0

Python produces the expected answer by dividing the time into tenequal 0.1-second intervals, and adding up the ten 0.1-meter segmentstraversed in them. Since the object moves at constant speed, itdoesn’t even matter whether we set n to 10, 1, or a million:

>>> print(dist(1))

1.0

Now let’s do an example where the answer isn’t obvious to peoplewho don’t know calculus: through what distance does an object fallin 1.0 s, starting from rest? By integrating a = g to find v = gtand the integrating again to get x = (1/2)gt2, we know that theexact answer is 4.9 m. Let’s see if we can reproduce that answernumerically. The main difference between this program and theprevious one is that now the velocity isn’t constant, so we need toupdate it as we go along.

1 import math

2 def dist2(n):

3 t = 1.0 # seconds

4 g=9.8 # strength of gravity, in m/s2

5 x=0 # Initialize the distance fallen.

6 v=0 # Initialize the velocity.

7 dt = t/n # Divide t into n equal parts.


9 dx = v*dt # tiny distance traveled during tiny time dt


11 dv = g*dt # tiny change in vel. during tiny time dt

12 v = v+dv

13 return x

With the drop split up into only 10 equal height intervals, the nu-merical technique provides a decent approximation:

>>> print(dist2(10))

4.41

By increasing n to ten thousand, we get an answer that’s as close aswe need, given the limited accuracy of the raw data:

>>> print(dist2(10000))

4.89951

Section 4.6 Numerical techniques 137

Now let’s use these techniques to solve the following somewhatwhimsical problem, which cannot be solved in closed form usingelementary functions such as sines, exponentials, roots, etc.

Ann E. Hodges of Sylacauga, Alabama is the only person everknown to have been injured by a meteorite. In 1954, she was struckin the hip by a meteorite that crashed through the roof of her houseand struck her in the hip while she was napping on the couch. SinceHodges was asleep, we do not have direct evidence on the followingsilly trivia question: if you’re going to be hit by a meteorite, willyou hear it coming, or will it approach at more than the speed ofsound? To answer this question, we start by constructing a physicalmodel that is as simple as possible. We take the meteor as enteringthe earth’s atmosphere directly along the vertical. The atmospheredoes not cut off suddenly at a certain height; its density can beapproximated as being proportional to e−x/H , where x is the altitudeand H ≈ 7.6 km is called the scale height. The force of air frictionis proportional to the density and to the square of the velocity, so

F = bv2e−x/H

where b is a constant and F is positive in the coordinate system we’vechosen, where +x is up. The constant b depends on the size of theobject, and its mass also affects the acceleration through Newton’asecond law, a = F/m. The answer to the question therefore dependson the size of the meteorite. However, it is reasonable to take theresults for the Sylacauga meteorite as constituting a general answerto our question, since larger ones are very rare, while the muchmore common pebble-sized smaller ones do not make it through theatmosphere before they disintegrate. The object’s initial velocityas it entered the atmosphere is not known, so we assume a typicalvalue of 20 km/s. The Sylacauga meteorite was seen breaking upinto three pieces, only two of which were recovered. The completeobject’s mass was probably about 7 kg and its radius about 9 cm.For an object with this radius, we expect b ≈ 1.5 × 10−3 kg/m.Using Newton’s second law, we find

a =Ftotalm

=bv2e−x/H −mg

m.

I don’t know of any way to solve this to find the function x(t) closedform, so let’s solve it numrically.

This problem is of a slightly different form than the ones above,where we knew we wanted to follow the motion up until a certaintime. This problem is more of an “Are we there yet?” We want tostop the calculation where the altitude reaches zero. If it startedat an initial position x at velocity v (v < 0) and maintained thatvelocity all the way down to sea level, the time interval would be


∆t = ∆x/v = (0 − x)/v = −x/v. Since it actually slows down, ∆twill be greater than that. We guess ten times that as a maximum,and then have the program check each time through the loop to seeif we’ve hit the ground yet. When this happens, we bail out of theloop before completing all n iterations.

1 import math

2 def meteor(n):

3 r = .09

4 m=7 # mass in kg

5 b=1.5e-3 # const. of prop. for friction, kg/m

6 x = 200.*1000. # start at 200 km altitude, far above air

7 v = -20.*1000. # 20 km/s

8 H = 7.6*1000. # scale height in meters

9 g = 9.8 # m/s2

10 t_max = -x/v*10. # guess the longest time it could take

11 dt = t_max/n # Divide t into n equal parts.


13 dx = v*dt


15 if x<0.:

16 return v

17 F = b*v**2*math.exp(-x/H)-m*g

18 a = F/m

19 dv = a*dt

20 v = v+dv

21 return -999. # If we get here, t_max was too short.

The result is:

>>> print(meteor(100000))

-3946.95754982

For comparison, the speed of sound is about 340 m/s. The an-swer is that if you are hit by a meteorite, you will not be able tohear its sound before it hits you.

Section 4.6 Numerical techniques 139

SummarySelected vocabularyweight . . . . . . . the force of gravity on an object, equal to mginertial frame . . a frame of reference that is not accelerating,

one in which Newton’s first law is truenoninertial frame an accelerating frame of reference, in which

Newton’s first law is violated

NotationFW . . . . . . . . weight

Other terminology and notationnet force . . . . . another way of saying “total force”

Summary

Newton’s first law of motion states that if all the forces on anobject cancel each other out, then the object continues in the samestate of motion. This is essentially a more refined version of Galileo’sprinciple of inertia, which did not refer to a numerical scale of force.

Newton’s second law of motion allows the prediction of an ob-ject’s acceleration given its mass and the total force on it, acm =Ftotal/m. This is only the one-dimensional version of the law; thefull-three dimensional treatment will come in chapter 8, Vectors.Without the vector techniques, we can still say that the situationremains unchanged by including an additional set of vectors thatcancel among themselves, even if they are not in the direction ofmotion.

Newton’s laws of motion are only true in frames of reference thatare not accelerating, known as inertial frames.

Even in one-dimensional motion, it is seldom possible to solvereal-world problems and predict the motion of an object in closedform. However, there are straightforward numerical techniques forsolving such problems.


Problem 4, part c.

ProblemsKey√



1 A car is accelerating forward along a straight road. If the forceof the road on the car’s wheels, pushing it forward, is a constant 3.0kN, and the car’s mass is 1000 kg, then how long will the car taketo go from 20 m/s to 50 m/s? . Solution, p. 499

2 (a) Let T be the maximum tension that an elevator’s cable canwithstand without breaking, i.e., the maximum force it can exert.If the motor is programmed to give the car an acceleration a, whatis the maximum mass that the car can have, including passengers,if the cable is not to break?

√

(b) Interpret the equation you derived in the special cases of a = 0and of a downward acceleration of magnitude g.

(“Interpret” means to analyze the behavior of the equation, andconnect that to reality, as in the self-check on page 129.)

3 An object is observed to be moving at constant speed in acertain direction. Can you conclude that no forces are acting on it?Explain. [Based on a problem by Serway and Faughn.]

4 You are given a large sealed box, and are not allowed to openit. Which of the following experiments measure its mass, and whichmeasure its weight? [Hint: Which experiments would give differentresults on the moon?](a) Put it on a frozen lake, throw a rock at it, and see how fast itscoots away after being hit.(b) Drop it from a third-floor balcony, and measure how loud thesound is when it hits the ground.(c) As shown in the figure, connect it with a spring to the wall, andwatch it vibrate.

. Solution, p. 499

5 While escaping from the palace of the evil Martian emperor,Sally Spacehound jumps from a tower of height h down to theground. Ordinarily the fall would be fatal, but she fires her blasterrifle straight down, producing an upward force of magnitude FB.This force is insufficient to levitate her, but it does cancel out someof the force of gravity. During the time t that she is falling, Sally isunfortunately exposed to fire from the emperor’s minions, and can’tdodge their shots. Let m be her mass, and g the strength of gravityon Mars.(a) Find the time t in terms of the other variables.(b) Check the units of your answer to part a.(c) For sufficiently large values of FB, your answer to part a becomesnonsense — explain what’s going on.

√

Problems 141

Problem 9.

6 At low speeds, every car’s acceleration is limited by traction,not by the engine’s power. Suppose that at low speeds, a certaincar is normally capable of an acceleration of 3 m/s2. If it is towinga trailer with half as much mass as the car itself, what accelerationcan it achieve? [Based on a problem from PSSC Physics.]

7 A helicopter of mass m is taking off vertically. The only forcesacting on it are the earth’s gravitational force and the force, Fair,of the air pushing up on the propeller blades.(a) If the helicopter lifts off at t = 0, what is its vertical speed attime t?(b) Check that the units of your answer to part a make sense.(c) Discuss how your answer to part a depends on all three variables,and show that it makes sense. That is, for each variable, discusswhat would happen to the result if you changed it while keeping theother two variables constant. Would a bigger value give a smallerresult, or a bigger result? Once you’ve figured out this mathematicalrelationship, show that it makes sense physically.(d) Plug numbers into your equation from part a, using m = 2300kg, Fair = 27000 N, and t = 4.0 s.

√

8 A uranium atom deep in the earth spits out an alpha particle.An alpha particle is a fragment of an atom. This alpha particle hasinitial speed v, and travels a distance d before stopping in the earth.(a) Find the force, F , that acted on the particle, in terms of v, d,and its mass, m. Don’t plug in any numbers yet. Assume that theforce was constant.

√

(b) Show that your answer has the right units.(c) Discuss how your answer to part a depends on all three variables,and show that it makes sense. That is, for each variable, discusswhat would happen to the result if you changed it while keeping theother two variables constant. Would a bigger value give a smallerresult, or a bigger result? Once you’ve figured out this mathematicalrelationship, show that it makes sense physically.(d) Evaluate your result for m = 6.7×10−27 kg, v = 2.0×104 km/s,and d = 0.71 mm.

√

9 A blimp is initially at rest, hovering, when at t = 0 the pilotturns on the motor of the propeller. The motor cannot instantlyget the propeller going, but the propeller speeds up steadily. Thesteadily increasing force between the air and the propeller is givenby the equation F = kt, where k is a constant. If the mass of theblimp is m, find its position as a function of time. (Assume thatduring the period of time you’re dealing with, the blimp is not yetmoving fast enough to cause a significant backward force due to airresistance.)

√

10 Some garden shears are like a pair of scissors: one sharp blade


slices past another. In the “anvil” type, however, a sharp bladepresses against a flat one rather than going past it. A gardeningbook says that for people who are not very physically strong, theanvil type can make it easier to cut tough branches, because itconcentrates the force on one side. Evaluate this claim based onNewton’s laws. [Hint: Consider the forces acting on the branch,and the motion of the branch.]

11 In the 1964 Olympics in Tokyo, the best men’s high jumpwas 2.18 m. Four years later in Mexico City, the gold medal in thesame event was for a jump of 2.24 m. Because of Mexico City’saltitude (2400 m), the acceleration of gravity there is lower thanthat in Tokyo by about 0.01 m/s2. Suppose a high-jumper has amass of 72 kg.(a) Compare his mass and weight in the two locations.(b) Assume that he is able to jump with the same initial verticalvelocity in both locations, and that all other conditions are the sameexcept for gravity. How much higher should he be able to jump inMexico City?

√

(Actually, the reason for the big change between ’64 and ’68 was theintroduction of the “Fosbury flop.”) ?

12 The factorial of an integer n, written n!, is defined as theproduct of all the positive integers less than or equal to n. Forexample, 3! = 1 × 2 × 3 = 6. Write a Python program to compute30!. (Python computes integer results with unlimited precision, soyou won’t get any problems with rounding or overflows.) Turn in aprintout of both your program and its output.

13 A ball falls from a height h. Without air resistance, the timeit takes to reach the floor is

√2h/g. Now suppose that air resistance

is not negligible. For a smooth sphere of radius r, moving at speedv through air of density ρ, the force of air resistance is (π/4)ρv2r2.Modify the program meteor on page 139 to handle this problem,and find the resulting change in the fall time in the case of a 21 gball of radius 1.0 cm, falling from a height of 1.0 m. The density ofair at sea level is about 1.2 kg/m3. You will need to use a very largevalue of n to achieve the required precision. Turn in a printout ofboth your program and its output. Answer: 0.34 ms.

Problems 143

Exercise 4: Force and motionEquipment:

2-meter pieces of butcher paper

wood blocks with hooks

string

masses to put on top of the blocks to increase friction

spring scales (preferably calibrated in Newtons)

Suppose a person pushes a crate, sliding it across the floor at a certain speed, and then repeatsthe same thing but at a higher speed. This is essentially the situation you will act out in thisexercise. What do you think is different about her force on the crate in the two situations?Discuss this with your group and write down your hypothesis:

1. First you will measure the amount of friction between the wood block and the butcher paperwhen the wood and paper surfaces are slipping over each other. The idea is to attach a springscale to the block and then slide the butcher paper under the block while using the scale tokeep the block from moving with it. Depending on the amount of force your spring scale wasdesigned to measure, you may need to put an extra mass on top of the block in order to increasethe amount of friction. It is a good idea to use long piece of string to attach the block to thespring scale, since otherwise one tends to pull at an angle instead of directly horizontally.

First measure the amount of friction force when sliding the butcher paper as slowly as possi-ble:

Now measure the amount of friction force at a significantly higher speed, say 1 meter per second.(If you try to go too fast, the motion is jerky, and it is impossible to get an accurate reading.)

Discuss your results. Why are we justified in assuming that the string’s force on the block (i.e.,the scale reading) is the same amount as the paper’s frictional force on the block?

2. Now try the same thing but with the block moving and the paper standing still. Try twodifferent speeds.

Do your results agree with your original hypothesis? If not, discuss what’s going on. How doesthe block “know” how fast to go?


What forces act on the girl?

Chapter 5

Analysis of forces

5.1 Newton’s third lawNewton created the modern concept of force starting from his insightthat all the effects that govern motion are interactions between twoobjects: unlike the Aristotelian theory, Newtonian physics has nophenomena in which an object changes its own motion.

145

a / Two magnets exert forceson each other.

b / Two people’s hands exertforces on each other.

c / Rockets work by pushingexhaust gases out the back.Newton’s third law says that if therocket exerts a backward forceon the gases, the gases mustmake an equal forward force onthe rocket. Rocket engines canfunction above the atmosphere,unlike propellers and jets, whichwork by pushing against thesurrounding air.

Is one object always the “order-giver” and the other the “order-follower”? As an example, consider a batter hitting a baseball. Thebat definitely exerts a large force on the ball, because the ball ac-celerates drastically. But if you have ever hit a baseball, you alsoknow that the ball makes a force on the bat — often with painfulresults if your technique is as bad as mine!

How does the ball’s force on the bat compare with the bat’sforce on the ball? The bat’s acceleration is not as spectacular asthe ball’s, but maybe we shouldn’t expect it to be, since the bat’smass is much greater. In fact, careful measurements of both objects’masses and accelerations would show that mballaball is very nearlyequal to −mbatabat, which suggests that the ball’s force on the batis of the same magnitude as the bat’s force on the ball, but in theopposite direction.

Figures a and b show two somewhat more practical laboratoryexperiments for investigating this issue accurately and without toomuch interference from extraneous forces.

In experiment a, a large magnet and a small magnet are weighedseparately, and then one magnet is hung from the pan of the topbalance so that it is directly above the other magnet. There is anattraction between the two magnets, causing the reading on the topscale to increase and the reading on the bottom scale to decrease.The large magnet is more “powerful” in the sense that it can pickup a heavier paperclip from the same distance, so many people havea strong expectation that one scale’s reading will change by a fardifferent amount than the other. Instead, we find that the twochanges are equal in magnitude but opposite in direction: the forceof the bottom magnet pulling down on the top one has the samestrength as the force of the top one pulling up on the bottom one.

In experiment b, two people pull on two spring scales. Regardlessof who tries to pull harder, the two forces as measured on the springscales are equal. Interposing the two spring scales is necessary inorder to measure the forces, but the outcome is not some artificialresult of the scales’ interactions with each other. If one person slapsanother hard on the hand, the slapper’s hand hurts just as muchas the slappee’s, and it doesn’t matter if the recipient of the slaptries to be inactive. (Punching someone in the mouth causes justas much force on the fist as on the lips. It’s just that the lips aremore delicate. The forces are equal, but not the levels of pain andinjury.)

Newton, after observing a series of results such as these, decidedthat there must be a fundamental law of nature at work:

146 Chapter 5 Analysis of forces

d / A swimmer doing the breaststroke pushes backward againstthe water. By Newton’s third law,the water pushes forward on him.

e / Newton’s third law doesnot mean that forces alwayscancel out so that nothing canever move. If these two figureskaters, initially at rest, pushagainst each other, they will bothmove.

Newton’s third lawForces occur in equal and opposite pairs: whenever object A exertsa force on object B, object B must also be exerting a force on objectA. The two forces are equal in magnitude and opposite in direction.

In one-dimensional situations, we can use plus and minus signs toindicate the directions of forces, and Newton’s third law can bewritten succinctly as FA on B = −FB on A.

self-check AFigure d analyzes swimming using Newton’s third law. Do a similaranalysis for a sprinter leaving the starting line. . Answer, p. 510

There is no cause and effect relationship between the two forcesin Newton’s third law. There is no “original” force, and neither oneis a response to the other. The pair of forces is a relationship, likemarriage, not a back-and-forth process like a tennis match. Newtoncame up with the third law as a generalization about all the types offorces with which he was familiar, such as frictional and gravitationalforces. When later physicists discovered a new type force, suchas the force that holds atomic nuclei together, they had to checkwhether it obeyed Newton’s third law. So far, no violation of thethird law has ever been discovered, whereas the first and secondlaws were shown to have limitations by Einstein and the pioneers ofatomic physics.

The English vocabulary for describing forces is unfortunatelyrooted in Aristotelianism, and often implies incorrectly that forcesare one-way relationships. It is unfortunate that a half-truth such as“the table exerts an upward force on the book” is so easily expressed,while a more complete and correct description ends up soundingawkward or strange: “the table and the book interact via a force,”or “the table and book participate in a force.”

To students, it often sounds as though Newton’s third law im-plies nothing could ever change its motion, since the two equal andopposite forces would always cancel. The two forces, however, arealways on two different objects, so it doesn’t make sense to addthem in the first place — we only add forces that are acting on thesame object. If two objects are interacting via a force and no otherforces are involved, then both objects will accelerate — in oppositedirections!

Section 5.1 Newton’s third law 147

f / It doesn’t make sense for theman to talk about using thewoman’s money to cancel out hisbar tab, because there is no goodreason to combine his debts andher assets. Similarly, it doesn’tmake sense to refer to the equaland opposite forces of Newton’sthird law as canceling. It onlymakes sense to add up forcesthat are acting on the same ob-ject, whereas two forces relatedto each other by Newton’s thirdlaw are always acting on two dif-ferent objects.

A mnemonic for using newton’s third law correctly

Mnemonics are tricks for memorizing things. For instance, themusical notes that lie between the lines on the treble clef spell theword FACE, which is easy to remember. Many people use themnemonic “SOHCAHTOA” to remember the definitions of the sine,cosine, and tangent in trigonometry. I have my own modest offering,POFOSTITO, which I hope will make it into the mnemonics hall offame. It’s a way to avoid some of the most common problems withapplying Newton’s third law correctly:

A book lying on a table example 1. A book is lying on a table. What force is the Newton’s-third-lawpartner of the earth’s gravitational force on the book?

Answer: Newton’s third law works like “B on A, A on B,” so thepartner must be the book’s gravitational force pulling upward onthe planet earth. Yes, there is such a force! No, it does not causethe earth to do anything noticeable.

Incorrect answer: The table’s upward force on the book is theNewton’s-third-law partner of the earth’s gravitational force on thebook.

This answer violates two out of three of the commandments ofPOFOSTITO. The forces are not of the same type, because the


Optional Topic: Newton’s ThirdLaw and Action at a DistanceNewton’s third law is completelysymmetric in the sense that nei-ther force constitutes a delayedresponse to the other. Newton’sthird law does not even mentiontime, and the forces are supposedto agree at any given instant. Thiscreates an interesting situationwhen it comes to noncontact forces.Suppose two people are hold-ing magnets, and when one per-son waves or wiggles her mag-net, the other person feels aneffect on his. In this way theycan send signals to each otherfrom opposite sides of a wall, andif Newton’s third law is correct, itwould seem that the signals aretransmitted instantly, with no timelag. The signals are indeed trans-mitted quite quickly, but experi-ments with electrically controlledmagnets show that the signalsdo not leap the gap instantly: theytravel at the same speed as light,which is an extremely high speedbut not an infinite one.

Is this a contradiction to New-ton’s third law? Not really. Ac-cording to current theories, thereare no true noncontact forces.Action at a distance does not ex-ist. Although it appears that thewiggling of one magnet affectsthe other with no need for any-thing to be in contact with any-thing, what really happens is thatwiggling a magnet creates a rip-ple in the magnetic field patternthat exists even in empty space.The magnet shoves the rippliesout with a kick and receives akick in return, in strict obedienceto Newton’s third law. The rip-ples spread out in all directions,and the ones that hit the othermagnet then interact with it, againobeying Newton’s third law.

table’s upward force on the book is not gravitational. Also, threeobjects are involved instead of two: the book, the table, and theplanet earth.

Pushing a box up a hill example 2. A person is pushing a box up a hill. What force is related byNewton’s third law to the person’s force on the box?

. The box’s force on the person.

Incorrect answer: The person’s force on the box is opposed byfriction, and also by gravity.

This answer fails all three parts of the POFOSTITO test, themost obvious of which is that three forces are referred to insteadof a pair.

. Solved problem: More about example 2 page 168, problem 2

. Solved problem: Why did it accelerate? page 168, problem 1


A When you fire a gun, the exploding gases push outward in alldirections, causing the bullet to accelerate down the barrel. What third-lawpairs are involved? [Hint: Remember that the gases themselves are anobject.]

B Tam Anh grabs Sarah by the hand and tries to pull her. She triesto remain standing without moving. A student analyzes the situation asfollows. “If Tam Anh’s force on Sarah is greater than her force on him,he can get her to move. Otherwise, she’ll be able to stay where she is.”What’s wrong with this analysis?

C You hit a tennis ball against a wall. Explain any and all incorrectideas in the following description of the physics involved: “According toNewton’s third law, there has to be a force opposite to your force on theball. The opposite force is the ball’s mass, which resists acceleration, andalso air resistance.”

5.2 Classification and behavior of forcesOne of the most basic and important tasks of physics is to classifythe forces of nature. I have already referred informally to “types” offorces such as friction, magnetism, gravitational forces, and so on.Classification systems are creations of the human mind, so there isalways some degree of arbitrariness in them. For one thing, the levelof detail that is appropriate for a classification system depends onwhat you’re trying to find out. Some linguists, the “lumpers,” like toemphasize the similarities among languages, and a few extremistshave even tried to find signs of similarities between words in lan-guages as different as English and Chinese, lumping the world’s lan-guages into only a few large groups. Other linguists, the “splitters,”might be more interested in studying the differences in pronuncia-tion between English speakers in New York and Connecticut. The

Section 5.2 Classification and behavior of forces 149

g / A scientific classificationsystem.

splitters call the lumpers sloppy, but the lumpers say that scienceisn’t worthwhile unless it can find broad, simple patterns within theseemingly complex universe.

Scientific classification systems are also usually compromises be-tween practicality and naturalness. An example is the question ofhow to classify flowering plants. Most people think that biologicalclassification is about discovering new species, naming them, andclassifying them in the class-order-family-genus-species system ac-cording to guidelines set long ago. In reality, the whole system is ina constant state of flux and controversy. One very practical way ofclassifying flowering plants is according to whether their petals areseparate or joined into a tube or cone — the criterion is so clear thatit can be applied to a plant seen from across the street. But herepracticality conflicts with naturalness. For instance, the begonia hasseparate petals and the pumpkin has joined petals, but they are sosimilar in so many other ways that they are usually placed withinthe same order. Some taxonomists have come up with classificationcriteria that they claim correspond more naturally to the apparentrelationships among plants, without having to make special excep-tions, but these may be far less practical, requiring for instance theexamination of pollen grains under an electron microscope.

In physics, there are two main systems of classification for forces.At this point in the course, you are going to learn one that is verypractical and easy to use, and that splits the forces up into a rel-atively large number of types: seven very common ones that we’lldiscuss explicitly in this chapter, plus perhaps ten less importantones such as surface tension, which we will not bother with rightnow.

Physicists, however, are obsessed with finding simple patterns,so recognizing as many as fifteen or twenty types of forces strikesthem as distasteful and overly complex. Since about the year 1900,physics has been on an aggressive program to discover ways in whichthese many seemingly different types of forces arise from a smallernumber of fundamental ones. For instance, when you press yourhands together, the force that keeps them from passing through eachother may seem to have nothing to do with electricity, but at theatomic level, it actually does arise from electrical repulsion betweenatoms. By about 1950, all the forces of nature had been explainedas arising from four fundamental types of forces at the atomic andnuclear level, and the lumping-together process didn’t stop there.By the 1960’s the length of the list had been reduced to three, andsome theorists even believe that they may be able to reduce it totwo or one. Although the unification of the forces of nature is one ofthe most beautiful and important achievements of physics, it makesmuch more sense to start this course with the more practical andeasy system of classification. The unified system of four forces willbe one of the highlights of the end of your introductory physics


sequence.

h / A practical classification scheme for forces.

The practical classification scheme which concerns us now canbe laid out in the form of the tree shown in figure h. The mostspecific types of forces are shown at the tips of the branches, andit is these types of forces that are referred to in the POFOSTITOmnemonic. For example, electrical and magnetic forces belong tothe same general group, but Newton’s third law would never relatean electrical force to a magnetic force.

The broadest distinction is that between contact and noncontactforces, which has been discussed in ch. 4. Among the contact forces,we distinguish between those that involve solids only and those thathave to do with fluids, a term used in physics to include both gasesand liquids.

It should not be necessary to memorize this diagram by rote.It is better to reinforce your memory of this system by calling tomind your commonsense knowledge of certain ordinary phenomena.For instance, we know that the gravitational attraction between us


and the planet earth will act even if our feet momentarily leave theground, and that although magnets have mass and are affected bygravity, most objects that have mass are nonmagnetic.

Hitting a wall example 3. A bullet, flying horizontally, hits a steel wall. What type of forceis there between the bullet and the wall?

. Starting at the bottom of the tree, we determine that the forceis a contact force, because it only occurs once the bullet touchesthe wall. Both objects are solid. The wall forms a vertical plane.If the nose of the bullet was some shape like a sphere, you mightimagine that it would only touch the wall at one point. Realisti-cally, however, we know that a lead bullet will flatten out a lot onimpact, so there is a surface of contact between the two, and itsorientation is vertical. The effect of the force on the bullet is tostop the horizontal motion of the bullet, and this horizontal ac-celeration must be produced by a horizontal force. The force istherefore perpendicular to the surface of contact, and it’s also re-pulsive (tending to keep the bullet from entering the wall), so itmust be a normal force.

Diagram h is meant to be as simple as possible while includingmost of the forces we deal with in everyday life. If you were aninsect, you would be much more interested in the force of surfacetension, which allowed you to walk on water. I have not includedthe nuclear forces, which are responsible for holding the nuclei ofatoms, because they are not evident in everyday life.

You should not be afraid to invent your own names for types offorces that do not fit into the diagram. For instance, the force thatholds a piece of tape to the wall has been left off of the tree, and ifyou were analyzing a situation involving scotch tape, you would beabsolutely right to refer to it by some commonsense name such as“sticky force.”

On the other hand, if you are having trouble classifying a certainforce, you should also consider whether it is a force at all. Forinstance, if someone asks you to classify the force that the earth hasbecause of its rotation, you would have great difficulty creating aplace for it on the diagram. That’s because it’s a type of motion,not a type of force!

Normal forces

A normal force, FN , is a force that keeps one solid object frompassing through another. “Normal” is simply a fancy word for “per-pendicular,” meaning that the force is perpendicular to the surfaceof contact. Intuitively, it seems the normal force magically adjustsitself to provide whatever force is needed to keep the objects fromoccupying the same space. If your muscles press your hands together


i / A model that correctly ex-plains many properties of friction.The microscopic bumps andholes in two surfaces dig intoeach other.

j / Static friction: the tray doesn’tslip on the waiter’s fingers.

k / Kinetic friction: the carskids.

gently, there is a gentle normal force. Press harder, and the normalforce gets stronger. How does the normal force know how strong tobe? The answer is that the harder you jam your hands together,the more compressed your flesh becomes. Your flesh is acting likea spring: more force is required to compress it more. The same istrue when you push on a wall. The wall flexes imperceptibly in pro-portion to your force on it. If you exerted enough force, would it bepossible for two objects to pass through each other? No, typicallythe result is simply to strain the objects so much that one of thembreaks.

Gravitational forces

As we’ll discuss in more detail later in the course, a gravitationalforce exists between any two things that have mass. In everyday life,the gravitational force between two cars or two people is negligible,so the only noticeable gravitational forces are the ones between theearth and various human-scale objects. We refer to these planet-earth-induced gravitational forces as weight forces, and as we havealready seen, their magnitude is given by |FW | = mg.

. Solved problem: Weight and mass page 168, problem 3

Static and kinetic friction

If you have pushed a refrigerator across a kitchen floor, you havefelt a certain series of sensations. At first, you gradually increasedyour force on the refrigerator, but it didn’t move. Finally, you sup-plied enough force to unstick the fridge, and there was a suddenjerk as the fridge started moving. Once the fridge was unstuck, youcould reduce your force significantly and still keep it moving.

While you were gradually increasing your force, the floor’s fric-tional force on the fridge increased in response. The two forces onthe fridge canceled, and the fridge didn’t accelerate. How did thefloor know how to respond with just the right amount of force? Fig-ure i shows one possible model of friction that explains this behavior.(A scientific model is a description that we expect to be incomplete,approximate, or unrealistic in some ways, but that nevertheless suc-ceeds in explaining a variety of phenomena.) Figure i/1 shows amicroscopic view of the tiny bumps and holes in the surfaces of thefloor and the refrigerator. The weight of the fridge presses the twosurfaces together, and some of the bumps in one surface will settleas deeply as possible into some of the holes in the other surface. Ini/2, your leftward force on the fridge has caused it to ride up a littlehigher on the bump in the floor labeled with a small arrow. Stillmore force is needed to get the fridge over the bump and allow it tostart moving. Of course, this is occurring simultaneously at millionsof places on the two surfaces.

Once you had gotten the fridge moving at constant speed, you


l / Many landfowl, even thosethat are competent fliers, preferto escape from a predator byrunning upward rather than byflying. This partridge is runningup a vertical tree trunk. Humanscan’t walk up walls because thereis no normal force and thereforeno frictional force; when FN = 0,the maximum force of staticfriction Fs,max = µsFN is alsozero. The partridge, however,has wings that it can flap in orderto create a force between it andthe air. Typically when a birdflaps its wings, the resulting forcefrom the air is in the directionthat would tend to lift the birdup. In this situation, however,the partridge changes its styleof flapping so that the directionis reversed. The normal forcebetween the feet and the treeallows a nonzero static frictionalforce. The mechanism is similarto that of a spoiler fin on a racingcar. Some evolutionary biologistsbelieve that when vertebrateflight first evolved, in dinosaurs,there was first a stage in whichthe wings were used only as anaid in running up steep inclines,and only later a transition toflight. (Redrawn from a figure byK.P. Dial.)

found that you needed to exert less force on it. Since zero total forceis needed to make an object move with constant velocity, the floor’srightward frictional force on the fridge has apparently decreasedsomewhat, making it easier for you to cancel it out. Our model alsogives a plausible explanation for this fact: as the surfaces slide pasteach other, they don’t have time to settle down and mesh with oneanother, so there is less friction.

Even though this model is intuitively appealing and fairly suc-cessful, it should not be taken too seriously, and in some situationsit is misleading. For instance, fancy racing bikes these days aremade with smooth tires that have no tread — contrary to whatwe’d expect from our model, this does not cause any decrease infriction. Machinists know that two very smooth and clean metalsurfaces may stick to each other firmly and be very difficult to slideapart. This cannot be explained in our model, but makes moresense in terms of a model in which friction is described as arisingfrom chemical bonds between the atoms of the two surfaces at theirpoints of contact: very flat surfaces allow more atoms to come incontact.

Since friction changes its behavior dramatically once the sur-faces come unstuck, we define two separate types of frictional forces.Static friction is friction that occurs between surfaces that are notslipping over each other. Slipping surfaces experience kinetic fric-tion. The forces of static and kinetic friction, notated Fs and Fk, arealways parallel to the surface of contact between the two objects.

self-check B1. When a baseball player slides in to a base, is the friction static, orkinetic?

2. A mattress stays on the roof of a slowly accelerating car. Is thefriction static, or kinetic?

3. Does static friction create heat? Kinetic friction? . Answer, p. 510

The maximum possible force of static friction depends on whatkinds of surfaces they are, and also on how hard they are beingpressed together. The approximate mathematical relationships canbe expressed as follows:

Fs,max = µsFN ,

where µs is a unitless number, called the coefficient of static friction,which depends on what kinds of surfaces they are. The maximumforce that static friction can supply, µsFN , represents the boundarybetween static and kinetic friction. It depends on the normal force,which is numerically equal to whatever force is pressing the twosurfaces together. In terms of our model, if the two surfaces arebeing pressed together more firmly, a greater sideways force will be


required in order to make the irregularities in the surfaces ride upand over each other.

Note that just because we use an adjective such as “applied” torefer to a force, that doesn’t mean that there is some special typeof force called the “applied force.” The applied force could be anytype of force, or it could be the sum of more than one force tryingto make an object move.

self-check CThe arrows in figure l show the forces of the tree trunk on the partridge.Describe the forces the bird makes on the tree. . Answer, p. 510

The force of kinetic friction on each of the two objects is in thedirection that resists the slippage of the surfaces. Its magnitude isusually well approximated as

Fk = µkFN

where µk is the coefficient of kinetic friction. Kinetic friction isusually more or less independent of velocity.

m / We choose a coordinate sys-tem in which the applied force,i.e., the force trying to move theobjects, is positive. The frictionforce is then negative, since it isin the opposite direction. As youincrease the applied force, theforce of static friction increases tomatch it and cancel it out, until themaximum force of static friction issurpassed. The surfaces then be-gin slipping past each other, andthe friction force becomes smallerin absolute value.

self-check DCan a frictionless surface exert a normal force? Can a frictional forceexist without a normal force? . Answer, p. 510

If you try to accelerate or decelerate your car too quickly, theforces between your wheels and the road become too great, and theybegin slipping. This is not good, because kinetic friction is weakerthan static friction, resulting in less control. Also, if this occurswhile you are turning, the car’s handling changes abruptly becausethe kinetic friction force is in a different direction than the staticfriction force had been: contrary to the car’s direction of motion,rather than contrary to the forces applied to the tire.

Most people respond with disbelief when told of the experimen-tal evidence that both static and kinetic friction are approximately


independent of the amount of surface area in contact. Even afterdoing a hands-on exercise with spring scales to show that it is true,many students are unwilling to believe their own observations, andinsist that bigger tires “give more traction.” In fact, the main rea-son why you would not want to put small tires on a big heavy caris that the tires would burst!

Although many people expect that friction would be propor-tional to surface area, such a proportionality would make predictionscontrary to many everyday observations. A dog’s feet, for example,have very little surface area in contact with the ground comparedto a human’s feet, and yet we know that a dog can often win atug-of-war with a person.

The reason a smaller surface area does not lead to less frictionis that the force between the two surfaces is more concentrated,causing their bumps and holes to dig into each other more deeply.

self-check EFind the direction of each of the forces in figure n. . Answer, p. 510

n / 1. The cliff’s normal force onthe climber’s feet. 2. The track’sstatic frictional force on the wheelof the accelerating dragster. 3.The ball’s normal force on thebat.

Locomotives example 4Looking at a picture of a locomotive, o, we notice two obviousthings that are different from an automobile. Where a car typi-cally has two drive wheels, a locomotive normally has many —ten in this example. (Some also have smaller, unpowered wheelsin front of and behind the drive wheels, but this example doesn’t.)Also, cars these days are generally built to be as light as possi-ble for their size, whereas locomotives are very massive, and noeffort seems to be made to keep their weight low. (The steamlocomotive in the photo is from about 1900, but this is true evenfor modern diesel and electric trains.)


p / Fluid friction depends onthe fluid’s pattern of flow, so it ismore complicated than frictionbetween solids, and there areno simple, universally applicableformulas to calculate it. Fromtop to bottom: supersonic windtunnel, vortex created by a cropduster, series of vortices createdby a single object, turbulence.

o / Example 4.

The reason locomotives are built to be so heavy is for traction.The upward normal force of the rails on the wheels, FN , cancelsthe downward force of gravity, FW , so ignoring plus and minussigns, these two forces are equal in absolute value, FN = FW .Given this amount of normal force, the maximum force of staticfriction is Fs = µsFN = µsFW . This static frictional force, of therails pushing forward on the wheels, is the only force that canaccelerate the train, pull it uphill, or cancel out the force of airresistance while cruising at constant speed. The coefficient ofstatic friction for steel on steel is about 1/4, so no locomotive canpull with a force greater than about 1/4 of its own weight. If theengine is capable of supplying more than that amount of force, theresult will be simply to break static friction and spin the wheels.

The reason this is all so different from the situation with a car isthat a car isn’t pulling something else. If you put extra weight ina car, you improve the traction, but you also increase the inertiaof the car, and make it just as hard to accelerate. In a train, theinertia is almost all in the cars being pulled, not in the locomotive.

The other fact we have to explain is the large number of driv-ing wheels. First, we have to realize that increasing the num-ber of driving wheels neither increases nor decreases the totalamount of static friction, because static friction is independent ofthe amount of surface area in contact. (The reason four-wheel-drive is good in a car is that if one or more of the wheels is slip-ping on ice or in mud, the other wheels may still have traction.This isn’t typically an issue for a train, since all the wheels experi-ence the same conditions.) The advantage of having more drivingwheels on a train is that it allows us to increase the weight of thelocomotive without crushing the rails, or damaging bridges.

Fluid friction

Try to drive a nail into a waterfall and you will be confrontedwith the main difference between solid friction and fluid friction.Fluid friction is purely kinetic; there is no static fluid friction. Thenail in the waterfall may tend to get dragged along by the waterflowing past it, but it does not stick in the water. The same is truefor gases such as air: recall that we are using the word “fluid” toinclude both gases and liquids.


q / What do the golf ball andthe shark have in common? Bothuse the same trick to reduce fluidfriction. The dimples on the golfball modify the pattern of flow ofthe air around it, counterintuitivelyreducing friction. Recent studieshave shown that sharks canaccomplish the same thing byraising, or “bristling,” the scaleson their skin at high speeds.

r / The wheelbases of theHummer H3 and the Toyota Priusare surprisingly similar, differingby only 10%. The main differencein shape is that the Hummer ismuch taller and wider. It presentsa much greater cross-sectionalarea to the wind, and this is themain reason that it uses about 2.5times more gas on the freeway.

Unlike kinetic friction between solids, fluid friction increasesrapidly with velocity. It also depends on the shape of the object,which is why a fighter jet is more streamlined than a Model T. Forobjects of the same shape but different sizes, fluid friction typicallyscales up with the cross-sectional area of the object, which is oneof the main reasons that an SUV gets worse mileage on the freewaythan a compact car.


A A student states that when he tries to push his refrigerator, thereason it won’t move is because Newton’s third law says there’s an equaland opposite frictional force pushing back. After all, the static friction forceis equal and opposite to the applied force. How would you convince himhe is wrong?

B Kinetic friction is usually more or less independent of velocity. How-ever, inexperienced drivers tend to produce a jerk at the last moment ofdeceleration when they stop at a stop light. What does this tell you aboutthe kinetic friction between the brake shoes and the brake drums?

C Some of the following are correct descriptions of types of forces thatcould be added on as new branches of the classification tree. Others arenot really types of forces, and still others are not force phenomena at all.In each case, decide what’s going on, and if appropriate, figure out howyou would incorporate them into the tree.

sticky force makes tape stick to thingsopposite force the force that Newton’s third law says relates to ev-

ery force you makeflowing force the force that water carries with it as it flows out of a

hosesurface tension lets insects walk on waterhorizontal force a force that is horizontalmotor force the force that a motor makes on the thing it is turningcanceled force a force that is being canceled out by some other

force

5.3 Analysis of forcesNewton’s first and second laws deal with the total of all the forcesexerted on a specific object, so it is very important to be able tofigure out what forces there are. Once you have focused your atten-tion on one object and listed the forces on it, it is also helpful todescribe all the corresponding forces that must exist according toNewton’s third law. We refer to this as “analyzing the forces” inwhich the object participates.


A barge example 5A barge is being pulled along a canal by teams of horses on the shores. Analyze all the forces in which thebarge participates.

force acting on barge force related to it by Newton’s third lawropes’ forward normal forces on barge barge’s backward normal force on ropeswater’s backward fluid friction force on barge barge’s forward fluid friction force on waterplanet earth’s downward gravitational forceon barge

barge’s upward gravitational force on earth

water’s upward “floating” force on barge barge’s downward “floating” force on water

Here I’ve used the word “floating” force as an example of a sensible invented term for a type of force notclassified on the tree in the previous section. A more formal technical term would be “hydrostatic force.”Note how the pairs of forces are all structured as “A’s force on B, B’s force on A”: ropes on barge and bargeon ropes; water on barge and barge on water. Because all the forces in the left column are forces acting onthe barge, all the forces in the right column are forces being exerted by the barge, which is why each entry inthe column begins with “barge.”

Often you may be unsure whether you have forgotten one of theforces. Here are three strategies for checking your list:

1. See what physical result would come from the forces you’vefound so far. Suppose, for instance, that you’d forgotten the“floating” force on the barge in the example above. Lookingat the forces you’d found, you would have found that therewas a downward gravitational force on the barge which wasnot canceled by any upward force. The barge isn’t supposedto sink, so you know you need to find a fourth, upward force.

2. Another technique for finding missing forces is simply to gothrough the list of all the common types of forces and see ifany of them apply.

3. Make a drawing of the object, and draw a dashed boundaryline around it that separates it from its environment. Look forpoints on the boundary where other objects come in contactwith your object. This strategy guarantees that you’ll findevery contact force that acts on the object, although it won’thelp you to find non-contact forces.

The following is another example in which we can profit by check-ing against our physical intuition for what should be happening.

Section 5.3 Analysis of forces 159


Rappelling example 6As shown in the figure below, Cindy is rappelling down a cliff. Her downward motion is at constant speed, andshe takes little hops off of the cliff, as shown by the dashed line. Analyze the forces in which she participatesat a moment when her feet are on the cliff and she is pushing off.

force acting on Cindy force related to it by Newton’s third lawplanet earth’s gravitational force on Cindy, ↓ Cindy’s gravitational force on earth, ↑rope’s frictional force on Cindy (her hand), ↑ Cindy’s frictional force on the rope, ↓cliff’s normal force on Cindy,→ Cindy’s normal force on the cliff,←

The two vertical forces cancel, which is what they should be doing if she is to go down at a constant rate. Theonly horizontal force on her is the cliff’s force, which is not canceled by any other force, and which thereforewill produce an acceleration of Cindy to the right. This makes sense, since she is hopping off. (This solutionis a little oversimplified, because the rope is slanting, so it also applies a small leftward force to Cindy. As sheflies out to the right, the slant of the rope will increase, pulling her back in more strongly.)

I believe that constructing the type of table described in thissection is the best method for beginning students. Most textbooks,however, prescribe a pictorial way of showing all the forces acting onan object. Such a picture is called a free-body diagram. It shouldnot be a big problem if a future physics professor expects you tobe able to draw such diagrams, because the conceptual reasoningis the same. You simply draw a picture of the object, with arrowsrepresenting the forces that are acting on it. Arrows representingcontact forces are drawn from the point of contact, noncontact forcesfrom the center of mass. Free-body diagrams do not show the equaland opposite forces exerted by the object itself.


A In the example of the barge going down the canal, I referred toa “floating” or “hydrostatic” force that keeps the boat from sinking. If youwere adding a new branch on the force-classification tree to represent thisforce, where would it go?

B A pool ball is rebounding from the side of the pool table. Analyzethe forces in which the ball participates during the short time when it is incontact with the side of the table.

C The earth’s gravitational force on you, i.e., your weight, is alwaysequal to mg, where m is your mass. So why can you get a shovel to godeeper into the ground by jumping onto it? Just because you’re jumping,that doesn’t mean your mass or weight is any greater, does it?


5.4 Transmission of forces by low-massobjects

You’re walking your dog. The dog wants to go faster than you do,and the leash is taut. Does Newton’s third law guarantee that yourforce on your end of the leash is equal and opposite to the dog’sforce on its end? If they’re not exactly equal, is there any reasonwhy they should be approximately equal?

If there was no leash between you, and you were in direct contactwith the dog, then Newton’s third law would apply, but Newton’sthird law cannot relate your force on the leash to the dog’s forceon the leash, because that would involve three separate objects.Newton’s third law only says that your force on the leash is equaland opposite to the leash’s force on you,

FyL = −FLy,

and that the dog’s force on the leash is equal and opposite to itsforce on the dog

FdL = −FLd.

Still, we have a strong intuitive expectation that whatever forcewe make on our end of the leash is transmitted to the dog, andvice-versa. We can analyze the situation by concentrating on theforces that act on the leash, FdL and FyL. According to Newton’ssecond law, these relate to the leash’s mass and acceleration:

FdL + FyL = mLaL.

The leash is far less massive then any of the other objects involved,and if mL is very small, then apparently the total force on the leashis also very small, FdL + FyL ≈ 0, and therefore

FdL ≈ −FyL .

Thus even though Newton’s third law does not apply directly tothese two forces, we can approximate the low-mass leash as if it wasnot intervening between you and the dog. It’s at least approximatelyas if you and the dog were acting directly on each other, in whichcase Newton’s third law would have applied.

In general, low-mass objects can be treated approximately as ifthey simply transmitted forces from one object to another. This canbe true for strings, ropes, and cords, and also for rigid objects suchas rods and sticks.

If you look at a piece of string under a magnifying glass as youpull on the ends more and more strongly, you will see the fibersstraightening and becoming taut. Different parts of the string are

Section 5.4 Transmission of forces by low-mass objects 161

t / The Golden Gate Bridge’sroadway is held up by the tensionin the vertical cables.

s / If we imagine dividing a taut rope up into small segments, thenany segment has forces pulling outward on it at each end. If the ropeis of negligible mass, then all the forces equal +T or −T , where T , thetension, is a single number.

apparently exerting forces on each other. For instance, if we think ofthe two halves of the string as two objects, then each half is exertinga force on the other half. If we imagine the string as consistingof many small parts, then each segment is transmitting a force tothe next segment, and if the string has very little mass, then allthe forces are equal in magnitude. We refer to the magnitude ofthe forces as the tension in the string, T . Although the tensionis measured in units of Newtons, it is not itself a force. There aremany forces within the string, some in one direction and some in theother direction, and their magnitudes are only approximately equal.The concept of tension only makes sense as a general, approximatestatement of how big all the forces are.

If a rope goes over a pulley or around some other object, then thetension throughout the rope is approximately equal so long as thereis not too much friction. A rod or stick can be treated in much thesame way as a string, but it is possible to have either compressionor tension.

Since tension is not a type of force, the force exerted by a ropeon some other object must be of some definite type such as staticfriction, kinetic friction, or a normal force. If you hold your dog’sleash with your hand through the loop, then the force exerted by theleash on your hand is a normal force: it is the force that keeps theleash from occupying the same space as your hand. If you grasp aplain end of a rope, then the force between the rope and your handis a frictional force.

A more complex example of transmission of forces is the waya car accelerates. Many people would describe the car’s engine asmaking the force that accelerates the car, but the engine is part ofthe car, so that’s impossible: objects can’t make forces on them-selves. What really happens is that the engine’s force is transmittedthrough the transmission to the axles, then through the tires to theroad. By Newton’s third law, there will thus be a forward force fromthe road on the tires, which accelerates the car.

Discussion question

A When you step on the gas pedal, is your foot’s force being transmittedin the sense of the word used in this section?


5.5 Objects under strainA string lengthens slightly when you stretch it. Similarly, we havealready discussed how an apparently rigid object such as a wall isactually flexing when it participates in a normal force. In othercases, the effect is more obvious. A spring or a rubber band visiblyelongates when stretched.

Common to all these examples is a change in shape of some kind:lengthening, bending, compressing, etc. The change in shape canbe measured by picking some part of the object and measuring itsposition, x. For concreteness, let’s imagine a spring with one endattached to a wall. When no force is exerted, the unfixed end of thespring is at some position xo. If a force acts at the unfixed end, itsposition will change to some new value of x. The more force, thegreater the departure of x from xo.

u / Defining the quantities F , x ,and xo in Hooke’s law.

Back in Newton’s time, experiments like this were consideredcutting-edge research, and his contemporary Hooke is rememberedtoday for doing them and for coming up with a simple mathematicalgeneralization called Hooke’s law:

F ≈ k(x− xo) . [force required to stretch a spring; valid

for small forces only]

Here k is a constant, called the spring constant, that depends onhow stiff the object is. If too much force is applied, the springexhibits more complicated behavior, so the equation is only a goodapproximation if the force is sufficiently small. Usually when theforce is so large that Hooke’s law is a bad approximation, the forceends up permanently bending or breaking the spring.

Although Hooke’s law may seem like a piece of trivia aboutsprings, it is actually far more important than that, because all

Section 5.5 Objects under strain 163

solid objects exert Hooke’s-law behavior over some range of suffi-ciently small forces. For example, if you push down on the hood ofa car, it dips by an amount that is directly proportional to the force.(But the car’s behavior would not be as mathematically simple ifyou dropped a boulder on the hood!)

. Solved problem: Combining springs page 173, problem 26

. Solved problem: Young’s modulus page 173, problem 28

Discussion question

A A car is connected to its axles through big, stiff springs called shockabsorbers, or “shocks.” Although we’ve discussed Hooke’s law above onlyin the case of stretching a spring, a car’s shocks are continually goingthrough both stretching and compression. In this situation, how wouldyou interpret the positive and negative signs in Hooke’s law?

5.6 Simple machines: the pulleyEven the most complex machines, such as cars or pianos, are builtout of certain basic units called simple machines. The following aresome of the main functions of simple machines:

transmitting a force: The chain on a bicycle transmits a forcefrom the crank set to the rear wheel.

changing the direction of a force: If you push down on a see-saw, the other end goes up.

changing the speed and precision of motion: When you makethe “come here” motion, your biceps only moves a couple ofcentimeters where it attaches to your forearm, but your armmoves much farther and more rapidly.

changing the amount of force: A lever or pulley can be usedto increase or decrease the amount of force.

You are now prepared to understand one-dimensional simple ma-chines, of which the pulley is the main example.

v / Example 7.

A pulley example 7. Farmer Bill says this pulley arrangement doubles the force ofhis tractor. Is he just a dumb hayseed, or does he know what he’sdoing?


. To use Newton’s first law, we need to pick an object and con-sider the sum of the forces on it. Since our goal is to relate thetension in the part of the cable attached to the stump to the ten-sion in the part attached to the tractor, we should pick an objectto which both those cables are attached, i.e., the pulley itself. Asdiscussed in section 5.4, the tension in a string or cable remainsapproximately constant as it passes around a pulley, provided thatthere is not too much friction. There are therefore two leftwardforces acting on the pulley, each equal to the force exerted by thetractor. Since the acceleration of the pulley is essentially zero, theforces on it must be canceling out, so the rightward force of thepulley-stump cable on the pulley must be double the force exertedby the tractor. Yes, Farmer Bill knows what he’s talking about.

Section 5.6 Simple machines: the pulley 165

SummarySelected vocabularyrepulsive . . . . . describes a force that tends to push the two

participating objects apartattractive . . . . describes a force that tends to pull the two

participating objects togetheroblique . . . . . . describes a force that acts at some other angle,

one that is not a direct repulsion or attractionnormal force . . . the force that keeps two objects from occupy-

ing the same spacestatic friction . . a friction force between surfaces that are not

slipping past each otherkinetic friction . a friction force between surfaces that are slip-

ping past each otherfluid . . . . . . . . a gas or a liquidfluid friction . . . a friction force in which at least one of the

object is is a fluidspring constant . the constant of proportionality between force

and elongation of a spring or other object un-der strain

NotationFN . . . . . . . . . a normal forceFs . . . . . . . . . a static frictional forceFk . . . . . . . . . a kinetic frictional forceµs . . . . . . . . . the coefficient of static friction; the constant of

proportionality between the maximum staticfrictional force and the normal force; dependson what types of surfaces are involved

µk . . . . . . . . . the coefficient of kinetic friction; the constantof proportionality between the kinetic fric-tional force and the normal force; depends onwhat types of surfaces are involved

k . . . . . . . . . . the spring constant; the constant of propor-tionality between the force exerted on an ob-ject and the amount by which the object islengthened or compressed

Summary

Newton’s third law states that forces occur in equal and oppositepairs. If object A exerts a force on object B, then object B mustsimultaneously be exerting an equal and opposite force on object A.Each instance of Newton’s third law involves exactly two objects,and exactly two forces, which are of the same type.

There are two systems for classifying forces. We are presentlyusing the more practical but less fundamental one. In this system,forces are classified by whether they are repulsive, attractive, oroblique; whether they are contact or noncontact forces; and whether


the two objects involved are solids or fluids.

Static friction adjusts itself to match the force that is trying tomake the surfaces slide past each other, until the maximum value isreached,

Fs,max = µsFN .

Once this force is exceeded, the surfaces slip past one another, andkinetic friction applies,

Fk = µkFN .

Both types of frictional force are nearly independent of surface area,and kinetic friction is usually approximately independent of thespeed at which the surfaces are slipping. The direction of the forceis in the direction that would tend to stop or prevent slipping.

A good first step in applying Newton’s laws of motion to anyphysical situation is to pick an object of interest, and then to listall the forces acting on that object. We classify each force by itstype, and find its Newton’s-third-law partner, which is exerted bythe object on some other object.

When two objects are connected by a third low-mass object,their forces are transmitted to each other nearly unchanged.

Objects under strain always obey Hooke’s law to a good approx-imation, as long as the force is small. Hooke’s law states that thestretching or compression of the object is proportional to the forceexerted on it,

F ≈ k(x− xo) .

Summary 167

Problem 6.

Problem 9.

ProblemsKey√



1 In each case, identify the force that causes the acceleration,and give its Newton’s-third-law partner. Describe the effect of thepartner force. (a) A swimmer speeds up. (b) A golfer hits the balloff of the tee. (c) An archer fires an arrow. (d) A locomotive slowsdown. . Solution, p. 499

2 Example 2 on page 149 involves a person pushing a box up ahill. The incorrect answer describes three forces. For each of thesethree forces, give the force that it is related to by Newton’s thirdlaw, and state the type of force. . Solution, p. 500

3 (a) Compare the mass of a one-liter water bottle on earth, onthe moon, and in interstellar space. . Solution, p. 500(b) Do the same for its weight.

In problems 4-8, analyze the forces using a table in the format shownin section 5.3. Analyze the forces in which the italicized object par-ticipates.

4 Some people put a spare car key in a little magnetic box thatthey stick under the chassis of their car. Let’s say that the box isstuck directly underneath a horizontal surface, and the car is parked.(See instructions above.)

5 Analyze two examples of objects at rest relative to the earththat are being kept from falling by forces other than the normalforce. Do not use objects in outer space, and do not duplicateproblem 4 or 8. (See instructions above.)

6 A person is rowing a boat, with her feet braced. She is doingthe part of the stroke that propels the boat, with the ends of theoars in the water (not the part where the oars are out of the water).(See instructions above.)

7 A farmer is in a stall with a cow when the cow decides to presshim against the wall, pinning him with his feet off the ground. An-alyze the forces in which the farmer participates. (See instructionsabove.)

8 A propeller plane is cruising east at constant speed and alti-tude. (See instructions above.)

9 A little old lady and a pro football player collide head-on.Compare their forces on each other, and compare their accelerations.Explain.


Problem 12.

10 The earth is attracted to an object with a force equal andopposite to the force of the earth on the object. If this is true,why is it that when you drop an object, the earth does not have anacceleration equal and opposite to that of the object?

11 When you stand still, there are two forces acting on you,the force of gravity (your weight) and the normal force of the floorpushing up on your feet. Are these forces equal and opposite? DoesNewton’s third law relate them to each other? Explain.

12 Today’s tallest buildings are really not that much taller thanthe tallest buildings of the 1940’s. One big problem with making aneven taller skyscraper is that every elevator needs its own shaft run-ning the whole height of the building. So many elevators are neededto serve the building’s thousands of occupants that the elevatorshafts start taking up too much of the space within the building.An alternative is to have elevators that can move both horizontallyand vertically: with such a design, many elevator cars can share afew shafts, and they don’t get in each other’s way too much becausethey can detour around each other. In this design, it becomes im-possible to hang the cars from cables, so they would instead have toride on rails which they grab onto with wheels. Friction would keepthem from slipping. The figure shows such a frictional elevator inits vertical travel mode. (The wheels on the bottom are for when itneeds to switch to horizontal motion.)(a) If the coefficient of static friction between rubber and steel isµs, and the maximum mass of the car plus its passengers is M ,how much force must there be pressing each wheel against the railin order to keep the car from slipping? (Assume the car is notaccelerating.)

√

(b) Show that your result has physically reasonable behavior withrespect to µs. In other words, if there was less friction, would thewheels need to be pressed more firmly or less firmly? Does yourequation behave that way?

13 An ice skater builds up some speed, and then coasts acrossthe ice passively in a straight line. (a) Analyze the forces.(b) If his initial speed is v, and the coefficient of kinetic friction isµk, find the maximum theoretical distance he can glide before com-ing to a stop. Ignore air resistance.

√

(c) Show that your answer to part b has the right units.(d) Show that your answer to part b depends on the variables in away that makes sense physically.(e) Evaluate your answer numerically for µk = 0.0046, and a world-record speed of 14.58 m/s. (The coefficient of friction was mea-sured by De Koning et al., using special skates worn by real speedskaters.)

√

(f) Comment on whether your answer in part e seems realistic. If itdoesn’t, suggest possible reasons why.

Problems 169

14 A cop investigating the scene of an accident measures thelength L of a car’s skid marks in order to find out its speed v atthe beginning of the skid. Express v in terms of L and any otherrelevant variables.

√

15 Someone tells you she knows of a certain type of CentralAmerican earthworm whose skin, when rubbed on polished dia-mond, has µk > µs. Why is this not just empirically unlikely butlogically suspect?

16 When I cook rice, some of the dry grains always stick tothe measuring cup. To get them out, I turn the measuring cupupside-down and hit the “roof” with my hand so that the grainscome off of the “ceiling.” (a) Explain why static friction is irrelevanthere. (b) Explain why gravity is negligible. (c) Explain why hittingthe cup works, and why its success depends on hitting the cup hardenough.

17 Pick up a heavy object such as a backpack or a chair, andstand on a bathroom scale. Shake the object up and down. Whatdo you observe? Interpret your observations in terms of Newton’sthird law.

18 The following reasoning leads to an apparent paradox; explainwhat’s wrong with the logic. A baseball player hits a ball. The balland the bat spend a fraction of a second in contact. During thattime they’re moving together, so their accelerations must be equal.Newton’s third law says that their forces on each other are alsoequal. But a = F/m, so how can this be, since their masses areunequal? (Note that the paradox isn’t resolved by considering theforce of the batter’s hands on the bat. Not only is this force verysmall compared to the ball-bat force, but the batter could have justthrown the bat at the ball.)

19 A tugboat of mass m pulls a ship of mass M , accelerating it.The speeds are low enough that you can ignore fluid friction actingon their hulls, although there will of course need to be fluid frictionacting on the tug’s propellers.(a) Analyze the forces in which the tugboat participates, using atable in the format shown in section 5.3. Don’t worry about verticalforces.(b) Do the same for the ship.(c) Assume now that water friction on the two vessels’ hulls is neg-ligible. If the force acting on the tug’s propeller is F , what is thetension, T , in the cable connecting the two ships? [Hint: Writedown two equations, one for Newton’s second law applied to eachobject. Solve these for the two unknowns T and a.]

√

(d) Interpret your answer in the special cases of M = 0 and M =∞.


Problem 20.

Problem 21

20 Unequal masses M and m are suspended from a pulley asshown in the figure.(a) Analyze the forces in which mass m participates, using a tablethe format shown in section 5.3. [The forces in which the other massparticipates will of course be similar, but not numerically the same.](b) Find the magnitude of the accelerations of the two masses.[Hints: (1) Pick a coordinate system, and use positive and nega-tive signs consistently to indicate the directions of the forces andaccelerations. (2) The two accelerations of the two masses have tobe equal in magnitude but of opposite signs, since one side eats uprope at the same rate at which the other side pays it out. (3) Youneed to apply Newton’s second law twice, once to each mass, andthen solve the two equations for the unknowns: the acceleration, a,and the tension in the rope, T .](c) Many people expect that in the special case of M = m, the twomasses will naturally settle down to an equilibrium position side byside. Based on your answer from part b, is this correct?(d) Find the tension in the rope, T .(e) Interpret your equation from part d in the special case where oneof the masses is zero. Here “interpret” means to figure out what hap-pens mathematically, figure out what should happen physically, andconnect the two.

21 The figure shows a stack of two blocks, sitting on top of a tablethat is bolted to the floor. All three objects are made from iden-tical wood, with their surfaces finished identically using the samesandpaper. We tap the middle block, giving it an initial velocity vto the right. The tap is executed so rapidly that almost no initialvelocity is imparted to the top block.(a) Find the time that will elapse until the slipping between the topand middle blocks stops. Express your answer in terms of v, m, M ,g, and the relevant coefficient of friction.

√

(b) Show that your answer makes sense in terms of units.(c) Check that your result has the correct behavior when you makem bigger or smaller. Explain. This means that you should discussthe mathematical behavior of the result, and then explain how thiscorresponds to what would really happen physically.(d) Similarly, discuss what happens when you make M bigger orsmaller.(e) Similarly, discuss what happens when you make g bigger orsmaller.

Problems 171

Problem 22.

Problem 23.

Problem 24.

22 Mountain climbers with masses m and M are roped togetherwhile crossing a horizontal glacier when a vertical crevasse opens upunder the climber with mass M . The climber with mass m dropsdown on the snow and tries to stop by digging into the snow withthe pick of an ice ax. Alas, this story does not have a happy ending,because this doesn’t provide enough friction to stop. Both m and Mcontinue accelerating, with M dropping down into the crevasse andm being dragged across the snow, slowed only by the kinetic frictionwith coefficient µk acting between the ax and the snow. There is nosignificant friction between the rope and the lip of the crevasse.(a) Find the acceleration a.

√

(b) Check the units of your result.(c) Check the dependence of your equation on the variables. Thatmeans that for each variable, you should determine what its effecton a should be physically, and then what your answer from part asays its effect would be mathematically.

23 Ginny has a plan. She is going to ride her sled while her dogFoo pulls her, and she holds on to his leash. However, Ginny hasn’ttaken physics, so there may be a problem: she may slide right offthe sled when Foo starts pulling.(a) Analyze all the forces in which Ginny participates, making atable as in section 5.3.(b) Analyze all the forces in which the sled participates.(c) The sled has mass m, and Ginny has mass M . The coefficientof static friction between the sled and the snow is µ1, and µ2 isthe corresponding quantity for static friction between the sled andher snow pants. Ginny must have a certain minimum mass so thatshe will not slip off the sled. Find this in terms of the other threevariables.

√

(d) Interpreting your equation from part c, under what conditionswill there be no physically realistic solution for M? Discuss whatthis means physically.

24 In the system shown in the figure, the pulleys on the left andright are fixed, but the pulley in the center can move to the left orright. The two masses are identical. Find the upward accelerationof the mass on the left, in terms of g only. Assume all the ropes andpulleys are massless and frictionless.

√

25 Example 7 on page 164 describes a force-doubling setupinvolving a pulley. Make up a more complicated arrangement, usingmore than one pulley, that would multiply the force by a factorgreater than two.


Problem 26.

Problem 29.

26 The figure shows two different ways of combining a pair ofidentical springs, each with spring constant k. We refer to the topsetup as parallel, and the bottom one as a series arrangement.(a) For the parallel arrangement, analyze the forces acting on theconnector piece on the left, and then use this analysis to determinethe equivalent spring constant of the whole setup. Explain whetherthe combined spring constant should be interpreted as being stifferor less stiff.(b) For the series arrangement, analyze the forces acting on eachspring and figure out the same things. . Solution, p. 500

27 Generalize the results of problem 26 to the case where thetwo spring constants are unequal.

28 (a) Using the solution of problem 26, which is given in theback of the book, predict how the spring constant of a fiber willdepend on its length and cross-sectional area.(b) The constant of proportionality is called the Young’s modu-lus, E, and typical values of the Young’s modulus are about 1010

to 1011. What units would the Young’s modulus have in the SI(meter-kilogram-second) system? . Solution, p. 500

29 This problem depends on the results of problems 26 and28, whose solutions are in the back of the book. When atoms formchemical bonds, it makes sense to talk about the spring constant ofthe bond as a measure of how “stiff” it is. Of course, there aren’treally little springs — this is just a mechanical model. The purposeof this problem is to estimate the spring constant, k, for a singlebond in a typical piece of solid matter. Suppose we have a fiber,like a hair or a piece of fishing line, and imagine for simplicity thatit is made of atoms of a single element stacked in a cubical manner,as shown in the figure, with a center-to-center spacing b. A typicalvalue for b would be about 10−10 m.(a) Find an equation for k in terms of b, and in terms of the Young’smodulus, E, defined in problem 16 and its solution.(b) Estimate k using the numerical data given in problem 28.(c) Suppose you could grab one of the atoms in a diatomic moleculelike H2 or O2, and let the other atom hang vertically below it. Doesthe bond stretch by any appreciable fraction due to gravity?

Problems 173

30 A cross-country skier is gliding on a level trail, with negligi-ble friction. Then, when he is at position x = 0, the tip of his skisenters a patch of dirt. As he rides onto the dirt, more and more ofhis weight is being supported by the dirt. The skis have length `,so if he reached x = ` without stopping, his weight would be com-pletely on the dirt. This problem deals with the motion for x < `.(a) Find the acceleration in terms of x, as well as any other relevantconstants.(b) This is a second-order differential equation. You should be ableto find the solution simply by thinking about some commonly oc-curing functions that you know about, and finding two that havethe right properties. If these functions are x = f(t) and x = g(t),then the most general solution to the equations of motion will be ofthe form x = af + bg, where a and b are constants to be determinedfrom the initial conditions.(c) Suppose that the initial velocity vo at x = 0 is such that he stopsat x < `. Find the time until he stops, and show that, counterintu-itively, this time is independent of vo. Explain physically why thisis true. ?


Motion in threedimensions

175

Chapter 6

Newton’s laws in threedimensions

6.1 Forces have no perpendicular effectsSuppose you could shoot a rifle and arrange for a second bullet tobe dropped from the same height at the exact moment when thefirst left the barrel. Which would hit the ground first? Nearlyeveryone expects that the dropped bullet will reach the dirt first,

177

and Aristotle would have agreed. Aristotle would have described itlike this. The shot bullet receives some forced motion from the gun.It travels forward for a split second, slowing down rapidly becausethere is no longer any force to make it continue in motion. Onceit is done with its forced motion, it changes to natural motion, i.e.falling straight down. While the shot bullet is slowing down, thedropped bullet gets on with the business of falling, so according toAristotle it will hit the ground first.

a / A bullet is shot from a gun, and another bullet is simultaneously dropped from the same height. 1.Aristotelian physics says that the horizontal motion of the shot bullet delays the onset of falling, so the droppedbullet hits the ground first. 2. Newtonian physics says the two bullets have the same vertical motion, regardlessof their different horizontal motions.

Luckily, nature isn’t as complicated as Aristotle thought! Toconvince yourself that Aristotle’s ideas were wrong and needlesslycomplex, stand up now and try this experiment. Take your keysout of your pocket, and begin walking briskly forward. Withoutspeeding up or slowing down, release your keys and let them fallwhile you continue walking at the same pace.

You have found that your keys hit the ground right next to yourfeet. Their horizontal motion never slowed down at all, and thewhole time they were dropping, they were right next to you. Thehorizontal motion and the vertical motion happen at the same time,and they are independent of each other. Your experiment provesthat the horizontal motion is unaffected by the vertical motion, butit’s also true that the vertical motion is not changed in any way bythe horizontal motion. The keys take exactly the same amount oftime to get to the ground as they would have if you simply droppedthem, and the same is true of the bullets: both bullets hit the ground

178 Chapter 6 Newton’s laws in three dimensions

simultaneously.

These have been our first examples of motion in more than onedimension, and they illustrate the most important new idea thatis required to understand the three-dimensional generalization ofNewtonian physics:

Forces have no perpendicular effects.When a force acts on an object, it has no effect on the part of theobject’s motion that is perpendicular to the force.

In the examples above, the vertical force of gravity had no effecton the horizontal motions of the objects. These were examples ofprojectile motion, which interested people like Galileo because ofits military applications. The principle is more general than that,however. For instance, if a rolling ball is initially heading straightfor a wall, but a steady wind begins blowing from the side, the balldoes not take any longer to get to the wall. In the case of projectilemotion, the force involved is gravity, so we can say more specificallythat the vertical acceleration is 9.8 m/s2, regardless of the horizontalmotion.

self-check AIn the example of the ball being blown sideways, why doesn’t the balltake longer to get there, since it has to travel a greater distance? .

Answer, p. 510

Relationship to relative motion

These concepts are directly related to the idea that motion is rel-ative. Galileo’s opponents argued that the earth could not possiblybe rotating as he claimed, because then if you jumped straight up inthe air you wouldn’t be able to come down in the same place. Theirargument was based on their incorrect Aristotelian assumption thatonce the force of gravity began to act on you and bring you backdown, your horizontal motion would stop. In the correct Newtoniantheory, the earth’s downward gravitational force is acting before,during, and after your jump, but has no effect on your motion inthe perpendicular (horizontal) direction.

If Aristotle had been correct, then we would have a handy wayto determine absolute motion and absolute rest: jump straight upin the air, and if you land back where you started, the surface fromwhich you jumped must have been in a state of rest. In reality, thistest gives the same result as long as the surface under you is aninertial frame. If you try this in a jet plane, you land back on thesame spot on the deck from which you started, regardless of whetherthe plane is flying at 500 miles per hour or parked on the runway.The method would in fact only be good for detecting whether the

Section 6.1 Forces have no perpendicular effects 179

c / The shadow on the wallshows the ball’s y motion, theshadow on the floor its x motion.

plane was accelerating.


A The following is an incorrect explanation of a fact about targetshooting:

“Shooting a high-powered rifle with a high muzzle velocity is different fromshooting a less powerful gun. With a less powerful gun, you have to aimquite a bit above your target, but with a more powerful one you don’t haveto aim so high because the bullet doesn’t drop as fast.”

What is the correct explanation?

B You have thrown a rock, and it is flying through the air in an arc. Ifthe earth’s gravitational force on it is always straight down, why doesn’t itjust go straight down once it leaves your hand?

C Consider the example of the bullet that is dropped at the samemoment another bullet is fired from a gun. What would the motion of thetwo bullets look like to a jet pilot flying alongside in the same direction asthe shot bullet and at the same horizontal speed?

6.2 Coordinates and components’Cause we’re allBold as love,Just ask the axis.

Jimi Hendrix

How do we convert these ideas into mathematics? Figure b showsa good way of connecting the intuitive ideas to the numbers. In onedimension, we impose a number line with an x coordinate on acertain stretch of space. In two dimensions, we imagine a grid ofsquares which we label with x and y values, as shown in figure b.

But of course motion doesn’t really occur in a series of discretehops like in chess or checkers. Figure c shows a way of conceptual-izing the smooth variation of the x and y coordinates. The ball’sshadow on the wall moves along a line, and we describe its positionwith a single coordinate, y, its height above the floor. The wallshadow has a constant acceleration of -9.8 m/s2. A shadow on thefloor, made by a second light source, also moves along a line, and wedescribe its motion with an x coordinate, measured from the wall.


b / This object experiences a force that pulls it down toward thebottom of the page. In each equal time interval, it moves three units tothe right. At the same time, its vertical motion is making a simple patternof +1, 0, −1, −2, −3, −4, . . . units. Its motion can be described by an xcoordinate that has zero acceleration and a y coordinate with constantacceleration. The arrows labeled x and y serve to explain that we aredefining increasing x to the right and increasing y as upward.

The velocity of the floor shadow is referred to as the x componentof the velocity, written vx. Similarly we can notate the accelerationof the floor shadow as ax. Since vx is constant, ax is zero.

Similarly, the velocity of the wall shadow is called vy, its accel-eration ay. This example has ay = −9.8 m/s2.

Because the earth’s gravitational force on the ball is acting alongthe y axis, we say that the force has a negative y component, Fy,but Fx = Fz = 0.

The general idea is that we imagine two observers, each of whomperceives the entire universe as if it was flattened down to a singleline. The y-observer, for instance, perceives y, vy, and ay, and willinfer that there is a force, Fy, acting downward on the ball. Thatis, a y component means the aspect of a physical phenomenon, suchas velocity, acceleration, or force, that is observable to someone whocan only see motion along the y axis.

All of this can easily be generalized to three dimensions. In theexample above, there could be a z-observer who only sees motiontoward or away from the back wall of the room.

Section 6.2 Coordinates and components 181

d / Example 1.

A car going over a cliff example 1. The police find a car at a distance w = 20 m from the base of acliff of height h = 100 m. How fast was the car going when it wentover the edge? Solve the problem symbolically first, then plug inthe numbers.

. Let’s choose y pointing up and x pointing away from the cliff.The car’s vertical motion was independent of its horizontal mo-tion, so we know it had a constant vertical acceleration of a =−g = −9.8 m/s2. The time it spent in the air is therefore relatedto the vertical distance it fell by the constant-acceleration equa-tion

∆y =12

ay∆t2 ,

or

−h =12

(−g)∆t2 .

Solving for ∆t gives

∆t =

√2hg

.

Since the vertical force had no effect on the car’s horizontal mo-tion, it had ax = 0, i.e., constant horizontal velocity. We can applythe constant-velocity equation

vx =∆x∆t

,

i.e.,

vx =w∆t

.

We now substitute for ∆t to find

vx = w/

√2hg

,

which simplifies to

vx = w√

g2h

.

Plugging in numbers, we find that the car’s speed when it wentover the edge was 4 m/s, or about 10 mi/hr.


e / A parabola can be defined asthe shape made by cutting a coneparallel to its side. A parabola isalso the graph of an equation ofthe form y ∝ x2.

f / Each water droplet followsa parabola. The faster drops’parabolas are bigger.

Projectiles move along parabolas.

What type of mathematical curve does a projectile follow throughspace? To find out, we must relate x to y, eliminating t. The rea-soning is very similar to that used in the example above. Arbitrarilychoosing x = y = t = 0 to be at the top of the arc, we convenientlyhave x = ∆x, y = ∆y, and t = ∆t, so

y =1

2ayt

2 (ay < 0)

x = vxt

We solve the second equation for t = x/vx and eliminate t in thefirst equation:

y =1

2ay

(x

vx

)2

.

Since everything in this equation is a constant except for x and y,we conclude that y is proportional to the square of x. As you mayor may not recall from a math class, y ∝ x2 describes a parabola.

Discussion question

A At the beginning of this section I represented the motion of a projec-tile on graph paper, breaking its motion into equal time intervals. Supposeinstead that there is no force on the object at all. It obeys Newton’s first lawand continues without changing its state of motion. What would the corre-sponding graph-paper diagram look like? If the time interval representedby each arrow was 1 second, how would you relate the graph-paper dia-gram to the velocity components vx and vy ?

B Make up several different coordinate systems oriented in differentways, and describe the ax and ay of a falling object in each one.

6.3 Newton’s laws in three dimensionsIt is now fairly straightforward to extend Newton’s laws to threedimensions:

Newton’s first lawIf all three components of the total force on an object are zero,then it will continue in the same state of motion.

Newton’s second lawThe components of an object’s acceleration are predicted bythe equations

ax = Fx,total/m ,

ay = Fy,total/m , and

az = Fz,total/m .

Newton’s third lawIf two objects A and B interact via forces, then the compo-

Section 6.3 Newton’s laws in three dimensions 183

g / Example 2.

nents of their forces on each other are equal and opposite:

FA on B,x = −FB on A,x ,

FA on B,y = −FB on A,y , and

FA on B,z = −FB on A,z .

Forces in perpendicular directions on the same objectexample 2. An object is initially at rest. Two constant forces begin acting onit, and continue acting on it for a while. As suggested by the twoarrows, the forces are perpendicular, and the rightward force isstronger. What happens?

. Aristotle believed, and many students still do, that only one forcecan “give orders” to an object at one time. They therefore thinkthat the object will begin speeding up and moving in the directionof the stronger force. In fact the object will move along a diagonal.In the example shown in the figure, the object will respond to thelarge rightward force with a large acceleration component to theright, and the small upward force will give it a small accelerationcomponent upward. The stronger force does not overwhelm theweaker force, or have any effect on the upward motion at all. Theforce components simply add together:

Fx ,total = F1,x +*0

F2,x

Fy ,total =>

0F1,y + F2,y

Discussion question

A The figure shows two trajectories, made by splicing together linesand circular arcs, which are unphysical for an object that is only beingacted on by gravity. Prove that they are impossible based on Newton’slaws.


SummarySelected vocabularycomponent . . . . the part of a velocity, acceleration, or force

that would be perceptible to an observer whocould only see the universe projected along acertain one-dimensional axis

parabola . . . . . the mathematical curve whose graph has yproportional to x2

Notationx, y, z . . . . . . an object’s positions along the x, y, and z axesvx, vy, vz . . . . . the x, y, and z components of an object’s ve-

locity; the rates of change of the object’s x, y,and z coordinates

ax, ay, az . . . . . the x, y, and z components of an object’s ac-celeration; the rates of change of vx, vy, andvz

Summary

A force does not produce any effect on the motion of an objectin a perpendicular direction. The most important application ofthis principle is that the horizontal motion of a projectile has zeroacceleration, while the vertical motion has an acceleration equal to g.That is, an object’s horizontal and vertical motions are independent.The arc of a projectile is a parabola.

Motion in three dimensions is measured using three coordinates,x, y, and z. Each of these coordinates has its own correspondingvelocity and acceleration. We say that the velocity and accelerationboth have x, y, and z components

Newton’s second law is readily extended to three dimensions byrewriting it as three equations predicting the three components ofthe acceleration,

ax = Fx,total/m ,

ay = Fy,total/m ,

az = Fz,total/m ,

and likewise for the first and third laws.

Summary 185

ProblemsKey√



1 Two daredevils, Wendy and Bill, go over Niagara Falls. Wendysits in an inner tube, and lets the 30 km/hr velocity of the river throwher out horizontally over the falls. Bill paddles a kayak, adding anextra 10 km/hr to his velocity. They go over the edge of the fallsat the same moment, side by side. Ignore air friction. Explain yourreasoning.(a) Who hits the bottom first?(b) What is the horizontal component of Wendy’s velocity on im-pact?(c) What is the horizontal component of Bill’s velocity on impact?(d) Who is going faster on impact?

2 At the 2010 Salinas Lettuce Festival Parade, the Lettuce Queendrops her bouquet while riding on a float moving toward the right.Sketch the shape of its trajectory in her frame of reference, andcompare with the shape seen by one of her admirers standing onthe sidewalk.

Problem 3.

3 A baseball pitcher throws a pitch clocked at vx=73.3 mi/h.He throws horizontally. By what amount, d, does the ball drop bythe time it reaches home plate, L=60.0 ft away?(a) First find a symbolic answer in terms of L, vx, and g.

√

(b) Plug in and find a numerical answer. Express your answer inunits of ft. (Note: 1 ft=12 in, 1 mi=5280 ft, and 1 in=2.54 cm)√

4 Two cars go over the same bump in the road, Maria’s Maseratiat 25 miles per hour and Park’s Porsche at 37. How many timesgreater is the vertical acceleration of the Porsche? Hint: Rememberthat acceleration depends both on how much the velocity changesand on how much time it takes to change.

√


5 A batter hits a baseball at speed v, at an angle θ abovehorizontal.(a) Find an equation for the range (horizontal distance to wherethe ball falls), R, in terms of the relevant variables. Neglect airfriction and the height of the ball above the ground when it is hit.

. Answer, p. 514(b) Interpret your equation in the cases of θ=0 and θ = 90 .(c) Find the angle that gives the maximum range.

. Answer, p. 514

6 (a) A ball is thrown straight up with velocity v. Find anequation for the height to which it rises.

√

(b) Generalize your equation for a ball thrown at an angle θ abovehorizontal, in which case its initial velocity components are vx =v cos θ and vy = v sin θ.

√

Problems 187


a / Vectors are used in aerial nav-igation.

Chapter 7

Vectors

7.1 Vector notationThe idea of components freed us from the confines of one-dimensionalphysics, but the component notation can be unwieldy, since everyone-dimensional equation has to be written as a set of three separateequations in the three-dimensional case. Newton was stuck with thecomponent notation until the day he died, but eventually someonesufficiently lazy and clever figured out a way of abbreviating threeequations as one.

(a)−→F A on B = −

−→F B on A stands for

FA on B,x = −FB on A,x

FA on B,y = −FB on A,y

FA on B,z = −FB on A,z

(b)−→F total =

−→F 1 +

−→F 2 + . . . stands for

Ftotal,x = F1,x + F2,x + . . .Ftotal,y = F1,y + F2,y + . . .Ftotal,z = F1,z + F2,z + . . .

(c) −→a = ∆−→v∆t stands for

ax = ∆vx/∆tay = ∆vy/∆taz = ∆vz/∆t

Example (a) shows both ways of writing Newton’s third law. Whichwould you rather write?

The idea is that each of the algebra symbols with an arrow writ-

189

ten on top, called a vector, is actually an abbreviation for threedifferent numbers, the x, y, and z components. The three compo-nents are referred to as the components of the vector, e.g., Fx is the

x component of the vector−→F . The notation with an arrow on top

is good for handwritten equations, but is unattractive in a printedbook, so books use boldface, F, to represent vectors. After thispoint, I’ll use boldface for vectors throughout this book.

In general, the vector notation is useful for any quantity thathas both an amount and a direction in space. Even when you arenot going to write any actual vector notation, the concept itself is auseful one. We say that force and velocity, for example, are vectors.A quantity that has no direction in space, such as mass or time,is called a scalar. The amount of a vector quantity is called itsmagnitude. The notation for the magnitude of a vector A is |A|,like the absolute value sign used with scalars.

Often, as in example (b), we wish to use the vector notation torepresent adding up all the x components to get a total x component,etc. The plus sign is used between two vectors to indicate this typeof component-by-component addition. Of course, vectors are reallytriplets of numbers, not numbers, so this is not the same as the useof the plus sign with individual numbers. But since we don’t want tohave to invent new words and symbols for this operation on vectors,we use the same old plus sign, and the same old addition-relatedwords like “add,” “sum,” and “total.” Combining vectors this wayis called vector addition.

Similarly, the minus sign in example (a) was used to indicatenegating each of the vector’s three components individually. Theequals sign is used to mean that all three components of the vectoron the left side of an equation are the same as the correspondingcomponents on the right.

Example (c) shows how we abuse the division symbol in a similarmanner. When we write the vector ∆v divided by the scalar ∆t,we mean the new vector formed by dividing each one of the velocitycomponents by ∆t.

It’s not hard to imagine a variety of operations that would com-bine vectors with vectors or vectors with scalars, but only four ofthem are required in order to express Newton’s laws:

operation definitionvector + vector Add component by component to

make a new set of three numbers.vector− vector Subtract component by component

to make a new set of three numbers.vector · scalar Multiply each component of the vec-

tor by the scalar.vector/scalar Divide each component of the vector

by the scalar.

190 Chapter 7 Vectors

b / The x an y componentsof a vector can be thought of asthe shadows it casts onto the xand y axes.

c / Self-check B.

As an example of an operation that is not useful for physics, therejust aren’t any useful physics applications for dividing a vector byanother vector component by component. In optional section 7.5,we discuss in more detail the fundamental reasons why some vectoroperations are useful and others useless.

We can do algebra with vectors, or with a mixture of vectorsand scalars in the same equation. Basically all the normal rules ofalgebra apply, but if you’re not sure if a certain step is valid, youshould simply translate it into three component-based equations andsee if it works.

Order of addition example 1. If we are adding two force vectors, F + G, is it valid to assumeas in ordinary algebra that F + G is the same as G + F?

. To tell if this algebra rule also applies to vectors, we simplytranslate the vector notation into ordinary algebra notation. Interms of ordinary numbers, the components of the vector F + Gwould be Fx + Gx , Fy + Gy , and Fz + Gz , which are certainly thesame three numbers as Gx + Fx , Gy + Fy , and Gz + Fz . Yes, F + Gis the same as G + F.

It is useful to define a symbol r for the vector whose componentsare x, y, and z, and a symbol ∆r made out of ∆x, ∆y, and ∆z.

Although this may all seem a little formidable, keep in mind thatit amounts to nothing more than a way of abbreviating equations!Also, to keep things from getting too confusing the remainder of thischapter focuses mainly on the ∆r vector, which is relatively easy tovisualize.

self-check ATranslate the equations vx = ∆x/∆t , vy = ∆y/∆t , and vz = ∆z/∆t formotion with constant velocity into a single equation in vector notation.. Answer, p. 510

Drawing vectors as arrows

A vector in two dimensions can be easily visualized by drawingan arrow whose length represents its magnitude and whose directionrepresents its direction. The x component of a vector can then bevisualized as the length of the shadow it would cast in a beam oflight projected onto the x axis, and similarly for the y component.Shadows with arrowheads pointing back against the direction of thepositive axis correspond to negative components.

In this type of diagram, the negative of a vector is the vectorwith the same magnitude but in the opposite direction. Multiplyinga vector by a scalar is represented by lengthening the arrow by thatfactor, and similarly for division.

self-check B

Section 7.1 Vector notation 191

d / Example 2.

Given vector Q represented by an arrow in figure c, draw arrows repre-senting the vectors 1.5Q and −Q. . Answer, p.511


A Would it make sense to define a zero vector? Discuss what thezero vector’s components, magnitude, and direction would be; are thereany issues here? If you wanted to disqualify such a thing from being avector, consider whether the system of vectors would be complete. Forcomparison, can you think of a simple arithmetic problem with ordinarynumbers where you need zero as the result? Does the same reasoningapply to vectors, or not? From your group, choose one person to act aszero’s advocate and one to argue against letting zero in the club. The restof the group should act as jurors.

B You drive to your friend’s house. How does the magnitude of your ∆rvector compare with the distance you’ve added to the car’s odometer?

7.2 Calculations with magnitude and directionIf you ask someone where Las Vegas is compared to Los Angeles,they are unlikely to say that the ∆x is 290 km and the ∆y is 230km, in a coordinate system where the positive x axis is east and they axis points north. They will probably say instead that it’s 370 kmto the northeast. If they were being precise, they might specify thedirection as 38 counterclockwise from east. In two dimensions, wecan always specify a vector’s direction like this, using a single angle.A magnitude plus an angle suffice to specify everything about thevector. The following two examples show how we use trigonometryand the Pythagorean theorem to go back and forth between the x−yand magnitude-angle descriptions of vectors.

Finding magnitude and angle from components example 2. Given that the ∆r vector from LA to Las Vegas has ∆x = 290 kmand ∆y = 230 km, how would we find the magnitude and directionof ∆r?

. We find the magnitude of ∆r from the Pythagorean theorem:

|∆r| =√∆x2 + ∆y2

= 370 km

We know all three sides of the triangle, so the angle θ can befound using any of the inverse trig functions. For example, weknow the opposite and adjacent sides, so

θ = tan−1 ∆y∆x

= 38 .


e / Example 4.

Finding components from magnitude and angle example 3. Given that the straight-line distance from Los Angeles to LasVegas is 370 km, and that the angle θ in the figure is 38 , howcan the xand y components of the ∆r vector be found?

. The sine and cosine of θ relate the given information to theinformation we wish to find:

cos θ =∆x|∆r|

sin θ =∆y|∆r|

Solving for the unknowns gives

∆x = |∆r| cos θ= 290 km and

∆y = |∆r| sin θ= 230 km .

The following example shows the correct handling of the plusand minus signs, which is usually the main cause of mistakes.

Negative components example 4. San Diego is 120 km east and 150 km south of Los Angeles. Anairplane pilot is setting course from San Diego to Los Angeles. Atwhat angle should she set her course, measured counterclock-wise from east, as shown in the figure?

. If we make the traditional choice of coordinate axes, with xpointing to the right and y pointing up on the map, then her ∆x isnegative, because her final x value is less than her initial x value.Her ∆y is positive, so we have

∆x = −120 km∆y = 150 km .

If we work by analogy with example 2, we get

θ = tan−1 ∆y∆x

= tan−1(−1.25)= −51 .

According to the usual way of defining angles in trigonometry,a negative result means an angle that lies clockwise from the xaxis, which would have her heading for the Baja California. Whatwent wrong? The answer is that when you ask your calculator to

Section 7.2 Calculations with magnitude and direction 193

f / Example 5.

g / Vectors can be added graph-ically by placing them tip to tail,and then drawing a vector fromthe tail of the first vector to the tipof the second vector.

take the arctangent of a number, there are always two valid pos-sibilities differing by 180 . That is, there are two possible angleswhose tangents equal -1.25:

tan 129 = −1.25tan−51 = −1.25

You calculator doesn’t know which is the correct one, so it justpicks one. In this case, the one it picked was the wrong one, andit was up to you to add 180 to it to find the right answer.

Discussion question

A In the example above, we dealt with components that were negative.Does it make sense to talk about positive and negative vectors?

7.3 Techniques for adding vectorsAddition of vectors given their components

The easiest type of vector addition is when you are in possessionof the components, and want to find the components of their sum.

Adding components example 5. Given the ∆x and ∆y values from the previous examples, findthe ∆x and ∆y from San Diego to Las Vegas.

.

∆xtotal = ∆x1 + ∆x2

= −120 km + 290 km= 170 km

∆ytotal = ∆y1 + ∆y2

= 150 km + 230 km= 380

Note how the signs of the x components take care of the west-ward and eastward motions, which partially cancel.

Addition of vectors given their magnitudes and directions

In this case, you must first translate the magnitudes and direc-tions into components, and the add the components.

Graphical addition of vectors

Often the easiest way to add vectors is by making a scale drawingon a piece of paper. This is known as graphical addition, as opposedto the analytic techniques discussed previously. (It has nothing todo with x − y graphs or graph paper. “Graphical” here simplymeans drawing. It comes from the Greek verb “grapho,” to write,like related English words including “graphic.”)


LA to Vegas, graphically example 6. Given the magnitudes and angles of the ∆r vectors from SanDiego to Los Angeles and from Los Angeles to Las Vegas, findthe magnitude and angle of the ∆r vector from San Diego to LasVegas.

. Using a protractor and a ruler, we make a careful scale draw-ing, as shown in figure h. The protractor can be convenientlyaligned with the blue rules on the notebook paper. A scale of1 mm→ 2 km was chosen for this solution because it was as bigas possible (for accuracy) without being so big that the drawingwouldn’t fit on the page. With a ruler, we measure the distancefrom San Diego to Las Vegas to be 206 mm, which correspondsto 412 km. With a protractor, we measure the angle θ to be 65 .

h / Example 6.

Even when we don’t intend to do an actual graphical calculationwith a ruler and protractor, it can be convenient to diagram theaddition of vectors in this way. With ∆r vectors, it intuitively makessense to lay the vectors tip-to-tail and draw the sum vector from thetail of the first vector to the tip of the second vector. We can dothe same when adding other vectors such as force vectors.

self-check CHow would you subtract vectors graphically? . Answer, p. 511

Section 7.3 Techniques for adding vectors 195


A If you’re doing graphical addition of vectors, does it matter whichvector you start with and which vector you start from the other vector’stip?

B If you add a vector with magnitude 1 to a vector of magnitude 2,what magnitudes are possible for the vector sum?

C Which of these examples of vector addition are correct, and whichare incorrect?

7.4 ? Unit vector notationWhen we want to specify a vector by its components, it can be cum-bersome to have to write the algebra symbol for each component:

∆x = 290 km, ∆y = 230 km

A more compact notation is to write

∆r = (290 km)x + (230 km)y ,

where the vectors x, y, and z, called the unit vectors, are definedas the vectors that have magnitude equal to 1 and directions lyingalong the x, y, and z axes. In speech, they are referred to as “x-hat”and so on.

A slightly different, and harder to remember, version of thisnotation is unfortunately more prevalent. In this version, the unitvectors are called i, j, and k:

∆r = (290 km)i + (230 km)j .

7.5 ? Rotational invarianceLet’s take a closer look at why certain vector operations are use-ful and others are not. Consider the operation of multiplying twovectors component by component to produce a third vector:

Rx = PxQx

Ry = PyQy

Rz = PzQz

As a simple example, we choose vectors P and Q to have length1, and make them perpendicular to each other, as shown in figure


i / Component-by-componentmultiplication of the vectors in 1would produce different vectorsin coordinate systems 2 and 3.

i/1. If we compute the result of our new vector operation using thecoordinate system in i/2, we find:

Rx = 0

Ry = 0

Rz = 0

The x component is zero because Px = 0, the y component is zerobecause Qy = 0, and the z component is of course zero because bothvectors are in the x − y plane. However, if we carry out the sameoperations in coordinate system i/3, rotated 45 degrees with respectto the previous one, we find

Rx = 1/2

Ry = −1/2

Rz = 0

The operation’s result depends on what coordinate system we use,and since the two versions of R have different lengths (one being zeroand the other nonzero), they don’t just represent the same answerexpressed in two different coordinate systems. Such an operationwill never be useful in physics, because experiments show physicsworks the same regardless of which way we orient the laboratorybuilding! The useful vector operations, such as addition and scalarmultiplication, are rotationally invariant, i.e., come out the sameregardless of the orientation of the coordinate system.

Section 7.5 ? Rotational invariance 197

SummarySelected vocabularyvector . . . . . . . a quantity that has both an amount (magni-

tude) and a direction in spacemagnitude . . . . the “amount” associated with a vectorscalar . . . . . . . a quantity that has no direction in space, only

an amount

NotationA . . . . . . . . . a vector with components Ax, Ay, and Az−→A . . . . . . . . . handwritten notation for a vector|A| . . . . . . . . the magnitude of vector Ar . . . . . . . . . . the vector whose components are x, y, and z∆r . . . . . . . . . the vector whose components are ∆x, ∆y, and

∆zx, y, z . . . . . . (optional topic) unit vectors; the vectors with

magnitude 1 lying along the x, y, and z axes

i, j, k . . . . . . . a harder to remember notation for the unitvectors

Other terminology and notationdisplacement vec-tor . . . . . . . . .

a name for the symbol ∆r

speed . . . . . . . the magnitude of the velocity vector, i.e., thevelocity stripped of any information about itsdirection

Summary

A vector is a quantity that has both a magnitude (amount) anda direction in space, as opposed to a scalar, which has no direction.The vector notation amounts simply to an abbreviation for writingthe vector’s three components.

In two dimensions, a vector can be represented either by its twocomponents or by its magnitude and direction. The two ways ofdescribing a vector can be related by trigonometry.

The two main operations on vectors are addition of a vector toa vector, and multiplication of a vector by a scalar.

Vector addition means adding the components of two vectorsto form the components of a new vector. In graphical terms, thiscorresponds to drawing the vectors as two arrows laid tip-to-tail anddrawing the sum vector from the tail of the first vector to the tipof the second one. Vector subtraction is performed by negating thevector to be subtracted and then adding.

Multiplying a vector by a scalar means multiplying each of itscomponents by the scalar to create a new vector. Division by ascalar is defined similarly.

Differentiation and integration of vectors is defined component


by component.

Summary 199

Problem 1.

Problem 4.

ProblemsKey√



1 The figure shows vectors A and B. Graphically calculate thefollowing:

A + B, A−B, B−A, −2B, A− 2B

No numbers are involved.

2 Phnom Penh is 470 km east and 250 km south of Bangkok.Hanoi is 60 km east and 1030 km north of Phnom Penh.(a) Choose a coordinate system, and translate these data into ∆xand ∆y values with the proper plus and minus signs.(b) Find the components of the ∆r vector pointing from Bangkokto Hanoi.

√

3 If you walk 35 km at an angle 25 counterclockwise from east,and then 22 km at 230 counterclockwise from east, find the distanceand direction from your starting point to your destination.

√

4 A machinist is drilling holes in a piece of aluminum accordingto the plan shown in the figure. She starts with the top hole, thenmoves to the one on the left, and then to the one on the right. Sincethis is a high-precision job, she finishes by moving in the directionand at the angle that should take her back to the top hole, andchecks that she ends up in the same place. What are the distanceand direction from the right-hand hole to the top one?

√


5 Suppose someone proposes a new operation in which a vectorA and a scalar B are added together to make a new vector C likethis:

Cx = Ax +B

Cy = Ay +B

Cz = Az +B

Prove that this operation won’t be useful in physics, because it’snot rotationally invariant.

6 In this problem you’ll extend the analysis in problem 5 onp. 187 to include air friction by writing a computer program. Fora game played at sea level, the force due to air friction is approxi-mately (7×10−4 N·s2/m2)v2, in the direction opposite to the motionof the ball.1 The mass of a baseball is 0.146 kg.(a) For a ball hit at a speed of 45.0 m/s from a height of 1.0 m, findthe optimal angle and the resulting range. . Answer, p. 514(b) How much farther would the ball fly at the Colorado Rockies’stadium, where the thinner air gives 18 percent less air friction?

. Answer, p. 514

1A standard baseball is supposed to have a circumference of 9 18

inches. Astandard way of paramerizing the force of fluid friction is F = (1/2)ρACDv

2,where ρ is the density of the fluid, A is the object’s cross-sectional area, andCD is a unitless constant. E. Meyer and J. Bohn, in a 2008 paper published atarxiv.org, survey existing data on CD for baseballs and estimate it to be in therange from about 0.13 to 0.5. This leads to a figure something like the one given.

Problems 201


Chapter 8

Vectors and motion

In 1872, capitalist and former California governor Leland Stanfordasked photographer Eadweard Muybridge if he would work for himon a project to settle a $25,000 bet (a princely sum at that time).Stanford’s friends were convinced that a galloping horse always hadat least one foot on the ground, but Stanford claimed that there wasa moment during each cycle of the motion when all four feet werein the air. The human eye was simply not fast enough to settle thequestion. In 1878, Muybridge finally succeeded in producing whatamounted to a motion picture of the horse, showing conclusivelythat all four feet did leave the ground at one point. (Muybridge wasa colorful figure in San Francisco history, and his acquittal for themurder of his wife’s lover was considered the trial of the century inCalifornia.)

The losers of the bet had probably been influenced by Aris-totelian reasoning, for instance the expectation that a leaping horsewould lose horizontal velocity while in the air with no force to pushit forward, so that it would be more efficient for the horse to runwithout leaping. But even for students who have converted whole-

203

heartedly to Newtonianism, the relationship between force and ac-celeration leads to some conceptual difficulties, the main one beinga problem with the true but seemingly absurd statement that anobject can have an acceleration vector whose direction is not thesame as the direction of motion. The horse, for instance, has nearlyconstant horizontal velocity, so its ax is zero. But as anyone can tellyou who has ridden a galloping horse, the horse accelerates up anddown. The horse’s acceleration vector therefore changes back andforth between the up and down directions, but is never in the samedirection as the horse’s motion. In this chapter, we will examinemore carefully the properties of the velocity, acceleration, and forcevectors. No new principles are introduced, but an attempt is madeto tie things together and show examples of the power of the vectorformulation of Newton’s laws.

8.1 The velocity vectorFor motion with constant velocity, the velocity vector is

v = ∆r/∆t . [only for constant velocity]

The ∆r vector points in the direction of the motion, and dividingit by the scalar ∆t only changes its length, not its direction, so thevelocity vector points in the same direction as the motion. Whenthe velocity vector is not constant, we form it from the componentsvx = dx/dt, vy = dy/dt, and vz = dz/dt. This set of three equationscan be notated more compactly as

v = dr/dt .

This is an example of a more general rule about differentiating vec-tors: to differentiate a vector, take the derivative component bycomponent. Even when the velocity vector is not constant, it stillpoints along the direction of motion.

A car bouncing on its shock absorbers example 1. A car bouncing on its shock absorbers has a position as a func-tion of time given by

r = bt x + (c sinωt)y ,

where b, c, and ω (Greek letter omega) are constants. Infer theunits of the constants, find the velocity, and check the units of theresult.

. The components of the position vector are bt and c sinωt , andif these are both to have units of meters, then b must have units ofm/s and c units of meters. The sine function requires a unitlessinput, so ω must have units of s−1 (interpreted as radians persecond, e.g., if c = 2π rad/s, then the car completes one cycle ofvertical oscillation in one second).

204 Chapter 8 Vectors and motion

a / Example 2.

Differentiating component by component, we find

v = bx + (cω cosωt)y .

The units of the first component are simply the units of b, m/s,which makes sense. The units of the second component arem · s−1, which also checks out.

Vector addition is the correct way to generalize the one-dimensionalconcept of adding velocities in relative motion, as shown in the fol-lowing example:

Velocity vectors in relative motion example 2. You wish to cross a river and arrive at a dock that is directlyacross from you, but the river’s current will tend to carry youdownstream. To compensate, you must steer the boat at an an-gle. Find the angle θ, given the magnitude, |vWL|, of the water’svelocity relative to the land, and the maximum speed, |vBW |, ofwhich the boat is capable relative to the water.

. The boat’s velocity relative to the land equals the vector sum ofits velocity with respect to the water and the water’s velocity withrespect to the land,

vBL = vBW + vWL .

If the boat is to travel straight across the river, i.e., along the yaxis, then we need to have vBL,x = 0. This x component equalsthe sum of the x components of the other two vectors,

vBL,x = vBW ,x + vWL,x ,

or0 = −|vBW | sin θ + |vWL| .

Solving for θ, we find

sin θ = |vWL|/|vBW | ,

so

θ = sin−1 |vWL||vBW |

.

. Solved problem: Annie Oakley page 214, problem 3


A Is it possible for an airplane to maintain a constant velocity vectorbut not a constant |v|? How about the opposite – a constant |v| but not aconstant velocity vector? Explain.

B New York and Rome are at about the same latitude, so the earth’srotation carries them both around nearly the same circle. Do the two citieshave the same velocity vector (relative to the center of the earth)? If not,is there any way for two cities to have the same velocity vector?

Section 8.1 The velocity vector 205

b / A change in the magni-tude of the velocity vector impliesan acceleration.

c / A change in the directionof the velocity vector also pro-duces a nonzero ∆v vector, andthus a nonzero accelerationvector, ∆v/∆t .

8.2 The acceleration vector

When the acceleration is constant, we have

a = ∆v/∆t , [only for constant acceleration]

which can be written in terms of initial and final velocities as

a = (vf − vi)/∆t . [only for constant acceleration]

In general, we define the acceleration vector as the derivative

a = dv/dt .

Now there are two ways in which we could have a nonzero accel-eration. Either the magnitude or the direction of the velocity vectorcould change. This can be visualized with arrow diagrams as shownin figures b and c. Both the magnitude and direction can changesimultaneously, as when a car accelerates while turning. Only whenthe magnitude of the velocity changes while its direction stays con-stant do we have a ∆v vector and an acceleration vector along thesame line as the motion.

self-check A(1) In figure b, is the object speeding up, or slowing down? (2) Whatwould the diagram look like if vi was the same as vf ? (3) Describe howthe ∆v vector is different depending on whether an object is speedingup or slowing down. . Answer, p. 511

If this all seems a little strange and abstract to you, you’re notalone. It doesn’t mean much to most physics students the firsttime someone tells them that acceleration is a vector, and that theacceleration vector does not have to be in the same direction as thevelocity vector. One way to understand those statements better isto imagine an object such as an air freshener or a pair of fuzzy dicehanging from the rear-view mirror of a car. Such a hanging object,called a bob, constitutes an accelerometer. If you watch the bobas you accelerate from a stop light, you’ll see it swing backward.The horizontal direction in which the bob tilts is opposite to thedirection of the acceleration. If you apply the brakes and the car’sacceleration vector points backward, the bob tilts forward.

After accelerating and slowing down a few times, you thinkyou’ve put your accelerometer through its paces, but then you makea right turn. Surprise! Acceleration is a vector, and needn’t pointin the same direction as the velocity vector. As you make a rightturn, the bob swings outward, to your left. That means the car’sacceleration vector is to your right, perpendicular to your velocityvector. A useful definition of an acceleration vector should relatein a systematic way to the actual physical effects produced by theacceleration, so a physically reasonable definition of the acceleration


vector must allow for cases where it is not in the same direction asthe motion.

self-check BIn projectile motion, what direction does the acceleration vector have?. Answer, p. 511

d / Example 3.

Rappelling example 3In figure d, the rappeller’s velocity has long periods of gradualchange interspersed with short periods of rapid change. Thesecorrespond to periods of small acceleration and force, and peri-ods of large acceleration and force.

The galloping horse example 4Figure e on page 208 shows outlines traced from the first, third,fifth, seventh, and ninth frames in Muybridge’s series of pho-tographs of the galloping horse. The estimated location of thehorse’s center of mass is shown with a circle, which bobs aboveand below the horizontal dashed line.

If we don’t care about calculating velocities and accelerations inany particular system of units, then we can pretend that the timebetween frames is one unit. The horse’s velocity vector as itmoves from one point to the next can then be found simply bydrawing an arrow to connect one position of the center of mass tothe next. This produces a series of velocity vectors which alter-nate between pointing above and below horizontal.

Section 8.2 The acceleration vector 207

e / Example 4.

The ∆v vector is the vector which we would have to add onto onevelocity vector in order to get the next velocity vector in the series.The ∆v vector alternates between pointing down (around the timewhen the horse is in the air, B) and up (around the time when thehorse has two feet on the ground, D).


A When a car accelerates, why does a bob hanging from the rearviewmirror swing toward the back of the car? Is it because a force throws itbackward? If so, what force? Similarly, describe what happens in theother cases described above.

B Superman is guiding a crippled spaceship into port. The ship’sengines are not working. If Superman suddenly changes the direction ofhis force on the ship, does the ship’s velocity vector change suddenly? Itsacceleration vector? Its direction of motion?


f / Example 5.

g / The applied force FA pushesthe block up the frictionless ramp.

h / Three forces act on theblock. Their vector sum is zero.

i / If the block is to move atconstant velocity, Newton’s firstlaw says that the three forcevectors acting on it must addup to zero. To perform vectoraddition, we put the vectors tipto tail, and in this case we areadding three vectors, so eachone’s tail goes against the tip ofthe previous one. Since they aresupposed to add up to zero, thethird vector’s tip must come backto touch the tail of the first vector.They form a triangle, and sincethe applied force is perpendicularto the normal force, it is a righttriangle.

8.3 The force vector and simple machinesForce is relatively easy to intuit as a vector. The force vector pointsin the direction in which it is trying to accelerate the object it isacting on.

Since force vectors are so much easier to visualize than accel-eration vectors, it is often helpful to first find the direction of the(total) force vector acting on an object, and then use that informa-tion to determine the direction of the acceleration vector. Newton’ssecond law, Ftotal = ma, tells us that the two must be in the samedirection.

A component of a force vector example 5Figure f, redrawn from a classic 1920 textbook, shows a boypulling another child on a sled. His force has both a horizontalcomponent and a vertical one, but only the horizontal one accel-erates the sled. (The vertical component just partially cancels theforce of gravity, causing a decrease in the normal force betweenthe runners and the snow.) There are two triangles in the figure.One triangle’s hypotenuse is the rope, and the other’s is the mag-nitude of the force. These triangles are similar, so their internalangles are all the same, but they are not the same triangle. Oneis a distance triangle, with sides measured in meters, the othera force triangle, with sides in newtons. In both cases, the hori-zontal leg is 93% as long as the hypotenuse. It does not makesense, however, to compare the sizes of the triangles — the forcetriangle is not smaller in any meaningful sense.

Pushing a block up a ramp example 6. Figure (a) shows a block being pushed up a frictionless rampat constant speed by an applied force FA. How much force isrequired, in terms of the block’s mass, m, and the angle of theramp, θ?

. Figure (b) shows the other two forces acting on the block: anormal force, FN , created by the ramp, and the weight force, FW ,created by the earth’s gravity. Because the block is being pushedup at constant speed, it has zero acceleration, and the total forceon it must be zero. From figure (c), we find

|FA| = |FW | sin θ= mg sin θ .

Since the sine is always less than one, the applied force is alwaysless than mg, i.e., pushing the block up the ramp is easier thanlifting it straight up. This is presumably the principle on which thepyramids were constructed: the ancient Egyptians would havehad a hard time applying the forces of enough slaves to equal thefull weight of the huge blocks of stone.

Essentially the same analysis applies to several other simple ma-

Section 8.3 The force vector and simple machines 209


j / Discussion question B.

chines, such as the wedge and the screw.

. Solved problem: A cargo plane page 216, problem 9

. Solved problem: The angle of repose page 217, problem 13

. Solved problem: A wagon page 217, problem 14


A The figure shows a block being pressed diagonally upward against awall, causing it to slide up the wall. Analyze the forces involved, includingtheir directions.

B The figure shows a roller coaster car rolling down and then up underthe influence of gravity. Sketch the car’s velocity vectors and accelerationvectors. Pick an interesting point in the motion and sketch a set of forcevectors acting on the car whose vector sum could have resulted in theright acceleration vector.

8.4 More about calculus with vectorsOur definition of the derivative of a vector implies the familiar prop-erties

d(cf)

dt= c

d(f)

dt[c is a constant]

and

d(f + g)

dt=

d(f)

dt+

d(g)

dt.

The integral of a vector is likewise defined as integrating componentby component.

The second derivative of a vector example 7. Two objects have positions as functions of time given by theequations

r1 = 3t2x + t y

and

r2 = 3t4x + t y .

Find both objects’ accelerations using calculus. Could either an-swer have been found without calculus?

. Taking the first derivative of each component, we find

v1 = 6t x + y

v2 = 12t3x + y ,

and taking the derivatives again gives acceleration,

a1 = 6x

a2 = 36t2x .


The first object’s acceleration could have been found without cal-culus, simply by comparing the x and y coordinates with theconstant-acceleration equation ∆x = vo∆t + 1

2a∆t2. The secondequation, however, isn’t just a second-order polynomial in t , sothe acceleration isn’t constant, and we really did need calculus tofind the corresponding acceleration.

The integral of a vector example 8. Starting from rest, a flying saucer of mass m is observed tovary its propulsion with mathematical precision according to theequation

F = bt42x + ct137y .

(The aliens inform us that the numbers 42 and 137 have a specialreligious significance for them.) Find the saucer’s velocity as afunction of time.

. From the given force, we can easily find the acceleration

a =Fm

=bm

t42x +cm

t137y .

The velocity vector v is the integral with respect to time of theacceleration,

v =∫

a dt

=∫ (

bm

t42x +cm

t137y)

dt ,

and integrating component by component gives

=(∫

bm

t42 dt)

x +(∫

cm

t137 dt)

y

=b

43mt43x +

c138m

t138y ,

where we have omitted the constants of integration, since thesaucer was starting from rest.

A fire-extinguisher stunt on ice example 9. Prof. Puerile smuggles a fire extinguisher into a skating rink.Climbing out onto the ice without any skates on, he sits down andpushes off from the wall with his feet, acquiring an initial veloc-ity voy. At t = 0, he then discharges the fire extinguisher at a45-degree angle so that it applies a force to him that is backwardand to the left, i.e., along the negative y axis and the positive xaxis. The fire extinguisher’s force is strong at first, but then diesdown according to the equation |F| = b − ct , where b and c areconstants. Find the professor’s velocity as a function of time.

Section 8.4 More about calculus with vectors 211

. Measured counterclockwise from the x axis, the angle of theforce vector becomes 315 . Breaking the force down into x andy components, we have

Fx = |F| cos 315

= (b − ct)Fy = |F| sin 315

= (−b + ct) .

In unit vector notation, this is

F = (b − ct)x + (−b + ct)y .

Newton’s second law gives

a = F/m

=b − ct√

2mx +−b + ct√

2my .

To find the velocity vector as a function of time, we need to inte-grate the acceleration vector with respect to time,

v =∫

a dt

=∫ (

b − ct√2m

x +−b + ct√

2my)

dt

=1√2m

∫ [(b − ct) x + (−b + ct) y

]dt

A vector function can be integrated component by component, sothis can be broken down into two integrals,

v =x√2m

∫(b − ct) dt +

y√2m

∫(−b + ct) dt

=

(bt − 1

2ct2√

2m+ constant #1

)x +

(−bt + 1

2ct2√

2m+ constant #2

)y

Here the physical significance of the two constants of integrationis that they give the initial velocity. Constant #1 is therefore zero,and constant #2 must equal vo. The final result is

v =

(bt − 1

2ct2√

2m

)x +

(−bt + 1

2ct2√

2m+ vo

)y .


SummaryThe velocity vector points in the direction of the object’s motion.Relative motion can be described by vector addition of velocities.

The acceleration vector need not point in the same direction asthe object’s motion. We use the word “acceleration” to describe anychange in an object’s velocity vector, which can be either a changein its magnitude or a change in its direction.

An important application of the vector addition of forces is theuse of Newton’s first law to analyze mechanical systems.

Summary 213

Problem 2.

ProblemsKey√



1 Is it possible for a helicopter to have an acceleration due eastand a velocity due west? If so, what would be going on? If not, whynot?

2 The figure shows the path followed by Hurricane Irene in 2005as it moved north. The dots show the location of the center of thestorm at six-hour intervals, with lighter dots at the time when thestorm reached its greatest intensity. Find the time when the storm’scenter had a velocity vector to the northeast and an accelerationvector to the southeast. Explain.

3 Annie Oakley, riding north on horseback at 30 mi/hr, shootsher rifle, aiming horizontally and to the northeast. The muzzle speedof the rifle is 140 mi/hr. When the bullet hits a defenseless fuzzyanimal, what is its speed of impact? Neglect air resistance, andignore the vertical motion of the bullet. . Solution, p. 501

Problem 4.

4 A dinosaur fossil is slowly moving down the slope of a glacierunder the influence of wind, rain and gravity. At the same time,the glacier is moving relative to the continent underneath. Thedashed lines represent the directions but not the magnitudes of thevelocities. Pick a scale, and use graphical addition of vectors to findthe magnitude and the direction of the fossil’s velocity relative tothe continent. You will need a ruler and protractor.

√

5 A bird is initially flying horizontally east at 21.1 m/s, but onesecond later it has changed direction so that it is flying horizontallyand 7 north of east, at the same speed. What are the magnitudeand direction of its acceleration vector during that one second timeinterval? (Assume its acceleration was roughly constant.)

√


Problem 7.

6 A gun is aimed horizontally to the west, and fired at t = 0. Thebullet’s position vector as a function of time is r = bx + cty + dt2z,where b, c, and d are positive constants.(a) What units would b, c, and d need to have for the equation tomake sense?(b) Find the bullet’s velocity and acceleration as functions of time.(c) Give physical interpretations of b, c, d, x, y, and z.

7 The figure shows an experiment in which a cart is released fromrest at A, and accelerates down the slope through a distance x untilit passes through a sensor’s light beam. The point of the experimentis to determine the cart’s acceleration. At B, a cardboard vanemounted on the cart enters the light beam, blocking the light beam,and starts an electronic timer running. At C, the vane emerges fromthe beam, and the timer stops.(a) Find the final velocity of the cart in terms of the width w ofthe vane and the time tb for which the sensor’s light beam wasblocked.

√

(b) Find the magnitude of the cart’s acceleration in terms of themeasurable quantities x, tb, and w.

√

(c) Analyze the forces in which the cart participates, using a table inthe format introduced in section 5.3. Assume friction is negligible.(d) Find a theoretical value for the acceleration of the cart, whichcould be compared with the experimentally observed value extractedin part b. Express the theoretical value in terms of the angle θ ofthe slope, and the strength g of the gravitational field.

√

8 A person of mass M stands in the middle of a tightrope,which is fixed at the ends to two buildings separated by a horizontaldistance L. The rope sags in the middle, stretching and lengtheningthe rope slightly.(a) If the tightrope walker wants the rope to sag vertically by nomore than a height h, find the minimum tension, T , that the ropemust be able to withstand without breaking, in terms of h, g, M ,and L.

√

(b) Based on your equation, explain why it is not possible to geth = 0, and give a physical interpretation.

Problem 8.

Problems 215

Problem 9.

Problem 11.

9 A cargo plane has taken off from a tiny airstrip in the Andes,and is climbing at constant speed, at an angle of θ=17 with respectto horizontal. Its engines supply a thrust of Fthrust = 200 kN, andthe lift from its wings is Flift = 654 kN. Assume that air resistance(drag) is negligible, so the only forces acting are thrust, lift, andweight. What is its mass, in kg? . Solution, p. 501

10 A skier of mass m is coasting down a slope inclined atan angle θ compared to horizontal. Assume for simplicity that thetreatment of kinetic friction given in chapter 5 is appropriate here,although a soft and wet surface actually behaves a little differently.The coefficient of kinetic friction acting between the skis and thesnow is µk, and in addition the skier experiences an air frictionforce of magnitude bv2, where b is a constant.(a) Find the maximum speed that the skier will attain, in terms ofthe variables m, g, θ, µk, and b.

√

(b) For angles below a certain minimum angle θmin, the equationgives a result that is not mathematically meaningful. Find an equa-tion for θmin, and give a physical explanation of what is happeningfor θ < θmin.

√

11 Your hand presses a block of mass m against a wall witha force FH acting at an angle θ, as shown in the figure. Find theminimum and maximum possible values of |FH | that can keep theblock stationary, in terms of m, g, θ, and µs, the coefficient of staticfriction between the block and the wall. Check both your answersin the case of θ = 90 , and interpret the case where the maximumforce is infinite.

√

12 Driving down a hill inclined at an angle θ with respect tohorizontal, you slam on the brakes to keep from hitting a deer. Yourantilock brakes kick in, and you don’t skid.(a) Analyze the forces. (Ignore rolling resistance and air friction.)(b) Find the car’s maximum possible deceleration, a (expressed asa positive number), in terms of g, θ, and the relevant coefficient offriction.

√

(c) Explain physically why the car’s mass has no effect on youranswer.(d) Discuss the mathematical behavior and physical interpretationof your result for negative values of θ.(e) Do the same for very large positive values of θ.


Problem 14.

Problem 15 (Millikan and Gale,1920).

Problem 16.

13 The angle of repose is the maximum slope on which an objectwill not slide. On airless, geologically inert bodies like the moon oran asteroid, the only thing that determines whether dust or rubblewill stay on a slope is whether the slope is less steep than the angleof repose.(a) Find an equation for the angle of repose, deciding for yourselfwhat are the relevant variables.(b) On an asteroid, where g can be thousands of times lower thanon Earth, would rubble be able to lie at a steeper angle of repose?

. Solution, p. 501

14 A wagon is being pulled at constant speed up a slope θ by arope that makes an angle φ with the vertical.(a) Assuming negligible friction, show that the tension in the ropeis given by the equation

FT =sin θ

sin(θ + φ)FW ,

where FW is the weight force acting on the wagon.(b) Interpret this equation in the special cases of φ = 0 and φ =180 − θ. . Solution, p. 502

15 The figure shows a boy hanging in three positions: (1) withhis arms straight up, (2) with his arms at 45 degrees, and (3) withhis arms at 60 degrees with respect to the vertical. Compare thetension in his arms in the three cases.

16 For safety, mountain climbers often wear a climbing harnessand tie in to other climbers on a rope team or to anchors such aspitons or snow anchors. When using anchors, the climber usuallywants to tie in to more than one, both for extra strength and forredundancy in case one fails. The figure shows such an arrangement,with the climber hanging from a pair of anchors forming a “Y’ atan angle θ. The usual advice is to make θ < 90 ; for large values ofθ, the stress placed on the anchors can be many times greater thanthe actual load L, so that two anchors are actually less safe thanone.(a) Find the force S at each anchor in terms of L and θ.

√

(b) Verify that your answer makes sense in the case of θ = 0.(c) Interpret your answer in the case of θ = 180 .(d) What is the smallest value of θ for which S exceeds L, so that afailure of at least one anchor is more likely than it would have beenwith a single anchor?

√

Problems 217

Problem 17.

Problem 18.

Problem 19.

Problem 20.

17 (a) A block is sitting on a wedge inclined at an angle θwith respect to horizontal. Someone grabs the wedge and movesit horizontally with acceleration a. The motion is in the directionshown by the arrow in the figure. Find the maximum accelerationthat can be applied without causing the block to slide downhill.(b) Show that your answer to part a has the right units.

√

(c) Show that it also has the right dependence on θ, by comparingits mathematical behavior to its physically expected behavior.

18 The two blocks shown in the figure have equal mass, m, andthe surface is frictionless. (a) What is the tension in the masslessrope? . Hint, p. 494

√

(b) Show that the units of your answer make sense.(c) Check the physical behavior of your answer in the special casesof φ ≤ θ and θ = 0, φ = 90 .

19 The photo shows a coil of rope wound around a smooth metalpost. A large amount of tension is applied at the bottom of the coil,but only a tiny force, supplied by a piece of sticky tape, is neededat the top to keep the rope from slipping. Show that the ratio ofthese two forces increases exponentially with the number of turns ofrope, and find an expression for that ratio. . Hint, p. 494

√?

20 Two wheels of radius r rotate in the same vertical plane withangular velocities +Ω and −Ω about axes that are parallel and atthe same height. The wheels touch one another at a point on theircircumferences, so that their rotations mesh like gears in a gear train.A board is laid on top of the wheels, so that two friction forces actupon it, one from each wheel. Characterize the three qualitativelydifferent types of motion that the board can exhibit, depending onthe initial conditions. ?


Problems 219

Exercise 8: Vectors and motionEach diagram on page 221 shows the motion of an object in an x − y plane. Each dot is onelocation of the object at one moment in time. The time interval from one dot to the next isalways the same, so you can think of the vector that connects one dot to the next as a v vector,and subtract to find ∆v vectors.

1. Suppose the object in diagram 1 is moving from the top left to the bottom right. Deducewhatever you can about the force acting on it. Does the force always have the same magnitude?The same direction?

Invent a physical situation that this diagram could represent.

What if you reinterpret the diagram, and reverse the object’s direction of motion?

2. What can you deduce about the force that is acting in diagram 2?

Invent a physical situation that diagram 2 could represent.

3. What can you deduce about the force that is acting in diagram 3?

Invent a physical situation.


Exercise 8: Vectors and motion 221


Chapter 9

Circular motion

9.1 Conceptual frameworkI now live fifteen minutes from Disneyland, so my friends and familyin my native Northern California think it’s a little strange that I’venever visited the Magic Kingdom again since a childhood trip to thesouth. The truth is that for me as a preschooler, Disneyland wasnot the Happiest Place on Earth. My mother took me on a ride inwhich little cars shaped like rocket ships circled rapidly around acentral pillar. I knew I was going to die. There was a force trying tothrow me outward, and the safety features of the ride would surelyhave been inadequate if I hadn’t screamed the whole time to makesure Mom would hold on to me. Afterward, she seemed surprisinglyindifferent to the extreme danger we had experienced.

Circular motion does not produce an outward force

My younger self’s understanding of circular motion was partlyright and partly wrong. I was wrong in believing that there was aforce pulling me outward, away from the center of the circle. Theeasiest way to understand this is to bring back the parable of thebowling ball in the pickup truck from chapter 4. As the truck makesa left turn, the driver looks in the rearview mirror and thinks thatsome mysterious force is pulling the ball outward, but the truckis accelerating, so the driver’s frame of reference is not an inertialframe. Newton’s laws are violated in a noninertial frame, so the ballappears to accelerate without any actual force acting on it. Becausewe are used to inertial frames, in which accelerations are caused by

223

b / This crane fly’s haltereshelp it to maintain its orientationin flight.

forces, the ball’s acceleration creates a vivid illusion that there mustbe an outward force.

a / 1. In the turning truck’s frameof reference, the ball appearsto violate Newton’s laws, dis-playing a sideways accelerationthat is not the result of a force-interaction with any other object.2. In an inertial frame of refer-ence, such as the frame fixed tothe earth’s surface, the ball obeysNewton’s first law. No forces areacting on it, and it continues mov-ing in a straight line. It is the truckthat is participating in an interac-tion with the asphalt, the truck thataccelerates as it should accordingto Newton’s second law. In an inertial frame everything makes more sense. The ball has

no force on it, and goes straight as required by Newton’s first law.The truck has a force on it from the asphalt, and responds to itby accelerating (changing the direction of its velocity vector) asNewton’s second law says it should.

The halteres example 1Another interesting example is an insect organ called the hal-teres, a pair of small knobbed limbs behind the wings, which vi-brate up and down and help the insect to maintain its orientationin flight. The halteres evolved from a second pair of wings pos-sessed by earlier insects. Suppose, for example, that the halteresare on their upward stroke, and at that moment an air currentcauses the fly to pitch its nose down. The halteres follow New-ton’s first law, continuing to rise vertically, but in the fly’s rotatingframe of reference, it seems as though they have been subjectedto a backward force. The fly has special sensory organs that per-ceive this twist, and help it to correct itself by raising its nose.

Circular motion does not persist without a force

I was correct, however, on a different point about the Disneylandride. To make me curve around with the car, I really did need someforce such as a force from my mother, friction from the seat, or anormal force from the side of the car. (In fact, all three forces wereprobably adding together.) One of the reasons why Galileo failed to

224 Chapter 9 Circular motion

c / 1. An overhead view of a per-son swinging a rock on a rope. Aforce from the string is requiredto make the rock’s velocity vectorkeep changing direction. 2. If thestring breaks, the rock will followNewton’s first law and go straightinstead of continuing around thecircle.

refine the principle of inertia into a quantitative statement like New-ton’s first law is that he was not sure whether motion without a forcewould naturally be circular or linear. In fact, the most impressiveexamples he knew of the persistence of motion were mostly circular:the spinning of a top or the rotation of the earth, for example. New-ton realized that in examples such as these, there really were forcesat work. Atoms on the surface of the top are prevented from flyingoff straight by the ordinary force that keeps atoms stuck together insolid matter. The earth is nearly all liquid, but gravitational forcespull all its parts inward.

Uniform and nonuniform circular motion

Circular motion always involves a change in the direction of thevelocity vector, but it is also possible for the magnitude of the ve-locity to change at the same time. Circular motion is referred to asuniform if |v| is constant, and nonuniform if it is changing.

Your speedometer tells you the magnitude of your car’s velocityvector, so when you go around a curve while keeping your speedome-ter needle steady, you are executing uniform circular motion. If yourspeedometer reading is changing as you turn, your circular motionis nonuniform. Uniform circular motion is simpler to analyze math-ematically, so we will attack it first and then pass to the nonuniformcase.

self-check AWhich of these are examples of uniform circular motion and which arenonuniform?

(1) the clothes in a clothes dryer (assuming they remain against theinside of the drum, even at the top)

(2) a rock on the end of a string being whirled in a vertical circle .

Answer, p. 511

Section 9.1 Conceptual framework 225

d / To make the brick go in acircle, I had to exert an inwardforce on the rope.

f / When a car is going straightat constant speed, the forwardand backward forces on it arecanceling out, producing a totalforce of zero. When it movesin a circle at constant speed,there are three forces on it, butthe forward and backward forcescancel out, so the vector sum isan inward force.

Only an inward force is required for uniform circular motion.

Figure c showed the string pulling in straight along a radius ofthe circle, but many people believe that when they are doing thisthey must be “leading” the rock a little to keep it moving along.That is, they believe that the force required to produce uniformcircular motion is not directly inward but at a slight angle to theradius of the circle. This intuition is incorrect, which you can easilyverify for yourself now if you have some string handy. It is onlywhile you are getting the object going that your force needs to be atan angle to the radius. During this initial period of speeding up, themotion is not uniform. Once you settle down into uniform circularmotion, you only apply an inward force.

If you have not done the experiment for yourself, here is a theo-retical argument to convince you of this fact. We have discussed inchapter 6 the principle that forces have no perpendicular effects. Tokeep the rock from speeding up or slowing down, we only need tomake sure that our force is perpendicular to its direction of motion.We are then guaranteed that its forward motion will remain unaf-fected: our force can have no perpendicular effect, and there is noother force acting on the rock which could slow it down. The rockrequires no forward force to maintain its forward motion, any morethan a projectile needs a horizontal force to “help it over the top”of its arc.

e / A series of three hammer taps makes the rolling ball trace a tri-angle, seven hammers a heptagon. If the number of hammers was largeenough, the ball would essentially be experiencing a steady inward force,and it would go in a circle. In no case is any forward force necessary.


g / Example 2.

Why, then, does a car driving in circles in a parking lot stopexecuting uniform circular motion if you take your foot off the gas?The source of confusion here is that Newton’s laws predict an ob-ject’s motion based on the total force acting on it. A car driving incircles has three forces on it

(1) an inward force from the asphalt, controlled with the steeringwheel;

(2) a forward force from the asphalt, controlled with the gaspedal; and

(3) backward forces from air resistance and rolling resistance.

You need to make sure there is a forward force on the car so thatthe backward forces will be exactly canceled out, creating a vectorsum that points directly inward.

A motorcycle making a turn example 2The motorcyclist in figure g is moving along an arc of a circle. Itlooks like he’s chosen to ride the slanted surface of the dirt at aplace where it makes just the angle he wants, allowing him to getthe force he needs on the tires as a normal force, without needingany frictional force. The dirt’s normal force on the tires points upand to our left. The vertical component of that force is canceledby gravity, while its horizontal component causes him to curve.

In uniform circular motion, the acceleration vector is inward

Since experiments show that the force vector points directlyinward, Newton’s second law implies that the acceleration vectorpoints inward as well. This fact can also be proven on purely kine-matical grounds, and we will do so in the next section.

Section 9.1 Conceptual framework 227

Discussion questions A-D

Discussion question E.


A In the game of crack the whip, a line of people stand holding hands,and then they start sweeping out a circle. One person is at the center, androtates without changing location. At the opposite end is the person whois running the fastest, in a wide circle. In this game, someone always endsup losing their grip and flying off. Suppose the person on the end losesher grip. What path does she follow as she goes flying off? (Assume sheis going so fast that she is really just trying to put one foot in front of theother fast enough to keep from falling; she is not able to get any significanthorizontal force between her feet and the ground.)

B Suppose the person on the outside is still holding on, but feels thatshe may loose her grip at any moment. What force or forces are actingon her, and in what directions are they? (We are not interested in thevertical forces, which are the earth’s gravitational force pulling down, andthe ground’s normal force pushing up.)

C Suppose the person on the outside is still holding on, but feels thatshe may loose her grip at any moment. What is wrong with the followinganalysis of the situation? “The person whose hand she’s holding exertsan inward force on her, and because of Newton’s third law, there’s anequal and opposite force acting outward. That outward force is the oneshe feels throwing her outward, and the outward force is what might makeher go flying off, if it’s strong enough.”

D If the only force felt by the person on the outside is an inward force,why doesn’t she go straight in?

E In the amusement park ride shown in the figure, the cylinder spinsfaster and faster until the customer can pick her feet up off the floor with-out falling. In the old Coney Island version of the ride, the floor actuallydropped out like a trap door, showing the ocean below. (There is also aversion in which the whole thing tilts up diagonally, but we’re discussingthe version that stays flat.) If there is no outward force acting on her, whydoes she stick to the wall? Analyze all the forces on her.

F What is an example of circular motion where the inward force is anormal force? What is an example of circular motion where the inwardforce is friction? What is an example of circular motion where the inwardforce is the sum of more than one force?

G Does the acceleration vector always change continuously in circularmotion? The velocity vector?


h / This figure shows an intu-itive justification for the factproved mathematically in thissection, that the direction of theforce and acceleration in circularmotion is inward. The heptagon,2, is a better approximation toa circle than the triangle, 1. Tomake an infinitely good approx-imation to circular motion, wewould need to use an infinitelylarge number of infinitesimal taps,which would amount to a steadyinward force.

i / The total force in the forward-backward direction is zero in bothcases.

9.2 Uniform circular motionIn this section I derive some convenient results, which you will usefrequently, for the acceleration of an object performing uniform cir-cular motion.

An object moving in a circle of radius r in the x-y plane has

x = r cos ωt and

y = r sin ωt ,

where ω is the number of radians traveled per second, and the pos-itive or negative sign indicates whether the motion is clockwise orcounterclockwise.

Differentiating, we find that the components of the velocity are

vx = −ωr sin ωt and

vy = ωr cos ωt ,

and for the acceleration we have

ax = −ω2r cos ωt and

ay = −ω2r sin ωt .

The acceleration vector has cosines and sines in the same placesas the r vector, but with minus signs in front, so it points in theopposite direction, i.e., toward the center of the circle. By Newton’ssecond law, a=F/m, this shows that the force must be inward aswell; without this force, the object would fly off straight.

The magnitude of the acceleration is

|a| =√a2x + a2

y

= ω2r .

It makes sense that ω is squared, since reversing the sign of ω cor-responds to reversing the direction of motion, but the accelerationis toward the center of the circle, regardless of whether the motionis clockwise or counterclockwise. This result can also be rewrittenin the form

|a| = |v|2

r.

These results are counterintuitive. Until Newton, physicists andlaypeople alike had assumed that the planets would need a forceto push them forward in their orbits. Figure h may help to makeit more plausible that only an inward force is required. A forwardforce might be needed in order to cancel out a backward force suchas friction, i, but the total force in the forward-backward directionneeds to be exactly zero for constant-speed motion. When you are in

Section 9.2 Uniform circular motion 229

j / There is no outward forceon the bowling ball, but in thenoninertial frame it seems likeone exists.

a car undergoing circular motion, there is also a strong illusion of anoutward force. But what object could be making such a force? Thecar’s seat makes an inward force on you, not an outward one. Thereis no object that could be exerting an outward force on your body.In reality, this force is an illusion that comes from our brain’s intu-itive efforts to interpret the situation within a noninertial frame ofreference. As shown in figure j, we can describe everything perfectlywell in an inertial frame of reference, such as the frame attached tothe sidewalk. In such a frame, the bowling ball goes straight becausethere is no force on it. The wall of the truck’s bed hits the ball, notthe other way around.

Force required to turn on a bike example 3. A bicyclist is making a turn along an arc of a circle with radius20 m, at a speed of 5 m/s. If the combined mass of the cyclistplus the bike is 60 kg, how great a static friction force must theroad be able to exert on the tires?

. Taking the magnitudes of both sides of Newton’s second lawgives

|F| = |ma|= m|a| .

Substituting |a| = |v|2/r gives

|F| = m|v|2/r≈ 80 N

(rounded off to one sig fig).

Don’t hug the center line on a curve! example 4. You’re driving on a mountain road with a steep drop on yourright. When making a left turn, is it safer to hug the center line orto stay closer to the outside of the road?

. You want whichever choice involves the least acceleration, be-cause that will require the least force and entail the least risk ofexceeding the maximum force of static friction. Assuming thecurve is an arc of a circle and your speed is constant, your caris performing uniform circular motion, with |a| = |v|2/r . The de-pendence on the square of the speed shows that driving slowlyis the main safety measure you can take, but for any given speedyou also want to have the largest possible value of r . Even thoughyour instinct is to keep away from that scary precipice, you are ac-tually less likely to skid if you keep toward the outside, becausethen you are describing a larger circle.

Acceleration related to radius and period of rotation example 5. How can the equation for the acceleration in uniform circularmotion be rewritten in terms of the radius of the circle and theperiod, T , of the motion, i.e., the time required to go around once?


k / Example 6.

. The period can be related to the speed as follows:

|v| = circumferenceT

= 2πr/T .

Substituting into the equation |a| = |v|2/r gives

|a| = 4π2rT 2 .

A clothes dryer example 6. My clothes dryer has a drum with an inside radius of 35 cm, andit spins at 48 revolutions per minute. What is the acceleration ofthe clothes inside?

. We can solve this by finding the period and plugging in to theresult of the previous example. If it makes 48 revolutions in oneminute, then the period is 1/48 of a minute, or 1.25 s. To get anacceleration in mks units, we must convert the radius to 0.35 m.Plugging in, the result is 8.8 m/s2.

More about clothes dryers! example 7. In a discussion question in the previous section, we made theassumption that the clothes remain against the inside of the drumas they go over the top. In light of the previous example, is this acorrect assumption?

. No. We know that there must be some minimum speed at whichthe motor can run that will result in the clothes just barely stay-ing against the inside of the drum as they go over the top. If theclothes dryer ran at just this minimum speed, then there would beno normal force on the clothes at the top: they would be on theverge of losing contact. The only force acting on them at the topwould be the force of gravity, which would give them an acceler-ation of g = 9.8 m/s2. The actual dryer must be running slowerthan this minimum speed, because it produces an acceleration ofonly 8.8 m/s2. My theory is that this is done intentionally, to makethe clothes mix and tumble.

. Solved problem: The tilt-a-whirl page 235, problem 3

. Solved problem: An off-ramp page 236, problem 5


A A certain amount of force is needed to provide the acceleration ofcircular motion. What if were are exerting a force perpendicular to thedirection of motion in an attempt to make an object trace a circle of radiusr , but the force isn’t as big as m|v|2/r?

B Suppose a rotating space station, as in figure l on page 232, is built.It gives its occupants the illusion of ordinary gravity. What happens when

Section 9.2 Uniform circular motion 231

m / 1. Moving in a circle whilespeeding up. 2. Uniform circularmotion. 3. Slowing down.

a person in the station lets go of a ball? What happens when she throwsa ball straight “up” in the air (i.e., towards the center)?

l / Discussion question B. An artist’s conception of a rotating space colonyin the form of a giant wheel. A person living in this noninertial frame ofreference has an illusion of a force pulling her outward, toward the deck,for the same reason that a person in the pickup truck has the illusionof a force pulling the bowling ball. By adjusting the speed of rotation, thedesigners can make an acceleration |v|2/r equal to the usual accelerationof gravity on earth. On earth, your acceleration standing on the groundis zero, and a falling rock heads for your feet with an acceleration of 9.8m/s2. A person standing on the deck of the space colony has an upwardacceleration of 9.8 m/s2, and when she lets go of a rock, her feet head upat the nonaccelerating rock. To her, it seems the same as true gravity.

9.3 Nonuniform circular motion

What about nonuniform circular motion? Although so far wehave been discussing components of vectors along fixed x and yaxes, it now becomes convenient to discuss components of the accel-eration vector along the radial line (in-out) and the tangential line(along the direction of motion). For nonuniform circular motion,the radial component of the acceleration obeys the same equationas for uniform circular motion,

ar = v2/r ,

where v = |v|, but the acceleration vector also has a tangentialcomponent,

at =dv

dt.

The latter quantity has a simple interpretation. If you are goingaround a curve in your car, and the speedometer needle is mov-ing, the tangential component of the acceleration vector is simplywhat you would have thought the acceleration was if you saw thespeedometer and didn’t know you were going around a curve.

Slow down before a turn, not during it. example 8. When you’re making a turn in your car and you’re afraid youmay skid, isn’t it a good idea to slow down?

. If the turn is an arc of a circle, and you’ve already completedpart of the turn at constant speed without skidding, then the roadand tires are apparently capable of enough static friction to sup-ply an acceleration of |v|2/r . There is no reason why you wouldskid out now if you haven’t already. If you get nervous and brake,however, then you need to have a tangential acceleration com-ponent in addition to the radial component you were already ableto produce successfully. This would require an acceleration vec-tor with a greater magnitude, which in turn would require a larger


force. Static friction might not be able to supply that much force,and you might skid out. As in the previous example on a similartopic, the safe thing to do is to approach the turn at a comfortablylow speed.

. Solved problem: A bike race page 237, problem 10

Section 9.3 Nonuniform circular motion 233

SummarySelected vocabularyuniform circularmotion . . . . . .

circular motion in which the magnitude of thevelocity vector remains constant

nonuniform circu-lar motion . . . .

circular motion in which the magnitude of thevelocity vector changes

radial . . . . . . . parallel to the radius of a circle; the in-outdirection

tangential . . . . tangent to the circle, perpendicular to the ra-dial direction

Notationar . . . . . . . . . radial acceleration; the component of the ac-

celeration vector along the in-out directionat . . . . . . . . . tangential acceleration; the component of the

acceleration vector tangent to the circle

Summary

If an object is to have circular motion, a force must be exerted onit toward the center of the circle. There is no outward force on theobject; the illusion of an outward force comes from our experiencesin which our point of view was rotating, so that we were viewingthings in a noninertial frame.

An object undergoing uniform circular motion has an inwardacceleration vector of magnitude

|a| = v2/r ,

where v = |v|. In nonuniform circular motion, the radial and tan-gential components of the acceleration vector are

ar = v2/r

at =dv

dt.


Problem 3.

ProblemsKey√



1 Show that the expression |v|2/r has the units of acceleration.

2 A plane is flown in a loop-the-loop of radius 1.00 km. Theplane starts out flying upside-down, straight and level, then beginscurving up along the circular loop, and is right-side up when itreaches the top. (The plane may slow down somewhat on the wayup.) How fast must the plane be going at the top if the pilot is toexperience no force from the seat or the seatbelt while at the top ofthe loop?

√

3 The amusement park ride shown in the figure consists of acylindrical room that rotates about its vertical axis. When the ro-tation is fast enough, a person against the wall can pick his or herfeet up off the floor and remain “stuck” to the wall without falling.(a) Suppose the rotation results in the person having a speed v. Theradius of the cylinder is r, the person’s mass is m, the downwardacceleration of gravity is g, and the coefficient of static friction be-tween the person and the wall is µs. Find an equation for the speed,v, required, in terms of the other variables. (You will find that oneof the variables cancels out.)(b) Now suppose two people are riding the ride. Huy is wearingdenim, and Gina is wearing polyester, so Huy’s coefficient of staticfriction is three times greater. The ride starts from rest, and as itbegins rotating faster and faster, Gina must wait longer before beingable to lift her feet without sliding to the floor. Based on your equa-tion from part a, how many times greater must the speed be beforeGina can lift her feet without sliding down? . Solution, p. 503 ?

4 The bright star Sirius has a mass of 4.02 × 1030 kg and liesat a distance of 8.1× 1016 m from our solar system. Suppose you’restanding on a merry-go-round carousel rotating with a period of 10seconds. You adopt a rotating, noninertial frame of reference, inwhich the carousel is at rest, and the universe is spinning around it.If you drop a corndog, you see it accelerate horizontally away fromthe axis, and you interpret this is the result of some horizontal force.This force does not actually exist; it only seems to exist becauseyou’re insisting on using a noninertial frame. Similarly, calculatethe force that seems to act on Sirius in this frame of reference.Comment on the physical plausibility of this force, and on whatobject could be exerting it.

√

Problems 235

Problem 5.

Problem 7.

5 An engineer is designing a curved off-ramp for a freeway.Since the off-ramp is curved, she wants to bank it to make it lesslikely that motorists going too fast will wipe out. If the radius ofthe curve is r, how great should the banking angle, θ, be so thatfor a car going at a speed v, no static friction force whatsoever isrequired to allow the car to make the curve? State your answer interms of v, r, and g, and show that the mass of the car is irrelevant.

. Solution, p. 503

6 Lionel brand toy trains come with sections of track in standardlengths and shapes. For circular arcs, the most commonly usedsections have diameters of 662 and 1067 mm at the inside of the outerrail. The maximum speed at which a train can take the broadercurve without flying off the tracks is 0.95 m/s. At what speed mustthe train be operated to avoid derailing on the tighter curve?

√

7 Psychology professor R.O. Dent requests funding for an exper-iment on compulsive thrill-seeking behavior in guinea pigs, in whichthe subject is to be attached to the end of a spring and whirledaround in a horizontal circle. The spring has equilibrium length b,and obeys Hooke’s law with spring constant k. It is stiff enough tokeep from bending significantly under the guinea pig’s weight.(a) Calculate the length of the spring when it is undergoing steadycircular motion in which one rotation takes a time T . Express yourresult in terms of k, b, T , and the guinea pig’s mass m.

√

(b) The ethics committee somehow fails to veto the experiment, butthe safety committee expresses concern. Why? Does your equa-tion do anything unusual, or even spectacular, for any particularvalue of T? What do you think is the physical significance of thismathematical behavior?

8 The acceleration of an object in uniform circular motion can begiven either by |a| = |v|2/r or, equivalently, by |a| = 4π2r/T 2, whereT is the time required for one cycle (example 5 on page 230). PersonA says based on the first equation that the acceleration in circularmotion is greater when the circle is smaller. Person B, arguing fromthe second equation, says that the acceleration is smaller when thecircle is smaller. Rewrite the two statements so that they are lessmisleading, eliminating the supposed paradox. [Based on a problemby Arnold Arons.]


Problem 10.

Problem 11.

Problem 12.

9 When you’re done using an electric mixer, you can get mostof the batter off of the beaters by lifting them out of the batter withthe motor running at a high enough speed. Let’s imagine, to makethings easier to visualize, that we instead have a piece of tape stuckto one of the beaters.(a) Explain why static friction has no effect on whether or not thetape flies off.(b) Suppose you find that the tape doesn’t fly off when the motoris on a low speed, but at a greater speed, the tape won’t stay on.Why would the greater speed change things?

10 Three cyclists in a race are rounding a semicircular curve.At the moment depicted, cyclist A is using her brakes to apply aforce of 375 N to her bike. Cyclist B is coasting. Cyclist C ispedaling, resulting in a force of 375 N on her bike Each cyclist,with her bike, has a mass of 75 kg. At the instant shown, theinstantaneous speed of all three cyclists is 10 m/s. On the diagram,draw each cyclist’s acceleration vector with its tail on top of herpresent position, indicating the directions and lengths reasonablyaccurately. Indicate approximately the consistent scale you are usingfor all three acceleration vectors. Extreme precision is not necessaryas long as the directions are approximately right, and lengths ofvectors that should be equal appear roughly equal, etc. Assume allthree cyclists are traveling along the road all the time, not wanderingacross their lane or wiping out and going off the road.

. Solution, p. 503

11 The figure shows a ball on the end of a string of length Lattached to a vertical rod which is spun about its vertical axis by amotor. The period (time for one rotation) is P .(a) Analyze the forces in which the ball participates.(b) Find how the angle θ depends on P , g, and L. [Hints: (1)Write down Newton’s second law for the vertical and horizontalcomponents of force and acceleration. This gives two equations,which can be solved for the two unknowns, θ and the tension inthe string. (2) If you introduce variables like v and r, relate themto the variables your solution is supposed to contain, and eliminatethem.]

√

(c) What happens mathematically to your solution if the motor isrun very slowly (very large values of P )? Physically, what do youthink would actually happen in this case?

12 The figure shows two blocks of masses m1 and m2 slidingin circles on a frictionless table. Find the tension in the strings ifthe period of rotation (time required for one complete rotation) isP .

√

Problems 237

Problem 14.

Problem 15.

Problem 16.

13 A little kid in my neighborhood came home from shoppingwith his mother. They live on a hill, with their driveway orientedperpendicular to the slope. Their minivan was parked in the drive-way, and while she was bringing groceries inside, he unlocked theparking brake and put the car in neutral. The steering wheel waslocked with the wheels banked. The car rolled downhill in a circulararc with the driveway at its top, eventually crashing through thewall of a neighbor’s living room. (Nobody was hurt.) Suppose theneighbor’s house hadn’t intervened. The car just rolls freely, and wewant to know whether it will ever skid. Static friction acts betweenthe asphalt and the tires with coefficient µs, the radius of the circleis r, the slope of the hill is θ, and the gravitational field has strengthg. Find the maximum value of θ such that the car will never skid.√

?

14 The figure shows an old-fashioned device called a flyballgovernor, used for keeping an engine running at the correct speed.The whole thing rotates about the vertical shaft, and the mass Mis free to slide up and down. This mass would have a connection(not shown) to a valve that controlled the engine. If, for instance,the engine ran too fast, the mass would rise, causing the engine toslow back down.(a) Show that in the special case of a = 0, the angle θ is given by

θ = cos−1

(g(m+M)P 2

4π2mL

),

where P is the period of rotation (time required for one completerotation).(b) There is no closed-form solution for θ in the general case wherea is not zero. However, explain how the undesirable low-speed be-havior of the a = 0 device would be improved by making a nonzero.

?

15 The vertical post rotates at frequency ω. The bead slidesfreely along the string, reaching an equilibrium in which its distancefrom the axis is r and the angles θ and φ have some particular values.Find φ in terms of θ, g, ω, and r.

√?

16 A bead slides down along a piece of wire that is in the shapeof a helix. The helix lies on the surface of a vertical cylinder ofradius r, and the vertical distance between turns is d.(a) Ordinarily when an object slides downhill under the influence ofkinetic friction, the velocity-independence of kinetic friction impliesthat the acceleration is constant, and therefore there is no limit tothe object’s velocity. Explain the physical reason why this argumentfails here, so that the bead will in fact have some limiting velocity.(b) Find the limiting velocity.(c) Show that your result has the correct behavior in the limit ofr →∞. [Problem by B. Korsunsky.]

√?


17 In a well known stunt from circuses and carnivals, a motor-cyclist rides around inside a big bowl, gradually speeding up andrising higher. Eventually the cyclist can get up to where the wallsof the bowl are vertical. Let’s estimate the conditions under whicha running human could do the same thing. (a) If the runner canrun at speed v, and her shoes have a coefficient of static friction µs,what is the maximum radius of the circle? (b) Evaluate your resultnumerically for v = 10 m/s (the speed of an olympic sprinter) andµs = 5. (This is roughly the highest coefficient of static friction everachieved for surfaces that are not sticky. The surface has an arrayof microscopic fibers like a hair brush, and is inspired by the hairson the feet of a gecko.)

Problems 239


a / Johannes Kepler found amathematical description of themotion of the planets, which ledto Newton’s theory of gravity.

Gravity is the only really important force on the cosmic scale. This false-color representation of saturn’s rings was made from an image sent backby the Voyager 2 space probe. The rings are composed of innumerabletiny ice particles orbiting in circles under the influence of saturn’s gravity.

Chapter 10

Gravity

Cruise your radio dial today and try to find any popular song thatwould have been imaginable without Louis Armstrong. By introduc-ing solo improvisation into jazz, Armstrong took apart the jigsawpuzzle of popular music and fit the pieces back together in a dif-ferent way. In the same way, Newton reassembled our view of theuniverse. Consider the titles of some recent physics books writtenfor the general reader: The God Particle, Dreams of a Final The-ory. When the subatomic particle called the neutrino was recentlyproven for the first time to have mass, specialists in cosmology be-gan discussing seriously what effect this would have on calculationsof the ultimate fate of the universe: would the neutrinos’ mass causeenough extra gravitational attraction to make the universe eventu-ally stop expanding and fall back together? Without Newton, suchattempts at universal understanding would not merely have seemeda little pretentious, they simply would not have occurred to anyone.

This chapter is about Newton’s theory of gravity, which he usedto explain the motion of the planets as they orbited the sun. Whereas

241

b / Tycho Brahe made his nameas an astronomer by showing thatthe bright new star, today calleda supernova, that appeared inthe skies in 1572 was far beyondthe Earth’s atmosphere. This,along with Galileo’s discovery ofsunspots, showed that contraryto Aristotle, the heavens werenot perfect and unchanging.Brahe’s fame as an astronomerbrought him patronage from KingFrederick II, allowing him to carryout his historic high-precisionmeasurements of the planets’motions. A contradictory charac-ter, Brahe enjoyed lecturing othernobles about the evils of dueling,but had lost his own nose in ayouthful duel and had it replacedwith a prosthesis made of analloy of gold and silver. Willing toendure scandal in order to marrya peasant, he nevertheless usedthe feudal powers given to him bythe king to impose harsh forcedlabor on the inhabitants of hisparishes. The result of their work,an Italian-style palace with anobservatory on top, surely ranksas one of the most luxuriousscience labs ever built. Keplerdescribed Brahe as dying of aruptured bladder after falling froma wagon on the way home froma party, but other contemporaryaccounts and modern medicalanalysis suggest mercury poison-ing, possibly as a result of courtintrigue.

this book has concentrated on Newton’s laws of motion, leavinggravity as a dessert, Newton tosses off the laws of motion in thefirst 20 pages of the Principia Mathematica and then spends thenext 130 discussing the motion of the planets. Clearly he saw thisas the crucial scientific focus of his work. Why? Because in it heshowed that the same laws of motion applied to the heavens as tothe earth, and that the gravitational force that made an apple fallwas the same as the force that kept the earth’s motion from carryingit away from the sun. What was radical about Newton was not hislaws of motion but his concept of a universal science of physics.

10.1 Kepler’s laws

Newton wouldn’t have been able to figure out why the planetsmove the way they do if it hadn’t been for the astronomer TychoBrahe (1546-1601) and his protege Johannes Kepler (1571-1630),who together came up with the first simple and accurate descriptionof how the planets actually do move. The difficulty of their task issuggested by figure c, which shows how the relatively simple orbitalmotions of the earth and Mars combine so that as seen from earthMars appears to be staggering in loops like a drunken sailor.

c / As the Earth and Mars revolve around the sun at different rates,the combined effect of their motions makes Mars appear to trace astrange, looped path across the background of the distant stars.

242 Chapter 10 Gravity

Brahe, the last of the great naked-eye astronomers, collected ex-tensive data on the motions of the planets over a period of manyyears, taking the giant step from the previous observations’ accuracyof about 10 minutes of arc (10/60 of a degree) to an unprecedented1 minute. The quality of his work is all the more remarkable consid-ering that his observatory consisted of four giant brass protractorsmounted upright in his castle in Denmark. Four different observerswould simultaneously measure the position of a planet in order tocheck for mistakes and reduce random errors.

With Brahe’s death, it fell to his former assistant Kepler to tryto make some sense out of the volumes of data. Kepler, in con-tradiction to his late boss, had formed a prejudice, a correct oneas it turned out, in favor of the theory that the earth and planetsrevolved around the sun, rather than the earth staying fixed andeverything rotating about it. Although motion is relative, it is notjust a matter of opinion what circles what. The earth’s rotationand revolution about the sun make it a noninertial reference frame,which causes detectable violations of Newton’s laws when one at-tempts to describe sufficiently precise experiments in the earth-fixedframe. Although such direct experiments were not carried out untilthe 19th century, what convinced everyone of the sun-centered sys-tem in the 17th century was that Kepler was able to come up witha surprisingly simple set of mathematical and geometrical rules fordescribing the planets’ motion using the sun-centered assumption.After 900 pages of calculations and many false starts and dead-endideas, Kepler finally synthesized the data into the following threelaws:

Kepler’s elliptical orbit lawThe planets orbit the sun in elliptical orbits with the sun atone focus.

Kepler’s equal-area lawThe line connecting a planet to the sun sweeps out equal areasin equal amounts of time.

Kepler’s law of periodsThe time required for a planet to orbit the sun, called itsperiod, is proportional to the long axis of the ellipse raised tothe 3/2 power. The constant of proportionality is the samefor all the planets.

Although the planets’ orbits are ellipses rather than circles, mostare very close to being circular. The earth’s orbit, for instance, isonly flattened by 1.7% relative to a circle. In the special case of aplanet in a circular orbit, the two foci (plural of “focus”) coincideat the center of the circle, and Kepler’s elliptical orbit law thus saysthat the circle is centered on the sun. The equal-area law impliesthat a planet in a circular orbit moves around the sun with constantspeed. For a circular orbit, the law of periods then amounts to a

Section 10.1 Kepler’s laws 243

d / An ellipse is a circle thathas been distorted by shrinkingand stretching along perpendicu-lar axes.

e / An ellipse can be con-structed by tying a string to twopins and drawing like this with thepencil stretching the string taut.Each pin constitutes one focus ofthe ellipse.

f / If the time interval takenby the planet to move from P to Qis equal to the time interval fromR to S, then according to Kepler’sequal-area law, the two shadedareas are equal. The planetis moving faster during intervalRS than it did during PQ, whichNewton later determined was dueto the sun’s gravitational forceaccelerating it. The equal-arealaw predicts exactly how much itwill speed up.

statement that the time for one orbit is proportional to r3/2, wherer is the radius. If all the planets were moving in their orbits at thesame speed, then the time for one orbit would simply depend onthe circumference of the circle, so it would only be proportional tor to the first power. The more drastic dependence on r3/2 meansthat the outer planets must be moving more slowly than the innerplanets.

10.2 Newton’s law of gravityThe sun’s force on the planets obeys an inverse square law.

Kepler’s laws were a beautifully simple explanation of what theplanets did, but they didn’t address why they moved as they did.Did the sun exert a force that pulled a planet toward the center ofits orbit, or, as suggested by Descartes, were the planets circulatingin a whirlpool of some unknown liquid? Kepler, working in theAristotelian tradition, hypothesized not just an inward force exertedby the sun on the planet, but also a second force in the directionof motion to keep the planet from slowing down. Some speculatedthat the sun attracted the planets magnetically.

Once Newton had formulated his laws of motion and taughtthem to some of his friends, they began trying to connect themto Kepler’s laws. It was clear now that an inward force would beneeded to bend the planets’ paths. This force was presumably anattraction between the sun and each planet. (Although the sun doesaccelerate in response to the attractions of the planets, its mass is sogreat that the effect had never been detected by the prenewtonianastronomers.) Since the outer planets were moving slowly alongmore gently curving paths than the inner planets, their accelerationswere apparently less. This could be explained if the sun’s force wasdetermined by distance, becoming weaker for the farther planets.Physicists were also familiar with the noncontact forces of electricityand magnetism, and knew that they fell off rapidly with distance,so this made sense.

In the approximation of a circular orbit, the magnitude of thesun’s force on the planet would have to be

[1] F = ma = mv2/r .

Now although this equation has the magnitude, v, of the velocityvector in it, what Newton expected was that there would be a morefundamental underlying equation for the force of the sun on a planet,and that that equation would involve the distance, r, from the sunto the object, but not the object’s speed, v — motion doesn’t makeobjects lighter or heavier.

self-check AIf eq. [1] really was generally applicable, what would happen to an


g / The moon’s accelerationis 602 = 3600 times smaller thanthe apple’s.

object released at rest in some empty region of the solar system? .

Answer, p. 511

Equation [1] was thus a useful piece of information which couldbe related to the data on the planets simply because the planetshappened to be going in nearly circular orbits, but Newton wantedto combine it with other equations and eliminate v algebraically inorder to find a deeper truth.

To eliminate v, Newton used the equation

[2] v =circumference

T=

2πr

T.

Of course this equation would also only be valid for planets in nearlycircular orbits. Plugging this into eq. [1] to eliminate v gives

[3] F =4π2mr

T 2.

This unfortunately has the side-effect of bringing in the period, T ,which we expect on similar physical grounds will not occur in thefinal answer. That’s where the circular-orbit case, T ∝ r3/2, ofKepler’s law of periods comes in. Using it to eliminate T gives aresult that depends only on the mass of the planet and its distancefrom the sun:

F ∝ m/r2 . [force of the sun on a planet of mass

m at a distance r from the sun; same

proportionality constant for all the planets]

(Since Kepler’s law of periods is only a proportionality, the finalresult is a proportionality rather than an equation, so there is nopoint in hanging on to the factor of 4π2.)

As an example, the “twin planets” Uranus and Neptune havenearly the same mass, but Neptune is about twice as far from thesun as Uranus, so the sun’s gravitational force on Neptune is aboutfour times smaller.

self-check BFill in the steps leading from equation [3] to F ∝ m/r2. . Answer, p.511

The forces between heavenly bodies are the same type offorce as terrestrial gravity.

OK, but what kind of force was it? It probably wasn’t magnetic,since magnetic forces have nothing to do with mass. Then cameNewton’s great insight. Lying under an apple tree and looking upat the moon in the sky, he saw an apple fall. Might not the earthalso attract the moon with the same kind of gravitational force?The moon orbits the earth in the same way that the planets orbit

Section 10.2 Newton’s law of gravity 245

the sun, so maybe the earth’s force on the falling apple, the earth’sforce on the moon, and the sun’s force on a planet were all the sametype of force.

There was an easy way to test this hypothesis numerically. If itwas true, then we would expect the gravitational forces exerted bythe earth to follow the same F ∝ m/r2 rule as the forces exerted bythe sun, but with a different constant of proportionality appropriateto the earth’s gravitational strength. The issue arises now of how todefine the distance, r, between the earth and the apple. An applein England is closer to some parts of the earth than to others, butsuppose we take r to be the distance from the center of the earth tothe apple, i.e., the radius of the earth. (The issue of how to measurer did not arise in the analysis of the planets’ motions because thesun and planets are so small compared to the distances separatingthem.) Calling the proportionality constant k, we have

Fearth on apple = k mapple/r2earth

Fearth on moon = k mmoon/d2earth-moon .

Newton’s second law says a = F/m, so

aapple = k / r2earth

amoon = k / d2earth-moon .

The Greek astronomer Hipparchus had already found 2000 yearsbefore that the distance from the earth to the moon was about 60times the radius of the earth, so if Newton’s hypothesis was right,the acceleration of the moon would have to be 602 = 3600 times lessthan the acceleration of the falling apple.

Applying a = v2/r to the acceleration of the moon yielded anacceleration that was indeed 3600 times smaller than 9.8 m/s2, andNewton was convinced he had unlocked the secret of the mysteriousforce that kept the moon and planets in their orbits.

Newton’s law of gravity

The proportionality F ∝ m/r2 for the gravitational force on anobject of mass m only has a consistent proportionality constant forvarious objects if they are being acted on by the gravity of the sameobject. Clearly the sun’s gravitational strength is far greater thanthe earth’s, since the planets all orbit the sun and do not exhibitany very large accelerations caused by the earth (or by one another).What property of the sun gives it its great gravitational strength?Its great volume? Its great mass? Its great temperature? Newtonreasoned that if the force was proportional to the mass of the objectbeing acted on, then it would also make sense if the determiningfactor in the gravitational strength of the object exerting the forcewas its own mass. Assuming there were no other factors affectingthe gravitational force, then the only other thing needed to make


h / Students often have ahard time understanding thephysical meaning of G. It’s justa proportionality constant thattells you how strong gravitationalforces are. If you could change it,all the gravitational forces all overthe universe would get strongeror weaker. Numerically, thegravitational attraction betweentwo 1-kg masses separated by adistance of 1 m is 6.67×10−11 N,and this is what G is in SI units.

i / Example 3. Computer-enhanced images of Pluto andCharon, taken by the HubbleSpace Telescope.

quantitative predictions of gravitational forces would be a propor-tionality constant. Newton called that proportionality constant G,so here is the complete form of the law of gravity he hypothesized.

Newton’s law of gravity

F =Gm1m2

r2[gravitational force between objects of mass

m1 and m2, separated by a distance r; r is not

the radius of anything ]

Newton conceived of gravity as an attraction between any twomasses in the universe. The constant G tells us how many newtonsthe attractive force is for two 1-kg masses separated by a distanceof 1 m. The experimental determination of G in ordinary units(as opposed to the special, nonmetric, units used in astronomy)is described in section 10.5. This difficult measurement was notaccomplished until long after Newton’s death.

The units of G example 1. What are the units of G?

. Solving for G in Newton’s law of gravity gives

G =Fr2

m1m2,

so the units of G must be N·m2/kg2. Fully adorned with units, thevalue of G is 6.67× 10−11 N·m2/kg2.

Newton’s third law example 2. Is Newton’s law of gravity consistent with Newton’s third law?

. The third law requires two things. First, m1’s force on m2 shouldbe the same as m2’s force on m1. This works out, because theproduct m1m2 gives the same result if we interchange the labels 1and 2. Second, the forces should be in opposite directions. Thiscondition is also satisfied, because Newton’s law of gravity refersto an attraction: each mass pulls the other toward itself.

Pluto and Charon example 3. Pluto’s moon Charon is unusually large considering Pluto’s size,giving them the character of a double planet. Their masses are1.25×1022 and 1.9x1021 kg, and their average distance from oneanother is 1.96× 104 km. What is the gravitational force betweenthem?

. If we want to use the value of G expressed in SI (meter-kilogram-second) units, we first have to convert the distance to 1.96 ×


j / The conic sections are thecurves made by cutting thesurface of an infinite cone with aplane.

k / An imaginary cannon ableto shoot cannonballs at very highspeeds is placed on top of animaginary, very tall mountainthat reaches up above the at-mosphere. Depending on thespeed at which the ball is fired,it may end up in a tightly curvedelliptical orbit, 1, a circular orbit,2, a bigger elliptical orbit, 3, or anearly straight hyperbolic orbit, 4.

107 m. The force is(6.67× 10−11 N·m2/kg2

) (1.25× 1022 kg

) (1.9× 1021 kg

)(1.96× 107 m

)2

= 4.1× 1018 N

The proportionality to 1/r2 in Newton’s law of gravity was notentirely unexpected. Proportionalities to 1/r2 are found in manyother phenomena in which some effect spreads out from a point.For instance, the intensity of the light from a candle is proportionalto 1/r2, because at a distance r from the candle, the light has tobe spread out over the surface of an imaginary sphere of area 4πr2.The same is true for the intensity of sound from a firecracker, or theintensity of gamma radiation emitted by the Chernobyl reactor. It’simportant, however, to realize that this is only an analogy. Forcedoes not travel through space as sound or light does, and force isnot a substance that can be spread thicker or thinner like butter ontoast.

Although several of Newton’s contemporaries had speculatedthat the force of gravity might be proportional to 1/r2, none ofthem, even the ones who had learned Newton’s laws of motion, hadhad any luck proving that the resulting orbits would be ellipses, asKepler had found empirically. Newton did succeed in proving thatelliptical orbits would result from a 1/r2 force, but we postponethe proof until the chapter 15 because it can be accomplished muchmore easily using the concepts of energy and angular momentum.

Newton also predicted that orbits in the shape of hyperbolasshould be possible, and he was right. Some comets, for instance,orbit the sun in very elongated ellipses, but others pass throughthe solar system on hyperbolic paths, never to return. Just as thetrajectory of a faster baseball pitch is flatter than that of a moreslowly thrown ball, so the curvature of a planet’s orbit depends onits speed. A spacecraft can be launched at relatively low speed,resulting in a circular orbit about the earth, or it can be launchedat a higher speed, giving a more gently curved ellipse that reachesfarther from the earth, or it can be launched at a very high speedwhich puts it in an even less curved hyperbolic orbit. As you govery far out on a hyperbola, it approaches a straight line, i.e., itscurvature eventually becomes nearly zero.

Newton also was able to prove that Kepler’s second law (sweep-ing out equal areas in equal time intervals) was a logical consequenceof his law of gravity. Newton’s version of the proof is moderatelycomplicated, but the proof becomes trivial once you understand theconcept of angular momentum, which will be covered later in thecourse. The proof will therefore be deferred until section section15.9.


self-check CWhich of Kepler’s laws would it make sense to apply to hyperbolic or-bits? . Answer, p.511

. Solved problem: Visiting Ceres page 261, problem 1

. Solved problem: Why a equals g page 265, problem 20

. Solved problem: Ida and Dactyl page 265, problem 21

. Solved problem: Another solar system page 262, problem 6

. Solved problem: Weight loss page 261, problem 3

. Solved problem: The receding moon page 265, problem 22


A How could Newton find the speed of the moon to plug in to a =v2/r?

B Two projectiles of different mass shot out of guns on the surface ofthe earth at the same speed and angle will follow the same trajectories,assuming that air friction is negligible. (You can verify this by throwing twoobjects together from your hand and seeing if they separate or stay sideby side.) What corresponding fact would be true for satellites of the earthhaving different masses?

C What is wrong with the following statement? “A comet in an ellipticalorbit speeds up as it approaches the sun, because the sun’s force on it isincreasing.”

D Why would it not make sense to expect the earth’s gravitational forceon a bowling ball to be inversely proportional to the square of the distancebetween their surfaces rather than their centers?

E Does the earth accelerate as a result of the moon’s gravitationalforce on it? Suppose two planets were bound to each other gravitationallythe way the earth and moon are, but the two planets had equal masses.What would their motion be like?

F Spacecraft normally operate by firing their engines only for a fewminutes at a time, and an interplanetary probe will spend months or yearson its way to its destination without thrust. Suppose a spacecraft is in acircular orbit around Mars, and it then briefly fires its engines in reverse,causing a sudden decrease in speed. What will this do to its orbit? Whatabout a forward thrust?


10.3 Apparent weightlessnessIf you ask somebody at the bus stop why astronauts are weightless,you’ll probably get one of the following two incorrect answers:

(1) They’re weightless because they’re so far from the earth.

(2) They’re weightless because they’re moving so fast.

The first answer is wrong, because the vast majority of astro-nauts never get more than a thousand miles from the earth’s surface.The reduction in gravity caused by their altitude is significant, butnot 100%. The second answer is wrong because Newton’s law ofgravity only depends on distance, not speed.

The correct answer is that astronauts in orbit around the earthare not really weightless at all. Their weightlessness is only appar-ent. If there was no gravitational force on the spaceship, it wouldobey Newton’s first law and move off on a straight line, rather thanorbiting the earth. Likewise, the astronauts inside the spaceship arein orbit just like the spaceship itself, with the earth’s gravitationalforce continually twisting their velocity vectors around. The reasonthey appear to be weightless is that they are in the same orbit asthe spaceship, so although the earth’s gravity curves their trajectorydown toward the deck, the deck drops out from under them at thesame rate.

Apparent weightlessness can also be experienced on earth. Anytime you jump up in the air, you experience the same kind of ap-parent weightlessness that the astronauts do. While in the air, youcan lift your arms more easily than normal, because gravity does notmake them fall any faster than the rest of your body, which is fallingout from under them. The Russian air force now takes rich foreigntourists up in a big cargo plane and gives them the feeling of weight-lessness for a short period of time while the plane is nose-down anddropping like a rock.

10.4 Vector addition of gravitational forcesPick a flower on earth and you move the farthest star.

Paul Dirac

When you stand on the ground, which part of the earth is pullingdown on you with its gravitational force? Most people are temptedto say that the effect only comes from the part directly under you,since gravity always pulls straight down. Here are three observationsthat might help to change your mind:

• If you jump up in the air, gravity does not stop affecting youjust because you are not touching the earth: gravity is a non-contact force. That means you are not immune from the grav-


l / Gravity only appears topull straight down because thenear perfect symmetry of theearth makes the sideways com-ponents of the total force on anobject cancel almost exactly. Ifthe symmetry is broken, e.g., bya dense mineral deposit, the totalforce is a little off to the side.

ity of distant parts of our planet just because you are nottouching them.

• Gravitational effects are not blocked by intervening matter.For instance, in an eclipse of the moon, the earth is lined updirectly between the sun and the moon, but only the sun’s lightis blocked from reaching the moon, not its gravitational force— if the sun’s gravitational force on the moon was blocked inthis situation, astronomers would be able to tell because themoon’s acceleration would change suddenly. A more subtlebut more easily observable example is that the tides are causedby the moon’s gravity, and tidal effects can occur on the sideof the earth facing away from the moon. Thus, far-off partsof the earth are not prevented from attracting you with theirgravity just because there is other stuff between you and them.

• Prospectors sometimes search for underground deposits of denseminerals by measuring the direction of the local gravitationalforces, i.e., the direction things fall or the direction a plumbbob hangs. For instance, the gravitational forces in the regionto the west of such a deposit would point along a line slightlyto the east of the earth’s center. Just because the total grav-itational force on you points down, that doesn’t mean thatonly the parts of the earth directly below you are attractingyou. It’s just that the sideways components of all the forcevectors acting on you come very close to canceling out.

A cubic centimeter of lava in the earth’s mantle, a grain of silicainside Mt. Kilimanjaro, and a flea on a cat in Paris are all attractingyou with their gravity. What you feel is the vector sum of all thegravitational forces exerted by all the atoms of our planet, and forthat matter by all the atoms in the universe.

When Newton tested his theory of gravity by comparing theorbital acceleration of the moon to the acceleration of a falling appleon earth, he assumed he could compute the earth’s force on theapple using the distance from the apple to the earth’s center. Washe wrong? After all, it isn’t just the earth’s center attracting theapple, it’s the whole earth. A kilogram of dirt a few feet under hisbackyard in England would have a much greater force on the applethan a kilogram of molten rock deep under Australia, thousands ofmiles away. There’s really no obvious reason why the force shouldcome out right if you just pretend that the earth’s whole mass isconcentrated at its center. Also, we know that the earth has someparts that are more dense, and some parts that are less dense. Thesolid crust, on which we live, is considerably less dense than themolten rock on which it floats. By all rights, the computation of thevector sum of all the forces exerted by all the earth’s parts shouldbe a horrendous mess.

Section 10.4 Vector addition of gravitational forces 251

m / An object outside a sphericalshell of mass will feel gravitationalforces from every part of the shell— stronger forces from the closerparts, and weaker ones from theparts farther away. The shelltheorem states that the vectorsum of all the forces is the sameas if all the mass had beenconcentrated at the center of theshell.

Actually, Newton had sound mathematical reasons for treatingthe earth’s mass as if it was concentrated at its center. First, al-though Newton no doubt suspected the earth’s density was nonuni-form, he knew that the direction of its total gravitational force wasvery nearly toward the earth’s center. That was strong evidencethat the distribution of mass was very symmetric, so that we canthink of the earth as being made of many layers, like an onion,with each layer having constant density throughout. (Today thereis further evidence for symmetry based on measurements of how thevibrations from earthquakes and nuclear explosions travel throughthe earth.) Newton then concentrated on the gravitational forcesexerted by a single such thin shell, and proved the following math-ematical theorem, known as the shell theorem:

If an object lies outside a thin, spherical shell of mass, thenthe vector sum of all the gravitational forces exerted by all theparts of the shell is the same as if the shell’s mass had beenconcentrated at its center. If the object lies inside the shell,then all the gravitational forces cancel out exactly.

For terrestrial gravity, each shell acts as though its mass was con-centrated at the earth’s center, so the final result is the same as ifthe earth’s whole mass was concentrated at its center.

The second part of the shell theorem, about the gravitationalforces canceling inside the shell, is a little surprising. Obviously theforces would all cancel out if you were at the exact center of a shell,but why should they still cancel out perfectly if you are inside theshell but off-center? The whole idea might seem academic, since wedon’t know of any hollow planets in our solar system that astronautscould hope to visit, but actually it’s a useful result for understandinggravity within the earth, which is an important issue in geology. Itdoesn’t matter that the earth is not actually hollow. In a mine shaftat a depth of, say, 2 km, we can use the shell theorem to tell us thatthe outermost 2 km of the earth has no net gravitational effect, andthe gravitational force is the same as what would be produced if theremaining, deeper, parts of the earth were all concentrated at itscenter.


self-check DSuppose you’re at the bottom of a deep mineshaft, which means you’restill quite far from the center of the earth. The shell theorem says thatthe shell of mass you’ve gone inside exerts zero total force on you.Discuss which parts of the shell are attracting you in which directions,and how strong these forces are. Explain why it’s at least plausible thatthey cancel. . Answer, p. 511


A If you hold an apple, does the apple exert a gravitational force onthe earth? Is it much weaker than the earth’s gravitational force on theapple? Why doesn’t the earth seem to accelerate upward when you dropthe apple?

B When astronauts travel from the earth to the moon, how does thegravitational force on them change as they progress?

C How would the gravity in the first-floor lobby of a massive skyscrapercompare with the gravity in an open field outside of the city?

D In a few billion years, the sun will start undergoing changes that willeventually result in its puffing up into a red giant star. (Near the beginningof this process, the earth’s oceans will boil off, and by the end, the sunwill probably swallow the earth completely.) As the sun’s surface starts toget closer and close to the earth, how will the earth’s orbit be affected?

10.5 Weighing the earthLet’s look more closely at the application of Newton’s law of gravityto objects on the earth’s surface. Since the earth’s gravitationalforce is the same as if its mass was all concentrated at its center,the force on a falling object of mass m is given by

F = GMearth m / r2earth .

The object’s acceleration equals F/m, so the object’s mass cancelsout and we get the same acceleration for all falling objects, as weknew we should:

g = GMearth / r2earth .

Newton knew neither the mass of the earth nor a numerical valuefor the constant G. But if someone could measure G, then it wouldbe possible for the first time in history to determine the mass of theearth! The only way to measure G is to measure the gravitationalforce between two objects of known mass, but that’s an exceedinglydifficult task, because the force between any two objects of ordinary

Section 10.5 Weighing the earth 253

o / A simplified version ofCavendish’s apparatus.

n / Cavendish’s apparatus. The two large balls are fixed in place,but the rod from which the two small balls hang is free to twist under theinfluence of the gravitational forces.

size is extremely small. The English physicist Henry Cavendish wasthe first to succeed, using the apparatus shown in figures n and o.The two larger balls were lead spheres 8 inches in diameter, and eachone attracted the small ball near it. The two small balls hung fromthe ends of a horizontal rod, which itself hung by a thin thread. Theframe from which the larger balls hung could be rotated by handabout a vertical axis, so that for instance the large ball on the rightwould pull its neighboring small ball toward us and while the smallball on the left would be pulled away from us. The thread fromwhich the small balls hung would thus be twisted through a smallangle, and by calibrating the twist of the thread with known forces,the actual gravitational force could be determined. Cavendish setup the whole apparatus in a room of his house, nailing all the doorsshut to keep air currents from disturbing the delicate apparatus.The results had to be observed through telescopes stuck throughholes drilled in the walls. Cavendish’s experiment provided the firstnumerical values for G and for the mass of the earth. The presentlyaccepted value of G is 6.67× 10−11 N·m2/kg2.

Knowing G not only allowed the determination of the earth’smass but also those of the sun and the other planets. For instance,by observing the acceleration of one of Jupiter’s moons, we can inferthe mass of Jupiter. The following table gives the distances of theplanets from the sun and the masses of the sun and planets. (Otherdata are given in the back of the book.)


average distance fromthe sun, in units of theearth’s average distancefrom the sun

mass, in units of theearth’s mass

sun — 330,000

mercury 0.38 0.056

venus 0.72 0.82

earth 1 1

mars 1.5 0.11

jupiter 5.2 320

saturn 9.5 95

uranus 19 14

neptune 30 17

pluto 39 0.002

The following example applies the numerical techniques of sec-tion 4.6.

From the earth to the moon example 4The Apollo 11 mission landed the first humans on the moon in1969. In this example, we’ll estimate the time it took to get tothe moon, and compare our estimate with the actual time, whichwas 73.0708 hours from the engine burn that took the ship out ofearth orbit to the engine burn that inserted it into lunar orbit. Dur-ing this time, the ship was coasting with the engines off, exceptfor a small course-correction burn, which we neglect. More im-portantly, we do the calculation for a straight-line trajectory ratherthan the real S-shaped one, so the result can only be expectedto agree roughly with what really happened. The following datacome from the original press kit, which NASA has scanned andposted on the Web:

initial altitude 3.363× 105 minitial velocity 1.083× 104 m/s

The endpoint of the the straight-line trajectory is a free-fall im-pact on the lunar surface, which is also unrealistic (luckily for theastronauts).

The force acting on the ship is

F = −GMemr2 +

GMmm(rm − r )2 ,

but since everything is proportional to the mass of the ship, m, wecan divide it out

Fm

= −GMe

r2 +GMm

(rm − r )2 ,

and the variables F in the program is actually the force per unitmass F/m. The program is a straightforward modification of thefunction meteor on page 139.

Section 10.5 Weighing the earth 255

1 import math

2 def apollo(vi,n):

3 bigg=6.67e-11 # gravitational constant, SI

4 me=5.97e24 # mass of earth, kg

5 mm=7.35e22 # mass of moon, kg

6 em=3.84e8 # earth-moon distance, m

7 re=6.378e6 # radius of earth, m

8 rm=1.74e6 # radius of moon, m

9 v=vi

10 x=re+3.363e5 # re+initial altitude

11 xf=em-rm # surface of moon

12 dt = 360000./n # split 100 hours into n parts

13 t = 0.


15 dx = v*dt


17 if x>xf:

18 return t/3600.

19 a = -bigg*me/x**2+bigg*mm/(em-x)**2

20 t = t + dt

21 dv = a*dt

22 v = v+dv

>>> print apollo(1.083e4,1000000)

59.7488999991

>>> vi=1.083e4

This is pretty decent agreement with the real-world time of 73hours, considering the wildly inaccurate trajectory assumed. It’sinteresting to see how much the duration of the trip changes if weincrease the initial velocity by only ten percent:

>>> print apollo(1.2e4,1000000)

18.3682

The most important reason for using the lower speed was thatif something had gone wrong, the ship would have been able towhip around the moon and take a “free return” trajectory back tothe earth, without having to do any further burns. At a higherspeed, the ship would have had so much kinetic energy that inthe absence of any further engine burns, it would have escapedfrom the earth-moon system. The Apollo 13 mission had to take afree return trajectory after an explosion crippled the spacecraft.


A It would have been difficult for Cavendish to start designing anexperiment without at least some idea of the order of magnitude of G.How could he estimate it in advance to within a factor of 10?


B Fill in the details of how one would determine Jupiter’s mass byobserving the acceleration of one of its moons. Why is it only necessaryto know the acceleration of the moon, not the actual force acting on it?Why don’t we need to know the mass of the moon? What about a planetthat has no moons, such as Venus — how could its mass be found?

10.6 ? Dark energyUntil recently, physicists thought they understood gravity fairlywell. Einstein had modified Newton’s theory, but certain charac-teristrics of gravitational forces were firmly established. For onething, they were always attractive. If gravity always attracts, thenit is logical to ask why the universe doesn’t collapse. Newton hadanswered this question by saying that if the universe was infinite inall directions, then it would have no geometric center toward whichit would collapse; the forces on any particular star or planet ex-erted by distant parts of the universe would tend to cancel out bysymmetry. More careful calculations, however, show that Newton’suniverse would have a tendency to collapse on smaller scales: anypart of the universe that happened to be slightly more dense thanaverage would contract further, and this contraction would resultin stronger gravitational forces, which would cause even more rapidcontraction, and so on.

When Einstein overhauled gravity, the same problem reared itsugly head. Like Newton, Einstein was predisposed to believe in auniverse that was static, so he added a special repulsive term to hisequations, intended to prevent a collapse. This term was not associ-ated with any interaction of mass with mass, but represented merelyan overall tendency for space itself to expand unless restrained bythe matter that inhabited it. It turns out that Einstein’s solution,like Newton’s, is unstable. Furthermore, it was soon discoveredobservationally that the universe was expanding, and this was in-terpreted by creating the Big Bang model, in which the universe’scurrent expansion is the aftermath of a fantastically hot explosion.1

An expanding universe, unlike a static one, was capable of being ex-plained with Einstein’s equations, without any repulsion term. Theuniverse’s expansion would simply slow down over time due to theattractive gravitational forces. After these developments, Einsteinsaid woefully that adding the repulsive term, known as the cosmo-logical constant, had been the greatest blunder of his life.

This was the state of things until 1999, when evidence began toturn up that the universe’s expansion has been speeding up ratherthan slowing down! The first evidence came from using a telescopeas a sort of time machine: light from a distant galaxy may havetaken billions of years to reach us, so we are seeing it as it was farin the past. Looking back in time, astronomers saw the universe

1Book 3, section 3.5, presents some of the evidence for the Big Bang.

Section 10.6 ? Dark energy 257

p / The WMAP probe’s map of thecosmic microwave background islike a “baby picture” of the uni-verse.

expanding at speeds that were lower, rather than higher. At firstthey were mortified, since this was exactly the opposite of whathad been expected. The statistical quality of the data was also notgood enough to constitute ironclad proof, and there were worriesabout systematic errors. The case for an accelerating expansionhas however been supported by high-precision mapping of the dim,sky-wide afterglow of the Big Bang, known as the cosmic microwavebackground.

So now Einstein’s “greatest blunder” has been resurrected. Sincewe don’t actually know whether or not this self-repulsion of spacehas a constant strength, the term “cosmological constant” has lostcurrency. Nowadays physicists usually refer to the phenomenon as“dark energy.” Picking an impressive-sounding name for it shouldnot obscure the fact that we know absolutely nothing about thenature of the effect or why it exists.


SummarySelected vocabularyellipse . . . . . . . a flattened circle; one of the conic sectionsconic section . . . a curve formed by the intersection of a plane

and an infinite conehyperbola . . . . another conic section; it does not close back

on itselfperiod . . . . . . . the time required for a planet to complete one

orbit; more generally, the time for one repeti-tion of some repeating motion

focus . . . . . . . one of two special points inside an ellipse: theellipse consists of all points such that the sumof the distances to the two foci equals a certainnumber; a hyperbola also has a focus

NotationG . . . . . . . . . the constant of proportionality in Newton’s

law of gravity; the gravitational force of at-traction between two 1-kg spheres at a center-to-center distance of 1 m

Summary

Kepler deduced three empirical laws from data on the motion ofthe planets:

Kepler’s elliptical orbit law: The planets orbit the sun in ellip-tical orbits with the sun at one focus.

Kepler’s equal-area law: The line connecting a planet to the sunsweeps out equal areas in equal amounts of time.

Kepler’s law of periods: The time required for a planet to orbitthe sun is proportional to the long axis of the ellipse raised tothe 3/2 power. The constant of proportionality is the samefor all the planets.

Newton was able to find a more fundamental explanation for theselaws. Newton’s law of gravity states that the magnitude of theattractive force between any two objects in the universe is given by

F = Gm1m2/r2 .

Weightlessness of objects in orbit around the earth is only appar-ent. An astronaut inside a spaceship is simply falling along withthe spaceship. Since the spaceship is falling out from under the as-tronaut, it appears as though there was no gravity accelerating theastronaut down toward the deck.

Gravitational forces, like all other forces, add like vectors. Agravitational force such as we ordinarily feel is the vector sum of all

Summary 259

the forces exerted by all the parts of the earth. As a consequence ofthis, Newton proved the shell theorem for gravitational forces:

If an object lies outside a thin, uniform shell of mass, then thevector sum of all the gravitational forces exerted by all the parts ofthe shell is the same as if all the shell’s mass was concentrated at itscenter. If the object lies inside the shell, then all the gravitationalforces cancel out exactly.


Problem 4.

ProblemsKey√



1 Ceres, the largest asteroid in our solar system, is a sphericalbody with a mass 6000 times less than the earth’s, and a radiuswhich is 13 times smaller. If an astronaut who weighs 400 N onearth is visiting the surface of Ceres, what is her weight?

. Solution, p. 503

2 Roy has a mass of 60 kg. Laurie has a mass of 65 kg. Theyare 1.5 m apart.(a) What is the magnitude of the gravitational force of the earth onRoy?(b) What is the magnitude of Roy’s gravitational force on the earth?(c) What is the magnitude of the gravitational force between Royand Laurie?(d) What is the magnitude of the gravitational force between Laurieand the sun?

√

3 (a) A certain vile alien gangster lives on the surface of anasteroid, where his weight is 0.20 N. He decides he needs to loseweight without reducing his consumption of princesses, so he’s goingto move to a different asteroid where his weight will be 0.10 N. Thereal estate agent’s database has asteroids listed by mass, however,not by surface gravity. Assuming that all asteroids are sphericaland have the same density, how should the mass of his new asteroidcompare with that of his old one?(b) Jupiter’s mass is 318 times the Earth’s, and its gravity is abouttwice Earth’s. Is this consistent with the results of part a? If not,how do you explain the discrepancy? . Solution, p. 504

4 The planet Uranus has a mass of 8.68× 1025 kg and a radiusof 2.56× 104 km. The figure shows the relative sizes of Uranus andEarth.(a) Compute the ratio gU/gE , where gU is the strength of the grav-itational field at the surface of Uranus and gE is the correspondingquantity at the surface of the Earth.

√

(b) What is surprising about this result? How do you explain it?

5 How high above the Earth’s surface must a rocket be in orderto have 1/100 the weight it would have at the surface? Express youranswer in units of the radius of the Earth.

√

Problems 261

Problem 8.

6 Astronomers have detected a solar system consisting of threeplanets orbiting the star Upsilon Andromedae. The planets havebeen named b, c, and d. Planet b’s average distance from the staris 0.059 A.U., and planet c’s average distance is 0.83 A.U., where anastronomical unit or A.U. is defined as the distance from the Earthto the sun. For technical reasons, it is possible to determine theratios of the planets’ masses, but their masses cannot presently bedetermined in absolute units. Planet c’s mass is 3.0 times that ofplanet b. Compare the star’s average gravitational force on planetc with its average force on planet b. [Based on a problem by ArnoldArons.] . Solution, p. 504

7 The star Lalande 21185 was found in 1996 to have two planetsin roughly circular orbits, with periods of 6 and 30 years. What isthe ratio of the two planets’ orbital radii?

√

8 You are considering going on a space voyage to Mars, in whichyour route would be half an ellipse, tangent to the Earth’s orbit atone end and tangent to Mars’ orbit at the other. Your spacecraft’sengines will only be used at the beginning and end, not during thevoyage. How long would the outward leg of your trip last? (Assumethe orbits of Earth and Mars are circular.)

√

9 Where would an object have to be located so that it wouldexperience zero total gravitational force from the earth and moon?√

10 In a Star Trek episode, the Enterprise is in a circular orbitaround a planet when something happens to the engines. Spockthen tells Kirk that the ship will spiral into the planet’s surfaceunless they can fix the engines. Is this scientifically correct? Why?

11 Astronomers have recently observed stars orbiting at veryhigh speeds around an unknown object near the center of our galaxy.For stars orbiting at distances of about 1014 m from the object,the orbital velocities are about 106 m/s. Assuming the orbits arecircular, estimate the mass of the object, in units of the mass ofthe sun, 2 × 1030 kg. If the object was a tightly packed cluster ofnormal stars, it should be a very bright source of light. Since novisible light is detected coming from it, it is instead believed to bea supermassive black hole.

√

12 During a solar eclipse, the moon, earth and sun all lie onthe same line, with the moon between the earth and sun. Defineyour coordinates so that the earth and moon lie at greater x valuesthan the sun. For each force, give the correct sign as well as themagnitude. (a) What force is exerted on the moon by the sun? (b)On the moon by the earth? (c) On the earth by the sun? (d) Whattotal force is exerted on the sun? (e) On the moon? (f) On theearth?

√


13 Suppose that on a certain day there is a crescent moon,and you can tell by the shape of the crescent that the earth, sunand moon form a triangle with a 135 interior angle at the moon’scorner. What is the magnitude of the total gravitational force ofthe earth and the sun on the moon? (If you haven’t done problem12 already, you might want to try it first, since it’s easier, and someof its results can be recycled in this problem.)

√

Problem 13.

14 On Feb. 28, 2007, the New Horizons space probe, on itsway to a 2015 flyby of Pluto, passed by the planet Jupiter for agravity-assisted maneuver that increased its speed and changed itscourse. The dashed line in the figure shows the spacecraft’s tra-jectory, which is curved because of three forces: the force of theexhaust gases from the probe’s own engines, the sun’s gravitationalforce, and Jupiter’s gravitational force. Find the magnitude of thetotal gravitational force acting on the probe. You will find that thesun’s force is much smaller than Jupiter’s, so that the magnitudeof the total force is determined almost entirely by Jupiter’s force.However, this is a high-precision problem, and you will find that thetotal force is slightly different from Jupiter’s force.

√

Problem 14: New Horizons atits closest approach to Jupiter.(Jupiter’s four largest moons areshown for illustrative purposes.)The masses are:sun: 1.9891× 1030 kgJupiter: 1.8986× 1027 kgNew Horizons: 465.0 kg

15 The International Space Station orbits at an average altitudeof about 370 km above sea level. Compute the value of g at thataltitude.

√

Problems 263

16 (a) If the earth was of uniform density, would your weightbe increased or decreased at the bottom of a mine shaft? Explain.(b) In real life, objects weigh slightly more at the bottom of a mineshaft. What does that allow us to infer about the Earth? ?

17 (a) A geosynchronous orbit is one in which the satellite orbitsabove the equator, and has an orbital period of 24 hours, so that itis always above the same point on the spinning earth. Calculate thealtitude of such a satellite.

√

(b) What is the gravitational field experienced by the satellite? Giveyour answer as a percentage in relation to the gravitational field atthe earth’s surface.

√

18 If a bullet is shot straight up at a high enough velocity, it willnever return to the earth. This is known as the escape velocity. Wewill discuss escape velocity using the concept of energy later in thecourse, but it can also be gotten at using straightforward calculus.In this problem, you will analyze the motion of an object of mass mwhose initial velocity is exactly equal to escape velocity. We assumethat it is starting from the surface of a spherically symmetric planetof mass M and radius b. The trick is to guess at the general formof the solution, and then determine the solution in more detail. As-sume (as is true) that the solution is of the form r = ktp, where r isthe object’s distance from the center of the planet at time t, and kand p are constants.(a) Find the acceleration, and use Newton’s second law and New-ton’s law of gravity to determine k and p. You should find that theresult is independent of m.

√

(b) What happens to the velocity as t approaches infinity?(c) Determine escape velocity from the Earth’s surface.

√

19 (a) Suppose a rotating spherical body such as a planet hasa radius r and a uniform density ρ, and the time required for onerotation is T . At the surface of the planet, the apparent accelerationof a falling object is reduced by the acceleration of the ground outfrom under it. Derive an equation for the apparent acceleration ofgravity, g, at the equator in terms of r, ρ, T , and G.

√

(b) Applying your equation from a, by what fraction is your appar-ent weight reduced at the equator compared to the poles, due to theEarth’s rotation?

√

(c) Using your equation from a, derive an equation giving the valueof T for which the apparent acceleration of gravity becomes zero,i.e., objects can spontaneously drift off the surface of the planet.Show that T only depends on ρ, and not on r.

√

(d) Applying your equation from c, how long would a day have tobe in order to reduce the apparent weight of objects at the equatorof the Earth to zero? [Answer: 1.4 hours](e) Astronomers have discovered objects they called pulsars, whichemit bursts of radiation at regular intervals of less than a second.


Problem 21.

If a pulsar is to be interpreted as a rotating sphere beaming out anatural “searchlight” that sweeps past the earth with each rotation,use your equation from c to show that its density would have to bemuch greater than that of ordinary matter.(f) Astrophysicists predicted decades ago that certain stars that usedup their sources of energy could collapse, forming a ball of neutronswith the fantastic density of ∼ 1017 kg/m3. If this is what pulsarsreally are, use your equation from c to explain why no pulsar hasever been observed that flashes with a period of less than 1 ms orso.

20 Prove, based on Newton’s laws of motion and Newton’s lawof gravity, that all falling objects have the same acceleration if theyare dropped at the same location on the earth and if other forcessuch as friction are unimportant. Do not just say, “g = 9.8 m/s2 –it’s constant.” You are supposed to be proving that g should be thesame number for all objects. . Solution, p. 504

21 The figure shows an image from the Galileo space probetaken during its August 1993 flyby of the asteroid Ida. Astronomerswere surprised when Galileo detected a smaller object orbiting Ida.This smaller object, the only known satellite of an asteroid in oursolar system, was christened Dactyl, after the mythical creatureswho lived on Mount Ida, and who protected the infant Zeus. Forscale, Ida is about the size and shape of Orange County, and Dactylthe size of a college campus. Galileo was unfortunately unable tomeasure the time, T , required for Dactyl to orbit Ida. If it had,astronomers would have been able to make the first accurate deter-mination of the mass and density of an asteroid. Find an equationfor the density, ρ, of Ida in terms of Ida’s known volume, V , theknown radius, r, of Dactyl’s orbit, and the lamentably unknownvariable T . (This is the same technique that was used successfullyfor determining the masses and densities of the planets that havemoons.) . Solution, p. 504

22 As is discussed in more detail in example 3 on p. 386, tidalinteractions with the earth are causing the moon’s orbit to growgradually larger. Laser beams bounced off of a mirror left on themoon by astronauts have allowed a measurement of the moon’s rateof recession, which is about 1 cm per year. This means that thegravitational force acting between earth and moon is decreasing. Bywhat fraction does the force decrease with each 27-day orbit? [Hint:If you try to calculate the two forces and subtract, your calculatorwill probably give a result of zero due to rounding. Instead, reasonabout the fractional amount by which the quantity 1/r2 will change.As a warm-up, you may wish to observe the percentage change in1/r2 that results from changing r from 1 to 1.01. Based on a problemby Arnold Arons.] . Solution, p. 504

Problems 265

23 Astronomers calculating orbits of planets often work in anonmetric system of units, in which the unit of time is the year,the unit of mass is the sun’s mass, and the unit of distance is theastronomical unit (A.U.), defined as half the long axis of the earth’sorbit. In these units, find an exact expression for the gravitationalconstant, G.

√

24 Suppose that we inhabited a universe in which, instead ofNewton’s law of gravity, we had F = k

√m1m2/r

2, where k is someconstant with different units than G. (The force is still attrac-tive.) However, we assume that a = F/m and the rest of Newtonianphysics remains true, and we use a = F/m to define our mass scale,so that, e.g., a mass of 2 kg is one which exhibits half the accelera-tion when the same force is applied to it as to a 1 kg mass.(a) Is this new law of gravity consistent with Newton’s third law?(b) Suppose you lived in such a universe, and you dropped two un-equal masses side by side. What would happen?(c) Numerically, suppose a 1.0-kg object falls with an accelerationof 10 m/s2. What would be the acceleration of a rain drop with amass of 0.1 g? Would you want to go out in the rain?(d) If a falling object broke into two unequal pieces while it fell,what would happen?(e) Invent a law of gravity that results in behavior that is the op-posite of what you found in part b. [Based on a problem by ArnoldArons.]

25 The structures that we see in the universe, such as solarsystems, galaxies, and clusters of galaxies, are believed to have con-densed from clumps that formed, due to gravitational attraction,in preexisting clouds of gas and dust. Observations of the cosmicmicrowave background radiation (p. 258) suggest that the mixtureof hot hydrogen and helium that existed soon after the Big Bangwas extremely uniform, but not perfectly so. We can imagine thatany region that started out a little more dense would form a naturalcenter for the collapse of a clump. Suppose that we have a spheri-cal region with density ρ and radius r, and for simplicity let’s justassume that it’s surrounded by vacuum. (a) Find the accelerationof the material at the edge of the cloud. To what power of r is itproportional? (b) The cloud will take a time t to collapse to somefraction of its original size. Show that t is independent of r.

Remark: This result suggests that structures would get a chance to form at allscales in the universe. That is, solar systems would not form before galaxies gotto, or vice versa. It is therefore physically natural that when we look at theuniverse at essentially all scales less than a billion light-years, we see structure.

?

26 You have a fixed amount of material with a fixed density. Ifthe material is formed into some shape S, then there will be somepoint in space at which the resulting gravitational field attains itsmaximum value gS . What shape maximizes gS? ?


Exercise 10: The shell theoremThis exercise is an approximate numerical test of the shell theorem. There are seven massesA-G, each being one kilogram. Masses A-E, each one meter from the center, form a shape liketwo Egyptian pyramids joined at their bases; this is a rough approximation to a six-kilogramspherical shell of mass. Mass G is five meters from the center of the main group. The class willdivide into six groups and split up the work required in order to calculate the vector sum of thesix gravitational forces exerted on mass G. Depending on the size of the class, more than onegroup may be assigned to deal with the contribution of the same mass to the total force, andthe redundant groups can check each other’s results.

1. Discuss as a class what can be done to simplify the task of calculating the vector sum, andhow to organize things so that each group can work in parallel with the others.

2. Each group should write its results on the board in units of piconewtons, retaining fivesignificant figures of precision. Everyone will need to use the same value for the gravitationalconstant, G = 6.6743× 10−11 N·m2/kg2.

3. The class will determine the vector sum and compare with the result that would be obtainedwith the shell theorem.

Exercise 10: The shell theorem 267

Conservation laws

268

270

In July of 1994, Comet Shoemaker-Levy struck the planet Jupiter, de-positing 7 × 1022 joules of energy, and incidentally giving rise to a seriesof Hollywood movies in which our own planet is threatened by an impactby a comet or asteroid. There is evidence that such an impact causedthe extinction of the dinosaurs. Left: Jupiter’s gravitational force on thenear side of the comet was greater than on the far side, and this differ-ence in force tore up the comet into a string of fragments. Two separatetelescope images have been combined to create the illusion of a point ofview just behind the comet. (The colored fringes at the edges of Jupiterare artifacts of the imaging system.) Top: A series of images of the plumeof superheated gas kicked up by the impact of one of the fragments. Theplume is about the size of North America. Bottom: An image after all theimpacts were over, showing the damage done.

Chapter 11

Conservation of energy

11.1 The search for a perpetual motionmachine

Don’t underestimate greed and laziness as forces for progress. Mod-ern chemistry was born from the collision of lust for gold with dis-taste for the hard work of finding it and digging it up. Failed effortsby generations of alchemists to turn lead into gold led finally to theconclusion that it could not be done: certain substances, the chem-

271

a / The magnet draws theball to the top of the ramp, whereit falls through the hole and rollsback to the bottom.

b / As the wheel spins clock-wise, the flexible arms sweeparound and bend and unbend. Bydropping off its ball on the ramp,the arm is supposed to makeitself lighter and easier to lift overthe top. Picking its own ball backup again on the right, it helps topull the right side down.

ical elements, are fundamental, and chemical reactions can neitherincrease nor decrease the amount of an element such as gold.

Now flash forward to the early industrial age. Greed and lazinesshave created the factory, the train, and the ocean liner, but in eachof these is a boiler room where someone gets sweaty shoveling thecoal to fuel the steam engine. Generations of inventors have tried tocreate a machine, called a perpetual motion machine, that would runforever without fuel. Such a machine is not forbidden by Newton’slaws of motion, which are built around the concepts of force andinertia. Force is free, and can be multiplied indefinitely with pulleys,gears, or levers. The principle of inertia seems even to encouragethe belief that a cleverly constructed machine might not ever rundown.

Figures a and b show two of the innumerable perpetual motionmachines that have been proposed. The reason these two examplesdon’t work is not much different from the reason all the others havefailed. Consider machine a. Even if we assume that a properlyshaped ramp would keep the ball rolling smoothly through eachcycle, friction would always be at work. The designer imagined thatthe machine would repeat the same motion over and over again, sothat every time it reached a given point its speed would be exactlythe same as the last time. But because of friction, the speed wouldactually be reduced a little with each cycle, until finally the ballwould no longer be able to make it over the top.

Friction has a way of creeping into all moving systems. Therotating earth might seem like a perfect perpetual motion machine,since it is isolated in the vacuum of outer space with nothing to exertfrictional forces on it. But in fact our planet’s rotation has sloweddrastically since it first formed, and the earth continues to slowits rotation, making today just a little longer than yesterday. Thevery subtle source of friction is the tides. The moon’s gravity raisesbulges in the earth’s oceans, and as the earth rotates the bulgesprogress around the planet. Where the bulges encounter land, thereis friction, which slows the earth’s rotation very gradually.

11.2 EnergyThe analysis based on friction is somewhat superficial, however. Onecould understand friction perfectly well and yet imagine the follow-ing situation. Astronauts bring back a piece of magnetic ore fromthe moon which does not behave like ordinary magnets. A normalbar magnet, c/1, attracts a piece of iron essentially directly towardit, and has no left- or right-handedness. The moon rock, however,exerts forces that form a whirlpool pattern around it, 2. NASAgoes to a machine shop and has the moon rock put in a lathe andmachined down to a smooth cylinder, 3. If we now release a ballbearing on the surface of the cylinder, the magnetic force whips it

272 Chapter 11 Conservation of energy

c / A mysterious moon rockmakes a perpetual motionmachine.

d / Example 1.

around and around at ever higher speeds. Of course there is somefriction, but there is a net gain in speed with each revolution.

Physicists would lay long odds against the discovery of such amoon rock, not just because it breaks the rules that magnets nor-mally obey but because, like the alchemists, they have discovereda very deep and fundamental principle of nature which forbids cer-tain things from happening. The first alchemist who deserved tobe called a chemist was the one who realized one day, “In all theseattempts to create gold where there was none before, all I’ve beendoing is shuffling the same atoms back and forth among differenttest tubes. The only way to increase the amount of gold in my lab-oratory is to bring some in through the door.” It was like havingsome of your money in a checking account and some in a savings ac-count. Transferring money from one account into the other doesn’tchange the total amount.

We say that the number of grams of gold is a conserved quan-tity. In this context, the word “conserve” does not have its usualmeaning of trying not to waste something. In physics, a conservedquantity is something that you wouldn’t be able to get rid of evenif you wanted to. Conservation laws in physics always refer to aclosed system, meaning a region of space with boundaries throughwhich the quantity in question is not passing. In our example, thealchemist’s laboratory is a closed system because no gold is comingin or out through the doors.

Conservation of mass example 1In figure d, the stream of water is fatter near the mouth of thefaucet, and skinnier lower down. This is because the water speedsup as it falls. If the cross-sectional area of the stream was equalall along its length, then the rate of flow through a lower cross-section would be greater than the rate of flow through a cross-section higher up. Since the flow is steady, the amount of wa-ter between the two cross-sections stays constant. The cross-sectional area of the stream must therefore shrink in inverse pro-portion to the increasing speed of the falling water. This is anexample of conservation of mass.

In general, the amount of any particular substance is not con-served. Chemical reactions can change one substance into another,and nuclear reactions can even change one element into another.The total mass of all substances is however conserved:

the law of conservation of massThe total mass of a closed system always remains constant. Energycannot be created or destroyed, but only transferred from one systemto another.

A similar lightbulb eventually lit up in the heads of the people

Section 11.2 Energy 273

who had been frustrated trying to build a perpetual motion machine.In perpetual motion machine a, consider the motion of one of theballs. It performs a cycle of rising and falling. On the way down itgains speed, and coming up it slows back down. Having a greaterspeed is like having more money in your checking account, and beinghigh up is like having more in your savings account. The device issimply shuffling funds back and forth between the two. Having moreballs doesn’t change anything fundamentally. Not only that, butfriction is always draining off money into a third “bank account:”heat. The reason we rub our hands together when we’re cold is thatkinetic friction heats things up. The continual buildup in the “heataccount” leaves less and less for the “motion account” and “heightaccount,” causing the machine eventually to run down.

These insights can be distilled into the following basic principleof physics:

the law of conservation of energyIt is possible to give a numerical rating, called energy, to the stateof a physical system. The total energy is found by adding up contri-butions from characteristics of the system such as motion of objectsin it, heating of the objects, and the relative positions of objectsthat interact via forces. The total energy of a closed system alwaysremains constant. Energy cannot be created or destroyed, but onlytransferred from one system to another.

The moon rock story violates conservation of energy because therock-cylinder and the ball together constitute a closed system. Oncethe ball has made one revolution around the cylinder, its positionrelative to the cylinder is exactly the same as before, so the numer-ical energy rating associated with its position is the same as before.Since the total amount of energy must remain constant, it is im-possible for the ball to have a greater speed after one revolution. Ifit had picked up speed, it would have more energy associated withmotion, the same amount of energy associated with position, and alittle more energy associated with heating through friction. Therecannot be a net increase in energy.

Converting one form of energy to another example 2Dropping a rock: The rock loses energy because of its changingposition with respect to the earth. Nearly all that energy is trans-formed into energy of motion, except for a small amount lost toheat created by air friction.

Sliding in to home base: The runner’s energy of motion is nearlyall converted into heat via friction with the ground.

Accelerating a car: The gasoline has energy stored in it, whichis released as heat by burning it inside the engine. Perhaps 10%


e / Example 3.

of this heat energy is converted into the car’s energy of motion.The rest remains in the form of heat, which is carried away by theexhaust.

Cruising in a car: As you cruise at constant speed in your car, allthe energy of the burning gas is being converted into heat. Thetires and engine get hot, and heat is also dissipated into the airthrough the radiator and the exhaust.

Stepping on the brakes: All the energy of the car’s motion is con-verted into heat in the brake shoes.

Stevin’s machine example 3The Dutch mathematician and engineer Simon Stevin proposed

the imaginary machine shown in figure e, which he had inscribedon his tombstone. This is an interesting example, because itshows a link between the force concept used earlier in this course,and the energy concept being developed now.

The point of the imaginary machine is to show the mechanicaladvantage of an inclined plane. In this example, the triangle hasthe proportions 3-4-5, but the argument works for any right trian-gle. We imagine that the chain of balls slides without friction, sothat no energy is ever converted into heat. If we were to slidethe chain clockwise by one step, then each ball would take theplace of the one in front of it, and the over all configuration wouldbe exactly the same. Since energy is something that only de-pends on the state of the system, the energy would have to bethe same. Similarly for a counterclockwise rotation, no energy ofposition would be released by gravity. This means that if we placethe chain on the triangle, and release it at rest, it can’t start mov-ing, because there would be no way for it to convert energy ofposition into energy of motion. Thus the chain must be perfectlybalanced. Now by symmetry, the arc of the chain hanging under-neath the triangle has equal tension at both ends, so removingthis arc wouldn’t affect the balance of the rest of the chain. Thismeans that a weight of three units hanging vertically balances aweight of five units hanging diagonally along the hypotenuse.

The mechanical advantage of the inclined plane is therefore 5/3,which is exactly the same as the result, 1/ sin θ, that we goton p. 209 by analyzing force vectors. What this shows is thatNewton’s laws and conservation laws are not logically separate,but rather are very closely related descriptions of nature. In thecases where Newton’s laws are true, they give the same answersas the conservation laws. This is an example of a more gen-eral idea, called the correspondence principle, about how scienceprogresses over time. When a newer, more general theory is pro-posed to replace an older theory, the new theory must agree withthe old one in the realm of applicability of the old theory, since theold theory only became accepted as a valid theory by being ver-

Section 11.2 Energy 275

Discussion question A. Thewater behind the Hoover Damhas energy because of its posi-tion relative to the planet earth,which is attracting it with a gravi-tational force. Letting water downto the bottom of the dam convertsthat energy into energy of motion.When the water reaches thebottom of the dam, it hits turbineblades that drive generators, andits energy of motion is convertedinto electrical energy.

ified experimentally in a variety of experiments. In other words,the new theory must be backward-compatible with the old one.Even though conservation laws can prove things that Newton’slaws can’t (that perpetual motion is impossible, for example), theyaren’t going to disprove Newton’s laws when applied to mechani-cal systems where we already knew Newton’s laws were valid.

Discussion question

A Hydroelectric power (water flowing over a dam to spin turbines)appears to be completely free. Does this violate conservation of energy?If not, then what is the ultimate source of the electrical energy producedby a hydroelectric plant?

B How does the proof in example 3 fail if the assumption of a frictionlesssurface doesn’t hold?

11.3 A numerical scale of energyEnergy comes in a variety of forms, and physicists didn’t discover allof them right away. They had to start somewhere, so they pickedone form of energy to use as a standard for creating a numericalenergy scale. (In fact the history is complicated, and several differentenergy units were defined before it was realized that there was asingle general energy concept that deserved a single consistent unitof measurement.) One practical approach is to define an energyunit based on heating water. The SI unit of energy is the joule,J, (rhymes with “cool”), named after the British physicist JamesJoule. One Joule is the amount of energy required in order to heat0.24 g of water by 1 C. The number 0.24 is not worth memorizing.

Note that heat, which is a form of energy, is completely differ-ent from temperature, which is not. Twice as much heat energyis required to prepare two cups of coffee as to make one, but twocups of coffee mixed together don’t have double the temperature.In other words, the temperature of an object tells us how hot it is,but the heat energy contained in an object also takes into accountthe object’s mass and what it is made of.1

Later we will encounter other quantities that are conserved inphysics, such as momentum and angular momentum, and the methodfor defining them will be similar to the one we have used for energy:pick some standard form of it, and then measure other forms bycomparison with this standard. The flexible and adaptable natureof this procedure is part of what has made conservation laws such adurable basis for the evolution of physics.

1In standard, formal terminology, there is another, finer distinction. Theword “heat” is used only to indicate an amount of energy that is transferred,whereas “thermal energy” indicates an amount of energy contained in an object.I’m informal on this point, and refer to both as heat, but you should be awareof the distinction.


Heating a swimming pool example 4. If electricity costs 3.9 cents per MJ (1 MJ = 1 megajoule = 106

J), how much does it cost to heat a 26000-gallon swimming poolfrom 10 C to 18 C?

. Converting gallons to cm3 gives

26000 gallons× 3780 cm3

1 gallon= 9.8× 107 cm3 .

Water has a density of 1 gram per cubic centimeter, so the massof the water is 9.8 × 107 g. One joule is sufficient to heat 0.24 gby 1 C, so the energy needed to heat the swimming pool is

1 J× 9.8× 107 g0.24 g

× 8 C1 C

= 3.3× 109 J

= 3.3× 103 MJ .

The cost of the electricity is (3.3× 103 MJ)($0.039/MJ)=$130.

Irish coffee example 5. You make a cup of Irish coffee out of 300 g of coffee at 100 Cand 30 g of pure ethyl alcohol at 20 C. One Joule is enoughenergy to produce a change of 1 C in 0.42 g of ethyl alcohol (i.e.,alcohol is easier to heat than water). What temperature is thefinal mixture?

. Adding up all the energy after mixing has to give the same resultas the total before mixing. We let the subscript i stand for theinitial situation, before mixing, and f for the final situation, and usesubscripts c for the coffee and a for the alcohol. In this notation,we have

total initial energy = total final energyEci + Eai = Ecf + Eaf .

We assume coffee has the same heat-carrying properties as wa-ter. Our information about the heat-carrying properties of the twosubstances is stated in terms of the change in energy required fora certain change in temperature, so we rearrange the equation toexpress everything in terms of energy differences:

Eaf − Eai = Eci − Ecf .

Using the given ratios of temperature change to energy change,we have

Eci − Ecf = (Tci − Tcf )(mc)/(0.24 g)Eaf − Eai = (Taf − Tai )(ma)/(0.42 g)

Setting these two quantities to be equal, we have

(Taf − Tai )(ma)/(0.42 g) = (Tci − Tcf )(mc)/(0.24 g) .

Section 11.3 A numerical scale of energy 277

In the final mixture the two substances must be at the same tem-perature, so we can use a single symbol Tf = Tcf = Taf for thetwo quantities previously represented by two different symbols,

(Tf − Tai )(ma)/(0.42 g) = (Tci − Tf )(mc)/(0.24 g) .

Solving for Tf gives

Tf =Tci

mc0.24 + Tai

ma0.42

mc0.24 + ma

0.42

= 96 C .

Once a numerical scale of energy has been established for someform of energy such as heat, it can easily be extended to other typesof energy. For instance, the energy stored in one gallon of gasolinecan be determined by putting some gasoline and some water in aninsulated chamber, igniting the gas, and measuring the rise in thewater’s temperature. (The fact that the apparatus is known as a“bomb calorimeter” will give you some idea of how dangerous theseexperiments are if you don’t take the right safety precautions.) Hereare some examples of other types of energy that can be measuredusing the same units of joules:

type of energy example

chemical energyreleased by burning

About 50 MJ are released by burninga kg of gasoline.

energy required tobreak an object

When a person suffers a spiral frac-ture of the thighbone (a commontype in skiing accidents), about 2 Jof energy go into breaking the bone.

energy required tomelt a solid substance

7 MJ are required to melt 1 kg of tin.

chemical energyreleased by digestingfood

A bowl of Cheeries with milk providesus with about 800 kJ of usable en-ergy.

raising a mass againstthe force of gravity

Lifting 1.0 kg through a height of 1.0m requires 9.8 J.

nuclear energyreleased in fission

1 kg of uranium oxide fuel consumedby a reactor releases 2 × 1012 J ofstored nuclear energy.

It is interesting to note the disproportion between the megajouleenergies we consume as food and the joule-sized energies we expendin physical activities. If we could perceive the flow of energy aroundus the way we perceive the flow of water, eating a bowl of cerealwould be like swallowing a bathtub’s worth of energy, the continualloss of body heat to one’s environment would be like an energy-hoseleft on all day, and lifting a bag of cement would be like flickingit with a few tiny energy-drops. The human body is tremendously


f / Example 6.

inefficient. The calories we “burn” in heavy exercise are almost alldissipated directly as body heat.

You take the high road and I’ll take the low road. example 6. Figure f shows two ramps which two balls will roll down. Com-pare their final speeds, when they reach point B. Assume frictionis negligible.

. Each ball loses some energy because of its decreasing heightabove the earth, and conservation of energy says that it must gainan equal amount of energy of motion (minus a little heat createdby friction). The balls lose the same amount of height, so theirfinal speeds must be equal.

It’s impressive to note the complete impossibility of solving thisproblem using only Newton’s laws. Even if the shape of the trackhad been given mathematically, it would have been a formidabletask to compute the balls’ final speed based on vector addition ofthe normal force and gravitational force at each point along the way.

How new forms of energy are discovered

Textbooks often give the impression that a sophisticated physicsconcept was created by one person who had an inspiration one day,but in reality it is more in the nature of science to rough out an ideaand then gradually refine it over many years. The idea of energywas tinkered with from the early 1800’s on, and new types of energykept getting added to the list.

To establish the existence of a new form of energy, a physicisthas to

(1) show that it could be converted to and from other forms ofenergy; and

(2) show that it related to some definite measurable property ofthe object, for example its temperature, motion, position relative toanother object, or being in a solid or liquid state.

For example, energy is released when a piece of iron is soaked inwater, so apparently there is some form of energy already stored inthe iron. The release of this energy can also be related to a definitemeasurable property of the chunk of metal: it turns reddish-orange.There has been a chemical change in its physical state, which wecall rusting.

Although the list of types of energy kept getting longer andlonger, it was clear that many of the types were just variations ona theme. There is an obvious similarity between the energy neededto melt ice and to melt butter, or between the rusting of iron andmany other chemical reactions. The topic of the next chapter ishow this process of simplification reduced all the types of energyto a very small number (four, according to the way I’ve chosen to

Section 11.3 A numerical scale of energy 279

count them).

It might seem that if the principle of conservation of energy everappeared to be violated, we could fix it up simply by inventing somenew type of energy to compensate for the discrepancy. This wouldbe like balancing your checkbook by adding in an imaginary depositor withdrawal to make your figures agree with the bank’s statements.Step (2) above guards against this kind of chicanery. In the 1920sthere were experiments that suggested energy was not conserved inradioactive processes. Precise measurements of the energy releasedin the radioactive decay of a given type of atom showed inconsistentresults. One atom might decay and release, say, 1.1 × 10−10 J ofenergy, which had presumably been stored in some mysterious formin the nucleus. But in a later measurement, an atom of exactly thesame type might release 1.2× 10−10 J. Atoms of the same type aresupposed to be identical, so both atoms were thought to have startedout with the same energy. If the amount released was random, thenapparently the total amount of energy was not the same after thedecay as before, i.e., energy was not conserved.

Only later was it found that a previously unknown particle,which is very hard to detect, was being spewed out in the decay.The particle, now called a neutrino, was carrying off some energy,and if this previously unsuspected form of energy was added in,energy was found to be conserved after all. The discovery of theenergy discrepancies is seen with hindsight as being step (1) in theestablishment of a new form of energy, and the discovery of the neu-trino was step (2). But during the decade or so between step (1)and step (2) (the accumulation of evidence was gradual), physicistshad the admirable honesty to admit that the cherished principle ofconservation of energy might have to be discarded.

self-check AHow would you carry out the two steps given above in order to estab-lish that some form of energy was stored in a stretched or compressedspring? . Answer, p. 512

Mass Into EnergyEinstein showed that mass itself could be converted to and from energy,according to his celebrated equation E = mc2, in which c is the speedof light. We thus speak of mass as simply another form of energy, andit is valid to measure it in units of joules. The mass of a 15-gram pencilcorresponds to about 1.3× 1015 J. The issue is largely academic in thecase of the pencil, because very violent processes such as nuclear re-actions are required in order to convert any significant fraction of an ob-ject’s mass into energy. Cosmic rays, however, are continually strikingyou and your surroundings and converting part of their energy of motioninto the mass of newly created particles. A single high-energy cosmicray can create a “shower” of millions of previously nonexistent particleswhen it strikes the atmosphere. Einstein’s theories are discussed laterin this book.


Even today, when the energy concept is relatively mature and sta-ble, a new form of energy has been proposed based on observationsof distant galaxies whose light began its voyage to us billions of yearsago. Astronomers have found that the universe’s continuing expansion,resulting from the Big Bang, has not been decelerating as rapidly in thelast few billion years as would have been expected from gravitationalforces. They suggest that a new form of energy may be at work.

Discussion question

A I’m not making this up. XS Energy Drink has ads that read like this:All the “Energy” ... Without the Sugar! Only 8 Calories! Comment onthis.

11.4 Kinetic energyThe technical term for the energy associated with motion is kineticenergy, from the Greek word for motion. (The root is the same asthe root of the word “cinema” for a motion picture, and in Frenchthe term for kinetic energy is “energie cinetique.”) To find howmuch kinetic energy is possessed by a given moving object, we mustconvert all its kinetic energy into heat energy, which we have chosenas the standard reference type of energy. We could do this, forexample, by firing projectiles into a tank of water and measuring theincrease in temperature of the water as a function of the projectile’smass and velocity. Consider the following data from a series of threesuch experiments:

m (kg) v (m/s) energy (J)

1.00 1.00 0.50

1.00 2.00 2.00

2.00 1.00 1.00

Comparing the first experiment with the second, we see that dou-bling the object’s velocity doesn’t just double its energy, it quadru-ples it. If we compare the first and third lines, however, we findthat doubling the mass only doubles the energy. This suggests thatkinetic energy is proportional to mass and to the square of veloc-ity, KE ∝ mv2, and further experiments of this type would indeedestablish such a general rule. The proportionality factor equals 0.5because of the design of the metric system, so the kinetic energy ofa moving object is given by

KE =1

2mv2 .

The metric system is based on the meter, kilogram, and second,with other units being derived from those. Comparing the units onthe left and right sides of the equation shows that the joule can bereexpressed in terms of the basic units as kg·m2/s2.

Students are often mystified by the occurrence of the factor of1/2, but it is less obscure than it looks. The metric system was

Section 11.4 Kinetic energy 281

designed so that some of the equations relating to energy wouldcome out looking simple, at the expense of some others, which hadto have inconvenient conversion factors in front. If we were usingthe old British Engineering System of units in this course, then we’dhave the British Thermal Unit (BTU) as our unit of energy. Inthat system, the equation you’d learn for kinetic energy would havean inconvenient proportionality constant, KE =

(1.29× 10−3

)mv2,

with KE measured in units of BTUs, v measured in feet per second,and so on. At the expense of this inconvenient equation for kineticenergy, the designers of the British Engineering System got a simplerule for calculating the energy required to heat water: one BTUper degree Fahrenheit per pound. The inventor of kinetic energy,Thomas Young, actually defined it as KE = mv2, which meant thatall his other equations had to be different from ours by a factor oftwo. All these systems of units work just fine as long as they arenot combined with one another in an inconsistent way.

Energy released by a comet impact example 7.Comet Shoemaker-Levy, which struck the planet Jupiter in 1994,had a mass of roughly 4 × 1013 kg, and was moving at a speedof 60 km/s. Compare the kinetic energy released in the impact tothe total energy in the world’s nuclear arsenals, which is 2× 1019

J. Assume for the sake of simplicity that Jupiter was at rest.

. Since we assume Jupiter was at rest, we can imagine that thecomet stopped completely on impact, and 100% of its kinetic en-ergy was converted to heat and sound. We first convert the speedto mks units, v = 6 × 104 m/s, and then plug in to the equationto find that the comet’s kinetic energy was roughly 7 × 1022 J, orabout 3000 times the energy in the world’s nuclear arsenals.

Is there any way to derive the equation KE = (1/2)mv2 math-ematically from first principles? No, it is purely empirical. Thefactor of 1/2 in front is definitely not derivable, since it is differentin different systems of units. The proportionality to v2 is not evenquite correct; experiments have shown deviations from the v2 rule athigh speeds, an effect that is related to Einstein’s theory of relativ-ity. Only the proportionality to m is inevitable. The whole energyconcept is based on the idea that we add up energy contributionsfrom all the objects within a system. Based on this philosophy, itis logically necessary that a 2-kg object moving at 1 m/s have thesame kinetic energy as two 1-kg objects moving side-by-side at thesame speed.

Energy and relative motion

Although I mentioned Einstein’s theory of relativity above, it’smore relevant right now to consider how conservation of energy re-lates to the simpler Galilean idea, which we’ve already studied, thatmotion is relative. Galileo’s Aristotelian enemies (and it is no ex-aggeration to call them enemies!) would probably have objected to


Discussion question B

conservation of energy. After all, the Galilean idea that an objectin motion will continue in motion indefinitely in the absence of aforce is not so different from the idea that an object’s kinetic energystays the same unless there is a mechanism like frictional heatingfor converting that energy into some other form.

More subtly, however, it’s not immediately obvious that whatwe’ve learned so far about energy is strictly mathematically consis-tent with the principle that motion is relative. Suppose we verifythat a certain process, say the collision of two pool balls, conservesenergy as measured in a certain frame of reference: the sum of theballs’ kinetic energies before the collision is equal to their sum afterthe collision. (In reality we’d need to add in other forms of energy,like heat and sound, that are liberated by the collision, but let’s keepit simple.) But what if we were to measure everything in a frame ofreference that was in a different state of motion? A particular poolball might have less kinetic energy in this new frame; for example, ifthe new frame of reference was moving right along with it, its kineticenergy in that frame would be zero. On the other hand, some otherballs might have a greater kinetic energy in the new frame. It’s notimmediately obvious that the total energy before the collision willstill equal the total energy after the collision. After all, the equationfor kinetic energy is fairly complicated, since it involves the squareof the velocity, so it would be surprising if everything still workedout in the new frame of reference. It does still work out. Homeworkproblem 13 in this chapter gives a simple numerical example, andthe general proof is taken up in problem 15 on p. 378 (with thesolution given in the back of the book).


A Suppose that, like Young or Einstein, you were trying out differentequations for kinetic energy to see if they agreed with the experimentaldata. Based on the meaning of positive and negative signs of velocity,why would you suspect that a proportionality to mv would be less likelythan mv2?

B The figure shows a pendulum that is released at A and caught by apeg as it passes through the vertical, B. To what height will the bob riseon the right?

11.5 PowerA car may have plenty of energy in its gas tank, but still may notbe able to increase its kinetic energy rapidly. A Porsche doesn’tnecessarily have more energy in its gas tank than a Hyundai, it isjust able to transfer it more quickly. The rate of transferring energyfrom one form to another is called power. The definition can bewritten as an equation,

P =∆E

∆t,

Section 11.5 Power 283

where the use of the delta notation in the symbol ∆E has the usualinterpretation: the final amount of energy in a certain form minusthe initial amount that was present in that form. Power has unitsof J/s, which are abbreviated as watts, W (rhymes with “lots”).

If the rate of energy transfer is not constant, the power at anyinstant can be defined as the slope of the tangent line on a graph ofE versus t. Likewise ∆E can be extracted from the area under theP -versus-t curve.

Converting kilowatt-hours to joules example 8. The electric company bills you for energy in units of kilowatt-hours (kilowatts multiplied by hours) rather than in SI units ofjoules. How many joules is a kilowatt-hour?

. 1 kilowatt-hour = (1 kW)(1 hour) = (1000 J/s)(3600 s) = 3.6 MJ.

Human wattage example 9. A typical person consumes 2000 kcal of food in a day, and con-verts nearly all of that directly to heat. Compare the person’s heatoutput to the rate of energy consumption of a 100-watt lightbulb.

. Looking up the conversion factor from calories to joules, we find

∆E = 2000 kcal× 1000 cal1 kcal

× 4.18 J1 cal

= 8× 106 J

for our daily energy consumption. Converting the time intervallikewise into mks,

∆t = 1 day× 24 hours1 day

× 60 min1 hour

× 60 s1 min

= 9× 104 s .

Dividing, we find that our power dissipated as heat is 90 J/s = 90W, about the same as a lightbulb.

Wind power density example 10Wind power is a renewable energy resource, but it is most prac-tical in areas where the wind is both strong and reliably strong.When a horizontal-axis wind turbine faces directly into a windflowing at speed v , the air it intercepts in time ∆t forms a cylin-der whose length is v∆t , and whose mass is proportional to thesame factor. The kinetic energy of this cylinder represents themaximum energy that can theoretically be extracted in this time.Since the mass is proportional to v , the kinetic energy is propor-tional to v × v2 = v3. That is, the “wind power density” varies asthe cube of the wind’s speed.

It is easy to confuse the concepts of force, energy, and power,especially since they are synonyms in ordinary speech. The table onthe following page may help to clear this up:


force energy power

conceptualdefinition

A force is an interactionbetween two objects thatcauses a push or a pull.A force can be defined asanything that is capableof changing an object’sstate of motion.

Heating an object, mak-ing it move faster, or in-creasing its distance fromanother object that is at-tracting it are all exam-ples of things that wouldrequire fuel or physical ef-fort. All these things canbe quantified using a sin-gle scale of measurement,and we describe them allas forms of energy.

Power is the rate atwhich energy is trans-formed from one formto another or transferredfrom one object to an-other.

operationaldefinition

A spring scale can be usedto measure force.

If we define a unit of en-ergy as the amount re-quired to heat a certainamount of water by a1 C, then we can mea-sure any other quantityof energy by transferringit into heat in water andmeasuring the tempera-ture increase.

Measure the change in theamount of some form ofenergy possessed by anobject, and divide by theamount of time requiredfor the change to occur.

scalar orvector?

vector — has a directionin space which is the di-rection in which it pulls orpushes

scalar — has no directionin space

scalar — has no directionin space

unit newtons (N) joules (J) watts (W) = joules/s

Can it runout? Does itcost money?

No. I don’t have topay a monthly bill forthe meganewtons of forcerequired to hold up myhouse.

Yes. We pay money forgasoline, electrical energy,batteries, etc., becausethey contain energy.

More power means youare paying money at ahigher rate. A 100-Wlightbulb costs a certainnumber of cents per hour.

Can it be aproperty ofan object?

No. A force is a rela-tionship between twointeracting objects.A home-run baseballdoesn’t “have” force.

Yes. What a home-runbaseball has is kinetic en-ergy, not force.

Not really. A 100-Wlightbulb doesn’t “have”100 W. 100 J/s is the rateat which it converts elec-trical energy into light.

11.6 ? Massless particlesFailure of newton’s laws

One of the main reasons for preferring conservation laws to New-ton’s laws as a foundation for physics is that conservation laws aremore general. For example, Newton’s laws apply only to matter,whereas conservation laws can handle light as well. No experimentin Newton’s day had ever shown anything but zero for the mass

Section 11.6 ? Massless particles 285

or weight of a ray of light, and substituting m = 0 into a = F/mresults in an infinite acceleration, which doesn’t make sense. Withhindsight, this is to be expected because of relativity (section 2.7).Newton’s laws are only a good approximation for velocities that aresmall compared to c, the maximum speed of cause and effect. Butlight travels at c, so Newton’s laws are not a good approximation tothe behavior of light.

For insight into the behavior of things that go at exactly c, let’sconsider a case where something goes very close to c. A typical22-caliber rifle shoots a bullet with a mass of about 3 g at a speedof about 400 m/s. Now consider the firing of such a rifle as seenthrough an ultra-powerful telescope by an alien in a distant galaxy.We happen to be firing in the direction away from the alien, whogets a view from over our shoulder. Since the universe is expanding,our two galaxies are receding from each other. In the alien’s frame,our own galaxy is the one that is moving — let’s say at c−(200 m/s).If the two velocities simply added, the bullet would be moving atc + (200 m/s). But velocities don’t simply add and subtract rela-tivistically (p. 86), and applying the correct equation for relativisticcombination of velocities, we find that in the alien’s frame, the bulletflies at only c− (199.9995 m/s). That is, according to the alien, theenergy in the gunpowder only succeeded in accelerating the bulletby 0.0005 m/s! If we insisted on believing in KE = (1/2)mv2, thiswould clearly violate conservation of energy in the alien’s frame ofreference. K must not only get bigger faster than (1/2)mv2 as vapproaches c, it must blow up to infinity. This gives a mechanicalexplanation for why no material object can ever reach or exceed c,which is reassuring because speeds greater than c lead to violationof causality.

Ultrarelativistic motion

The bullet as seen in the alien’s frame of reference is an exampleof an ultrarelativistic particle, meaning one moving very close to c.We can fairly easily infer quite a bit about how kinetic energy mustbehave at ultrarelativistic speeds. We know that it must get largerand larger, and the question is how large it is when the speed differsfrom c by some small amount.

A good way of thinking about an ultrarelativistic particle is thatit’s a particle with a very small mass. For example, the subatomicparticle called the neutrino has a very small mass, thousands of timessmaller than that of the electron. Neutrinos are emitted in radioac-tive decay, and because the neutrino’s mass is so small, the amountof energy available in these decays is always enough to accelerate itto very close to the speed of light. Nobody has ever succeeded in ob-serving a neutrino that was not ultrarelativistic. When a particle’smass is very small, it becomes difficult to measure. For almost 70years after the neutrino was discovered, its mass was thought to be


zero. Similarly, we currently believe that a ray of light has no mass,but it it always possible that its mass will be found to be nonzeroat some point in the future. A ray of light can be modeled as anultrarelativistic particle.

Let’s compare ultrarelativistic particles with train cars. A singlecar with kinetic energy E has different properties than a train of twocars each with kinetic energy E/2. The single car has half the massand a speed that is greater by a factor of

√2. But the same is not

true for ultrarelativistic particles. Since an idealized ultrarelativisticparticle has a mass too small to be detectable in any experiment,we can’t detect the difference between m and 2m. Furthermore,ultrarelativistic particles move at close to c, so there is no observabledifference in speed. Thus we expect that a single ultrarelativisticparticle with energy E compared with two such particles, each withenergy E/2, should have all the same properties as measured by amechanical detector.

An idealized zero-mass particle also has no frame in which itcan be at rest. It always travels at c, and no matter how fast wechase after it, we can never catch up. We can, however, observeit in different frames of reference, and we will find that its energyis different. For example, distant galaxies are receding from us atsubstantial fractions of c, and when we observe them through atelescope, they appear very dim not just because they are very faraway but also because their light has less energy in our frame thanin a frame at rest relative to the source. This effect must be suchthat changing frames of reference according to a specific Lorentztransformation always changes the energy of the particle by a fixedfactor, regardless of the particle’s original energy; for if not, thenthe effect of a Lorentz transformation on a single particle of energyE would be different from its effect on two particles of energy E/2.

How does this energy-shift factor depend on the velocity v of theLorentz transformation? Actually, it is more convenient to expressthis in terms of a different variable rather than v. In nonrelativisticphysics, we change frames of reference simply by adding a constantonto all our velocities, but this is only a low-velocity approximation.For this reason, it will be more convenient to work with a variable s,defined as the factor by which the long diagonal of a parallelogramlike the ones in section 2.7 stretches under a Lorentz transformation.For example, we found in problem 21 on p. 95 that a velocity of0.6c corresponds to a stretch factor s = 2. The convenient thingabout stretch factors is that when we change to a new frame ofreference, they simply multiply. For example, in problem 21 youfound the result of combining a velocity of 0.6c with another velocityof 0.6c by drawing a parallelogram with its long axis stretched bya factor of 2 × 2 = 4. The relation between s and v is given bys =

√(1 + v)/(1− v) (in units with c = 1; see problems 18 on p. 94

and 22 on p. 96).


A low-speed approximation example 11What happens when the the velocity is small compared to c? Inunits where c = 1, this means that v is small compared to 1. Thestretch factor s =

√(1 + v )/(1− v ) can then be approximated by

taking 1/(1−v ) ≈ 1+v and√

1 + ε ≈ 1+ε/2, so that s ≈ 1+v .

Let’s write f(s) for the energy-shift factor that results from agiven Lorentz transformation. Since a Lorentz transformation s1 fol-lowed by a second transformation s2 is equivalent to a single trans-formation by s1s2, we must have f(s1s2) = f(s1)f(s2). This tightlyconstrains the form of the function f ; it must be something likef(s) = sn, where n is a constant. The interpretation of n is thatunder a Lorentz transformation corresponding to 1% of c, energiesof ultrarelativistic particles change by about n% (making the ap-proximation that v = .01 gives s ≈ 1.01). We postpone until p. 372the proof that n = 1, which is also in agreement with experimentswith rays of light.

Our final result is that the energy of an ultrarelativistic particleis simply proportional to its Lorentz “stretch factor” s. Even inthe case where the particle is truly massless, so that s doesn’t haveany finite value, we can still find how the energy differs accordingto different observers by finding the s of the Lorentz transformationbetween the two observers’ frames of reference.

An astronomical energy shift example 12. For quantum-mechanical reasons, a hydrogen atom can onlyexist in states with certain specific energies. By conservationof energy, the atom can therefore only absorb or emit light thathas an energy equal to the difference between two such atomicenergies. The outer atmosphere of a star is mostly made ofmonoatomic hydrogen, and one of the energies that a hydrogenatom can absorb or emit is 3.0276 × 10−19 J. When we observelight from stars in the Andromeda Galaxy, it has an energy of3.0306 × 10−19 J. If this is assumed to be due entirely to themotion of the Milky Way and Andromeda Galaxy relative to oneanother, along the line connecting them, find the direction andmagnitude of this velocity.

. The energy is shifted upward, which means that the AndromedaGalaxy is moving toward us. (Galaxies at cosmological distancesare always observed to be receding from one another, but thisdoesn’t necessarily hold for galaxies as close as these.) Relatingthe energy shift to the velocity, we have

E ′

E= s =

√(1 + v )/(1− v ) .

Since the shift is only about one part per thousand, the velocityis small compared to c — or small compared to 1 in units wherec = 1. Therefore we can approximate as in example 11, s ≈ 1+v ,


and we find

v ≈ s − 1 =E ′

E− 1 = 1.0× 10−3 .

This is in units where c = 1. Converting to SI units, where c 6= 1,we have v = (1.0×10−3)c = 300 km/s. Although the AndromedaGalaxy’s tangential motion is not accurately known, it is consid-ered likely that it will collide with the Milky Way in a few billionyears.

A symmetry property of the energy shift example 13Suppose that A and B are at rest relative to one another, but C ismoving along the line between A and B. A sends a pulse of laserlight to C, who then measures its energy and transmits anotherpulse to B having the same energy. The pulse accumulates twoenergy shifts, and the result is their product s(v )s(−v ). But Cdidn’t actually need to absorb the original pulse and retransmit it;the results would have been the same if C had just stayed out ofthe way. Therefore this product must equal 1, so we must haves(−v )s(v ) = 1, which can be verified directly from the equation.

The Ives-Stilwell experiment example 14The result of example 13 was the basis of one of the earliest labo-ratory tests of special relativity, by Ives and Stilwell in 1938. Theyobserved the light emitted by excited by a beam of H+

2 and H+3 ions

with speeds of a few tenths of a percent of c. Measuring the lightfrom both ahead of and behind the beams, they found that theproduct s(v )s(−v ) was equal to 1, as predicted by relativity. If rel-ativity had been false, then one would have expected the productto differ from 1 by an amount that would have been detectable intheir experiment. In 2003, Saathoff et al. carried out an extremelyprecise version of the Ives-Stilwell technique with Li+ ions movingat 6.4% of c. The energies observed, in units of 10−28 J, were:

Eo = 3620927488± 3(unshifted energy)

Eos(v ) = 3859620256± 0.6(shifted energy, forward)

Eos(−v ) = 3396996334± 3(shifted energy, backward)√

Eos(v ) · Eos(−v ) = 3620927487± 2

The results show incredibly precise agreement between Eo and√Eos(v ) · Eos(−v ), as expected relativistically because s(v )s(−v )

is supposed to equal 1. The agreement extends to 9 significantfigures, whereas if relativity had been false there should havebeen a relative disagreement of about v2 = .004, i.e., a discrep-ancy in the third significant figure. The spectacular agreementwith theory has made this experiment a lightning rod for anti-relativity kooks.


SummarySelected vocabularyenergy . . . . . . A numerical scale used to measure the heat,

motion, or other properties that would requirefuel or physical effort to put into an object; ascalar quantity with units of joules (J).

power . . . . . . . The rate of transferring energy; a scalar quan-tity with units of watts (W).

kinetic energy . . The energy an object possesses because of itsmotion.

heat . . . . . . . . A form of energy that relates to temperature.Heat is different from temperature because anobject with twice as much mass requires twiceas much heat to increase its temperature bythe same amount. Heat is measured in joules,temperature in degrees. (In standard termi-nology, there is another, finer distinction be-tween heat and thermal energy, which is dis-cussed below. In this book, I informally referto both as heat.)

temperature . . . What a thermometer measures. Objects left incontact with each other tend to reach the sametemperature. Cf. heat. As discussed in moredetail in chapter 2, temperature is essentiallya measure of the average kinetic energy permolecule.

NotationE . . . . . . . . . energyJ . . . . . . . . . . joules, the SI unit of energyKE . . . . . . . . kinetic energyP . . . . . . . . . powerW . . . . . . . . . watts, the SI unit of power; equivalent to J/s

Other terminology and notationQ or ∆Q . . . . . the amount of heat transferred into or out of

an objectK or T . . . . . . alternative symbols for kinetic energy, used in

the scientific literature and in most advancedtextbooks

thermal energy . Careful writers make a distinction betweenheat and thermal energy, but the distinctionis often ignored in casual speech, even amongphysicists. Properly, thermal energy is usedto mean the total amount of energy possessedby an object, while heat indicates the amountof thermal energy transferred in or out. Theterm heat is used in this book to include bothmeanings.


Summary

Heating an object, making it move faster, or increasing its dis-tance from another object that is attracting it are all examples ofthings that would require fuel or physical effort. All these things canbe quantified using a single scale of measurement, and we describethem all as forms of energy. The SI unit of energy is the Joule.The reason why energy is a useful and important quantity is thatit is always conserved. That is, it cannot be created or destroyedbut only transferred between objects or changed from one form toanother. Conservation of energy is the most important and broadlyapplicable of all the laws of physics, more fundamental and generaleven than Newton’s laws of motion.

Heating an object requires a certain amount of energy per degreeof temperature and per unit mass, which depends on the substanceof which the object consists. Heat and temperature are completelydifferent things. Heat is a form of energy, and its SI unit is the joule(J). Temperature is not a measure of energy. Heating twice as muchof something requires twice as much heat, but double the amountof a substance does not have double the temperature.

The energy that an object possesses because of its motion iscalled kinetic energy. Kinetic energy is related to the mass of theobject and the magnitude of its velocity vector by the equation

KE =1

2mv2 .

Power is the rate at which energy is transformed from one formto another or transferred from one object to another,

P =dE

dt

The SI unit of power is the watt (W).

The equation KE = (1/2)mv2 is a nonrelativistic approxima-tion, valid at speeds that are small compared to c. In the oppo-site limit, of a particle with a speed very close to c, the energy isproportional to the “stretch factor” of the Lorentz transformation,s =

√(1 + v)/(1− v) (in units with c = 1), for v → +c and 1/s for

v → −c. This gives a mechanical explanation for why no materialobject can ever reach or exceed c, which is reassuring because speedsgreater than c lead to violation of causality.

Summary 291

ProblemsKey√



1 Can kinetic energy ever be less than zero? Explain. [Basedon a problem by Serway and Faughn.]

2 Estimate the kinetic energy of an Olympic sprinter.

3 You are driving your car, and you hit a brick wall head on,at full speed. The car has a mass of 1500 kg. The kinetic energyreleased is a measure of how much destruction will be done to the carand to your body. Calculate the energy released if you are travelingat (a) 40 mi/hr, and again (b) if you’re going 80 mi/hr. What iscounterintuitive about this, and what implication does this have fordriving at high speeds?

√

4 The following table gives the amount of energy required inorder to heat, melt, or boil a gram of water.heat 1 g of ice by 1 C 2.05 Jmelt 1 g of ice 333 Jheat 1 g of liquid by 1 C 4.19 Jboil 1 g of water 2500 Jheat 1 g of steam by 1 C 2.01 J

(a) How much energy is required in order to convert 1.00 g of ice at-20 C into steam at 137 C?

√

(b) What is the minimum amount of hot water that could melt 1.00g of ice?

√

5 A closed system can be a bad thing — for an astronautsealed inside a space suit, getting rid of body heat can be difficult.Suppose a 60-kg astronaut is performing vigorous physical activity,expending 200 W of power. If none of the heat can escape from herspace suit, how long will it take before her body temperature risesby 6 C(11 F), an amount sufficient to kill her? Assume that theamount of heat required to raise her body temperature by 1 C isthe same as it would be for an equal mass of water. Express youranswer in units of minutes.

√


6 A bullet flies through the air, passes through a paperbackbook, and then continues to fly through the air beyond the book.When is there a force? When is there energy?

. Solution, p. 505

7 Experiments show that the power consumed by a boat’s en-gine is approximately proportional to third power of its speed. (Weassume that it is moving at constant speed.) (a) When a boat is crus-ing at constant speed, what type of energy transformation do youthink is being performed? (b) If you upgrade to a motor with doublethe power, by what factor is your boat’s crusing speed increased?[Based on a problem by Arnold Arons.] . Solution, p. 505

8 Object A has a kinetic energy of 13.4 J. Object B has a massthat is greater by a factor of 3.77, but is moving more slowly bya factor of 2.34. What is object B’s kinetic energy? [Based on aproblem by Arnold Arons.] . Solution, p. 505

9 Example 10 on page 284 showed that the power produced bya wind turbine is proportional to the cube of the wind speed v. VonKarman found empirically that when a fluid flows turbulently over asurface, the speed of the fluid is often well approximated by v ∝ z1/7,where z is the distance from the surface. Wind turbine towers areoften constructed at heights of 50 m, but surveys of wind speeds areusually conducted at heights of about 3 m. By what factor shouldthe predicted wind power density be scaled up relative to the surveydata?

√

10 The moon doesn’t really just orbit the Earth. By Newton’sthird law, the moon’s gravitational force on the earth is the same asthe earth’s force on the moon, and the earth must respond to themoon’s force by accelerating. If we consider the earth in moon inisolation and ignore outside forces, then Newton’s first law says theircommon center of mass doesn’t accelerate, i.e., the earth wobblesaround the center of mass of the earth-moon system once per month,and the moon also orbits around this point. The moon’s mass is 81times smaller than the earth’s. Compare the kinetic energies of theearth and moon. (We know that the center of mass is a kind ofbalance point, so it must be closer to the earth than to the moon.In fact, the distance from the earth to the center of mass is 1/81of the distance from the moon to the center of mass, which makessense intuitively, and can be proved rigorously using the equationon page 360.)

11 My 1.25 kW microwave oven takes 126 seconds to bring 250g of water from room temperature to a boil. What percentage ofthe power is being wasted? Where might the rest of the energy begoing? . Solution, p. 505

Problems 293

12 The multiflash photograph shows a collision between twopool balls. The ball that was initially at rest shows up as a darkimage in its initial position, because its image was exposed severaltimes before it was struck and began moving. By making measure-ments on the figure, determine numerically whether or not energyappears to have been conserved in the collision. What systematiceffects would limit the accuracy of your test? [From an example inPSSC Physics.]

Problem 12.

13 This problem is a numerical example of the imaginary exper-iment discussed on p. 282 regarding the relationship between energyand relative motion. Let’s say that the pool balls both have massesof 1.00 kg. Suppose that in the frame of reference of the pool table,the cue ball moves at a speed of 1.00 m/s toward the eight ball,which is initially at rest. The collision is head-on, and as you canverify for yourself the next time you’re playing pool, the result ofsuch a collision is that the incoming ball stops dead and the ball thatwas struck takes off with the same speed originally possessed by theincoming ball. (This is actually a bit of an idealization. To keepthings simple, we’re ignoring the spin of the balls, and we assumethat no energy is liberated by the collision as heat or sound.) (a)Calculate the total initial kinetic energy and the total final kineticenergy, and verify that they are equal. (b) Now carry out the wholecalculation again in the frame of reference that is moving in the samedirection that the cue ball was initially moving, but at a speed of0.50 m/s. In this frame of reference, both balls have nonzero initialand final velocities, which are different from what they were in thetable’s frame. [See also problem 15 on p. 378.]


14 One theory about the destruction of the space shuttle Columbiain 2003 is that one of its wings had been damaged on liftoff by achunk of foam insulation that fell off of one of its external fuel tanks.The New York Times reported on June 5, 2003, that NASA engi-neers had recreated the impact to see if it would damage a mock-upof the shuttle’s wing. “Before last week’s test, many engineers atNASA said they thought lightweight foam could not harm the seem-ingly tough composite panels, and privately predicted that the foamwould bounce off harmlessly, like a Nerf ball.” In fact, the 1.7-poundpiece of foam, moving at 531 miles per hour, did serious damage.A member of the board investigating the disaster said this demon-strated that “people’s intuitive sense of physics is sometimes wayoff.” (a) Compute the kinetic energy of the foam, and (b) com-pare with the energy of a 170-pound boulder moving at 5.3 milesper hour (the speed it would have if you dropped it from aboutknee-level).

√

(c) The boulder is a hundred times more massive, but its speedis a hundred times smaller, so what’s counterintuitive about yourresults?

15 The figure above is from a classic 1920 physics textbookby Millikan and Gale. It represents a method for raising the waterfrom the pond up to the water tower, at a higher level, withoutusing a pump. Water is allowed into the drive pipe, and once it isflowing fast enough, it forces the valve at the bottom closed. Explainhow this works in terms of conservation of mass and energy. (Cf.example 1 on page 273.)

Problems 295

16 All stars, including our sun, show variations in their lightoutput to some degree. Some stars vary their brightness by a factorof two or even more, but our sun has remained relatively steady dur-ing the hundred years or so that accurate data have been collected.Nevertheless, it is possible that climate variations such as ice agesare related to long-term irregularities in the sun’s light output. Ifthe sun was to increase its light output even slightly, it could meltenough Antarctic ice to flood all the world’s coastal cities. The totalsunlight that falls on Antarctica amounts to about 1 × 1016 watts.Presently, this heat input to the poles is balanced by the loss ofheat via winds, ocean currents, and emission of infrared light, sothat there is no net melting or freezing of ice at the poles from yearto year. Suppose that the sun changes its light output by some smallpercentage, but there is no change in the rate of heat loss by thepolar caps. Estimate the percentage by which the sun’s light outputwould have to increase in order to melt enough ice to raise the levelof the oceans by 10 meters over a period of 10 years. (This would beenough to flood New York, London, and many other cities.) Melting1 kg of ice requires 3× 103 J.


Do these forms of energy have anything in common?

Chapter 12

Simplifying the energy zoo

Variety is the spice of life, not of science. The figure shows a fewexamples from the bewildering array of forms of energy that sur-rounds us. The physicist’s psyche rebels against the prospect of along laundry list of types of energy, each of which would requireits own equations, concepts, notation, and terminology. The pointat which we’ve arrived in the study of energy is analogous to theperiod in the 1960’s when a half a dozen new subatomic particleswere being discovered every year in particle accelerators. It was anembarrassment. Physicists began to speak of the “particle zoo,”and it seemed that the subatomic world was distressingly complex.The particle zoo was simplified by the realization that most of the

297

new particles being whipped up were simply clusters of a previouslyunsuspected set of more fundamental particles (which were whimsi-cally dubbed quarks, a made-up word from a line of poetry by JamesJoyce, “Three quarks for Master Mark.”) The energy zoo can alsobe simplified, and it is the purpose of this chapter to demonstratethe hidden similarities between forms of energy as seemingly differ-ent as heat and motion.

a / A vivid demonstration thatheat is a form of motion. A smallamount of boiling water is pouredinto the empty can, which rapidlyfills up with hot steam. The canis then sealed tightly, and sooncrumples. This can be explainedas follows. The high tempera-ture of the steam is interpreted asa high average speed of randommotions of its molecules. Beforethe lid was put on the can, therapidly moving steam moleculespushed their way out of the can,forcing the slower air moleculesout of the way. As the steam in-side the can thinned out, a sta-ble situation was soon achieved,in which the force from the lessdense steam molecules movingat high speed balanced againstthe force from the more dense butslower air molecules outside. Thecap was put on, and after a whilethe steam inside the can reachedthe same temperature as the airoutside. The force from the cool,thin steam no longer matched theforce from the cool, dense air out-side, and the imbalance of forcescrushed the can.

12.1 Heat is kinetic energyWhat is heat really? Is it an invisible fluid that your bare feet soakup from a hot sidewalk? Can one ever remove all the heat from anobject? Is there a maximum to the temperature scale?

The theory of heat as a fluid seemed to explain why colder ob-jects absorbed heat from hotter ones, but once it became clear thatheat was a form of energy, it began to seem unlikely that a materialsubstance could transform itself into and out of all those other formsof energy like motion or light. For instance, a compost pile gets hot,and we describe this as a case where, through the action of bacteria,chemical energy stored in the plant cuttings is transformed into heatenergy. The heating occurs even if there is no nearby warmer objectthat could have been leaking “heat fluid” into the pile.

An alternative interpretation of heat was suggested by the theorythat matter is made of atoms. Since gases are thousands of times lessdense than solids or liquids, the atoms (or clusters of atoms calledmolecules) in a gas must be far apart. In that case, what is keepingall the air molecules from settling into a thin film on the floor of theroom in which you are reading this book? The simplest explanationis that they are moving very rapidly, continually ricocheting off of

298 Chapter 12 Simplifying the energy zoo

b / Random motion of atomsin a gas, a liquid, and a solid.

the floor, walls, and ceiling. Though bizarre, the cloud-of-bulletsimage of a gas did give a natural explanation for the surprisingability of something as tenuous as a gas to exert huge forces. Yourcar’s tires can hold it up because you have pumped extra moleculesinto them. The inside of the tire gets hit by molecules more oftenthan the outside, forcing it to stretch and stiffen.

The outward forces of the air in your car’s tires increase evenfurther when you drive on the freeway for a while, heating up therubber and the air inside. This type of observation leads naturallyto the conclusion that hotter matter differs from colder in that itsatoms’ random motion is more rapid. In a liquid, the motion couldbe visualized as people in a milling crowd shoving past each othermore quickly. In a solid, where the atoms are packed together, themotion is a random vibration of each atom as it knocks against itsneighbors.

We thus achieve a great simplification in the theory of heat. Heatis simply a form of kinetic energy, the total kinetic energy of randommotion of all the atoms in an object. With this new understanding,it becomes possible to answer at one stroke the questions posed atthe beginning of the section. Yes, it is at least theoretically possibleto remove all the heat from an object. The coldest possible temper-ature, known as absolute zero, is that at which all the atoms havezero velocity, so that their kinetic energies, (1/2)mv2, are all zero.No, there is no maximum amount of heat that a certain quantity ofmatter can have, and no maximum to the temperature scale, sincearbitrarily large values of v can create arbitrarily large amounts ofkinetic energy per atom.

The kinetic theory of heat also provides a simple explanation ofthe true nature of temperature. Temperature is a measure of theamount of energy per molecule, whereas heat is the total amount ofenergy possessed by all the molecules in an object.

There is an entire branch of physics, called thermodynamics,that deals with heat and temperature and forms the basis for tech-nologies such as refrigeration.

Thermodynamics is not covered in this book, and I have pro-vided here only a brief overview of the thermodynamic conceptsthat relate directly to energy, glossing over at least one point thatwould be dealt with more carefully in a thermodynamics course: itis really only true for a gas that all the heat is in the form of ki-netic energy. In solids and liquids, the atoms are close enough toeach other to exert intense electrical forces on each other, and thereis therefore another type of energy involved, the energy associatedwith the atoms’ distances from each other. Strictly speaking, heatenergy is defined not as energy associated with random motion ofmolecules but as any form of energy that can be conducted betweenobjects in contact, without any force.

Section 12.1 Heat is kinetic energy 299

c / The skater has convertedall his kinetic energy into potentialenergy on the way up the side ofthe pool. Photo by J.D. Rogge,www.sonic.net/∼shawn.

12.2 Potential energy: energy of distance orcloseness

We have already seen many examples of energy related to the dis-tance between interacting objects. When two objects participate inan attractive noncontact force, energy is required to bring them far-ther apart. In both of the perpetual motion machines that startedoff the previous chapter, one of the types of energy involved was theenergy associated with the distance between the balls and the earth,which attract each other gravitationally. In the perpetual motionmachine with the magnet on the pedestal, there was also energyassociated with the distance between the magnet and the iron ball,which were attracting each other.

The opposite happens with repulsive forces: two socks with thesame type of static electric charge will repel each other, and cannotbe pushed closer together without supplying energy.

In general, the term potential energy, with algebra symbol PE, isused for the energy associated with the distance between two objectsthat attract or repel each other via a force that depends on thedistance between them. Forces that are not determined by distancedo not have potential energy associated with them. For instance,the normal force acts only between objects that have zero distancebetween them, and depends on other factors besides the fact thatthe distance is zero. There is no potential energy associated withthe normal force.

The following are some commonplace examples of potential en-ergy:

gravitational potential energy: The skateboarder in the photohas risen from the bottom of the pool, converting kinetic en-ergy into gravitational potential energy. After being at restfor an instant, he will go back down, converting PE back intoKE.

magnetic potential energy: When a magnetic compass needle isallowed to rotate, the poles of the compass change their dis-tances from the earth’s north and south magnetic poles, con-verting magnetic potential energy into kinetic energy. (Even-tually the kinetic energy is all changed into heat by friction,and the needle settles down in the position that minimizes itspotential energy.)

electrical potential energy: Socks coming out of the dryer clingtogether because of attractive electrical forces. Energy is re-quired in order to separate them.

potential energy of bending or stretching: The force betweenthe two ends of a spring depends on the distance between


d / As the skater free-falls,his PE is converted into KE. (Thenumbers would be equally validas a description of his motion onthe way up.)

them, i.e., on the length of the spring. If a car is presseddown on its shock absorbers and then released, the potentialenergy stored in the spring is transformed into kinetic andgravitational potential energy as the car bounces back up.

I have deliberately avoided introducing the term potential en-ergy up until this point, because it tends to produce unfortunateconnotations in the minds of students who have not yet been inoc-ulated with a careful description of the construction of a numericalenergy scale. Specifically, there is a tendency to generalize the terminappropriately to apply to any situation where there is the “poten-tial” for something to happen: “I took a break from digging, butI had potential energy because I knew I’d be ready to work hardagain in a few minutes.”

An equation for gravitational potential energy

All the vital points about potential energy can be made by focus-ing on the example of gravitational potential energy. For simplicity,we treat only vertical motion, and motion close to the surface of theearth, where the gravitational force is nearly constant. (The gener-alization to the three dimensions and varying forces is more easilyaccomplished using the concept of work, which is the subject thenext chapter.)

To find an equation for gravitational PE, we examine the caseof free fall, in which energy is transformed between kinetic energyand gravitational PE. Whatever energy is lost in one form is gainedin an equal amount in the other form, so using the notation ∆KEto stand for KEf −KEi and a similar notation for PE, we have

[1] ∆KE = −∆PEgrav .

It will be convenient to refer to the object as falling, so that PEis being changed into KE, but the math applies equally well to anobject slowing down on its way up. We know an equation for kineticenergy,

[2] KE =1

2mv2 ,

so if we can relate v to height, y, we will be able to relate ∆PE to y,which would tell us what we want to know about potential energy.The y component of the velocity can be connected to the height viathe constant acceleration equation

[3] v2f = v2

i + 2a∆y ,

and Newton’s second law provides the acceleration,

[4] a = F/m ,

Section 12.2 Potential energy: energy of distance or closeness 301

in terms of the gravitational force.

The algebra is simple because both equation [2] and equation [3]have velocity to the second power. Equation [2] can be solved forv2 to give v2 = 2KE/m, and substituting this into equation [3], wefind

2KEfm

= 2KEim

+ 2a∆y .

Making use of equations [1] and [4] gives the simple result

∆PEgrav = −F∆y . [change in gravitational PE

resulting from a change in height ∆y;

F is the gravitational force on the object,

i.e., its weight; valid only near the surface

of the earth, where F is constant]

Dropping a rock example 1. If you drop a 1-kg rock from a height of 1 m, how many joulesof KE does it have on impact with the ground? (Assume that anyenergy transformed into heat by air friction is negligible.)

. If we choose the y axis to point up, then Fy is negative, andequals −(1 kg)(g) = −9.8 N. A decrease in y is represented by anegative value of ∆y , ∆y = −1 m, so the change in potential en-ergy is −(−9.8 N)(−1 m) ≈ −10 J. (The proof that newtons mul-tiplied by meters give units of joules is left as a homework prob-lem.) Conservation of energy says that the loss of this amount ofPE must be accompanied by a corresponding increase in KE of10 J.

It may be dismaying to note how many minus signs had to behandled correctly even in this relatively simple example: a totalof four. Rather than depending on yourself to avoid any mistakeswith signs, it is better to check whether the final result make sensephysically. If it doesn’t, just reverse the sign.

Although the equation for gravitational potential energy was de-rived by imagining a situation where it was transformed into kineticenergy, the equation can be used in any context, because all thetypes of energy are freely convertible into each other.

Gravitational PE converted directly into heat example 2. A 50-kg firefighter slides down a 5-m pole at constant velocity.How much heat is produced?

. Since she slides down at constant velocity, there is no changein KE. Heat and gravitational PE are the only forms of energy thatchange. Ignoring plus and minus signs, the gravitational force on


her body equals mg, and the amount of energy transformed is

(mg)(5 m) = 2500 J .

On physical grounds, we know that there must have been an in-crease (positive change) in the heat energy in her hands and inthe flagpole.

Here are some questions and answers about the interpretation ofthe equation ∆PEgrav = −F∆y for gravitational potential energy.

Question: In a nutshell, why is there a minus sign in the equation?Answer: It is because we increase the PE by moving the object inthe opposite direction compared to the gravitational force.

Question: Why do we only get an equation for the change in po-tential energy? Don’t I really want an equation for the potentialenergy itself?Answer: No, you really don’t. This relates to a basic fact aboutpotential energy, which is that it is not a well defined quantity inthe absolute sense. Only changes in potential energy are unambigu-ously defined. If you and I both observe a rock falling, and agreethat it deposits 10 J of energy in the dirt when it hits, then we willbe forced to agree that the 10 J of KE must have come from a lossof 10 joules of PE. But I might claim that it started with 37 J of PEand ended with 27, while you might swear just as truthfully that ithad 109 J initially and 99 at the end. It is possible to pick somespecific height as a reference level and say that the PE is zero there,but it’s easier and safer just to work with changes in PE and avoidabsolute PE altogether.

Question: You referred to potential energy as the energy that twoobjects have because of their distance from each other. If a rockfalls, the object is the rock. Where’s the other object?Answer: Newton’s third law guarantees that there will always betwo objects. The other object is the planet earth.

Question: If the other object is the earth, are we talking about thedistance from the rock to the center of the earth or the distancefrom the rock to the surface of the earth?Answer: It doesn’t matter. All that matters is the change in dis-tance, ∆y, not y. Measuring from the earth’s center or its surfaceare just two equally valid choices of a reference point for definingabsolute PE.

Question: Which object contains the PE, the rock or the earth?Answer: We may refer casually to the PE of the rock, but techni-cally the PE is a relationship between the earth and the rock, andwe should refer to the earth and the rock together as possessing thePE.

Question: How would this be any different for a force other thangravity?

Section 12.2 Potential energy: energy of distance or closeness 303

e / All these energy transfor-mations turn out at the atomiclevel to be changes in potentialenergy resulting from changes inthe distances between atoms.

Answer: It wouldn’t. The result was derived under the assumptionof constant force, but the result would be valid for any other situa-tion where two objects interacted through a constant force. Gravityis unusual, however, in that the gravitational force on an object isso nearly constant under ordinary conditions. The magnetic forcebetween a magnet and a refrigerator, on the other hand, changesdrastically with distance. The math is a little more complex for avarying force, but the concepts are the same.

Question: Suppose a pencil is balanced on its tip and then fallsover. The pencil is simultaneously changing its height and rotating,so the height change is different for different parts of the object.The bottom of the pencil doesn’t lose any height at all. What doyou do in this situation?Answer: The general philosophy of energy is that an object’s en-ergy is found by adding up the energy of every little part of it.You could thus add up the changes in potential energy of all thelittle parts of the pencil to find the total change in potential en-ergy. Luckily there’s an easier way! The derivation of the equationfor gravitational potential energy used Newton’s second law, whichdeals with the acceleration of the object’s center of mass (i.e., itsbalance point). If you just define ∆y as the height change of thecenter of mass, everything works out. A huge Ferris wheel can berotated without putting in or taking out any PE, because its centerof mass is staying at the same height.

self-check AA ball thrown straight up will have the same speed on impact with theground as a ball thrown straight down at the same speed. How can thisbe explained using potential energy? . Answer, p. 512

Discussion question

A You throw a steel ball up in the air. How can you prove based onconservation of energy that it has the same speed when it falls back intoyour hand? What if you throw a feather up — is energy not conserved inthis case?

12.3 All energy is potential or kinetic

In the same way that we found that a change in temperatureis really only a change in kinetic energy at the atomic level, wenow find that every other form of energy turns out to be a formof potential energy. Boiling, for instance, means knocking some ofthe atoms (or molecules) out of the liquid and into the space above,where they constitute a gas. There is a net attractive force betweenessentially any two atoms that are next to each other, which is whymatter always prefers to be packed tightly in the solid or liquid stateunless we supply enough potential energy to pull it apart into a gas.This explains why water stops getting hotter when it reaches the


f / This figure looks similar tothe previous ones, but the scaleis a million times smaller. Thelittle balls are the neutrons andprotons that make up the tiny nu-cleus at the center of the uraniumatom. When the nucleus splits(fissions), the potential energychange is partly electrical andpartly a change in the potentialenergy derived from the forcethat holds atomic nuclei together(known as the strong nuclearforce).

g / A pellet of plutonium-238glows with its own heat. Itsnuclear potential energy is beingconverted into heat, a form ofkinetic energy. Pellets of this typeare used as power supplies onsome space probes.

boiling point: the power being pumped into the water by your stovebegins going into potential energy rather than kinetic energy.

As shown in figure e, every stored form of energy that we en-counter in everyday life turns out to be a form of potential energyat the atomic level. The forces between atoms are electrical andmagnetic in nature, so these are actually electrical and magneticpotential energies.

Even if we wish to include nuclear reactions in the picture, therestill turn out to be only four fundamental types of energy:

kinetic energy (including heat)

gravitational potential energy

electrical and magnetic potential energy

nuclear potential energy

How does light fit into this picture? Optional section 11.6 dis-cussed the idea of modeling a ray of light as a stream of masslessparticles. But the way in which we described the energy of such par-ticles was completely different from the use of KE = (1/2)mv2 forobjects made of atoms. Since the purpose of this chapter has beento bring every form of energy under the same roof, this inconsistencyfeels unsatisfying. Section 12.4 eliminates this inconsistency.

Discussion question

A Referring back to the pictures at the beginning of the chapter, howdo all these forms of energy fit into the shortened list of categories givenabove?

Section 12.3 All energy is potential or kinetic 305

12.4 ? E=mc2

In section 11.6 we found the relativistic expression for kinetic energyin the limiting case of an ultrarelativistic particle, i.e., one with aspeed very close to c: its energy is proportional to the “stretchfactor” of the Lorentz transformation, s =

√(1 + v)/(1− v) (in

units with c = 1), for v → +c and 1/s for v → −c. What aboutintermediate cases, like v = c/2?

h / The match is lit inside the belljar. It burns, and energy escapesfrom the jar in the form of light. Af-ter it stops burning, all the sameatoms are still in the jar: nonehave entered or escaped. The fig-ure shows the outcome expectedbefore relativity, which was thatthe mass measured on the bal-ance would remain exactly thesame.

When we are forced to tinker with a time-honored theory, ourfirst instinct should always be to tinker as conservatively as possible.Although we’ve been forced to admit that kinetic energy doesn’tvary as v2/2 at relativistic speeds, the next most conservative thingwe could do would be to assume that the only change necessary is toreplace the factor of v2/2 in the nonrelativistic expression for kineticenergy with some other function, which would have to act like s or1/s for v → ±c. I suspect that this is what Einstein thought when hecompleted his original paper on relativity in 1905, because it wasn’tuntil later that year that he published a second paper showing thatthis still wasn’t enough of a change to produce a working theory. Wenow know that there is something more that needs to be changedabout prerelativistic physics, and this is the assumption that mass isonly a property of material particles such as atoms (figure h). Callthis the “atoms-only hypothesis.”

Now that we know the correct relativistic way of finding theenergy of a ray of light, it turns out that we can use that to findwhat we were originally seeking, which was the energy of a materialobject. The following discussion closely follows Einstein’s.

Suppose that a material object O of mass m, initially at rest ina certain frame A, emits two rays of light, each with energy E/2.By conservation of energy, the object must have lost an amount ofenergy equal to E. By symmetry, O remains at rest.

We now switch to a different frame of reference B moving at somearbitrary speed corresponding to a stretch factor S. The change offrames means that we’re chasing one ray, so that its energy is scaleddown to (E/2)S−1, while running away from the other, whose energy


gets boosted to (E/2)S. In frame B, as in A, O retains the samespeed after emission of the light. But observers in frames A and Bdisagree on how much energy O has lost, the discrepancy being

E

[1

2(S + S−1)− 1

].

Let’s consider the case where B’s velocity relative to A is small. Forsmall velocities, the stretch factor is S ≈ 1+v/c, and the discrepancyin O’s energy loss is approximately

1

2Ev2/c2 .

The interpretation is that when O reduced its energy by E in orderto make the light rays, it reduced its mass from mo to mo − m,where m = E/c2. Rearranging factors, we have Einstein’s famous

E = mc2 .

We find that mass is not simply a built-in property of the parti-cles that make up an object, with the object’s mass being the sum ofthe masses of its particles. Rather, mass and energy are equivalent,so that if the experiment of figure h is carried out with a sufficientlyprecise balance, the reading will drop because of the mass equivalentof the energy emitted as light.

The equation E = mc2 tells us how much energy is equivalentto how much mass: the conversion factor is the square of the speedof light, c. Since c a big number, you get a really really big numberwhen you multiply it by itself to get c2. This means that evena small amount of mass is equivalent to a very large amount ofenergy. Conversely, an ordinary amount of energy corresponds to anextremely small mass, and this is why nobody detected the non-nullresult of experiments like the one in figure h hundreds of years ago.

The big event here is mass-energy equivalence, but we can alsoharvest a result for the energy of a material particle moving at acertain speed. Plugging in S =

√(1 + v)/(1− v) to the equation

above for the energy discrepancy of object O between frames A andB, we find m(γ − 1)c2. This is the difference between O’s energy inframe B and its energy when it is at rest, but since mass and energyare equivalent, we assign it energy mc2 when it is at rest. The resultis that the energy is

E = mγc2 .

Section 12.4 ? E=mc2 307

i / Top: A PET scanner. Mid-dle: Each positron annihilateswith an electron, producing twogamma-rays that fly off back-to-back. When two gamma raysare observed simultaneously inthe ring of detectors, they areassumed to come from the sameannihilation event, and the pointat which they were emitted mustlie on the line connecting thetwo detectors. Bottom: A scanof a person’s torso. The bodyhas concentrated the radioactivetracer around the stomach,indicating an abnormal medicalcondition.

Electron-positron annihilation example 3Natural radioactivity in the earth produces positrons, which arelike electrons but have the opposite charge. A form of antimat-ter, positrons annihilate with electrons to produce gamma rays, aform of high-frequency light. Such a process would have beenconsidered impossible before Einstein, because conservation ofmass and energy were believed to be separate principles, andthis process eliminates 100% of the original mass. The amountof energy produced by annihilating 1 kg of matter with 1 kg ofantimatter is

E = mc2

= (2 kg)(

3.0× 108 m/s)2

= 2× 1017 J ,

which is on the same order of magnitude as a day’s energy con-sumption for the entire world’s population!

Positron annihilation forms the basis for the medical imaging tech-nique called a PET (positron emission tomography) scan, in whicha positron-emitting chemical is injected into the patient and map-ped by the emission of gamma rays from the parts of the bodywhere it accumulates.

A rusting nail example 4. An iron nail is left in a cup of water until it turns entirely to rust.The energy released is about 0.5 MJ. In theory, would a suffi-ciently precise scale register a change in mass? If so, how much?

. The energy will appear as heat, which will be lost to the envi-ronment. The total mass-energy of the cup, water, and iron willindeed be lessened by 0.5 MJ. (If it had been perfectly insulated,there would have been no change, since the heat energy wouldhave been trapped in the cup.) The speed of light is c = 3 × 108

meters per second, so converting to mass units, we have

m =Ec2

=0.5× 106 J(

3× 108 m/s)2

= 6× 10−12 kilograms .

The change in mass is too small to measure with any practicaltechnique. This is because the square of the speed of light issuch a large number.

Gravity bending light example 5Gravity is a universal attraction between things that have mass,and since the energy in a beam of light is equivalent to a somevery small amount of mass, we expect that light will be affected


j / A New York Times head-line from November 10, 1919,describing the observationsdiscussed in example 5.

by gravity, although the effect should be very small. The first im-portant experimental confirmation of relativity came in 1919 whenstars next to the sun during a solar eclipse were observed to haveshifted a little from their ordinary position. (If there was no eclipse,the glare of the sun would prevent the stars from being observed.)Starlight had been deflected by the sun’s gravity. Figure k is aphotographic negative, so the circle that appears bright is actu-ally the dark face of the moon, and the dark area is really thebright corona of the sun. The stars, marked by lines above andbelow then, appeared at positions slightly different than their nor-mal ones.

k / Example 5.

Black holes example 6A star with sufficiently strong gravity can prevent light from leav-ing. Quite a few black holes have been detected via their gravita-tional forces on neighboring stars or clouds of gas and dust.

Section 12.4 ? E=mc2 309

SummarySelected vocabularypotential energy the energy having to do with the distance be-

tween two objects that interact via a noncon-tact force

NotationPE . . . . . . . . . potential energy

Other terminology and notationU or V . . . . . . symbols used for potential energy in the scien-

tific literature and in most advanced textbooks

Summary

Historically, the energy concept was only invented to include afew phenomena, but it was later generalized more and more to applyto new situations, for example nuclear reactions. This generalizingprocess resulted in an undesirably long list of types of energy, eachof which apparently behaved according to its own rules.

The first step in simplifying the picture came with the realizationthat heat was a form of random motion on the atomic level, i.e., heatwas nothing more than the kinetic energy of atoms.

A second and even greater simplification was achieved with therealization that all the other apparently mysterious forms of energyactually had to do with changing the distances between atoms (orsimilar processes in nuclei). This type of energy, which relates tothe distance between objects that interact via a force, is thereforeof great importance. We call it potential energy.

Most of the important ideas about potential energy can be un-derstood by studying the example of gravitational potential energy.The change in an object’s gravitational potential energy is given by

∆PEgrav = −Fgrav∆y , [if Fgrav is constant, i.e., the

the motion is all near the Earth’s surface]

The most important thing to understand about potential energyis that there is no unambiguous way to define it in an absolute sense.The only thing that everyone can agree on is how much the potentialenergy has changed from one moment in time to some later momentin time.

An implication of Einstein’s theory of special relativity is thatmass and energy are equivalent, as expressed by the famous E =mc2. The energy of a material object is given by E = mγc2.


ProblemsKey√



1 A ball rolls up a ramp, turns around, and comes back down.When does it have the greatest gravitational potential energy? Thegreatest kinetic energy? [Based on a problem by Serway and Faughn.]

2 Anya and Ivan lean over a balcony side by side. Anya throws apenny downward with an initial speed of 5 m/s. Ivan throws a pennyupward with the same speed. Both pennies end up on the groundbelow. Compare their kinetic energies and velocities on impact.

3 Can gravitational potential energy ever be negative? Notethat the question refers to PE, not ∆PE, so that you must thinkabout how the choice of a reference level comes into play. [Based ona problem by Serway and Faughn.]

4 (a) You release a magnet on a tabletop near a big piece ofiron, and the magnet slides across the table to the iron. Does themagnetic potential energy increase, or decrease? Explain.(b) Suppose instead that you have two repelling magnets. You givethem an initial push towards each other, so they decelerate while ap-proaching each other. Does the magnetic potential energy increaseor decrease? Explain.

5 Let Eb be the energy required to boil one kg of water. (a) Findan equation for the minimum height from which a bucket of watermust be dropped if the energy released on impact is to vaporize it.Assume that all the heat goes into the water, not into the dirt itstrikes, and ignore the relatively small amount of energy required toheat the water from room temperature to 100 C. [Numerical check,not for credit: Plugging in Eb = 2.3 MJ/kg should give a result of230 km.]

√

(b) Show that the units of your answer in part a come out rightbased on the units given for Eb.

6 A grasshopper with a mass of 110 mg falls from rest from aheight of 310 cm. On the way down, it dissipates 1.1 mJ of heat dueto air resistance. At what speed, in m/s, does it hit the ground?

. Solution, p. 505

Problems 311

7 At a given temperature, the average kinetic energy per moleculeis a fixed value, so for instance in air, the more massive oxygenmolecules are moving more slowly on the average than the nitrogenmolecules. The ratio of the masses of oxygen and nitrogen moleculesis 16.00 to 14.01. Now suppose a vessel containing some air is sur-rounded by a vacuum, and the vessel has a tiny hole in it, whichallows the air to slowly leak out. The molecules are bouncing aroundrandomly, so a given molecule will have to “try” many times beforeit gets lucky enough to head out through the hole. How many timesmore rapidly does the nitrogen escape?

√

8 A person on a bicycle is to coast down a ramp of height h andthen pass through a circular loop of radius r. What is the small-est value of h for which the cyclist will complete the loop withoutfalling? (Ignore the kinetic energy of the spinning wheels.)

√

9 Problem 9 has been deleted. ?

10 Students are often tempted to think of potential energy andkinetic energy as if they were always related to each other, likeyin and yang. To show this is incorrect, give examples of physicalsituations in which (a) PE is converted to another form of PE, and(b) KE is converted to another form of KE. . Solution, p. 505

11 Lord Kelvin, a physicist, told the story of how he encounteredJames Joule when Joule was on his honeymoon. As he traveled,Joule would stop with his wife at various waterfalls, and measurethe difference in temperature between the top of the waterfall andthe still water at the bottom. (a) It would surprise most peopleto learn that the temperature increased. Why should there be anysuch effect, and why would Joule care? How would this relate to theenergy concept, of which he was the principal inventor? (b) Howmuch of a gain in temperature should there be between the topand bottom of a 50-meter waterfall? (c) What assumptions did youhave to make in order to calculate your answer to part b? In reality,would the temperature change be more than or less than what youcalculated? [Based on a problem by Arnold Arons.]

√

12 Make an order-of-magnitude estimate of the power repre-sented by the loss of gravitational energy of the water going overNiagara Falls. If the hydroelectric plant at the bottom of the fallscould convert 100% of this to electrical power, roughly how manyhouseholds could be powered? . Solution, p. 505


Problem 16.

13 When you buy a helium-filled balloon, the seller has to inflateit from a large metal cylinder of the compressed gas. The heliuminside the cylinder has energy, as can be demonstrated for exampleby releasing a little of it into the air: you hear a hissing sound,and that sound energy must have come from somewhere. The totalamount of energy in the cylinder is very large, and if the valve isinadvertently damaged or broken off, the cylinder can behave likebomb or a rocket.

Suppose the company that puts the gas in the cylinders preparescylinder A with half the normal amount of pure helium, and cylinderB with the normal amount. Cylinder B has twice as much energy,and yet the temperatures of both cylinders are the same. Explain, atthe atomic level, what form of energy is involved, and why cylinderB has twice as much.

14 Explain in terms of conservation of energy why sweatingcools your body, even though the sweat is at the same temperatureas your body. Describe the forms of energy involved in this energytransformation. Why don’t you get the same cooling effect if youwipe the sweat off with a towel? Hint: The sweat is evaporating.

15 (a) A circular hoop of mass m and radius r spins like a wheelwhile its center remains at rest. Let ω (Greek letter omega) be thenumber of radians it covers per unit time, i.e., ω = 2π/T , wherethe period, T , is the time for one revolution. Show that its kineticenergy equals (1/2)mω2r2.(b) Show that the answer to part a has the right units. (Notethat radians aren’t really units, since the definition of a radian is aunitless ratio of two lengths.)(c) If such a hoop rolls with its center moving at velocity v, itskinetic energy equals (1/2)mv2, plus the amount of kinetic energyfound in part a. Show that a hoop rolls down an inclined plane withhalf the acceleration that a frictionless sliding block would have.

?

16 A skateboarder starts at nearly rest at the top of a giantcylinder, and begins rolling down its side. (If he started exactly atrest and exactly at the top, he would never get going!) Show that hisboard loses contact with the pipe after he has dropped by a heightequal to one third the radius of the pipe. . Solution, p. 506 ?

17 In example 5 on page 84, I remarked that accelerating amacroscopic (i.e., not microscopic) object to close to the speed oflight would require an unreasonable amount of energy. Suppose thatthe starship Enterprise from Star Trek has a mass of 8.0 × 107 kg,about the same as the Queen Elizabeth 2. Compute the kineticenergy it would have to have if it was moving at half the speed oflight. Compare with the total energy content of the world’s nucleararsenals, which is about 1021 J.

√

Problems 313

18 (a) A free neutron (as opposed to a neutron bound into anatomic nucleus) is unstable, and undergoes spontaneous radioactivedecay into a proton, an electron, and an antineutrino. The massesof the particles involved are as follows:

neutron 1.67495× 10−27 kgproton 1.67265× 10−27 kgelectron 0.00091× 10−27 kgantineutrino < 10−35 kg

Find the energy released in the decay of a free neutron.√

(b) Neutrons and protons make up essentially all of the mass of theordinary matter around us. We observe that the universe around ushas no free neutrons, but lots of free protons (the nuclei of hydrogen,which is the element that 90% of the universe is made of). We findneutrons only inside nuclei along with other neutrons and protons,not on their own.

If there are processes that can convert neutrons into protons, wemight imagine that there could also be proton-to-neutron conver-sions, and indeed such a process does occur sometimes in nucleithat contain both neutrons and protons: a proton can decay into aneutron, a positron, and a neutrino. A positron is a particle withthe same properties as an electron, except that its electrical chargeis positive. A neutrino, like an antineutrino, has negligible mass.

Although such a process can occur within a nucleus, explain whyit cannot happen to a free proton. (If it could, hydrogen would beradioactive, and you wouldn’t exist!)


Chapter 13

Work: the transfer ofmechanical energy

13.1 Work: the transfer of mechanical energyThe concept of work

The mass contained in a closed system is a conserved quantity,but if the system is not closed, we also have ways of measuring theamount of mass that goes in or out. The water company does thiswith a meter that records your water use.

Likewise, we often have a system that is not closed, and wouldlike to know how much energy comes in or out. Energy, however,is not a physical substance like water, so energy transfer cannotbe measured with the same kind of meter. How can we tell, forinstance, how much useful energy a tractor can “put out” on onetank of gas?

The law of conservation of energy guarantees that all the chem-

315

a / Work is a transfer of en-ergy.

b / The tractor raises the weightover the pulley, increasing itsgravitational potential energy.

c / The tractor acceleratesthe trailer, increasing its kineticenergy.

d / The tractor pulls a plow.Energy is expended in frictionalheating of the plow and the dirt,and in breaking dirt clods andlifting dirt up to the sides of thefurrow.

ical energy in the gasoline will reappear in some form, but not nec-essarily in a form that is useful for doing farm work. Tractors, likecars, are extremely inefficient, and typically 90% of the energy theyconsume is converted directly into heat, which is carried away bythe exhaust and the air flowing over the radiator. We wish to dis-tinguish the energy that comes out directly as heat from the energythat serves to accelerate a trailer or to plow a field, so we definea technical meaning of the ordinary word “work” to express thedistinction:

definition of workWork is the amount of energy transferred into or out of asystem, not counting energy transferred by heat conduction.

self-check ABased on this definition, is work a vector, or a scalar? What are itsunits? . Answer, p. 512

The conduction of heat is to be distinguished from heating byfriction. When a hot potato heats up your hands by conduction, theenergy transfer occurs without any force, but when friction heatsyour car’s brake shoes, there is a force involved. The transfer of en-ergy with and without a force are measured by completely differentmethods, so we wish to include heat transfer by frictional heatingunder the definition of work, but not heat transfer by conduction.The definition of work could thus be restated as the amount of en-ergy transferred by forces.

Calculating work as force multiplied by distance

The examples in figures b-d show that there are many differentways in which energy can be transferred. Even so, all these exampleshave two things in common:

1. A force is involved.

2. The tractor travels some distance as it does the work.

In b, the increase in the height of the weight, ∆y, is the same asthe distance the tractor travels, which we’ll call d. For simplicity,we discuss the case where the tractor raises the weight at constantspeed, so that there is no change in the kinetic energy of the weight,and we assume that there is negligible friction in the pulley, so thatthe force the tractor applies to the rope is the same as the rope’supward force on the weight. By Newton’s first law, these forces arealso of the same magnitude as the earth’s gravitational force on theweight. The increase in the weight’s potential energy is given byF∆y, so the work done by the tractor on the weight equals Fd, theproduct of the force and the distance moved:

W = Fd .

316 Chapter 13 Work: the transfer of mechanical energy

In example c, the tractor’s force on the trailer accelerates it, increas-ing its kinetic energy. If frictional forces on the trailer are negligible,then the increase in the trailer’s kinetic energy can be found usingthe same algebra that was used on page 301 to find the potentialenergy due to gravity. Just as in example b, we have

W = Fd .

Does this equation always give the right answer? Well, sort of.In example d, there are two quantities of work you might want tocalculate: the work done by the tractor on the plow and the workdone by the plow on the dirt. These two quantities can’t both equalFd. Most of the energy transmitted through the cable goes intofrictional heating of the plow and the dirt. The work done by theplow on the dirt is less than the work done by the tractor on theplow, by an amount equal to the heat absorbed by the plow. It turnsout that the equation W = Fd gives the work done by the tractor,not the work done by the plow. How are you supposed to know whenthe equation will work and when it won’t? The somewhat complexanswer is postponed until section 13.6. Until then, we will restrictourselves to examples in which W = Fd gives the right answer;essentially the reason the ambiguities come up is that when onesurface is slipping past another, d may be hard to define, becausethe two surfaces move different distances.

e / The baseball pitcher put ki-netic energy into the ball, so hedid work on it. To do the greatestpossible amount of work, he ap-plied the greatest possible forceover the greatest possible dis-tance.

We have also been using examples in which the force is in thesame direction as the motion, and the force is constant. (If the forcewas not constant, we would have to represent it with a function, nota symbol that stands for a number.) To summarize, we have:

rule for calculating work (simplest version)The work done by a force can be calculated as

W = Fd ,

if the force is constant and in the same direction as the motion.Some ambiguities are encountered in cases such as kinetic friction.

Section 13.1 Work: the transfer of mechanical energy 317

f / Example 1.

Mechanical work done in an earthquake example 1. In 1998, geologists discovered evidence for a big prehistoricearthquake in Pasadena, between 10,000 and 15,000 years ago.They found that the two sides of the fault moved 6.7 m relativeto one another, and estimated that the force between them was1.3× 1017 N. How much energy was released?

. Multiplying the force by the distance gives 9× 1017 J. For com-parison, the Northridge earthquake of 1994, which killed 57 peo-ple and did 40 billion dollars of damage, released 22 times lessenergy.

machines can increase force, but not work.

Figure g shows a pulley arrangement for doubling the force sup-plied by the tractor (book 1, section 5.6). The tension in the left-hand rope is equal throughout, assuming negligible friction, so thereare two forces pulling the pulley to the left, each equal to the origi-nal force exerted by the tractor on the rope. This doubled force istransmitted through the right-hand rope to the stump.

g / The pulley doubles the forcethe tractor can exert on thestump.

It might seem as though this arrangement would also double thework done by the tractor, but look again. As the tractor movesforward 2 meters, 1 meter of rope comes around the pulley, and thepulley moves 1 m to the left. Although the pulley exerts double theforce on the stump, the pulley and stump only move half as far, sothe work done on the stump is no greater that it would have beenwithout the pulley.

The same is true for any mechanical arrangement that increasesor decreases force, such as the gears on a ten-speed bike. You can’tget out more work than you put in, because that would violateconservation of energy. If you shift gears so that your force on thepedals is amplified, the result is that you just have to spin the pedalsmore times.

No work is done without motion.

It strikes most students as nonsensical when they are told thatif they stand still and hold a heavy bag of cement, they are doingno work on the bag. Even if it makes sense mathematically thatW = Fd gives zero when d is zero, it seems to violate commonsense. You would certainly become tired! The solution is simple.


h / Whenever energy is trans-ferred out of the spring, the sameamount has to be transferred intothe ball, and vice versa. As thespring compresses, the ball isdoing positive work on the spring(giving up its KE and transferringenergy into the spring as PE),and as it decompresses the ballis doing negative work (extractingenergy).

Physicists have taken over the common word “work” and given it anew technical meaning, which is the transfer of energy. The energyof the bag of cement is not changing, and that is what the physicistmeans by saying no work is done on the bag.

There is a transformation of energy, but it is taking place entirelywithin your own muscles, which are converting chemical energy intoheat. Physiologically, a human muscle is not like a tree limb, whichcan support a weight indefinitely without the expenditure of energy.Each muscle cell’s contraction is generated by zillions of little molec-ular machines, which take turns supporting the tension. When aparticular molecule goes on or off duty, it moves, and since it moveswhile exerting a force, it is doing work. There is work, but it is workdone by one molecule in a muscle cell on another.

Positive and negative work

When object A transfers energy to object B, we say that A doespositive work on B. B is said to do negative work on A. In otherwords, a machine like a tractor is defined as doing positive work.This use of the plus and minus signs relates in a logical and consis-tent way to their use in indicating the directions of force and motionin one dimension. In figure h, suppose we choose a coordinate sys-tem with the x axis pointing to the right. Then the force the springexerts on the ball is always a positive number. The ball’s motion,however, changes directions. The symbol d is really just a shorterway of writing the familiar quantity ∆x, whose positive and negativesigns indicate direction.

While the ball is moving to the left, we use d < 0 to representits direction of motion, and the work done by the spring, Fd, comesout negative. This indicates that the spring is taking kinetic energyout of the ball, and accepting it in the form of its own potentialenergy.

As the ball is reaccelerated to the right, it has d > 0, Fd ispositive, and the spring does positive work on the ball. Potentialenergy is transferred out of the spring and deposited in the ball askinetic energy.

In summary:

rule for calculating work (including cases of negativework)The work done by a force can be calculated as

W = Fd ,

if the force is constant and along the same line as the motion.The quantity d is to be interpreted as a synonym for ∆x, i.e.,positive and negative signs are used to indicate the directionof motion. Some ambiguities are encountered in cases such askinetic friction.

Section 13.1 Work: the transfer of mechanical energy 319

i / Because the force is inthe opposite direction comparedto the motion, the brake shoedoes negative work on the drum,i.e., accepts energy from it in theform of heat.

self-check BIn figure h, what about the work done by the ball on the spring?. Answer, p. 512

There are many examples where the transfer of energy out of anobject cancels out the transfer of energy in. When the tractor pullsthe plow with a rope, the rope does negative work on the tractorand positive work on the plow. The total work done by the rope iszero, which makes sense, since it is not changing its energy.

It may seem that when your arms do negative work by loweringa bag of cement, the cement is not really transferring energy intoyour body. If your body was storing potential energy like a com-pressed spring, you would be able to raise and lower a weight allday, recycling the same energy. The bag of cement does transferenergy into your body, but your body accepts it as heat, not as po-tential energy. The tension in the muscles that control the speed ofthe motion also results in the conversion of chemical energy to heat,for the same physiological reasons discussed previously in the casewhere you just hold the bag still.

One of the advantages of electric cars over gasoline-powered carsis that it is just as easy to put energy back in a battery as it is totake energy out. When you step on the brakes in a gas car, thebrake shoes do negative work on the rest of the car. The kineticenergy of the car is transmitted through the brakes and acceptedby the brake shoes in the form of heat. The energy cannot berecovered. Electric cars, however, are designed to use regenerativebraking. The brakes don’t use friction at all. They are electrical,and when you step on the brake, the negative work done by thebrakes means they accept the energy and put it in the battery forlater use. This is one of the reasons why an electric car is far betterfor the environment than a gas car, even if the ultimate source ofthe electrical energy happens to be the burning of oil in the electriccompany’s plant. The electric car recycles the same energy overand over, and only dissipates heat due to air friction and rollingresistance, not braking. (The electric company’s power plant canalso be fitted with expensive pollution-reduction equipment thatwould be prohibitively expensive or bulky for a passenger car.)


k / A force can do positive,negative, or zero work, depend-ing on its direction relative to thedirection of the motion.


A Besides the presence of a force, what other things differentiate theprocesses of frictional heating and heat conduction?

B Criticize the following incorrect statement: “A force doesn’t do anywork unless it’s causing the object to move.”

C To stop your car, you must first have time to react, and then it takessome time for the car to slow down. Both of these times contribute to thedistance you will travel before you can stop. The figure shows how theaverage stopping distance increases with speed. Because the stoppingdistance increases more and more rapidly as you go faster, the rule ofone car length per 10 m.p.h. of speed is not conservative enough at highspeeds. In terms of work and kinetic energy, what is the reason for themore rapid increase at high speeds?


13.2 Work in three dimensions

A force perpendicular to the motion does no work.

Suppose work is being done to change an object’s kinetic energy.A force in the same direction as its motion will speed it up, and aforce in the opposite direction will slow it down. As we have alreadyseen, this is described as doing positive work or doing negative workon the object. All the examples discussed up until now have beenof motion in one dimension, but in three dimensions the force canbe at any angle θ with respect to the direction of motion.

What if the force is perpendicular to the direction of motion? Wehave already seen that a force perpendicular to the motion resultsin circular motion at constant speed. The kinetic energy does notchange, and we conclude that no work is done when the force isperpendicular to the motion.

So far we have been reasoning about the case of a single forceacting on an object, and changing only its kinetic energy. The resultis more generally true, however. For instance, imagine a hockey pucksliding across the ice. The ice makes an upward normal force, but

Section 13.2 Work in three dimensions 321

l / Work is only done by thecomponent of the force parallel tothe motion.

m / Self-check. (BreakingTrail, by Walter E. Bohl.)

does not transfer energy to or from the puck.

Forces at other angles

Suppose the force is at some other angle with respect to themotion, say θ = 45 . Such a force could be broken down into twocomponents, one along the direction of the motion and the otherperpendicular to it. The force vector equals the vector sum of itstwo components, and the principle of vector addition of forces thustells us that the work done by the total force cannot be any differentthan the sum of the works that would be done by the two forces bythemselves. Since the component perpendicular to the motion doesno work, the work done by the force must be

W = F‖|d| , [work done by a constant force]

where the vector d is simply a less cumbersome version of the nota-tion ∆r. This result can be rewritten via trigonometry as

W = |F||d| cos θ . [work done by a constant force]

Even though this equation has vectors in it, it depends only ontheir magnitudes, and the magnitude of a vector is a scalar. Workis therefore still a scalar quantity, which only makes sense if it isdefined as the transfer of energy. Ten gallons of gasoline have theability to do a certain amount of mechanical work, and when youpull in to a full-service gas station you don’t have to say “Fill ’er upwith 10 gallons of south-going gas.”

Students often wonder why this equation involves a cosine ratherthan a sine, or ask if it would ever be a sine. In vector addition, thetreatment of sines and cosines seemed more equal and democratic,so why is the cosine so special now? The answer is that if we aregoing to describe, say, a velocity vector, we must give both thecomponent parallel to the x axis and the component perpendicularto the x axis (i.e., the y component). In calculating work, however,the force component perpendicular to the motion is irrelevant — itchanges the direction of motion without increasing or decreasing theenergy of the object on which it acts. In this context, it is only theparallel force component that matters, so only the cosine occurs.

self-check C(a) Work is the transfer of energy. According to this definition, is thehorse in the picture doing work on the pack? (b) If you calculate workby the method described in this section, is the horse in figure m doingwork on the pack? . Answer, p. 512

Pushing a broom example 2. If you exert a force of 21 N on a push broom, at an angle 35degrees below horizontal, and walk for 5.0 m, how much work doyou do? What is the physical significance of this quantity of work?

. Using the second equation above, the work done equals

(21 N)(5.0 m)(cos 35 ) = 86 J .


The form of energy being transferred is heat in the floor and thebroom’s bristles. This comes from the chemical energy stored inyour body. (The majority of the calories you burn are dissipateddirectly as heat inside your body rather than doing any work onthe broom. The 86 J is only the amount of energy transferredthrough the broom’s handle.)

A violin example 3As a violinist draws the bow across a string, the bow hairs exertboth a normal force and a kinetic frictional force on the string. Thenormal force is perpendicular to the direction of motion, and doesno work. However, the frictional force is in the same direction asthe motion of the bow, so it does work: energy is transferred tothe string, causing it to vibrate.

One way of playing a violin more loudly is to use longer strokes.Since W = Fd , the greater distance results in more work.

A second way of getting a louder sound is to press the bow morefirmly against the strings. This increases the normal force, andalthough the normal force itself does no work, an increase in thenormal force has the side effect of increasing the frictional force,thereby increasing W = Fd .

The violinist moves the bow back and forth, and sound is pro-duced on both the “up-bow” (the stroke toward the player’s left)and the “down-bow” (to the right). One may, for example, play aseries of notes in alternation between up-bows and down-bows.However, if the notes are of unequal length, the up and down mo-tions tend to be unequal, and if the player is not careful, she canrun out of bow in the middle of a note! To keep this from hap-pening, one can move the bow more quickly on the shorter notes,but the resulting increase in d will make the shorter notes louderthan they should be. A skilled player compensates by reducingthe force.

13.3 The dot productUp until now, we have not found any physically useful way to definethe multiplication of two vectors. It would be possible, for instance,to multiply two vectors component by component to form a thirdvector, but there are no physical situations where such a multipli-cation would be useful.

The equation W = |F||d| cos θ is an example of a sort of mul-tiplication of vectors that is useful. The result is a scalar, not avector, and this is therefore often referred to as the scalar productof the vectors F and d. There is a standard shorthand notation for

Section 13.3 The dot product 323

this operation,

A ·B = |A||B| cos θ , [definition of the notation A ·B;

θ is the angle between vectors A and B]

and because of this notation, a more common term for this operationis the dot product. In dot product notation, the equation for workis simply

W = F · d .

The dot product has the following geometric interpretation:

A ·B = |A|(component of B parallel to A)

= |B|(component of A parallel to B)

The dot product has some of the properties possessed by ordinarymultiplication of numbers,

A ·B = B ·AA · (B + C) = A ·B + A ·C

(cA) ·B = c (A ·B) ,

but it lacks one other: the ability to undo multiplication by dividing.

If you know the components of two vectors, you can easily cal-culate their dot product as follows:

A ·B = AxBx +AyBy +AzBz .

(This can be proved by first analyzing the special case where eachvector has only an x component, and the similar cases for y and z.We can then use the rule A · (B + C) = A · B + A · C to make ageneralization by writing each vector as the sum of its x, y, and zcomponents. See homework problem 23.)

Magnitude expressed with a dot product example 4If we take the dot product of any vector b with itself, we find

b · b =(bx x + by y + bz z

)·(bx x + by y + bz z

)= b2

x + b2y + b2

z ,

so its magnitude can be expressed as

|b| =√

b · b .

We will often write b2 to mean b · b, when the context makesit clear what is intended. For example, we could express kineticenergy as (1/2)m|v|2, (1/2)mv·v, or (1/2)mv2. In the third version,nothing but context tells us that v really stands for the magnitudeof some vector v.


n / The spring does work onthe cart. (Unlike the ball insection 13.1, the cart is attachedto the spring.)

Towing a barge example 5. A mule pulls a barge with a force F=(1100 N)x + (400 N)y, andthe total distance it travels is (1000 m)x. How much work does itdo?

. The dot product is 1.1× 106 N·m = 1.1× 106 J.

13.4 Varying forceUp until now we have done no actual calculations of work in caseswhere the force was not constant. The question of how to treatsuch cases is mathematically analogous to the issue of how to gener-alize the equation (distance) = (velocity)(time) to cases where thevelocity was not constant. We have to make the equation into anintegral:

W =

∫Fdx

The examples in this section are ones in which the force is varying,but is always along the same line as the motion.

self-check DIn which of the following examples would it be OK to calculate workusing Fd , and in which ones would you have to integrate?

(a) A fishing boat cruises with a net dragging behind it.

(b) A magnet leaps onto a refrigerator from a distance.

(c) Earth’s gravity does work on an outward-bound space probe. .

Answer, p. 512

Work done by a spring example 6An important and straightforward example is the calculation of thework done by a spring that obeys Hooke’s law,

F ≈ −k (x − xo) ,

where xo is the equilibrium position and the minus sign is becausethis is the force being exerted by the spring, not the force thatwould have to act on the spring to keep it at this position. That is,if the position of the cart in figure n is to the right of equilibrium,the spring pulls back to the left, and vice-versa. Integrating, wefind that the work done between x1 and x2 is

W = −12

k (x − xo)2∣∣∣∣x2

x1

.

Work done by gravity example 7Another important example is the work done by gravity when thechange in height is not small enough to assume a constant force.Newton’s law of gravity is

F =GMm

r2 ,

Section 13.4 Varying force 325

which can be integrated to give

W =∫ r2

r1

GMmr2 dr

= GMm(

1r2− 1

r1

).

13.5 Work and potential energyThe techniques for calculating work can also be applied to the cal-culation of potential energy. If a certain force depends only onthe distance between the two participating objects, then the energyreleased by changing the distance between them is defined as the po-tential energy, and the amount of potential energy lost equals minusthe work done by the force,

∆PE = −W .

The minus sign occurs because positive work indicates that the po-tential energy is being expended and converted to some other form.

It is sometimes convenient to pick some arbitrary position as areference position, and derive an equation for once and for all thatgives the potential energy relative to this position

PEx = −Wref→x . [potential energy at a point x]

To find the energy transferred into or out of potential energy, onethen subtracts two different values of this equation.

These equations might almost make it look as though work andenergy were the same thing, but they are not. First, potential energymeasures the energy that a system has stored in it, while workmeasures how much energy is transferred in or out. Second, thetechniques for calculating work can be used to find the amount ofenergy transferred in many situations where there is no potentialenergy involved, as when we calculate the amount of kinetic energytransformed into heat by a car’s brake shoes.

A toy gun example 8. A toy gun uses a spring with a spring constant of 10 N/m toshoot a ping-pong ball of mass 5 g. The spring is compressed to10 cm shorter than its equilibrium length when the gun is loaded.At what speed is the ball released?

. The equilibrium point is the natural choice for a reference point.Using the equation found previously for the work, we have

PEx =12

k (x − xo)2 .


The spring loses contact with the ball at the equilibrium point, sothe final potential energy is

PEf = 0 .

The initial potential energy is

PEi =12

(10 N/m)(0.10 m)2 .

= 0.05 J.

The loss in potential energy of 0.05 J means an increase in kineticenergy of the same amount. The velocity of the ball is found bysolving the equation K E = (1/2)mv2 for v ,

v =

√2K E

m

=

√(2)(0.05 J)0.005 kg

= 4 m/s .

Gravitational potential energy example 9. We have already found the equation ∆PE = −F∆y for the gravi-tational potential energy when the change in height is not enoughto cause a significant change in the gravitational force F . Whatif the change in height is enough so that this assumption is nolonger valid? Use the equation W = GMm(1/r2− 1/r1) derived inexample 7 to find the potential energy, using r =∞ as a referencepoint.

. The potential energy equals minus the work that would have tobe done to bring the object from r1 =∞ to r = r2, which is

PE = −GMmr

.

This is simpler than the equation for the work, which is an exam-ple of why it is advantageous to record an equation for potentialenergy relative to some reference point, rather than an equationfor work.

Although the equations derived in the previous two examplesmay seem arcane and not particularly useful except for toy design-ers and rocket scientists, their usefulness is actually greater thanit appears. The equation for the potential energy of a spring canbe adapted to any other case in which an object is compressed,stretched, twisted, or bent. While you are not likely to use theequation for gravitational potential energy for anything practical, itis directly analogous to an equation that is extremely useful in chem-istry, which is the equation for the potential energy of an electron

Section 13.5 Work and potential energy 327

at a distance r from the nucleus of its atom. As discussed in moredetail later in the course, the electrical force between the electronand the nucleus is proportional to 1/r2, just like the gravitationalforce between two masses. Since the equation for the force is of thesame form, so is the equation for the potential energy.

o / The twin Voyager spaceprobes were perhaps the great-est scientific successes of thespace program. Over a periodof decades, they flew by all theplanets of the outer solar system,probably accomplishing moreof scientific interest than theentire space shuttle program ata tiny fraction of the cost. BothVoyager probes completed theirfinal planetary flybys with speedsgreater than the escape velocityat that distance from the sun, andso headed on out of the solar sys-tem on hyperbolic orbits, neverto return. Radio contact hasbeen lost, and they are now likelyto travel interstellar space forbillions of years without collidingwith anything or being detectedby any intelligent species.


A What does the graph of PE = (1/2)k (x − xo)2 look like as a functionof x? Discuss the physical significance of its features.

B What does the graph of PE = −GMm/r look like as a function of r?Discuss the physical significance of its features. How would the equationand graph change if some other reference point was chosen rather thanr =∞?

C Starting at a distance r from a planet of mass M, how fast must anobject be moving in order to have a hyperbolic orbit, i.e., one that nevercomes back to the planet? This velocity is called the escape velocity. In-terpreting the result, does it matter in what direction the velocity is? Doesit matter what mass the object has? Does the object escape because it ismoving too fast for gravity to act on it?

D Does a spring have an “escape velocity?”

E If the form of energy being transferred is potential energy, thenthe equations F = dW/dx and W =

∫Fdx become F = −dPE/dx and

PE = −∫

Fdx . How would you then apply the following calculus con-cepts: zero derivative at minima and maxima, and the second derivativetest for concavity up or down.


13.6 ? When does work equal force timesdistance?

In the example of the tractor pulling the plow discussed on page317, the work did not equal Fd. The purpose of this section is toexplain more fully how the quantity Fd can and cannot be used.To simplify things, I write Fd throughout this section, but moregenerally everything said here would be true for the area under thegraph of F‖ versus d.

The following two theorems allow most of the ambiguity to becleared up.

the work-kinetic-energy theoremThe change in kinetic energy associated with the motion of anobject’s center of mass is related to the total force acting onit and to the distance traveled by its center of mass accordingto the equation ∆KEcm = Ftotaldcm.

This can be proved based on Newton’s second law and the equa-tion KE = (1/2)mv2. Note that despite the traditional name, itdoes not necessarily tell the amount of work done, since the forcesacting on the object could be changing other types of energy besidesthe KE associated with its center of mass motion.

The second theorem does relate directly to work:

When a contact force acts between two objects and the twosurfaces do not slip past each other, the work done equals Fd,where d is the distance traveled by the point of contact.

This one has no generally accepted name, so we refer to it simplyas the second theorem.

A great number of physical situations can be analyzed with thesetwo theorems, and often it is advantageous to apply both of themto the same situation.

An ice skater pushing off from a wall example 10The work-kinetic energy theorem tells us how to calculate theskater’s kinetic energy if we know the amount of force and thedistance her center of mass travels while she is pushing off.

The second theorem tells us that the wall does no work on theskater. This makes sense, since the wall does not have anysource of energy.

Absorbing an impact without recoiling? example 11. Is it possible to absorb an impact without recoiling? For in-stance, would a brick wall “give” at all if hit by a ping-pong ball?

. There will always be a recoil. In the example proposed, the wallwill surely have some energy transferred to it in the form of heatand vibration. The second theorem tells us that we can only have

Section 13.6 ? When does work equal force times distance? 329

nonzero work if the distance traveled by the point of contact isnonzero.

Dragging a refrigerator at constant velocity example 12Newton’s first law tells us that the total force on the refrigeratormust be zero: your force is canceling the floor’s kinetic frictionalforce. The work-kinetic energy theorem is therefore true but use-less. It tells us that there is zero total force on the refrigerator,and that the refrigerator’s kinetic energy doesn’t change.

The second theorem tells us that the work you do equals yourhand’s force on the refrigerator multiplied by the distance traveled.Since we know the floor has no source of energy, the only way forthe floor and refrigerator to gain energy is from the work you do.We can thus calculate the total heat dissipated by friction in therefrigerator and the floor.

Note that there is no way to find how much of the heat is dissi-pated in the floor and how much in the refrigerator.

Accelerating a cart example 13If you push on a cart and accelerate it, there are two forces actingon the cart: your hand’s force, and the static frictional force of theground pushing on the wheels in the opposite direction.

Applying the second theorem to your force tells us how to calcu-late the work you do.

Applying the second theorem to the floor’s force tells us that thefloor does no work on the cart. There is no motion at the pointof contact, because the atoms in the floor are not moving. (Theatoms in the surface of the wheel are also momentarily at restwhen they touch the floor.) This makes sense, since the floordoes not have any source of energy.

The work-kinetic energy theorem refers to the total force, and be-cause the floor’s backward force cancels part of your force, thetotal force is less than your force. This tells us that only part ofyour work goes into the kinetic energy associated with the forwardmotion of the cart’s center of mass. The rest goes into rotation ofthe wheels.

Up until now, we have not found any physically useful way todefine the multiplication of two vectors. It would be possible, forinstance, to multiply two vectors component by component to forma third vector, but there are no physical situations where such amultiplication would be useful.

The equation W = |F||d| cos θ is an example of a sort of mul-tiplication of vectors that is useful. The result is a scalar, not avector, and this is therefore often referred to as the scalar productof the vectors F and d. There is a standard shorthand notation for


this operation,

A ·B = |A||B| cos θ , [definition of the notation A ·B;

θ is the angle between vectors A and B]

and because of this notation, a more common term for this operationis the dot product. In dot product notation, the equation for workis simply

W = F · d .

The dot product has the following geometric interpretation:

A ·B = |A|(component of B parallel to A)

= |B|(component of A parallel to B)

The dot product has some of the properties possessed by ordinarymultiplication of numbers,

A ·B = B ·AA · (B + C) = A ·B + A ·C

(cA) ·B = c (A ·B) ,

but it lacks one other: the ability to undo multiplication by dividing.

If you know the components of two vectors, you can easily cal-culate their dot product as follows:

A ·B = AxBx +AyBy +AzBz .

(This can be proved by first analyzing the special case where eachvector has only an x component, and the similar cases for y and z.We can then use the rule A · (B + C) = A · B + A · C to make ageneralization by writing each vector as the sum of its x, y, and zcomponents. See homework problem 23.)

Magnitude expressed with a dot product example 14If we take the dot product of any vector b with itself, we find

b · b =(bx x + by y + bz z

)·(bx x + by y + bz z

)= b2

x + b2y + b2

z ,

so its magnitude can be expressed as

|b| =√

b · b .

We will often write b2 to mean b · b, when the context makesit clear what is intended. For example, we could express kineticenergy as (1/2)m|v|2, (1/2)mv·v, or (1/2)mv2. In the third version,nothing but context tells us that v really stands for the magnitudeof some vector v.

Section 13.6 ? When does work equal force times distance? 331

Towing a barge example 15. A mule pulls a barge with a force F=(1100 N)x + (400 N)y, andthe total distance it travels is (1000 m)x. How much work does itdo?

. The dot product is 1.1× 106 N·m = 1.1× 106 J.

13.7 ? Uniqueness of the dot productIn this section I prove that the vector dot product is unique, inthe sense that there is no other possible way to define it that isconsistent with rotational invariance and that reduces appropriatelyto ordinary multiplication in one dimension.

Suppose we want to find some way to multiply two vectors to geta scalar, and we don’t know how this operation should be defined?Let’s consider what we would get by performing this operation onvarious combinations of the unit vectors x, y, and z. Rotationalinvariance requires that we handle the three coordinate axes in thesame way, without giving special treatment to any of them, so wemust have x · x = y · y = z · z and x · y = y · z = z · x. This issupposed to be a way of generalizing ordinary multiplication, so forconsistency with the property 1 × 1 = 1 of ordinary numbers, theresult of multiplying a magnitude-one vector by itself had better bethe scalar 1, so x · x = y · y = z · z = 1. Furthermore, there is no wayto satisfy rotational invariance unless we define the mixed productsto be zero, x · y = y · z = z · x = 0; for example, a 90-degree rotationof our frame of reference about the z axis reverses the sign of x · y,but rotational invariance requires that x · y produce the same resulteither way, and zero is the only number that stays the same whenwe reverse its sign. Establishing these six products of unit vectorssuffices to define the operation in general, since any two vectorsthat we want to multiply can be broken down into components, e.g.,(2x+3z) · z = 2x · z+3z · z = 0+3 = 3. Thus by requiring rotationalinvariance and consistency with multiplication of ordinary numbers,we find that there is only one possible way to define a multiplicationoperation on two vectors that gives a scalar as the result. (Thereis, however, a different operation, discussed in chapter 15, whichmultiplies two vectors to give a vector.)

13.8 ? A dot product for relativity?In section 13.7 I showed that the dot product is the only physi-cally sensible way to multiply two vectors to get a scalar. This isessentially because the outcome of experiments shouldn’t dependon which way we rotate the laboratory. Dot products relate to thelengths of vectors and the angles between them, and rotations don’tchange lengths or angles.

Let’s consider how this would apply to relativity. Relativity


tells us that the length of a measuring rod is not absolute. Rotatingthe lab won’t change its length, but changing the lab’s state ofmotion will. The rod’s length is greatest in the frame that is atrest relative to the rod. This suggests that relativity requires somenew variation on the dot product: some slightly different way ofmultiplying two vectors to find a number that doesn’t depend onthe frame of reference.

Clock time

We do know of a number that stays the same in all frames of ref-erence. On p. 89 we proved that the Lorentz transformation doesn’tchange the area of a shape in the x-t plane. We used this only as astepping stone toward the Lorentz transformation, but it is naturalto wonder whether this kind of area has any physical interest of itsown.

The equal-area result is not relativistic, since the proof neverappeals to property 5 on page 79. Cases I and II on page 81 alsohave the equal-area property. We can see this clearly in a Galileantransformation like figure af on p. 79, where the distortion of therectangle could be accomplished by cutting it into vertical slices andthen displacing the slices upward without changing their areas.

But the area does have a nice interpretation in the relativisticcase. Suppose that we have events A (Charles VII is restored tothe throne) and B (Joan of Arc is executed). Now imagine thattechnologically advanced aliens want to be present at both A andB, but in the interim they wish to fly away in their spaceship, bepresent at some other event P (perhaps a news conference at whichthey give an update on the events taking place on earth), but getback in time for B. Since nothing can go faster than c (which wetake to equal 1 in appropriate units), P cannot be too far away. Theset of all possible events P forms a rectangle, figure p/1, in the x− tplane that has A and B at opposite corners and whose edges haveslopes equal to ±1. We call this type of rectangle a light-rectangle,because its sides could represent the motion of rays of light.

p / 1. The gray light-rectangle rep-resents the set of all events suchas P that could be visited after Aand before B.2. The rectangle becomes asquare in the frame in which Aand B occur at the same locationin space.3. The area of the dashed squareis τ2, so the area of the graysquare is τ2/2.The area of this rectangle will be the same regardless of one’s

frame of reference. In particular, we could choose a special frameof reference, panel 2 of the figure, such that A and B occur in the

Section 13.8 ? A dot product for relativity? 333

same place. (They do not occur at the same place, for example,in the sun’s frame, because the earth is spinning and going aroundthe sun.) Since the speed c, which equals 1 in our units, is thesame in all frames of reference, and the sides of the rectangle hadslopes ±1 in frame 1, they must still have slopes ±1 in frame 2. Therectangle becomes a square with its diagonals parallel to the x andt axes, and the length of these diagonals equals the time τ elapsedon a clock that is at rest in frame 2, i.e., a clock that glides throughspace at constant velocity from A to B, meeting up with the planetearth at the appointed time. As shown in panel 3 of the figure, thearea of the gray regions can be interpreted as half the square of thisgliding-clock time.

If events A and B are separated by a distance x and a time t,then in general t2 − x2 gives the square of the gliding-clock time.Proof: Based on units, the expression must have the form (. . .)t2 +(. . .)tx+(. . .)x2, where each (. . .) represents a unitless constant. Thetx coefficient must be zero by property 2 on p. 79. For consistencywith figure p/3, the t2 coefficient must equal 1. Since the areavanishes for x = t, the x2 coefficient must equal −1.

When |x| is greater than |t|, events A and are so far apart inspace and so close together in time that it would be impossible tohave a cause and effect relationship between them, since c = 1 isthe maximum speed of cause and effect. In this situation t2 − x2

is negative and cannot be interpreted as a clock time, but it canbe interpreted as minus the square of the distance bewteen A andB as measured by rulers at rest in a frame in which A and B aresimultaneous.

Four-vectors

No matter what, t2 − x2 is the same as measured in all framesof reference. Geometrically, it plays the same role in the x-t planethat ruler measurements play in the Euclidean plane. In Euclideangeometry, the ruler-distance between any two points stays the sameregardless of rotation, i.e., regardless of the angle from which weview the scene; according to the Pythagorean theorem, the squareof this distance is x2 + y2. In the x-t plane, t2 − x2 stays the sameregardless of the frame of reference. This suggests that by analogywith the dot product

x1x2 + y1y2

in the Euclidean x-y plane, we define a similar operation in the x-tplane,

t1t2 − x1x2 .

Putting in the other two spatial dimensions, we have

t1t2 − x1x2 − y1y2 − z1z2 .

A mathematical object like (t,x, y, z) is referred to as a four-vector,as opposed to a three-vector like (x, y, z). The term “dot product”


has connotations of referring only to three-vectors, so the operationof taking the scalar product of two four-vectors is usually referred toinstead as the “inner product.” There are various ways of notatingthe inner product of vectors a and b, such as a · b or < a, b >.

The magnitude of a three-vector is defined by taking the squareroot of its dot product with itself, and this square root is alwaysa real number, because a vector’s dot product with itself is alwayspositive. But the inner product of a four-vector with itself can bepositive, zero, or negative, and in these cases the vector is referred toas timelike, lightlike, spacelike, respectively. Since material objectscan never go as fast as c, the vector (∆t, ∆x, ∆y, ∆z) describing anobject’s motion from one event to another is always timelike.

The twin paradox example 16One of the classic paradoxes of relativity, known as the twin para-dox, is usually stated something like this. Alice and Betty are iden-tical twins. Betty goes on a space voyage at relativistic speeds,traveling away from the earth and then turning around and com-ing back. Meanwhile, Alice stays on earth. When Betty returns,she is younger than Alice because of relativistic time dilation. Butisn’t it valid to say that Betty’s spaceship is standing still and theearth moving? In that description, wouldn’t Alice end up youngerand Betty older?

The most common way of explaining the non-paradoxical natureof this paradox is that although special relativity says that inertialmotion is relative, it doesn’t say that noninertial motion is relative.In this respect it is the same as Newtonian mechanics. Betty ex-periences accelerations on her voyage, but Alice doesn’t. There-fore there is no doubt about who actually went on the trip and whodidn’t.

This resolution, however, may not be entirely satisfying because itmakes it sound as if relativistic time dilation is not occurring whileBetty’s ship cruises at constant velocity, but only while the ship isspeeding up or slowing down. This would appear to contradict ourearlier interpretation of relativistic time dilation, which was thata clock runs fastest according to an observer at rest relative tothe clock. Furthermore, if it’s acceleration that causes the effect,should we be looking for some new formula that computes timedilation based on acceleration?

The first thing to realize is that there is no unambiguous way todecide during which part of Betty’s journey the time dilation is oc-curring. To do this, we could need to be able to compare Alice andBetty’s clocks many times over the course of the trip. But eachtwin has no way of finding out what her sister’s clock reads “now,”except by exchanging radio signals, which travel at the speed oflight. The speed-of-light lag vanishes only at the beginning andend of the trip, when the twins are in the same place.

Section 13.8 ? A dot product for relativity? 335

Furthermore, we can use the inner product to show that the ac-cumulated difference in clock time doesn’t depend on the detailsof how Betty carries out her accelerations and decelerations. Infact, we can get the right answer simply by assuming that thesechanges in velocities occur instantaneously.

In Euclidean geometry, the triangle inequality |b + c| < |b| + |c|follows from

(|b| + |c|)2 − (b + c) · (b + c) = 2(|b||c| − b · c) ≥ 0 .

The reason this quantity always comes out positive is that for twovectors of fixed magnitude, the greatest dot product is alwaysachieved in the case where they lie along the same direction.

In the geometry of the x-t plane, the situation is different. Sup-pose that b and c are timelike vectors, so that they representpossible (∆t ,∆x , . . .) vectors for Betty on the outward and returnlegs of her trip. Then a = b+c describes the vector for Alice’s mo-tion. Alice goes by a direct route through the x-t plane while Bettytakes a detour. The magnitude of each timelike vector representsthe time elapsed on a clock carried by that twin. The triangleequality is now reversed, becoming |b + c| > |b| + |c|. The differ-ence from the Euclidean case arises because inner products areno longer necessarily maximized if vectors are in the same direc-tion. E.g., for two lightlike vectors, b · c vanishes entirely if b andc are parallel. For timelike vectors, parallelism actually minimizesthe inner product rather than maximizing it.1

1Proof: Let b and c be parallel and timelike, and directed forward in time.Adopt a frame of reference in which every spatial component of each vectorvanishes. This entails no loss of generality, since inner products are invariantunder such a transformation. Now let b and c be pulled away from parallelism,like opening a pair of scissors in the x− t plane. This reduces btct , while causingbx cx to become negative. Both effects increase the inner product.


SummarySelected vocabularywork . . . . . . . . the amount of energy transferred into or out

of a system, excluding energy transferred byheat conduction

NotationW . . . . . . . . . work

Summary

Work is a measure of the transfer of mechanical energy, i.e., thetransfer of energy by a force rather than by heat conduction. Whenthe force is constant, work can usually be calculated as

W = F‖|d| , [only if the force is constant]

where d is simply a less cumbersome notation for ∆r, the vectorfrom the initial position to the final position. Thus,

• A force in the same direction as the motion does positive work,i.e., transfers energy into the object on which it acts.

• A force in the opposite direction compared to the motion doesnegative work, i.e., transfers energy out of the object on whichit acts.

• When there is no motion, no mechanical work is done. Thehuman body burns calories when it exerts a force withoutmoving, but this is an internal energy transfer of energy withinthe body, and thus does not fall within the scientific definitionof work.

• A force perpendicular to the motion does no work.

When the force is not constant, the above equation should be gen-eralized as an integral,

∫F‖ dx.

There is only one meaningful (rotationally invariant) way ofdefining a multiplication of vectors whose result is a scalar, andit is known as the vector dot product:

b · c = bxcx + bycy + bzcz

= |b| |c| cos θbc .

The dot product has most of the usual properties associated withmultiplication, except that there is no “dot division.” The dot prod-uct can be used to compute mechanical work as W = F · d.

Machines such as pulleys, levers, and gears may increase or de-crease a force, but they can never increase or decrease the amountof work done. That would violate conservation of energy unless the

Summary 337

machine had some source of stored energy or some way to acceptand store up energy.

There are some situations in which the equation W = F‖ |d| isambiguous or not true, and these issues are discussed rigorously insection 13.6. However, problems can usually be avoided by analyzingthe types of energy being transferred before plunging into the math.In any case there is no substitute for a physical understanding ofthe processes involved.

The techniques developed for calculating work can also be ap-plied to the calculation of potential energy. We fix some positionas a reference position, and calculate the potential energy for someother position, x, as

PEx = −Wref→x .

The following two equations for potential energy have broadersignificance than might be suspected based on the limited situationsin which they were derived:

PE =1

2k (x− xo)2 .

[potential energy of a spring having spring constantk, when stretched or compressed from the equilibriumposition xo; analogous equations apply for the twisting,bending, compression, or stretching of any object.]

PE = −GMm

r

[gravitational potential energy of objects of masses Mand m, separated by a distance r; an analogous equationapplies to the electrical potential energy of an electronin an atom.]


ProblemsKey√



1 Two speedboats are identical, but one has more people aboardthan the other. Although the total masses of the two boats areunequal, suppose that they happen to have the same kinetic energy.In a boat, as in a car, it’s important to be able to stop in time toavoid hitting things. (a) If the frictional force from the water is thesame in both cases, how will the boats’ stopping distances compare?Explain. (b) Compare the times required for the boats to stop.

2 In each of the following situations, is the work being donepositive, negative, or zero? (a) a bull paws the ground; (b) a fishingboat pulls a net through the water behind it; (c) the water resiststhe motion of the net through it; (d) you stand behind a pickuptruck and lower a bale of hay from the truck’s bed to the ground.Explain. [Based on a problem by Serway and Faughn.]

3 (a) Suppose work is done in one-dimensional motion. Whathappens to the work if you reverse the direction of the positivecoordinate axis? Base your answer directly on the definition of work.(b) Now answer the question based on the W = Fd rule.

4 Does it make sense to say that work is conserved?. Solution, p. 506

5 A microwave oven works by twisting molecules one way andthen the other, counterclockwise and then clockwise about their owncenters, millions of times a second. If you put an ice cube or a stickof butter in a microwave, you’ll observe that the oven doesn’t heatthe solid very quickly, although eventually melting begins in onesmall spot. Once a melted spot forms, it grows rapidly, while therest of the solid remains solid. In other words, it appears based onthis experiment that a microwave oven heats a liquid much morerapidly than a solid. Explain why this should happen, based on theatomic-level description of heat, solids, and liquids. (See, e.g., figureb on page 299.)

Please don’t repeat the following common mistakes in your expla-nation:

In a solid, the atoms are packed more tightly and have lessspace between them. Not true. Ice floats because it’s lessdense than water.

In a liquid, the atoms are moving much faster. No, the differ-ence in average speed between ice at −1 C and water at 1 Cis only 0.4%.

Problems 339

Problem 6.

Problem 7: A cylinder fromthe 1965 Rambler’s engine. Thepiston is shown in its pushed outposition. The two bulges at thetop are for the valves that let freshair-gas mixture in. Based on afigure from Motor Service’s Au-tomotive Encyclopedia, Toboldtand Purvis.

6 Most modern bow hunters in the U.S. use a fancy mechanicalbow called a compound bow, which looks nothing like what mostpeople imagine when they think of a bow and arrow. It has a systemof pulleys designed to produce the force curve shown in the figure,where F is the force required to pull the string back, and x is thedistance between the string and the center of the bow’s body. It isnot a linear Hooke’s-law graph, as it would be for an old-fashionedbow. The big advantage of the design is that relatively little forceis required to hold the bow stretched to point B on the graph. Thisis the force required from the hunter in order to hold the bow readywhile waiting for a shot. Since it may be necessary to wait a longtime, this force can’t be too big. An old-fashioned bow, designedto require the same amount of force when fully drawn, would shootarrows at much lower speeds, since its graph would be a straight linefrom A to B. For the graph shown in the figure (taken from realisticdata), find the speed at which a 26 g arrow is released, assuming that70% of the mechanical work done by the hand is actually transmittedto the arrow. (The other 30% is lost to frictional heating inside thebow and kinetic energy of the recoiling and vibrating bow.)

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7 In the power stroke of a car’s gasoline engine, the fuel-air mix-ture is ignited by the spark plug, explodes, and pushes the pistonout. The exploding mixture’s force on the piston head is greatestat the beginning of the explosion, and decreases as the mixture ex-pands. It can be approximated by F = a/x, where x is the distancefrom the cylinder to the piston head, and a is a constant with unitsof N.m. (Actually a/x1.4 would be more accurate, but the problemworks out more nicely with a/x!) The piston begins its stroke atx = x1, and ends at x = x2. The 1965 Rambler had six cylinders,each with a = 220 N·m, x1 = 1.2 cm, and x2 = 10.2 cm.(a) Draw a neat, accurate graph of F vs x, on graph paper.(b) From the area under the curve, derive the amount of work donein one stroke by one cylinder.

√

(c) Assume the engine is running at 4800 r.p.m., so that duringone minute, each of the six cylinders performs 2400 power strokes.(Power strokes only happen every other revolution.) Find the en-gine’s power, in units of horsepower (1 hp=746 W).

√

(d) The compression ratio of an engine is defined as x2/x1. Explainin words why the car’s power would be exactly the same if x1 andx2 were, say, halved or tripled, maintaining the same compressionratio of 8.5. Explain why this would not quite be true with the morerealistic force equation F = a/x1.4.


Problem 8.

8 The figure, redrawn from Gray’s Anatomy, shows the tensionof which a muscle is capable. The variable x is defined as the con-traction of the muscle from its maximum length L, so that at x = 0the muscle has length L, and at x = L the muscle would theoreti-cally have zero length. In reality, the muscle can only contract tox = cL, where c is less than 1. When the muscle is extended to itsmaximum length, at x = 0, it is capable of the greatest tension, To.As the muscle contracts, however, it becomes weaker. Gray suggestsapproximating this function as a linear decrease, which would theo-retically extrapolate to zero at x = L. (a) Find the maximum workthe muscle can do in one contraction, in terms of c, L, and To.

√

(b) Show that your answer to part a has the right units.(c) Show that your answer to part a has the right behavior whenc = 0 and when c = 1.(d) Gray also states that the absolute maximum tension To hasbeen found to be approximately proportional to the muscle’s cross-sectional area A (which is presumably measured at x = 0), withproportionality constant k. Approximating the muscle as a cylin-der, show that your answer from part a can be reexpressed in termsof the volume, V , eliminating L and A.

√

(e) Evaluate your result numerically for a biceps muscle with a vol-ume of 200 cm3, with c = 0.8 and k = 100 N/cm2 as estimated byGray.

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9 In the earth’s atmosphere, the molecules are constantly movingaround. Because temperature is a measure of kinetic energy permolecule, the average kinetic energy of each type of molecule is thesame, e.g., the average KE of the O2 molecules is the same as theaverage KE of the N2 molecules. (a) If the mass of an O2 moleculeis eight times greater than that of a He atom, what is the ratio oftheir average speeds? Which way is the ratio, i.e., which is typicallymoving faster? (b) Use your result from part a to explain why anyhelium occurring naturally in the atmosphere has long since escapedinto outer space, never to return. (Helium is obtained commerciallyby extracting it from rocks.) You may want to do problem 12 first,for insight.

10 Weiping lifts a rock with a weight of 1.0 N through a heightof 1.0 m, and then lowers it back down to the starting point. Bubbapushes a table 1.0 m across the floor at constant speed, requiringa force of 1.0 N, and then pushes it back to where it started. (a)Compare the total work done by Weiping and Bubba. (b) Checkthat your answers to part a make sense, using the definition of work:work is the transfer of energy. In your answer, you’ll need to discusswhat specific type of energy is involved in each case.

Problems 341

11 In one of his more flamboyant moments, Galileo wrote “Whodoes not know that a horse falling from a height of three or fourcubits will break his bones, while a dog falling from the same heightor a cat from a height of eight or ten cubits will suffer no injury?Equally harmless would be the fall of a grasshopper from a tower orthe fall of an ant from the distance of the moon.” Find the speedof an ant that falls to earth from the distance of the moon at themoment when it is about to enter the atmosphere. Assume it isreleased from a point that is not actually near the moon, so themoon’s gravity is negligible.

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12 Starting at a distance r from a planet of mass M , how fastmust an object be moving in order to have a hyperbolic orbit, i.e.,one that never comes back to the planet? This velocity is calledthe escape velocity. Interpreting the result, does it matter in whatdirection the velocity is? Does it matter what mass the object has?Does the object escape because it is moving too fast for gravity toact on it?

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13 A projectile is moving directly away from a planet of massM at exactly escape velocity. (a) Find r, the distance from theprojectile to the center of the planet, as a function of time, t, andalso find v(t).(b) Check the units of your answer.(c) Does v show the correct behavior as t approaches infinity?

. Hint, p. 494

14 A car starts from rest at t = 0, and starts speeding up withconstant acceleration. (a) Find the car’s kinetic energy in terms ofits mass, m, acceleration, a, and the time, t. (b) Your answer inthe previous part also equals the amount of work, W , done fromt = 0 until time t. Take the derivative of the previous expressionto find the power expended by the car at time t. (c) Suppose twocars with the same mass both start from rest at the same time, butone has twice as much acceleration as the other. At any moment,how many times more power is being dissipated by the more quicklyaccelerating car? (The answer is not 2.)

√


15 A car accelerates from rest. At low speeds, its accelerationis limited by static friction, so that if we press too hard on thegas, we will “burn rubber” (or, for many newer cars, a computer-ized traction-control system will override the gas pedal). At higherspeeds, the limit on acceleration comes from the power of the engine,which puts a limit on how fast kinetic energy can be developed.(a) Show that if a force F is applied to an object moving at speedv, the power required is given by P = vF .(b) Find the speed v at which we cross over from the first regime de-scribed above to the second. At speeds higher than this, the enginedoes not have enough power to burn rubber. Express your resultin terms of the car’s power P , its mass m, the coefficient of staticfriction µs, and g.

√

(c) Show that your answer to part b has units that make sense.(d) Show that the dependence of your answer on each of the fourvariables makes sense physically.(e) The 2010 Maserati Gran Turismo Convertible has a maximumpower of 3.23 × 105 W (433 horsepower) and a mass (including a50-kg driver) of 2.03 × 103 kg. (This power is the maximum theengine can supply at its optimum frequency of 7600 r.p.m. Presum-ably the automatic transmission is designed so a gear is available inwhich the engine will be running at very nearly this frequency whenthe car is at moving at v.) Rubber on asphalt has µs ≈ 0.9. Find vfor this car. Answer: 18 m/s, or about 40 miles per hour.(f) Our analysis has neglected air friction, which can probably beapproximated as a force proportional to v2. The existence of thisforce is the reason that the car has a maximum speed, which is 176miles per hour. To get a feeling for how good an approximationit is to ignore air friction, find what fraction of the engine’s maxi-mum power is being used to overcome air resistance when the car ismoving at the speed v found in part e. Answer: 1%

16 In 1935, Yukawa proposed an early theory of the force thatheld the neutrons and protons together in the nucleus. His equa-tion for the potential energy of two such particles, at a center-to-center distance r, was PE(r) = gr−1e−r/a, where g parametrizes thestrength of the interaction, e is the base of natural logarithms, anda is about 10−15 m. Find the force between two nucleons that wouldbe consistent with this equation for the potential energy.

√

17 The magnitude of the force between two magnets separatedby a distance r can be approximated as kr−3 for large values of r.The constant k depends on the strengths of the magnets and therelative orientations of their north and south poles. Two magnetsare released on a slippery surface at an initial distance ri, and beginsliding towards each other. What will be the total kinetic energyof the two magnets when they reach a final distance rf? (Ignorefriction.)

√

Problems 343

18 A rail gun is a device like a train on a track, with the trainpropelled by a powerful electrical pulse. Very high speeds have beendemonstrated in test models, and rail guns have been proposed asan alternative to rockets for sending into outer space any objectthat would be strong enough to survive the extreme accelerations.Suppose that the rail gun capsule is launched straight up, and thatthe force of air friction acting on it is given by F = be−cx, where xis the altitude, b and c are constants, and e is the base of naturallogarithms. The exponential decay occurs because the atmospheregets thinner with increasing altitude. (In reality, the force wouldprobably drop off even faster than an exponential, because the cap-sule would be slowing down somewhat.) Find the amount of kineticenergy lost by the capsule due to air friction between when it islaunched and when it is completely beyond the atmosphere. (Grav-ity is negligible, since the air friction force is much greater than thegravitational force.)

√

19 A certain binary star system consists of two stars with massesm1 and m2, separated by a distance b. A comet, originally nearlyat rest in deep space, drops into the system and at a certain pointin time arrives at the midpoint between the two stars. For thatmoment in time, find its velocity, v, symbolically in terms of b, m1,m2, and fundamental constants.

√

20 Find the angle between the following two vectors:

x + 2y + 3z

4x + 5y + 6z

. Hint, p. 494√

21 An airplane flies in the positive direction along the x axis,through crosswinds that exert a force F = (a + bx)x + (c + dx)y.Find the work done by the wind on the plane, and by the plane onthe wind, in traveling from the origin to position x.

22 Prove that the dot product defined in section 13.3 is rota-tionally invariant in the sense of book 1, section 7.5.

23 Fill in the details of the proof of A·B = AxBx+AyBy+AzBzon page 331.


24 A space probe of mass m is dropped into a previously un-explored spherical cloud of gas and dust, and accelerates towardthe center of the cloud under the influence of the cloud’s gravity.Measurements of its velocity allow its potential energy, U , to bedetermined as a function of the distance r from the cloud’s center.The mass in the cloud is distributed in a spherically symmetric way,so its density, ρ(r), depends only on r and not on the angular coor-dinates. Show that by finding U(r), one can infer ρ(r) as follows:

ρ(r) =1

4πGmr2

d

dr

(r2 dU

dr

).

?

25 The purpose of this problem is to estimate the height of thetides. The main reason for the tides is the moon’s gravity, and we’llneglect the effect of the sun. Also, real tides are heavily influencedby landforms that channel the flow of water, but we’ll think of theearth as if it was completely covered with oceans. Under theseassumptions, the ocean surface should be a surface of constant U/m.That is, a thimbleful of water, m, should not be able to gain or loseany gravitational energy by moving from one point on the oceansurface to another. If only the spherical earth’s gravity was present,then we’d have U/m = −GMe/r, and a surface of constant U/mwould be a surface of constant r, i.e., the ocean’s surface would bespherical. Taking into account the moon’s gravity, the main effect isto shift the center of the sphere, but the sphere also becomes slightlydistorted into an approximately ellipsoidal shape. (The shift of thecenter is not physically related to the tides, since the solid part ofthe earth tends to be centered within the oceans; really, this effecthas to do with the motion of the whole earth through space, andthe way that it wobbles due to the moon’s gravity.) Determine theamount by which the long axis of the ellipsoid exceeds the shortaxis. . Hint, p. 494 ?

Problems 345


Pool balls exchange momentum.

Chapter 14

Conservation ofmomentum

In many subfields of physics these days, it is possible to read anentire issue of a journal without ever encountering an equation in-volving force or a reference to Newton’s laws of motion. In the lasthundred and fifty years, an entirely different framework has beendeveloped for physics, based on conservation laws.

The new approach is not just preferred because it is in fashion.It applies inside an atom or near a black hole, where Newton’s lawsdo not. Even in everyday situations the new approach can be supe-rior. We have already seen how perpetual motion machines could bedesigned that were too complex to be easily debunked by Newton’slaws. The beauty of conservation laws is that they tell us somethingmust remain the same, regardless of the complexity of the process.

So far we have discussed only two conservation laws, the laws ofconservation of mass and energy. Is there any reason to believe thatfurther conservation laws are needed in order to replace Newton’slaws as a complete description of nature? Yes. Conservation of massand energy do not relate in any way to the three dimensions of space,because both are scalars. Conservation of energy, for instance, doesnot prevent the planet earth from abruptly making a 90-degree turnand heading straight into the sun, because kinetic energy does notdepend on direction. In this chapter, we develop a new conservedquantity, called momentum, which is a vector.

347

14.1 MomentumA conserved quantity of motion

Your first encounter with conservation of momentum may havecome as a small child unjustly confined to a shopping cart. You spotsomething interesting to play with, like the display case of importedwine down at the end of the aisle, and decide to push the cart overthere. But being imprisoned by Dad in the cart was not the onlyinjustice that day. There was a far greater conspiracy to thwartyour young id, one that originated in the laws of nature. Pushingforward did nudge the cart forward, but it pushed you backward.If the wheels of the cart were well lubricated, it wouldn’t matterhow you jerked, yanked, or kicked off from the back of the cart.You could not cause any overall forward motion of the entire systemconsisting of the cart with you inside.

In the Newtonian framework, we describe this as arising fromNewton’s third law. The cart made a force on you that was equaland opposite to your force on it. In the framework of conservationlaws, we cannot attribute your frustration to conservation of energy.It would have been perfectly possible for you to transform some ofthe internal chemical energy stored in your body to kinetic energyof the cart and your body.

The following characteristics of the situation suggest that theremay be a new conservation law involved:

A closed system is involved. All conservation laws deal withclosed systems. You and the cart are a closed system, since thewell-oiled wheels prevent the floor from making any forward forceon you.

Something remains unchanged. The overall velocity of thesystem started out being zero, and you cannot change it. Thisvague reference to “overall velocity” can be made more precise:it is the velocity of the system’s center of mass that cannot bechanged.

Something can be transferred back and forth withoutchanging the total amount. If we define forward as positiveand backward as negative, then one part of the system can gainpositive motion if another part acquires negative motion. If wedon’t want to worry about positive and negative signs, we canimagine that the whole cart was initially gliding forward on itswell-oiled wheels. By kicking off from the back of the cart, youcould increase your own velocity, but this inevitably causes thecart to slow down.

348 Chapter 14 Conservation of momentum

It thus appears that there is some numerical measure of an object’squantity of motion that is conserved when you add up all the objectswithin a system.

Momentum

Although velocity has been referred to, it is not the total velocityof a closed system that remains constant. If it was, then firing agun would cause the gun to recoil at the same velocity as the bullet!The gun does recoil, but at a much lower velocity than the bullet.Newton’s third law tells us

Fgun on bullet = −Fbullet on gun ,

and assuming a constant force for simplicity, Newton’s second lawallows us to change this to

mbullet∆vbullet

∆t= −mgun

∆vgun∆t

.

Thus if the gun has 100 times more mass than the bullet, it willrecoil at a velocity that is 100 times smaller and in the oppositedirection, represented by the opposite sign. The quantity mv istherefore apparently a useful measure of motion, and we give it aname, momentum, and a symbol, p. (As far as I know, the letter“p” was just chosen at random, since “m” was already being used formass.) The situations discussed so far have been one-dimensional,but in three-dimensional situations it is treated as a vector.

definition of momentum for material objectsThe momentum of a material object, i.e., a piece of matter, is definedas

p = mv ,

the product of the object’s mass and its velocity vector.

The units of momentum are kg·m/s, and there is unfortunately noabbreviation for this clumsy combination of units.

The reasoning leading up to the definition of momentum was allbased on the search for a conservation law, and the only reason whywe bother to define such a quantity is that experiments show it isconserved:

the law of conservation of momentumIn any closed system, the vector sum of all the momenta remainsconstant,

p1i + p2i + . . . = p1f + p2f + . . . ,

where i labels the initial and f the final momenta. (A closed systemis one on which no external forces act.)

Section 14.1 Momentum 349

This chapter first addresses the one-dimensional case, in which thedirection of the momentum can be taken into account by using plusand minus signs. We then pass to three dimensions, necessitatingthe use of vector addition.

A subtle point about conservation laws is that they all refer to“closed systems,” but “closed” means different things in differentcases. When discussing conservation of mass, “closed” means a sys-tem that doesn’t have matter moving in or out of it. With energy,we mean that there is no work or heat transfer occurring acrossthe boundary of the system. For momentum conservation, “closed”means there are no external forces reaching into the system.

A cannon example 1. A cannon of mass 1000 kg fires a 10-kg shell at a velocity of200 m/s. At what speed does the cannon recoil?

. The law of conservation of momentum tells us that

pcannon,i + pshell ,i = pcannon,f + pshell ,f .

Choosing a coordinate system in which the cannon points in thepositive direction, the given information is

pcannon,i = 0pshell ,i = 0pshell ,f = 2000 kg·m/s .

We must have pcannon,f = −2000 kg·m/s, so the recoil velocity ofthe cannon is −2 m/s.

Ion drive for propelling spacecraft example 2. The experimental solar-powered ion drive of the Deep Space 1space probe expels its xenon gas exhaust at a speed of 30,000m/s, ten times faster than the exhaust velocity for a typical chem-ical-fuel rocket engine. Roughly how many times greater is themaximum speed this spacecraft can reach, compared with a chem-ical-fueled probe with the same mass of fuel (“reaction mass”)available for pushing out the back as exhaust?

. Momentum equals mass multiplied by velocity. Both spacecraftare assumed to have the same amount of reaction mass, and theion drive’s exhaust has a velocity ten times greater, so the mo-mentum of its exhaust is ten times greater. Before the enginestarts firing, neither the probe nor the exhaust has any momen-tum, so the total momentum of the system is zero. By conserva-tion of momentum, the total momentum must also be zero after


a / The ion drive engine of the NASA Deep Space 1 probe, shownunder construction (left) and being tested in a vacuum chamber (right)prior to its October 1998 launch. Intended mainly as a test vehicle for newtechnologies, the craft nevertheless carried out a successful scientificprogram that included a flyby of a comet.

all the exhaust has been expelled. If we define the positive di-rection as the direction the spacecraft is going, then the negativemomentum of the exhaust is canceled by the positive momen-tum of the spacecraft. The ion drive allows a final speed that isten times greater. (This simplified analysis ignores the fact thatthe reaction mass expelled later in the burn is not moving back-ward as fast, because of the forward speed of the already-movingspacecraft.)

Generalization of the momentum concept

As with all the conservation laws, the law of conservation of mo-mentum has evolved over time. In the 1800’s it was found that abeam of light striking an object would give it some momentum, eventhough light has no mass, and would therefore have no momentum


b / Steam and other gasesboiling off of the nucleus of Hal-ley’s comet. This close-up photowas taken by the European Giottospace probe, which passed within596 km of the nucleus on March13, 1986.

c / Halley’s comet, in a muchless magnified view from aground-based telescope.

according to the above definition. Rather than discarding the princi-ple of conservation of momentum, the physicists of the time decidedto see if the definition of momentum could be extended to includemomentum carried by light. The process is analogous to the processoutlined on page 279 for identifying new forms of energy. The firststep was the discovery that light could impart momentum to matter,and the second step was to show that the momentum possessed bylight could be related in a definite way to observable properties ofthe light. They found that conservation of momentum could be suc-cessfully generalized by attributing to a beam of light a momentumvector in the direction of the light’s motion and having a magnitudeproportional to the amount of energy the light possessed. The mo-mentum of light is negligible under ordinary circumstances, e.g., aflashlight left on for an hour would only absorb about 10−5 kg·m/sof momentum as it recoiled.

The tail of a comet example 3Momentum is not always equal to mv . Like many comets, Hal-ley’s comet has a very elongated elliptical orbit. About once percentury, its orbit brings it close to the sun. The comet’s head, ornucleus, is composed of dirty ice, so the energy deposited by theintense sunlight boils off steam and dust, b. The sunlight doesnot just carry energy, however — it also carries momentum. Themomentum of the sunlight impacting on the smaller dust particlespushes them away from the sun, forming a tail, c. By analogywith matter, for which momentum equals mv , you would expectthat massless light would have zero momentum, but the equationp = mv is not the correct one for light, and light does have mo-mentum. (The gases typically form a second, distinct tail whosemotion is controlled by the sun’s magnetic field.)

The reason for bringing this up is not so that you can plugnumbers into a formulas in these exotic situations. The point isthat the conservation laws have proven so sturdy exactly becausethey can easily be amended to fit new circumstances. Newton’slaws are no longer at the center of the stage of physics because theydid not have the same adaptability. More generally, the moral ofthis story is the provisional nature of scientific truth.

It should also be noted that conservation of momentum is nota consequence of Newton’s laws, as is often asserted in textbooks.Newton’s laws do not apply to light, and therefore could not pos-sibly be used to prove anything about a concept as general as theconservation of momentum in its modern form.

Momentum compared to kinetic energy

Momentum and kinetic energy are both measures of the quan-tity of motion, and a sideshow in the Newton-Leibnitz controversyover who invented calculus was an argument over whether mv (i.e.,momentum) or mv2 (i.e., kinetic energy without the 1/2 in front)


was the “true” measure of motion. The modern student can cer-tainly be excused for wondering why we need both quantities, whentheir complementary nature was not evident to the greatest mindsof the 1700’s. The following table highlights their differences.

kinetic energy . . . momentum . . .

is a scalar. is a vector

is not changed by a force perpendic-ular to the motion, which changesonly the direction of the velocityvector.

is changed by any force, since achange in either the magnitude orthe direction of the velocity vectorwill result in a change in the mo-mentum vector.

is always positive, and cannot cancelout.

cancels with momentum in the op-posite direction.

can be traded for other forms of en-ergy that do not involve motion. KEis not a conserved quantity by itself.

is always conserved in a closed sys-tem.

is quadrupled if the velocity is dou-bled.

is doubled if the velocity is doubled.

A spinning top example 4A spinning top has zero total momentum, because for every mov-ing point, there is another point on the opposite side that cancelsits momentum. It does, however, have kinetic energy.

Momentum and kinetic energy in firing a rifle example 5The rifle and bullet have zero momentum and zero kinetic energyto start with. When the trigger is pulled, the bullet gains some mo-mentum in the forward direction, but this is canceled by the rifle’sbackward momentum, so the total momentum is still zero. Thekinetic energies of the gun and bullet are both positive scalars,however, and do not cancel. The total kinetic energy is allowed toincrease, because kinetic energy is being traded for other formsof energy. Initially there is chemical energy in the gunpowder.This chemical energy is converted into heat, sound, and kineticenergy. The gun’s “backward” kinetic energy does not refrigeratethe shooter’s shoulder!

The wobbly earth example 6As the moon completes half a circle around the earth, its motionreverses direction. This does not involve any change in kineticenergy, and the earth’s gravitational force does not do any workon the moon. The reversed velocity vector does, however, implya reversed momentum vector, so conservation of momentum inthe closed earth-moon system tells us that the earth must alsochange its momentum. In fact, the earth wobbles in a little “or-bit” about a point below its surface on the line connecting it andthe moon. The two bodies’ momentum vectors always point inopposite directions and cancel each other out.


d / This Hubble Space Tele-scope photo shows a smallgalaxy (yellow blob in the lowerright) that has collided with alarger galaxy (spiral near thecenter), producing a wave of starformation (blue track) due to theshock waves passing throughthe galaxies’ clouds of gas. Thisis considered a collision in thephysics sense, even though it isstatistically certain that no star ineither galaxy ever struck a star inthe other. (This is because thestars are very small compared tothe distances between them.)

The earth and moon get a divorce example 7Why can’t the moon suddenly decide to fly off one way and theearth the other way? It is not forbidden by conservation of mo-mentum, because the moon’s newly acquired momentum in onedirection could be canceled out by the change in the momentumof the earth, supposing the earth headed the opposite directionat the appropriate, slower speed. The catastrophe is forbidden byconservation of energy, because both their energies would haveto increase greatly.

Momentum and kinetic energy of a glacier example 8A cubic-kilometer glacier would have a mass of about 1012 kg. Ifit moves at a speed of 10−5 m/s, then its momentum is 107 kg ·m/s. This is the kind of heroic-scale result we expect, perhapsthe equivalent of the space shuttle taking off, or all the cars in LAdriving in the same direction at freeway speed. Its kinetic energy,however, is only 50 J, the equivalent of the calories containedin a poppy seed or the energy in a drop of gasoline too smallto be seen without a microscope. The surprisingly small kineticenergy is because kinetic energy is proportional to the square ofthe velocity, and the square of a small number is an even smallernumber.


A If all the air molecules in the room settled down in a thin film on thefloor, would that violate conservation of momentum as well as conserva-tion of energy?

B A refrigerator has coils in back that get hot, and heat is molecularmotion. These moving molecules have both energy and momentum. Whydoesn’t the refrigerator need to be tied to the wall to keep it from recoilingfrom the momentum it loses out the back?

14.2 Collisions in one dimension

Physicists employ the term “collision” in a broader sense thanordinary usage, applying it to any situation where objects interactfor a certain period of time. A bat hitting a baseball, a radioactivelyemitted particle damaging DNA, and a gun and a bullet going theirseparate ways are all examples of collisions in this sense. Physicalcontact is not even required. A comet swinging past the sun on ahyperbolic orbit is considered to undergo a collision, even though itnever touches the sun. All that matters is that the comet and thesun exerted gravitational forces on each other.

The reason for broadening the term “collision” in this way isthat all of these situations can be attacked mathematically usingthe same conservation laws in similar ways. In the first example,conservation of momentum is all that is required.


Getting rear-ended example 9.Ms. Chang is rear-ended at a stop light by Mr. Nelson, and suesto make him pay her medical bills. He testifies that he was onlygoing 35 miles per hour when he hit Ms. Chang. She thinks hewas going much faster than that. The cars skidded together afterthe impact, and measurements of the length of the skid marksand the coefficient of friction show that their joint velocity immedi-ately after the impact was 19 miles per hour. Mr. Nelson’s Nissanweighs 3100 pounds, and Ms. Chang ’s Cadillac weighs 5200pounds. Is Mr. Nelson telling the truth?

. Since the cars skidded together, we can write down the equationfor conservation of momentum using only two velocities, v for Mr.Nelson’s velocity before the crash, and v ′ for their joint velocityafterward:

mNv = mNv ′ + mCv ′ .

Solving for the unknown, v , we find

v =(

1 +mC

mN

)v ′ .

Although we are given the weights in pounds, a unit of force, theratio of the masses is the same as the ratio of the weights, andwe find v = 51 miles per hour. He is lying.

The above example was simple because both cars had the samevelocity afterward. In many one-dimensional collisions, however, thetwo objects do not stick. If we wish to predict the result of such acollision, conservation of momentum does not suffice, because bothvelocities after the collision are unknown, so we have one equationin two unknowns.

Conservation of energy can provide a second equation, but itsapplication is not as straightforward, because kinetic energy is onlythe particular form of energy that has to do with motion. In manycollisions, part of the kinetic energy that was present before thecollision is used to create heat or sound, or to break the objectsor permanently bend them. Cars, in fact, are carefully designed tocrumple in a collision. Crumpling the car uses up energy, and that’sgood because the goal is to get rid of all that kinetic energy in arelatively safe and controlled way. At the opposite extreme, a su-perball is “super” because it emerges from a collision with almost allits original kinetic energy, having only stored it briefly as potentialenergy while it was being squashed by the impact.

Collisions of the superball type, in which almost no kinetic en-ergy is converted to other forms of energy, can thus be analyzedmore thoroughly, because they have KEf = KEi, as opposed tothe less useful inequality KEf < KEi for a case like a tennis ballbouncing on grass.

Section 14.2 Collisions in one dimension 355

Gory Details of the Proof inExample 10

The equation A + B = C + D saysthat the change in one ball’s ve-locity is equal and opposite to thechange in the other’s. We invent asymbol x = C − A for the changein ball 1’s velocity. The secondequation can then be rewritten asA2+B2 = (A+x)2+(B−x)2. Squar-ing out the quantities in parenthe-ses and then simplifying, we get0 = Ax − Bx + x2. The equationhas the trivial solution x = 0, i.e.,neither ball’s velocity is changed,but this is physically impossible be-cause the balls can’t travel througheach other like ghosts. Assumingx 6= 0, we can divide by x andsolve for x = B − A. This meansthat ball 1 has gained an amountof velocity exactly right to matchball 2’s initial velocity, and vice-versa. The balls must have swap-ped velocities.

Pool balls colliding head-on example 10. Two pool balls collide head-on, so that the collision is restrictedto one dimension. Pool balls are constructed so as to lose as littlekinetic energy as possible in a collision, so under the assumptionthat no kinetic energy is converted to any other form of energy,what can we predict about the results of such a collision?

. Pool balls have identical masses, so we use the same symbolm for both. Conservation of momentum and no loss of kineticenergy give us the two equations

mv1i + mv2i = mv1f + mv2f

12

mv21i +

12

mv22i =

12

mv21f +

12

mv22f

The masses and the factors of 1/2 can be divided out, and weeliminate the cumbersome subscripts by replacing the symbolsv1i ,... with the symbols A, B, C, and D:

A + B = C + D

A2 + B2 = C2 + D2 .

A little experimentation with numbers shows that given values of Aand B, it is impossible to find C and D that satisfy these equationsunless C and D equal A and B, or C and D are the same as Aand B but swapped around. A formal proof of this fact is givenin the sidebar. In the special case where ball 2 is initially at rest,this tells us that ball 1 is stopped dead by the collision, and ball2 heads off at the velocity originally possessed by ball 1. Thisbehavior will be familiar to players of pool.

Often, as in the example above, the details of the algebra arethe least interesting part of the problem, and considerable physicalinsight can be gained simply by counting the number of unknownsand comparing to the number of equations. Suppose a beginner atpool notices a case where her cue ball hits an initially stationaryball and stops dead. “Wow, what a good trick,” she thinks. “Ibet I could never do that again in a million years.” But she triesagain, and finds that she can’t help doing it even if she doesn’twant to. Luckily she has just learned about collisions in her physicscourse. Once she has written down the equations for conservationof energy and no loss of kinetic energy, she really doesn’t have tocomplete the algebra. She knows that she has two equations intwo unknowns, so there must be a well-defined solution. Once shehas seen the result of one such collision, she knows that the samething must happen every time. The same thing would happen withcolliding marbles or croquet balls. It doesn’t matter if the masses orvelocities are different, because that just multiplies both equationsby some constant factor.


The discovery of the neutron

This was the type of reasoning employed by James Chadwick inhis 1932 discovery of the neutron. At the time, the atom was imag-ined to be made out of two types of fundamental particles, protonsand electrons. The protons were far more massive, and clusteredtogether in the atom’s core, or nucleus. Attractive electrical forcescaused the electrons to orbit the nucleus in circles, in much thesame way that gravitational forces kept the planets from cruisingout of the solar system. Experiments showed that the helium nu-cleus, for instance, exerted exactly twice as much electrical force onan electron as a nucleus of hydrogen, the smallest atom, and this wasexplained by saying that helium had two protons to hydrogen’s one.The trouble was that according to this model, helium would havetwo electrons and two protons, giving it precisely twice the mass ofa hydrogen atom with one of each. In fact, helium has about fourtimes the mass of hydrogen.

Chadwick suspected that the helium nucleus possessed two addi-tional particles of a new type, which did not participate in electricalforces at all, i.e., were electrically neutral. If these particles had verynearly the same mass as protons, then the four-to-one mass ratio ofhelium and hydrogen could be explained. In 1930, a new type ofradiation was discovered that seemed to fit this description. It waselectrically neutral, and seemed to be coming from the nuclei of lightelements that had been exposed to other types of radiation. At thistime, however, reports of new types of particles were a dime a dozen,and most of them turned out to be either clusters made of previ-ously known particles or else previously known particles with higherenergies. Many physicists believed that the “new” particle that hadattracted Chadwick’s interest was really a previously known particlecalled a gamma ray, which was electrically neutral. Since gammarays have no mass, Chadwick decided to try to determine the newparticle’s mass and see if it was nonzero and approximately equalto the mass of a proton.

Unfortunately a subatomic particle is not something you canjust put on a scale and weigh. Chadwick came up with an ingenioussolution. The masses of the nuclei of the various chemical elementswere already known, and techniques had already been developed formeasuring the speed of a rapidly moving nucleus. He therefore setout to bombard samples of selected elements with the mysteriousnew particles. When a direct, head-on collision occurred betweena mystery particle and the nucleus of one of the target atoms, thenucleus would be knocked out of the atom, and he would measureits velocity.

Suppose, for instance, that we bombard a sample of hydrogenatoms with the mystery particles. Since the participants in thecollision are fundamental particles, there is no way for kinetic energy

Section 14.2 Collisions in one dimension 357

e / Chadwick’s subatomic pool table. A disk of the naturally occur-ring metal polonium provides a source of radiation capable of kickingneutrons out of the beryllium nuclei. The type of radiation emitted bythe polonium is easily absorbed by a few mm of air, so the air has to bepumped out of the left-hand chamber. The neutrons, Chadwick’s mysteryparticles, penetrate matter far more readily, and fly out through the walland into the chamber on the right, which is filled with nitrogen or hydrogengas. When a neutron collides with a nitrogen or hydrogen nucleus, itkicks it out of its atom at high speed, and this recoiling nucleus then ripsapart thousands of other atoms of the gas. The result is an electricalpulse that can be detected in the wire on the right. Physicists had alreadycalibrated this type of apparatus so that they could translate the strengthof the electrical pulse into the velocity of the recoiling nucleus. Thewhole apparatus shown in the figure would fit in the palm of your hand, indramatic contrast to today’s giant particle accelerators.

to be converted into heat or any other form of energy, and Chadwickthus had two equations in three unknowns:

equation #1: conservation of momentum

equation #2: no loss of kinetic energy

unknown #1: mass of the mystery particle

unknown #2: initial velocity of the mystery particle

unknown #3: final velocity of the mystery particle

The number of unknowns is greater than the number of equa-tions, so there is no unique solution. But by creating collisions withnuclei of another element, nitrogen, he gained two more equationsat the expense of only one more unknown:

equation #3: conservation of momentum in the new collision

equation #4: no loss of kinetic energy in the new collision

unknown #4: final velocity of the mystery particle in the newcollision


g / Two hockey pucks collide.Their mutual center of masstraces the straight path shown bythe dashed line.

He was thus able to solve for all the unknowns, including themass of the mystery particle, which was indeed within 1% of themass of a proton. He named the new particle the neutron, since itis electrically neutral.

Discussion question

A Good pool players learn to make the cue ball spin, which can causeit not to stop dead in a head-on collision with a stationary ball. If this doesnot violate the laws of physics, what hidden assumption was there in theexample above?

14.3 ? Relationship of momentum to thecenter of mass

f / In this multiple-flash photo-graph, we see the wrench fromabove as it flies through the air,rotating as it goes. Its centerof mass, marked with the blackcross, travels along a straight line,unlike the other points on thewrench, which execute loops.

We have already discussed the idea of the center of mass onp. 59, but using the concept of momentum we can now find a math-ematical method for defining the center of mass, explain why themotion of an object’s center of mass usually exhibits simpler mo-tion than any other point, and gain a very simple and powerful wayof understanding collisions.

The first step is to realize that the center of mass concept canbe applied to systems containing more than one object. Even some-thing like a wrench, which we think of as one object, is really madeof many atoms. The center of mass is particularly easy to visualizein the case shown on the left, where two identical hockey pucks col-lide. It is clear on grounds of symmetry that their center of massmust be at the midpoint between them. After all, we previously de-fined the center of mass as the balance point, and if the two hockeypucks were joined with a very lightweight rod whose own mass wasnegligible, they would obviously balance at the midpoint. It doesn’tmatter that the hockey pucks are two separate objects. It is stilltrue that the motion of their center of mass is exceptionally simple,just like that of the wrench’s center of mass.

The x coordinate of the hockey pucks’ center of mass is thusgiven by xcm = (x1 + x2)/2, i.e., the arithmetic average of theirx coordinates. Why is its motion so simple? It has to do withconservation of momentum. Since the hockey pucks are not beingacted on by any net external force, they constitute a closed system,

Section 14.3 ? Relationship of momentum to the center of mass 359

and their total momentum is conserved. Their total momentum is

mv1 +mv2 = m(v1 + v2)

= m

(∆x1

∆t+

∆x2

∆t

)=

m

∆t∆ (x1 + x2)

= m2∆xcm

∆t= mtotalvcm

In other words, the total momentum of the system is the same asif all its mass was concentrated at the center of mass point. Sincethe total momentum is conserved, the x component of the center ofmass’s velocity vector cannot change. The same is also true for theother components, so the center of mass must move along a straightline at constant speed.

The above relationship between the total momentum and themotion of the center of mass applies to any system, even if it is notclosed.

total momentum related to center of mass motionThe total momentum of any system is related to its total massand the velocity of its center of mass by the equation

ptotal = mtotalvcm .

What about a system containing objects with unequal masses,or containing more than two objects? The reasoning above can begeneralized to a weighted average

xcm =m1x1 +m2x2 + . . .

m1 +m2 + . . .,

with similar equations for the y and z coordinates.

Momentum in different frames of reference

Absolute motion is supposed to be undetectable, i.e., the lawsof physics are supposed to be equally valid in all inertial framesof reference. If we first calculate some momenta in one frame ofreference and find that momentum is conserved, and then reworkthe whole problem in some other frame of reference that is movingwith respect to the first, the numerical values of the momenta willall be different. Even so, momentum will still be conserved. All thatmatters is that we work a single problem in one consistent frame ofreference.

One way of proving this is to apply the equation ptotal =mtotalvcm. If the velocity of frame B relative to frame A is vBA,then the only effect of changing frames of reference is to changevcm from its original value to vcm + vBA. This adds a constantonto the momentum vector, which has no effect on conservation ofmomentum.


h / Moving your head so thatyou are always looking downfrom right above the center ofmass, you observe the collisionof the two hockey pucks in thecenter of mass frame.

i / The slingshot effect viewedin the sun’s frame of reference.Jupiter is moving to the left, andthe collision is head-on.

j / The slingshot viewed inthe frame of the center of mass ofthe Jupiter-spacecraft system.

The center of mass frame of reference

A particularly useful frame of reference in many cases is theframe that moves along with the center of mass, called the centerof mass (c.m.) frame. In this frame, the total momentum is zero.The following examples show how the center of mass frame can bea powerful tool for simplifying our understanding of collisions.

A collision of pool balls viewed in the c.m. frame example 11If you move your head so that your eye is always above the pointhalfway in between the two pool balls, you are viewing things inthe center of mass frame. In this frame, the balls come toward thecenter of mass at equal speeds. By symmetry, they must there-fore recoil at equal speeds along the lines on which they entered.Since the balls have essentially swapped paths in the center ofmass frame, the same must also be true in any other frame. Thisis the same result that required laborious algebra to prove previ-ously without the concept of the center of mass frame.

The slingshot effect example 12It is a counterintuitive fact that a spacecraft can pick up speedby swinging around a planet, if arrives in the opposite directioncompared to the planet’s motion. Although there is no physicalcontact, we treat the encounter as a one-dimensional collision,and analyze it in the center of mass frame. Figure i shows sucha “collision,” with a space probe whipping around Jupiter. In thesun’s frame of reference, Jupiter is moving.

What about the center of mass frame? Since Jupiter is so muchmore massive than the spacecraft, the center of mass is essen-tially fixed at Jupiter’s center, and Jupiter has zero velocity in thecenter of mass frame, as shown in figure j. The c.m. frame ismoving to the left compared to the sun-fixed frame used in i, sothe spacecraft’s initial velocity is greater in this frame.

Things are simpler in the center of mass frame, because it is moresymmetric. In the complicated sun-fixed frame, the incoming legof the encounter is rapid, because the two bodies are rushing to-ward each other, while their separation on the outbound leg ismore gradual, because Jupiter is trying to catch up. In the c.m.frame, Jupiter is sitting still, and there is perfect symmetry be-tween the incoming and outgoing legs, so by symmetry we havev1f = −v1i . Going back to the sun-fixed frame, the spacecraft’sfinal velocity is increased by the frames’ motion relative to eachother. In the sun-fixed frame, the spacecraft’s velocity has in-creased greatly.

The result can also be understood in terms of work and energy.In Jupiter’s frame, Jupiter is not doing any work on the spacecraftas it rounds the back of the planet, because the motion is per-pendicular to the force. But in the sun’s frame, the spacecraft’s

Section 14.3 ? Relationship of momentum to the center of mass 361

k / Power and force are therates at which energy andmomentum are transferred.

velocity vector at the same moment has a large component to theleft, so Jupiter is doing work on it.


A Make up a numerical example of two unequal masses moving in onedimension at constant velocity, and verify the equation ptotal = mtotalvcmover a time interval of one second.

B A more massive tennis racquet or baseball bat makes the ball flyoff faster. Explain why this is true, using the center of mass frame. Forsimplicity, assume that the racquet or bat is simply sitting still before thecollision, and that the hitter’s hands do not make any force large enoughto have a significant effect over the short duration of the impact.

14.4 Momentum transferThe rate of change of momentum

As with conservation of energy, we need a way to measure andcalculate the transfer of momentum into or out of a system when thesystem is not closed. In the case of energy, the answer was rathercomplicated, and entirely different techniques had to be used formeasuring the transfer of mechanical energy (work) and the transferof heat by conduction. For momentum, the situation is far simpler.

In the simplest case, the system consists of a single object actedon by a constant external force. Since it is only the object’s velocitythat can change, not its mass, the momentum transferred is

∆p = m∆v ,

which with the help of a = F/m and the constant-acceleration equa-tion a = ∆v/∆t becomes

∆p = ma∆t

= F∆t .

Thus the rate of transfer of momentum, i.e., the number of kg·m/sabsorbed per second, is simply the external force,

F =∆p

∆t.

[relationship between the force on an object and therate of change of its momentum; valid only if the forceis constant]

This is just a restatement of Newton’s second law, and in fact New-ton originally stated it this way. As shown in figure k, the rela-tionship between force and momentum is directly analogous to thatbetween power and energy.

The situation is not materially altered for a system composedof many objects. There may be forces between the objects, but the


l / The airbag increases ∆tso as to reduce F = ∆p/∆t .

internal forces cannot change the system’s momentum. (If they did,then removing the external forces would result in a closed systemthat could change its own momentum, like the mythical man whocould pull himself up by his own bootstraps. That would violateconservation of momentum.) The equation above becomes

Ftotal =∆ptotal

∆t.

[relationship between the total external force on a sys-tem and the rate of change of its total momentum; validonly if the force is constant]

Walking into a lamppost example 13. Starting from rest, you begin walking, bringing your momentumup to 100 kg·m/s. You walk straight into a lamppost. Why is themomentum change of −100 kg ·m/s caused by the lamppost somuch more painful than the change of +100 kg ·m/s when youstarted walking?

. The situation is one-dimensional, so we can dispense with thevector notation. It probably takes you about 1 s to speed up ini-tially, so the ground’s force on you is F = ∆p/∆t ≈ 100 N. Yourimpact with the lamppost, however, is over in the blink of an eye,say 1/10 s or less. Dividing by this much smaller ∆t gives a muchlarger force, perhaps thousands of newtons. (The negative signsimply indicates that the force is in the opposite direction.)

This is also the principle of airbags in cars. The time required forthe airbag to decelerate your head is fairly long, the time requiredfor your face to travel 20 or 30 cm. Without an airbag, your facewould hit the dashboard, and the time interval would be the muchshorter time taken by your skull to move a couple of centimeterswhile your face compressed. Note that either way, the same amountof mechanical work has to be done on your head: enough to eliminateall its kinetic energy.

Ion drive for spacecraft example 14. The ion drive of the Deep Space 1 spacecraft, pictured on page351 and discussed in example 2, produces a thrust of 90 mN(millinewtons). It carries about 80 kg of reaction mass, which itejects at a speed of 30,000 m/s. For how long can the enginecontinue supplying this amount of thrust before running out ofreaction mass to shove out the back?

. Solving the equation F = ∆p/∆t for the unknown ∆t , and treat-ing force and momentum as scalars since the problem is one-

Section 14.4 Momentum transfer 363

m / Example 15.

n / The F − t graph for atennis racquet hitting a ball mightlook like this. The amount ofmomentum transferred equalsthe area under the curve.

dimensional, we find

∆t =∆pF

=mexhaust∆vexhaust

F

=(80 kg)(30, 000 m/s)

0.090 N= 2.7× 107 s= 300 days

A toppling box example 15If you place a box on a frictionless surface, it will fall over with avery complicated motion that is hard to predict in detail. We know,however, that its center of mass moves in the same direction asits momentum vector points. There are two forces, a normal forceand a gravitational force, both of which are vertical. (The grav-itational force is actually many gravitational forces acting on allthe atoms in the box.) The total force must be vertical, so themomentum vector must be purely vertical too, and the center ofmass travels vertically. This is true even if the box bounces andtumbles. [Based on an example by Kleppner and Kolenkow.]

The area under the force-time graph

Few real collisions involve a constant force. For example, whena tennis ball hits a racquet, the strings stretch and the ball flattensdramatically. They are both acting like springs that obey Hooke’slaw, which says that the force is proportional to the amount ofstretching or flattening. The force is therefore small at first, rampsup to a maximum when the ball is about to reverse directions, andramps back down again as the ball is on its way back out. Theequation F = ∆p/∆t, derived under the assumption of constantacceleration, does not apply here, and the force does not even havea single well-defined numerical value that could be plugged in to theequation.

As with similar-looking equations such as v = ∆p/∆t, the equa-tion F = ∆p/∆t is correctly generalized by saying that the force isthe slope of the p− t graph.

Conversely, if we wish to find ∆p from a graph such as the onein figure n, one approach would be to divide the force by the mass ofthe ball, rescaling the F axis to create a graph of acceleration versustime. The area under the acceleration-versus-time graph gives thechange in velocity, which can then be multiplied by the mass tofind the change in momentum. An unnecessary complication wasintroduced, however, because we began by dividing by the massand ended by multiplying by it. It would have made just as much


o / Example 16.

sense to find the area under the original F − t graph, which wouldhave given us the momentum change directly.

Discussion question

A Many collisions, like the collision of a bat with a baseball, appear tobe instantaneous. Most people also would not imagine the bat and ball asbending or being compressed during the collision. Consider the followingpossibilities:

1. The collision is instantaneous.

2. The collision takes a finite amount of time, during which the ball andbat retain their shapes and remain in contact.

3. The collision takes a finite amount of time, during which the ball andbat are bending or being compressed.

How can two of these be ruled out based on energy or momentum con-siderations?

14.5 Momentum in three dimensionsIn this section we discuss how the concepts applied previously toone-dimensional situations can be used as well in three dimensions.Often vector addition is all that is needed to solve a problem:

An explosion example 16

. Astronomers observe the planet Mars as the Martians fight anuclear war. The Martian bombs are so powerful that they rip theplanet into three separate pieces of liquified rock, all having thesame mass. If one fragment flies off with velocity components

v1x = 0

v1y = 1.0× 104 km/hr ,

and the second with

v2x = 1.0× 104 km/hrv2y = 0 ,

(all in the center of mass frame) what is the magnitude of the thirdone’s velocity?

. In the center of mass frame, the planet initially had zero momen-tum. After the explosion, the vector sum of the momenta must stillbe zero. Vector addition can be done by adding components, so

mv1x + mv2x + mv3x = 0 , andmv1y + mv2y + mv3y = 0 ,

Section 14.5 Momentum in three dimensions 365

p / Example 17.

where we have used the same symbol m for all the terms, be-cause the fragments all have the same mass. The masses canbe eliminated by dividing each equation by m, and we find

v3x = −1.0× 104 km/hr

v3y = −1.0× 104 km/hr

which gives a magnitude of

|v3| =√

v23x + v2

3y

= 1.4× 104 km/hr

The center of mass

In three dimensions, we have the vector equations

Ftotal =∆ptotal

∆t

and


The following is an example of their use.

The bola example 17The bola, similar to the North American lasso, is used by SouthAmerican gauchos to catch small animals by tangling up theirlegs in the three leather thongs. The motion of the whirling bolathrough the air is extremely complicated, and would be a chal-lenge to analyze mathematically. The motion of its center ofmass, however, is much simpler. The only forces on it are gravi-tational, so

Ftotal = mtotalg .Using the equation Ftotal = ∆ptotal/∆t , we find

∆ptotal/∆t = mtotalg ,

and since the mass is constant, the equation ptotal = mtotalvcmallows us to change this to

mtotal∆vcm/∆t = mtotalg .

The mass cancels, and ∆vcm/∆t is simply the acceleration of thecenter of mass, so

acm = g .In other words, the motion of the system is the same as if all itsmass was concentrated at and moving with the center of mass.The bola has a constant downward acceleration equal to g, andflies along the same parabola as any other projectile thrown withthe same initial center of mass velocity. Throwing a bola with thecorrect rotation is presumably a difficult skill, but making it hit itstarget is no harder than it is with a ball or a single rock.

[Based on an example by Kleppner & Kolenkow.]


Counting equations and unknowns

Counting equations and unknowns is just as useful as in onedimension, but every object’s momentum vector has three compo-nents, so an unknown momentum vector counts as three unknowns.Conservation of momentum is a single vector equation, but it saysthat all three components of the total momentum vector stay con-stant, so we count it as three equations. Of course if the motionhappens to be confined to two dimensions, then we need only countvectors as having two components.

A two-car crash with sticking example 18Suppose two cars collide, stick together, and skid off together. Ifwe know the cars’ initial momentum vectors, we can count equa-tions and unknowns as follows:

unknown #1: x component of cars’ final, total momentum

unknown #2: y component of cars’ final, total momentum

equation #1: conservation of the total px

equation #2: conservation of the total py

Since the number of equations equals the number of unknowns,there must be one unique solution for their total momentum vectorafter the crash. In other words, the speed and direction at whichtheir common center of mass moves off together is unaffected byfactors such as whether the cars collide center-to-center or catcheach other a little off-center.

Shooting pool example 19Two pool balls collide, and as before we assume there is no de-crease in the total kinetic energy, i.e., no energy converted fromKE into other forms. As in the previous example, we assume weare given the initial velocities and want to find the final velocities.The equations and unknowns are:

unknown #1: x component of ball #1’s final momentum

unknown #2: y component of ball #1’s final momentum

unknown #3: x component of ball #2’s final momentum

unknown #4: y component of ball #2’s final momentum

equation #1: conservation of the total px

equation #2: conservation of the total py

equation #3: no decrease in total KE

Note that we do not count the balls’ final kinetic energies as un-knowns, because knowing the momentum vector, one can alwaysfind the velocity and thus the kinetic energy. The number of equa-tions is less than the number of unknowns, so no unique result isguaranteed. This is what makes pool an interesting game. By

Section 14.5 Momentum in three dimensions 367

q / Example 20.

aiming the cue ball to one side of the target ball you can havesome control over the balls’ speeds and directions of motion afterthe collision.

It is not possible, however, to choose any combination of finalspeeds and directions. For instance, a certain shot may give thecorrect direction of motion for the target ball, making it go into apocket, but may also have the undesired side-effect of making thecue ball go in a pocket.

Calculations with the momentum vector

The following example illustrates how a force is required tochange the direction of the momentum vector, just as one wouldbe required to change its magnitude.

A turbine example 20. In a hydroelectric plant, water flowing over a dam drives a tur-bine, which runs a generator to make electric power. The figureshows a simplified physical model of the water hitting the turbine,in which it is assumed that the stream of water comes in at a45 angle with respect to the turbine blade, and bounces off at a90 angle at nearly the same speed. The water flows at a rate R,in units of kg/s, and the speed of the water is v . What are themagnitude and direction of the water’s force on the turbine?

. In a time interval ∆t, the mass of water that strikes the blade isR∆t, and the magnitude of its initial momentum is mv = vR∆t .The water’s final momentum vector is of the same magnitude, butin the perpendicular direction. By Newton’s third law, the water’sforce on the blade is equal and opposite to the blade’s force onthe water. Since the force is constant, we can use the equation

Fblade on water =∆pwater

∆t.

Choosing the x axis to be to the right and the y axis to be up, thiscan be broken down into components as

Fblade on water,x =∆pwater,x

∆t

=−vR∆t − 0

∆t= −vR

and

Fblade on water,y =∆pwater,y

∆t

=0− (−vR∆t)

∆t= vR .


The water’s force on the blade thus has components

Fwater on blade,x = vRFwater on blade,y = −vR .

In situations like this, it is always a good idea to check that theresult makes sense physically. The x component of the water’sforce on the blade is positive, which is correct since we know theblade will be pushed to the right. The y component is negative,which also makes sense because the water must push the bladedown. The magnitude of the water’s force on the blade is

|Fwater on blade| =√

2vR

and its direction is at a 45-degree angle down and to the right.


A The figures show a jet of water striking two different objects. Howdoes the total downward force compare in the two cases? How could thisfact be used to create a better waterwheel? (Such a waterwheel is knownas a Pelton wheel.)


14.6 Applications of calculusBy now you will have learned to recognized the circumlocutions I usein the sections without calculus in order to introduce calculus-likeconcepts without using the notation, terminology, or techniques ofcalculus. It will therefore come as no surprise to you that the rateof change of momentum can be represented with a derivative,

Ftotal =dptotal

dt.

And of course the business about the area under the F − t curve isreally an integral, ∆ptotal =

∫Ftotaldt, which can be made into an

integral of a vector in the more general three-dimensional case:

∆ptotal =

∫Ftotaldt .

In the case of a material object that is neither losing nor picking upmass, these are just trivially rearranged versions of familiar equa-tions, e.g., F = mdv/dt rewritten as F = d(mv)/dt. The followingis a less trivial example, where F = ma alone would not have beenvery easy to work with.

Section 14.6 Applications of calculus 369

Rain falling into a moving cart example 21. If 1 kg/s of rain falls vertically into a 10-kg cart that is rollingwithout friction at an initial speed of 1.0 m/s, what is the effect onthe speed of the cart when the rain first starts falling?

. The rain and the cart make horizontal forces on each other, butthere is no external horizontal force on the rain-plus-cart system,so the horizontal motion obeys

F =d(mv )

dt= 0

We use the product rule to find

0 =dmdt

v + mdvdt

.

We are trying to find how v changes, so we solve for dv/dt ,

dvdt

= − vm

dmdt

= −(

1 m/s10 kg

)(1 kg/s)

= −0.1 m/s2 .

(This is only at the moment when the rain starts to fall.)

Finally we note that there are cases where F = ma is not justless convenient than F = dp/dt but in fact F = ma is wrong andF = dp/dt is right. A good example is the formation of a comet’stail by sunlight. We cannot use F = ma to describe this process,since we are dealing with a collision of light with matter, whereasNewton’s laws only apply to matter. The equation F = dp/dt, onthe other hand, allows us to find the force experienced by an atom ofgas in the comet’s tail if we know the rate at which the momentumvectors of light rays are being turned around by reflection from theatom.

14.7 ? Relativistic momentumHow does momentum behave in relativity?

Newtonian mechanics has two different measures of motion, ki-netic energy and momentum, and the relationship between them isnonlinear. Doubling your car’s momentum quadruples its kineticenergy.

But nonrelativistic mechanics can’t handle massless particles,which are always ultrarelativistic. We saw in section 11.6 that ul-trarelativistic particles are “generic,” in the sense that they have noindividual mechanical properties other than an energy and a direc-tion of motion. Therefore the relationship between kinetic energy


r / In the p-E plane, mass-less particles lie on the twodiagonals, while particles withmass lie to the right.

and momentum must be linear for ultrarelativistic particles. Indeed,experiments verify that light has momentum, and doubling the en-ergy of a ray of light doubles its momentum rather than quadruplingit.

How can we make sense of these energy-momentum relation-ships, which seem to take on two completely different forms in thelimiting cases of very low and very high velocities?

The first step is realize that since mass and energy are equivalent(section 12.4), we will get more of an apples-to-apples comparison ifwe stop talking about a material object’s kinetic energy and considerinstead its total energy E, which includes a contribution from itsmass.

On a graph of p versus E, massless particles, which have E ∝ |p|,lie on two diagonal lines that connect at the origin. If we like, wecan pick units such that the slopes of these lines are plus and minusone. Material particles lie to the right of these lines. For example,a car sitting in a parking lot has p = 0 and E = mc2.

Now what happens to such a graph when we change to a dif-ferent frame or reference that is in motion relative to the originalframe? A massless particle still has to act like a massless parti-cle, so the diagonals are simply stretched or contracted along theirown lengths. A transformation that always takes a line to a line isa linear transformation (p. 80), and if the transformation betweendifferent frames of reference preserves the linearity of the lines p = Eand p = −E, then it’s natural to suspect that it is actually somekind of linear transformation. By the same reasoning as on p. 89,the transformation must be area-preserving. We then have the samethree cases to consider as in figure ai on p. 81. Case I is ruled outbecause it would imply that particles keep the same energy whenwe change frames. (This is what would happen if c were infinite, sothat the mass-equivalent E/c2 of a given mass was zero, and there-fore E would be interpreted purely as the mass.) Case II can’t beright because it doesn’t preserve the E = |p| diagonals. We are leftwith case III, which establishes the following aesthetically appealingfact: the p-E plane transforms according to exactly the same kindof Lorentz transformation as the x-t plane. That is, (E, px, py, pz) isa four-vector (p. 334) just like (t,x, y, z). This is a highly desirableresult. If it were not true, it would be like having to learn differentmathematical rules for different kinds of three-vectors in Newtonianmechanics.

The only remaining issue to settle is whether the choice of unitsthat gives invariant 45-degree diagonals in the x-t plane is the sameas the choice of units that gives such diagonals in the p-E plane.That is, we need to establish that the c that applies to x and t isequal to the c′ needed for p and E, i.e., that the velocity scales of thetwo graphs are matched up. This is true because in the Newtonian

Section 14.7 ? Relativistic momentum 371

limit, the total mass-energy E is essentially just the particle’s mass,and then p/E ≈ p/m ≈ v. This establishes that the velocity scalesare matched at small velocities, which implies that they coincide forall velocities, since a large velocity, even one approaching c, can bebuilt up from many small increments. (This also establishes thatthe exponent n defined on p. 288 equals 1 as claimed.)

Suppose that a particle is at rest. Then it has p = 0 and mass-energy E equal to its mass m. Therefore the inner product of its(E, p) four-vector with itself equals m2. In other words, the “mag-nitude” of the energy-momentum four-vector is simply equal to theparticle’s mass. If we transform into a different frame of reference,the inner product stays the same. We can therefore always interpretthe magnitude of an energy-momentum four-vector as the mass. Insymbols,

m2 = E2 − p2 ,

or, in units with c 6= 1,

(mc2)2 = E2 − (pc)2 .

self-check AInterpret this relationship in the case where m = 0. . Answer, p. 512

Since we already have an equation E = mγ for the energy ofa material particle in terms of its velocity, we can find a similarequation for the momentum:

p =√E2 −m2

= m√γ2 − 1

= m

√1

1− v2− 1

= mγv .

As a material particle gets closer and closer to c, its momentumapproaches infinity, so that an infinite force would be required inorder to reach c.

Light rays don’t interact example 22We observe that when two rays of light cross paths, they continuethrough one another without bouncing like material objects. Thisbehavior follows directly from conservation of mass and energy.

Any two vectors can be contained in a single plane, so we canchoose our coordinates so that both rays have vanishing py andpz . Since only px is nonzero, we write it simply as p. In the result-ing p-E plane, there are two possibilities: either the rays both liealong the same diagonal, or they lie along different diagonals. Ifthey lie along the same diagonal, then there can’t be a collision,because the two rays are both moving in the same direction at


the same speed c, and the trailing one will never catch up withthe leading one.

Now suppose they lie along different diagonals. We add theirenergy-momentum vectors to get their total energy-momentum,which will lie in the gray area of figure r. That is, a pair of lightrays taken as a single system act sort of like a material objectwith a nonzero mass.1 By a Lorentz transformation, we can al-ways find a frame in which this total energy-momentum vectorlies along the E axis. This is a frame in which the momenta of thetwo rays cancel, and we have a symmetric head-on collision be-tween two rays of equal energy. It is the “center-of-mass” frame,although neither object has any mass on an individual basis. Forconvenience, let’s assume that the x-y -z coordinate system waschosen so that its origin was at rest in this frame.

Since the collision occurs along the x axis, by symmetry it is notpossible for the rays after the collision to depart from the x axis;for if they did, then there would be nothing to determine the ori-entation of the plane in which they emerged.2 Therefore we arejustified in continuing to use the same px -E plane to analyze thefour-vectors of the rays after the collision.

Let each ray have energy E in the frame described above. Giventhis total energy-momentum vector, how can we cook up twoenergy-momentum vectors for the final state such that energy andmomentum will have been conserved? Since there is zero totalmomentum, our only choice is two light rays, one with energy-momentum vector (E , E) and one with (E ,−E). But this is exactlythe same as our initial state, except that we can arbitrarily choosethe roles of the two rays to have been interchanged. Such an in-terchanging is only a matter of labeling, so there is no observablesense in which the rays have collided.3

1If you construct a box out of mirrors and put some light inside, it has weight,and theoretically even has a gravitational field! This is an example of the factthat mass is not additive in relativity. Two objects, each with zero mass, canhave an aggregate mass that is nonzero.

2In quantum mechanics, there is a loophole here. Quantum mechanics allowscertain kinds of randomness, so that the symmetry can be broken by letting theoutgoing rays be observed in a plane with some random orientation.

3There is a second loophole here, which is that a ray of light is actually awave, and the wave has a property called polarization. As a mechanical anal-ogy, consider a rope stretched taut. Side-to-side vibrations can propagate alongthe rope, and these vibrations can occur in any plane that coincides with therope. The orientation of this plane is referred to as the polarization of the wave.Returning to the case of the coliding light rays, it is possible to have nontrivialcollisions in the sense that the rays could affect one another’s polarizations.

Section 14.7 ? Relativistic momentum 373

SummarySelected vocabularymomentum . . . a measure of motion, equal to mv for material

objectscollision . . . . . an interaction between moving objects that

lasts for a certain timecenter of mass . . the balance point or average position of the

mass in a system

Notationp . . . . . . . . . . the momentum vectorcm . . . . . . . . . center of mass, as in xcm, acm, etc.

Other terminology and notationimpulse, I, J . . the amount of momentum transferred, ∆pelastic collision . one in which no KE is converted into other

forms of energyinelastic collision one in which some KE is converted to other

forms of energy

Summary

If two objects interact via a force, Newton’s third law guaran-tees that any change in one’s velocity vector will be accompaniedby a change in the other’s which is in the opposite direction. Intu-itively, this means that if the two objects are not acted on by anyexternal force, they cannot cooperate to change their overall state ofmotion. This can be made quantitative by saying that the quantitym1v1 + m2v2 must remain constant as long as the only forces arethe internal ones between the two objects. This is a conservationlaw, called the conservation of momentum, and like the conserva-tion of energy, it has evolved over time to include more and morephenomena unknown at the time the concept was invented. Themomentum of a material object is

p = mv ,

but this is more like a standard for comparison of momenta ratherthan a definition. For instance, light has momentum, but has nomass, and the above equation is not the right equation for light. Thelaw of conservation of momentum says that the total momentum ofany closed system, i.e., the vector sum of the momentum vectors ofall the things in the system, is a constant.

An important application of the momentum concept is to colli-sions, i.e., interactions between moving objects that last for a certainamount of time while the objects are in contact or near each other.Conservation of momentum tells us that certain outcomes of a col-lision are impossible, and in some cases may even be sufficient topredict the motion after the collision. In other cases, conservationof momentum does not provide enough equations to find all the un-knowns. In some collisions, such as the collision of a superball with


the floor, very little kinetic energy is converted into other forms ofenergy, and this provides one more equation, which may suffice topredict the outcome.

The total momentum of a system can be related to its total massand the velocity of its center of mass by the equation


The center of mass, introduced on an intuitive basis in book 1 asthe “balance point” of an object, can be generalized to any systemcontaining any number of objects, and is defined mathematicallyas the weighted average of the positions of all the parts of all theobjects,

xcm =m1x1 +m2x2 + . . .

m1 +m2 + . . .,

with similar equations for the y and z coordinates.

The frame of reference moving with the center of mass of a closedsystem is always a valid inertial frame, and many problems can begreatly simplified by working them in the inertial frame. For exam-ple, any collision between two objects appears in the c.m. frame asa head-on one-dimensional collision.

When a system is not closed, the rate at which momentum istransferred in or out is simply the total force being exerted externallyon the system,

Ftotal =dptotal

dt.

Summary 375

ProblemsKey√



1 Derive a formula expressing the kinetic energy of an object interms of its momentum and mass.

√

2 Two people in a rowboat wish to move around without causingthe boat to move. What should be true about their total momen-tum? Explain.

3 A firework shoots up into the air, and just before it explodesit has a certain momentum and kinetic energy. What can you sayabout the momenta and kinetic energies of the pieces immediatelyafter the explosion? [Based on a problem from PSSC Physics.]

. Solution, p. 506

4 A bullet leaves the barrel of a gun with a kinetic energy of 90J. The gun barrel is 50 cm long. The gun has a mass of 4 kg, thebullet 10 g.(a) Find the bullet’s final velocity.

√

(b) Find the bullet’s final momentum.√

(c) Find the momentum of the recoiling gun.(d) Find the kinetic energy of the recoiling gun, and explain whythe recoiling gun does not kill the shooter.

√

Problem 5

5 The graph shows the force, in meganewtons, exerted by arocket engine on the rocket as a function of time. If the rocket’smass is 4000 kg, at what speed is the rocket moving when the enginestops firing? Assume it goes straight up, and neglect the force ofgravity, which is much less than a meganewton.

√


Problem 8

6 Cosmic rays are particles from outer space, mostly protons andatomic nuclei, that are continually bombarding the earth. Most ofthem, although they are moving extremely fast, have no discernibleeffect even if they hit your body, because their masses are so small.Their energies vary, however, and a very small minority of themhave extremely large energies. In some cases the energy is as muchas several Joules, which is comparable to the KE of a well thrownrock! If you are in a plane at a high altitude and are so incrediblyunlucky as to be hit by one of these rare ultra-high-energy cosmicrays, what would you notice, the momentum imparted to your body,the energy dissipated in your body as heat, or both? Base your con-clusions on numerical estimates, not just random speculation. (Atthese high speeds, one should really take into account the devia-tions from Newtonian physics described by Einstein’s special theoryof relativity. Don’t worry about that, though.)

7 Show that for a body made up of many equal masses, theequation for the center of mass becomes a simple average of all thepositions of the masses.

8 The figure shows a view from above of a collision about tohappen between two air hockey pucks sliding without friction. Theyhave the same speed, vi, before the collision, but the big puck is 2.3times more massive than the small one. Their sides have sticky stuffon them, so when they collide, they will stick together. At whatangle will they emerge from the collision? In addition to giving anumerical answer, please indicate by drawing on the figure how yourangle is defined. . Solution, p. 506

9 A learjet traveling due east at 300 mi/hr collides with ajumbo jet which was heading southwest at 150 mi/hr. The jumbojet’s mass is 5.0 times greater than that of the learjet. When theycollide, the learjet sticks into the fuselage of the jumbo jet, and theyfall to earth together. Their engines stop functioning immediatelyafter the collision. On a map, what will be the direction from thelocation of the collision to the place where the wreckage hits theground? (Give an angle.)

√

10 A very massive object with velocity v collides head-on withan object at rest whose mass is very small. No kinetic energy isconverted into other forms. Prove that the low-mass object recoilswith velocity 2v. [Hint: Use the center-of-mass frame of reference.]

11 A mass m moving at velocity v collides with a stationarytarget having the same mass m. Find the maximum amount ofenergy that can be released as heat and sound.

√

Problems 377

12 When the contents of a refrigerator cool down, the changedmolecular speeds imply changes in both momentum and energy.Why, then, does a fridge transfer power through its radiator coils,but not force? . Solution, p. 507

13 A 10-kg bowling ball moving at 2.0 m/s hits a 1.0-kg bowlingpin, which is initially at rest. The other pins are all gone already,and the collision is head-on, so that the motion is one-dimensional.Assume that negligible amounts of heat and sound are produced.Find the velocity of the pin immediately after the collision.

14 A ball of mass 3m collides head-on with an initially stationaryball of mass m. No kinetic energy is transformed into heat or sound.In what direction is the mass-3m ball moving after the collision, andhow fast is it going compared to its original velocity?

√

15 Suppose a system consisting of pointlike particles has a totalkinetic energy Kcm measured in the center-of-mass frame of refer-ence. Since they are pointlike, they cannot have any energy due tointernal motion.(a) Prove that in a different frame of reference, moving with veloc-ity u relative to the center-of-mass frame, the total kinetic energyequals Kcm +M |u|2/2, where M is the total mass. [Hint: You cansave yourself a lot of writing if you express the total kinetic energyusing the dot product.] . Solution, p. 507(b) Use this to prove that if energy is conserved in one frame ofreference, then it is conserved in every frame of reference. The totalenergy equals the total kinetic energy plus the sum of the potentialenergies due to the particles’ interactions with each other, whichwe assume depends only on the distance between particles. [For asimpler numerical example, see problem 13 on p. 294.] ?

16 The big difference between the equations for momentum andkinetic energy is that one is proportional to v and one to v2. Both,however, are proportional to m. Suppose someone tells you thatthere’s a third quantity, funkosity, defined as f = m2v, and thatfunkosity is conserved. How do you know your leg is being pulled?

. Solution, p. 507

17 A rocket ejects exhaust with an exhaust velocity u. Therate at which the exhaust mass is used (mass per unit time) is b.We assume that the rocket accelerates in a straight line startingfrom rest, and that no external forces act on it. Let the rocket’sinitial mass (fuel plus the body and payload) be mi, and mf be itsfinal mass, after all the fuel is used up. (a) Find the rocket’s finalvelocity, v, in terms of u, mi, and mf . (b) A typical exhaust velocityfor chemical rocket engines is 4000 m/s. Estimate the initial massof a rocket that could accelerate a one-ton payload to 10% of thespeed of light, and show that this design won’t work. (For the sakeof the estimate, ignore the mass of the fuel tanks.)

√?


18 A flexible rope of mass m and length L slides without frictionover the edge of a table. Let x be the length of the rope that ishanging over the edge at a given moment in time.(a) Show that x satisfies the equation of motion d2x/dt2 = gx/L.[Hint: Use F = dp/dt, which allows you to handle the two parts ofthe rope separately even though mass is moving out of one part andinto the other.](b) Give a physical explanation for the fact that a larger value ofx on the right-hand side of the equation leads to a greater value ofthe acceleration on the left side.(c) When we take the second derivative of the function x(t) we aresupposed to get essentially the same function back again, exceptfor a constant out in front. The function ex has the property thatit is unchanged by differentiation, so it is reasonable to look forsolutions to this problem that are of the form x = bect, where b andc are constants. Show that this does indeed provide a solution fortwo specific values of c (and for any value of b).(d) Show that the sum of any two solutions to the equation of motionis also a solution.(e) Find the solution for the case where the rope starts at rest att = 0 with some nonzero value of x. ?

19 (a) Find a relativistic equation for the velocity of an objectin terms of its mass and momentum (eliminating γ).

√

(b) Show that your result is approximately the same as the classicalvalue, p/m, at low velocities.(c) Show that very large momenta result in speeds close to the speedof light. ?

Problems 379


A tornado touches down in Spring Hill, Kansas, May 20, 1957.

Chapter 15

Conservation of angularmomentum

“Sure, and maybe the sun won’t come up tomorrow.” Of course,the sun only appears to go up and down because the earth spins,so the cliche should really refer to the unlikelihood of the earth’sstopping its rotation abruptly during the night. Why can’t it stop?It wouldn’t violate conservation of momentum, because the earth’srotation doesn’t add anything to its momentum. While Californiaspins in one direction, some equally massive part of India goes theopposite way, canceling its momentum. A halt to Earth’s rotationwould entail a drop in kinetic energy, but that energy could simplybe converted into some other form, such as heat.

Other examples along these lines are not hard to find. A hydro-gen atom spins at the same rate for billions of years. A high-diverwho is rotating when he comes off the board does not need to make

381

any physical effort to continue rotating, and indeed would be unableto stop rotating before he hit the water.

These observations have the hallmarks of a conservation law:

A closed system is involved. Nothing is making an effort totwist the earth, the hydrogen atom, or the high-diver. They areisolated from rotation-changing influences, i.e., they are closedsystems.

Something remains unchanged. There appears to be a numer-ical quantity for measuring rotational motion such that the totalamount of that quantity remains constant in a closed system.

Something can be transferred back and forth withoutchanging the total amount. In figure a, the jumper wants toget his feet out in front of him so he can keep from doing a “faceplant” when he lands. Bringing his feet forward would involve acertain quantity of counterclockwise rotation, but he didn’t startout with any rotation when he left the ground. Suppose we con-sider counterclockwise as positive and clockwise as negative. Theonly way his legs can acquire some positive rotation is if some otherpart of his body picks up an equal amount of negative rotation.This is why he swings his arms up behind him, clockwise.

a / An early photograph of an old-fashioned long-jump.

What numerical measure of rotational motion is conserved? Carengines and old-fashioned LP records have speeds of rotation mea-sured in rotations per minute (r.p.m.), but the number of rota-tions per minute (or per second) is not a conserved quantity. Atwirling figure skater, for instance, can pull her arms in to increaseher r.p.m.’s. The first section of this chapter deals with the nu-merical definition of the quantity of rotation that results in a validconservation law.

382 Chapter 15 Conservation of angular momentum

b / An overhead view of apiece of putty being thrown ata door. Even though the puttyis neither spinning nor travelingalong a curve, we must define itas having some kind of “rotation”because it is able to make thedoor rotate.

c / As seen by someone standingat the axis, the putty changesits angular position. We there-fore define it as having angularmomentum.

15.1 Conservation of angular momentumWhen most people think of rotation, they think of a solid objectlike a wheel rotating in a circle around a fixed point. Examples ofthis type of rotation, called rigid rotation or rigid-body rotation, in-clude a spinning top, a seated child’s swinging leg, and a helicopter’sspinning propeller. Rotation, however, is a much more general phe-nomenon, and includes noncircular examples such as a comet inan elliptical orbit around the sun, or a cyclone, in which the corecompletes a circle more quickly than the outer parts.

If there is a numerical measure of rotational motion that is aconserved quantity, then it must include nonrigid cases like these,since nonrigid rotation can be traded back and forth with rigid ro-tation. For instance, there is a trick for finding out if an egg israw or hardboiled. If you spin a hardboiled egg and then stop itbriefly with your finger, it stops dead. But if you do the same witha raw egg, it springs back into rotation because the soft interior wasstill swirling around within the momentarily motionless shell. Thepattern of flow of the liquid part is presumably very complex andnonuniform due to the asymmetric shape of the egg and the differ-ent consistencies of the yolk and the white, but there is apparentlysome way to describe the liquid’s total amount of rotation with asingle number, of which some percentage is given back to the shellwhen you release it.

The best strategy is to devise a way of defining the amount ofrotation of a single small part of a system. The amount of rotationof a system such as a cyclone will then be defined as the total of allthe contributions from its many small parts.

The quest for a conserved quantity of rotation even requires usto broaden the rotation concept to include cases where the motiondoesn’t repeat or even curve around. If you throw a piece of puttyat a door, the door will recoil and start rotating. The putty wastraveling straight, not in a circle, but if there is to be a generalconservation law that can cover this situation, it appears that wemust describe the putty as having had some “rotation,” which itthen gave up to the door. The best way of thinking about it is toattribute rotation to any moving object or part of an object thatchanges its angle in relation to the axis of rotation. In the putty-and-door example, the hinge of the door is the natural point to thinkof as an axis, and the putty changes its angle as seen by someonestanding at the hinge. For this reason, the conserved quantity we areinvestigating is called angular momentum. The symbol for angularmomentum can’t be a or m, since those are used for accelerationand mass, so the symbol L is arbitrarily chosen instead.

Imagine a 1-kg blob of putty, thrown at the door at a speed of1 m/s, which hits the door at a distance of 1 m from the hinge.We define this blob to have 1 unit of angular momentum. When

Section 15.1 Conservation of angular momentum 383

d / A putty blob thrown di-rectly at the axis has no angularmotion, and therefore no angularmomentum. It will not cause thedoor to rotate.

e / Only the component ofthe velocity vector perpendicularto the dashed line should becounted into the definition ofangular momentum.

it hits the door, the door will recoil and start rotating. We canuse the speed at which the door recoils as a measure of the angularmomentum the blob brought in.1

Experiments show, not surprisingly, that a 2-kg blob thrown inthe same way makes the door rotate twice as fast, so the angularmomentum of the putty blob must be proportional to mass,

L ∝ m .

Similarly, experiments show that doubling the velocity of theblob will have a doubling effect on the result, so its angular momen-tum must be proportional to its velocity as well,

L ∝ mv .

You have undoubtedly had the experience of approaching a closeddoor with one of those bar-shaped handles on it and pushing on thewrong side, the side close to the hinges. You feel like an idiot, be-cause you have so little leverage that you can hardly budge the door.The same would be true with the putty blob. Experiments wouldshow that the amount of rotation the blob can give to the door isproportional to the distance, r, from the axis of rotation, so angularmomentum must also be proportional to r,

L ∝ mvr .

We are almost done, but there is one missing ingredient. Weknow on grounds of symmetry that a putty ball thrown directlyinward toward the hinge will have no angular momentum to giveto the door. After all, there would not even be any way to de-cide whether the ball’s rotation was clockwise or counterclockwisein this situation. It is therefore only the component of the blob’svelocity vector perpendicular to the door that should be counted inits angular momentum,

L = mv⊥r .

More generally, v⊥ should be thought of as the component of theobject’s velocity vector that is perpendicular to the line joining theobject to the axis of rotation.

We find that this equation agrees with the definition of the origi-nal putty blob as having one unit of angular momentum, and we cannow see that the units of angular momentum are (kg·m/s)·m, i.e.,kg·m2/s. This gives us a way of calculating the angular momentumof any material object or any system consisting of material objects:

1We assume that the door is much more massive than the blob. Under thisassumption, the speed at which the door recoils is much less than the originalspeed of the blob, so the blob has lost essentially all its angular momentum, andgiven it to the door.


f / A figure skater pulls in herarms so that she can execute aspin more rapidly.

angular momentum of a material objectThe angular momentum of a moving particle is

L = mv⊥r ,

where m is its mass, v⊥ is the component of its velocity vectorperpendicular to the line joining it to the axis of rotation, and r isits distance from the axis. Positive and negative signs are used todescribe opposite directions of rotation.

The angular momentum of a finite-sized object or a systemof many objects is found by dividing it up into many small parts,applying the equation to each part, and adding to find the totalamount of angular momentum.

Note that r is not necessarily the radius of a circle. (As impliedby the qualifiers, matter isn’t the only thing that can have angularmomentum. Light can also have angular momentum, and the aboveequation would not apply to light.)

Conservation of angular momentum has been verified over andover again by experiment, and is now believed to be one of the threemost fundamental principles of physics, along with conservation ofenergy and momentum.

A figure skater pulls her arms in example 1When a figure skater is twirling, there is very little friction betweenher and the ice, so she is essentially a closed system, and herangular momentum is conserved. If she pulls her arms in, she isdecreasing r for all the atoms in her arms. It would violate con-servation of angular momentum if she then continued rotating atthe same speed, i.e., taking the same amount of time for eachrevolution, because her arms’ contributions to her angular mo-mentum would have decreased, and no other part of her wouldhave increased its angular momentum. This is impossible be-cause it would violate conservation of angular momentum. If hertotal angular momentum is to remain constant, the decrease in rfor her arms must be compensated for by an overall increase inher rate of rotation. That is, by pulling her arms in, she substan-tially reduces the time for each rotation.

Section 15.1 Conservation of angular momentum 385

h / Example 3. A view of theearth-moon system from abovethe north pole. All distanceshave been highly distorted forlegibility. The earth’s rotation iscounterclockwise from this pointof view (arrow). The moon’s grav-ity creates a bulge on the sidenear it, because its gravitationalpull is stronger there, and an“anti-bulge” on the far side, sinceits gravity there is weaker. Forsimplicity, let’s focus on the tidalbulge closer to the moon. Itsfrictional force is trying to slowdown the earth’s rotation, so itsforce on the earth’s solid crust istoward the bottom of the figure.By Newton’s third law, the crustmust thus make a force on thebulge which is toward the top ofthe figure. This causes the bulgeto be pulled forward at a slightangle, and the bulge’s gravitytherefore pulls the moon forward,accelerating its orbital motionabout the earth and flinging itoutward.

g / Example 2.

Changing the axis example 2An object’s angular momentum can be different depending on theaxis about which it rotates. Figure g shows shows two double-exposure photographs a viola player tipping the bow in order tocross from one string to another. Much more angular momentumis required when playing near the bow’s handle, called the frog,as in the panel on the right; not only are most of the atoms in thebow at greater distances, r , from the axis of rotation, but the onesin the tip also have more momentum, p. It is difficult for the playerto quickly transfer a large angular momentum into the bow, andthen transfer it back out just as quickly. (In the language of section15.4, large torques are required.) This is one of the reasons thatstring players tend to stay near the middle of the bow as much aspossible.

Earth’s slowing rotation and the receding moon example 3As noted in chapter 1, the earth’s rotation is actually slowing downvery gradually, with the kinetic energy being dissipated as heat byfriction between the land and the tidal bulges raised in the seasby the earth’s gravity. Does this mean that angular momentum isnot really perfectly conserved? No, it just means that the earthis not quite a closed system by itself. If we consider the earthand moon as a system, then the angular momentum lost by theearth must be gained by the moon somehow. In fact very precisemeasurements of the distance between the earth and the moonhave been carried out by bouncing laser beams off of a mirrorleft there by astronauts, and these measurements show that themoon is receding from the earth at a rate of 4 centimeters peryear! The moon’s greater value of r means that it has a greater


angular momentum, and the increase turns out to be exactly theamount lost by the earth. In the days of the dinosaurs, the dayswere significantly shorter, and the moon was closer and appearedbigger in the sky.

But what force is causing the moon to speed up, drawing it outinto a larger orbit? It is the gravitational forces of the earth’s tidalbulges. The effect is described qualitatively in the caption of thefigure. The result would obviously be extremely difficult to calcu-late directly, and this is one of those situations where a conserva-tion law allows us to make precise quantitative statements aboutthe outcome of a process when the calculation of the processitself would be prohibitively complex.

Restriction to rotation in a plane

Is angular momentum a vector, or a scalar? It does have adirection in space, but it’s a direction of rotation, not a straight-linedirection like the directions of vectors such as velocity or force. Itturns out that there is a way of defining angular momentum as avector, but in this book the examples will be confined to a singleplane of rotation, i.e., effectively two-dimensional situations. In thisspecial case, we can choose to visualize the plane of rotation fromone side or the other, and to define clockwise and counterclockwiserotation as having opposite signs of angular momentum.

Discussion question

A Conservation of plain old momentum, p, can be thought of as thegreatly expanded and modified descendant of Galileo’s original principleof inertia, that no force is required to keep an object in motion. The princi-ple of inertia is counterintuitive, and there are many situations in which itappears superficially that a force is needed to maintain motion, as main-tained by Aristotle. Think of a situation in which conservation of angularmomentum, L, also seems to be violated, making it seem incorrectly thatsomething external must act on a closed system to keep its angular mo-mentum from “running down.”

15.2 Angular momentum in planetary motionWe now discuss the application of conservation of angular momen-tum to planetary motion, both because of its intrinsic importanceand because it is a good way to develop a visual intuition for angularmomentum.

Kepler’s law of equal areas states that the area swept out bya planet in a certain length of time is always the same. Angularmomentum had not been invented in Kepler’s time, and he did noteven know the most basic physical facts about the forces at work. Hethought of this law as an entirely empirical and unexpectedly simpleway of summarizing his data, a rule that succeeded in describingand predicting how the planets sped up and slowed down in their

Section 15.2 Angular momentum in planetary motion 387

i / The planet’s angular mo-mentum is related to the rate atwhich it sweeps out area.

elliptical paths. It is now fairly simple, however, to show that theequal area law amounts to a statement that the planet’s angularmomentum stays constant.

There is no simple geometrical rule for the area of a pie wedgecut out of an ellipse, but if we consider a very short time interval,as shown in figure i, the shaded shape swept out by the planet isvery nearly a triangle. We do know how to compute the area of atriangle. It is one half the product of the base and the height:

area =1

2bh .

We wish to relate this to angular momentum, which containsthe variables r and v⊥ . If we consider the sun to be the axis ofrotation, then the variable r is identical to the base of the triangle,r = b. Referring to the magnified portion of the figure, v⊥ can berelated to h, because the two right triangles are similar:

h

distance traveled=v⊥|v|

The area can thus be rewritten as

area =1

2rv⊥(distance traveled)

|v|.

The distance traveled equals |v|∆t, so this simplifies to

area =1

2rv⊥∆t .

We have found the following relationship between angular momen-tum and the rate at which area is swept out:

L = 2marea

∆t.

The factor of 2 in front is simply a matter of convention, since anyconserved quantity would be an equally valid conserved quantity ifyou multiplied it by a constant. The factor of m was not relevantto Kepler, who did not know the planets’ masses, and who was onlydescribing the motion of one planet at a time.

We thus find that Kepler’s equal-area law is equivalent to a state-ment that the planet’s angular momentum remains constant. Butwait, why should it remain constant? — the planet is not a closedsystem, since it is being acted on by the sun’s gravitational force.There are two valid answers. The first is that it is actually the to-tal angular momentum of the sun plus the planet that is conserved.The sun, however, is millions of times more massive than the typicalplanet, so it accelerates very little in response to the planet’s gravi-tational force. It is thus a good approximation to say that the sun



doesn’t move at all, so that no angular momentum is transferredbetween it and the planet.

The second answer is that to change the planet’s angular mo-mentum requires not just a force but a force applied in a certainway. In section 15.4 we discuss the transfer of angular momentumby a force, but the basic idea here is that a force directly in towardthe axis does not change the angular momentum.


A Suppose an object is simply traveling in a straight line at constantspeed. If we pick some point not on the line and call it the axis of rotation,is area swept out by the object at a constant rate? Would it matter if wechose a different axis?

B The figure is a strobe photo of a pendulum bob, taken from under-neath the pendulum looking straight up. The black string can’t be seenin the photograph. The bob was given a slight sideways push when itwas released, so it did not swing in a plane. The bright spot marks thecenter, i.e., the position the bob would have if it hung straight down at us.Does the bob’s angular momentum appear to remain constant if we con-sider the center to be the axis of rotation? What if we choose a differentaxis?


15.3 Two theorems about angular momentumWith plain old momentum, p, we had the freedom to work in anyinertial frame of reference we liked. The same object could havedifferent values of momentum in two different frames, if the frameswere not at rest with respect to each other. Conservation of mo-mentum, however, would be true in either frame. As long as weemployed a single frame consistently throughout a calculation, ev-erything would work.

The same is true for angular momentum, and in addition thereis an ambiguity that arises from the definition of an axis of rotation.For a wheel, the natural choice of an axis of rotation is obviouslythe axle, but what about an egg rotating on its side? The egg

Section 15.3 Two theorems about angular momentum 389

j / Example 4.

k / Everyone has a strongtendency to think of the diver asrotating about his own center ofmass. However, he is flying inan arc, and he also has angularmomentum because of thismotion.

l / This rigid object has angu-lar momentum both because it isspinning about its center of massand because it is moving throughspace.

has an asymmetric shape, and thus no clearly defined geometriccenter. A similar issue arises for a cyclone, which does not evenhave a sharply defined shape, or for a complicated machine withmany gears. The following theorem, the first of two presented inthis section without proof, explains how to deal with this issue.Although I have put descriptive titles above both theorems, theyhave no generally accepted names.

the choice of axis theoremIt is entirely arbitrary what point one defines as the axis forpurposes of calculating angular momentum. If a closed sys-tem’s angular momentum is conserved when calculated withone choice of axis, then it will also be conserved for any otherchoice. Likewise, any inertial frame of reference may be used.

Colliding asteroids described with different axes example 4Observers on planets A and B both see the two asteroids collid-ing. The asteroids are of equal mass and their impact speeds arethe same. Astronomers on each planet decide to define their ownplanet as the axis of rotation. Planet A is twice as far from the col-lision as planet B. The asteroids collide and stick. For simplicity,assume planets A and B are both at rest.

With planet A as the axis, the two asteroids have the same amountof angular momentum, but one has positive angular momentumand the other has negative. Before the collision, the total angularmomentum is therefore zero. After the collision, the two asteroidswill have stopped moving, and again the total angular momen-tum is zero. The total angular momentum both before and afterthe collision is zero, so angular momentum is conserved if youchoose planet A as the axis.

The only difference with planet B as axis is that r is smaller by afactor of two, so all the angular momenta are halved. Even thoughthe angular momenta are different than the ones calculated byplanet A, angular momentum is still conserved.

The earth spins on its own axis once a day, but simultaneouslytravels in its circular one-year orbit around the sun, so any givenpart of it traces out a complicated loopy path. It would seem difficultto calculate the earth’s angular momentum, but it turns out thatthere is an intuitively appealing shortcut: we can simply add up theangular momentum due to its spin plus that arising from its centerof mass’s circular motion around the sun. This is a special case ofthe following general theorem:

the spin theoremAn object’s angular momentum with respect to some outsideaxis A can be found by adding up two parts:(1) The first part is the object’s angular momentum foundby using its own center of mass as the axis, i.e., the angular


m / Energy, momentum, andangular momentum can be trans-ferred. The rates of transfer arecalled power, force, and torque.

momentum the object has because it is spinning.(2) The other part equals the angular momentum that theobject would have with respect to the axis A if it had all itsmass concentrated at and moving with its center of mass.

A system with its center of mass at rest example 5In the special case of an object whose center of mass is at rest,the spin theorem implies that the object’s angular momentum isthe same regardless of what axis we choose. (This is an evenstronger statement than the choice of axis theorem, which onlyguarantees that angular momentum is conserved for any givenchoice of axis, without specifying that it is the same for all suchchoices.)

Discussion question

A In the example of the colliding asteroids, suppose planet A was mov-ing toward the top of the page, at the same speed as the bottom asteroid.How would planet A’s astronomers describe the angular momenta of theasteroids? Would angular momentum still be conserved?

15.4 Torque: the rate of transfer of angularmomentum

Force can be interpreted as the rate of transfer of momentum. Theequivalent in the case of angular momentum is called torque (rhymeswith “fork”). Where force tells us how hard we are pushing orpulling on something, torque indicates how hard we are twisting onit. Torque is represented by the Greek letter tau, τ , and the rateof change of an object’s angular momentum equals the total torqueacting on it,

τtotal =∆L

∆t.

(If the angular momentum does not change at a constant rate, thetotal torque equals the slope of the tangent line on a graph of Lversus t.)

As with force and momentum, it often happens that angularmomentum recedes into the background and we focus our interest onthe torques. The torque-focused point of view is exemplified by thefact that many scientifically untrained but mechanically apt peopleknow all about torque, but none of them have heard of angularmomentum. Car enthusiasts eagerly compare engines’ torques, andthere is a tool called a torque wrench which allows one to apply adesired amount of torque to a screw and avoid overtightening it.

Torque distinguished from force

Of course a force is necessary in order to create a torque — youcan’t twist a screw without pushing on the wrench — but force andtorque are two different things. One distinction between them isdirection. We use positive and negative signs to represent forces in

Section 15.4 Torque: the rate of transfer of angular momentum 391

n / The plane’s four enginesproduce zero total torque but notzero total force.

o / The simple physical situa-tion we use to derive an equationfor torque. A force that pointsdirectly in at or out away from theaxis produces neither clockwisenor counterclockwise angularmomentum. A force in the per-pendicular direction does transferangular momentum.

the two possible directions along a line. The direction of a torque,however, is clockwise or counterclockwise, not a linear direction.

The other difference between torque and force is a matter ofleverage. A given force applied at a door’s knob will change thedoor’s angular momentum twice as rapidly as the same force appliedhalfway between the knob and the hinge. The same amount of forceproduces different amounts of torque in these two cases.

It is possible to have a zero total torque with a nonzero totalforce. An airplane with four jet engines, n, would be designed sothat their forces are balanced on the left and right. Their forces areall in the same direction, but the clockwise torques of two of theengines are canceled by the counterclockwise torques of the othertwo, giving zero total torque.

Conversely we can have zero total force and nonzero total torque.A merry-go-round’s engine needs to supply a nonzero torque on itto bring it up to speed, but there is zero total force on it. If therewas not zero total force on it, its center of mass would accelerate!

Relationship between force and torque

How do we calculate the amount of torque produced by a givenforce? Since it depends on leverage, we should expect it to dependon the distance between the axis and the point of application ofthe force. We’ll derive an equation relating torque to force for aparticular very simple situation, and state without proof that theequation actually applies to all situations.

Consider a pointlike object which is initially at rest at a distancer from the axis we have chosen for defining angular momentum.We first observe that a force directly inward or outward, along theline connecting the axis to the object, does not impart any angularmomentum to the object.

A force perpendicular to the line connecting the axis and theobject does, however, make the object pick up angular momentum.Newton’s second law gives

a =F

m,

and assuming for simplicity that the force is constant, the constantacceleration equation a = ∆v/∆t allows us to find the velocity theobject acquires after a time ∆t,

∆v =F∆t

m.


p / The geometric relationshipsrefered to in the relationshipbetween force and torque.

We are trying to relate force to a change in angular momentum, sowe multiply both sides of the equation by mr to give

m∆vr = F∆tr

∆L = F∆tr .

Dividing by ∆t gives the torque:

∆L

∆t= Fr

τ = Fr .

If a force acts at an angle other than 0 or 90 with respect to theline joining the object and the axis, it would be only the componentof the force perpendicular to the line that would produce a torque,

τ = F⊥r .

Although this result was proved under a simplified set of circum-stances, it is more generally valid:

relationship between force and torqueThe rate at which a force transfers angular momentum to anobject, i.e., the torque produced by the force, is given by

|τ | = r|F⊥| ,

where r is the distance from the axis to the point of applica-tion of the force, and F⊥ is the component of the force thatis perpendicular to the line joining the axis to the point ofapplication.

The equation is stated with absolute value signs because thepositive and negative signs of force and torque indicate differentthings, so there is no useful relationship between them. The signof the torque must be found by physical inspection of the case athand.

From the equation, we see that the units of torque can be writ-ten as newtons multiplied by meters. Metric torque wrenches arecalibrated in N·m, but American ones use foot-pounds, which is alsoa unit of distance multiplied by a unit of force. We know from ourstudy of mechanical work that newtons multiplied by meters equaljoules, but torque is a completely different quantity from work, andnobody writes torques with units of joules, even though it would betechnically correct.

self-check ACompare the magnitudes and signs of the four torques shown in thefigure. . Answer, p. 512



q / The quantity r⊥.

How torque depends on the direction of the force example 6. How can the torque applied to the wrench in the figure be ex-pressed in terms of r , |F |, and the angle θ?

. The force vector and its F⊥ component form the hypotenuseand one leg of a right triangle,

and the interior angle opposite to F⊥ equals θ. The absolute valueof F⊥ can thus be expressed as

F⊥ = |F| sin θ ,

leading to|τ| = r |F| sin θ .

Sometimes torque can be more neatly visualized in terms of thequantity r⊥ shown in figure q, which gives us a third way of express-ing the relationship between torque and force:

|τ | = r⊥|F| .

Of course you would not want to go and memorize all threeequations for torque. Starting from any one of them you could easilyderive the other two using trigonometry. Familiarizing yourself withthem can however clue you in to easier avenues of attack on certainproblems.

The torque due to gravity

Up until now we’ve been thinking in terms of a force that actsat a single point on an object, such as the force of your hand on thewrench. This is of course an approximation, and for an extremelyrealistic calculation of your hand’s torque on the wrench you mightneed to add up the torques exerted by each square millimeter whereyour skin touches the wrench. This is seldom necessary. But inthe case of a gravitational force, there is never any single point atwhich the force is applied. Our planet is exerting a separate tug onevery brick in the Leaning Tower of Pisa, and the total gravitationaltorque on the tower is the sum of the torques contributed by all thelittle forces. Luckily there is a trick that allows us to avoid sucha massive calculation. It turns out that for purposes of computingthe total gravitational torque on an object, you can get the rightanswer by just pretending that the whole gravitational force acts atthe object’s center of mass.


r / Example 7.

s / Example 8.

Gravitational torque on an outstretched arm example 7. Your arm has a mass of 3.0 kg, and its center of mass is 30cm from your shoulder. What is the gravitational torque on yourarm when it is stretched out horizontally to one side, taking theshoulder to be the axis?

. The total gravitational force acting on your arm is

|F | = (3.0 kg)(9.8 m/s2) = 29 N .

For the purpose of calculating the gravitational torque, we cantreat the force as if it acted at the arm’s center of mass. The forceis straight down, which is perpendicular to the line connecting theshoulder to the center of mass, so

F⊥ = |F | = 29 N .

Continuing to pretend that the force acts at the center of the arm,r equals 30 cm = 0.30 m, so the torque is

τ = rF⊥ = 9 N·m .

Cow tipping example 8In 2005, Dr. Margo Lillie and her graduate student Tracy Boech-ler published a study claiming to debunk cow tipping. Their claimwas based on an analysis of the torques that would be requiredto tip a cow, which showed that one person wouldn’t be able tomake enough torque to do it. A lively discussion ensued on thepopular web site slashdot.org (“news for nerds, stuff that mat-ters”) concerning the validity of the study. Personally, I had al-ways assumed that cow-tipping was a group sport anyway, but asa physicist, I also had some quibbles with their calculation. Here’smy own analysis.

There are three forces on the cow: the force of gravity FW , theground’s normal force FN , and the tippers’ force FA.

As soon as the cow’s left hooves (on the right from our point ofview) break contact with the ground, the ground’s force is beingapplied only to hooves on the other side. We don’t know theground’s force, and we don’t want to find it. Therefore we takethe axis to be at its point of application, so that its torque is zero.

For the purpose of computing torques, we can pretend that grav-ity acts at the cow’s center of mass, which I’ve placed a littlelower than the center of its torso, since its legs and head alsohave some mass, and the legs are more massive than the headand stick out farther, so they lower the c.m. more than the headraises it. The angle θW between the vertical gravitational forceand the line rW is about 14 . (An estimate by Matt Semke at the


University of Nebraska-Lincoln gives 20 , which is in the sameballpark.)

To generate the maximum possible torque with the least possibleforce, the tippers want to push at a point as far as possible fromthe axis, which will be the shoulder on the other side, and theywant to push at a 90 degree angle with respect to the radius linerA.

When the tippers are just barely applying enough force to raisethe cow’s hooves on one side, the total torque has to be justslightly more than zero. (In reality, they want to push a lot harderthan this — hard enough to impart a lot of angular momentum tothe cow fair in a short time, before it gets mad and hurts them.We’re just trying to calculate the bare minimum force they canpossibly use, which is the question that science can answer.) Set-ting the total torque equal to zero,

τN + τW + τA = 0 ,

and letting counterclockwise torques be positive, we have

0−mgrW sin θW + FArA sin 90 = 0

FA =rW

rAmg sin θW

≈ 11.5

(680 kg)(9.8 m/s2) sin 14

= 1100 N .

The 680 kg figure for the typical mass of a cow is due to Lillieand Boechler, who are veterinarians, so I assume it’s fairly accu-rate. My estimate of 1100 N comes out significantly lower thantheir 1400 N figure, mainly because their incorrect placement ofthe center of mass gives θW = 24 . I don’t think 1100 N is animpossible amount of force to require of one big, strong person(it’s equivalent to lifting about 110 kg, or 240 pounds), but giventhat the tippers need to impart a large angular momentum fairlyquickly, it’s probably true that several people would be required.

The main practical issue with cow tipping is that cows generallysleep lying down. Falling on its side can also seriously injure acow.



Discussion question E.


A This series of discussion questions deals with past students’ incorrectreasoning about the following problem.

Suppose a comet is at the point in its orbit shown in the figure. Theonly force on the comet is the sun’s gravitational force.

Throughout the question, define all torques and angular momentausing the sun as the axis.

(1) Is the sun producing a nonzero torque on the comet? Explain.(2) Is the comet’s angular momentum increasing, decreasing, orstaying the same? Explain.

Explain what is wrong with the following answers. In some cases, the an-swer is correct, but the reasoning leading up to it is wrong. (a) Incorrectanswer to part (1): “Yes, because the sun is exerting a force on the comet,and the comet is a certain distance from the sun.”(b) Incorrect answer to part (1): “No, because the torques cancel out.”(c) Incorrect answer to part (2): “Increasing, because the comet is speed-ing up.”

B Which claw hammer would make it easier to get the nail out of thewood if the same force was applied in the same direction?

C You whirl a rock over your head on the end of a string, and graduallypull in the string, eventually cutting the radius in half. What happens tothe rock’s angular momentum? What changes occur in its speed, the timerequired for one revolution, and its acceleration? Why might the stringbreak?

D A helicopter has, in addition to the huge fan blades on top, a smallerpropeller mounted on the tail that rotates in a vertical plane. Why?

E The photo shows an amusement park ride whose two cars rotate inopposite directions. Why is this a good design?


t / The windmills are not closedsystems, but angular momentumis being transferred out of themat the same rate it is transferredin, resulting in constant angularmomentum. To get an idea ofthe huge scale of the modernwindmill farm, note the sizes ofthe trucks and trailers.

15.5 StaticsEquilibrium

There are many cases where a system is not closed but maintainsconstant angular momentum. When a merry-go-round is running atconstant angular momentum, the engine’s torque is being canceledby the torque due to friction.

When an object has constant momentum and constant angularmomentum, we say that it is in equilibrium. This is a scientificredefinition of the common English word, since in ordinary speechnobody would describe a car spinning out on an icy road as beingin equilibrium.

Very commonly, however, we are interested in cases where an ob-ject is not only in equilibrium but also at rest, and this correspondsmore closely to the usual meaning of the word. Trees and bridgeshave been designed by evolution and engineers to stay at rest, andto do so they must have not just zero total force acting on them butzero total torque. It is not enough that they don’t fall down, theyalso must not tip over. Statics is the branch of physics concernedwith problems such as these.

Solving statics problems is now simply a matter of applying andcombining some things you already know:

• You know the behaviors of the various types of forces, forexample that a frictional force is always parallel to the surfaceof contact.

• You know about vector addition of forces. It is the vector sumof the forces that must equal zero to produce equilibrium.

• You know about torque. The total torque acting on an objectmust be zero if it is to be in equilibrium.

• You know that the choice of axis is arbitrary, so you can makea choice of axis that makes the problem easy to solve.

In general, this type of problem could involve four equations in fourunknowns: three equations that say the force components add upto zero, and one equation that says the total torque is zero. Mostcases you’ll encounter will not be this complicated. In the followingexample, only the equation for zero total torque is required in orderto get an answer.

Section 15.5 Statics 399

u / Example 9.

A flagpole example 9. A 10-kg flagpole is being held up by a lightweight horizontalcable, and is propped against the foot of a wall as shown in thefigure. If the cable is only capable of supporting a tension of 70N, how great can the angle α be without breaking the cable?

. All three objects in the figure are supposed to be in equilibrium:the pole, the cable, and the wall. Whichever of the three objectswe pick to investigate, all the forces and torques on it have tocancel out. It is not particularly helpful to analyze the forces andtorques on the wall, since it has forces on it from the ground thatare not given and that we don’t want to find. We could study theforces and torques on the cable, but that doesn’t let us use thegiven information about the pole. The object we need to analyzeis the pole.

The pole has three forces on it, each of which may also result ina torque: (1) the gravitational force, (2) the cable’s force, and (3)the wall’s force.

We are free to define an axis of rotation at any point we wish, andit is helpful to define it to lie at the bottom end of the pole, sinceby that definition the wall’s force on the pole is applied at r = 0and thus makes no torque on the pole. This is good, because wedon’t know what the wall’s force on the pole is, and we are nottrying to find it.

With this choice of axis, there are two nonzero torques on thepole, a counterclockwise torque from the cable and a clockwisetorque from gravity. Choosing to represent counterclockwise torquesas positive numbers, and using the equation |τ| = r |F | sin θ, wehave

rcable|Fcable| sin θcable − rgrav |Fgrav | sin θgrav = 0 .

A little geometry gives θcable = 90 − α and θgrav = α, so

rcable|Fcable| sin(90 − α)− rgrav |Fgrav | sinα = 0 .

The gravitational force can be considered as acting at the pole’scenter of mass, i.e., at its geometrical center, so rcable is twicergrav , and we can simplify the equation to read

2|Fcable| sin(90 − α)− |Fgrav | sinα = 0 .

These are all quantities we were given, except for α, which is theangle we want to find. To solve for α we need to use the trigidentity sin(90 − x) = cos x ,

2|Fcable| cosα− |Fgrav | sinα = 0 ,


v / Example 10.

which allows us to find

tanα = 2|Fcable||Fgrav |

α = tan−1(

2|Fcable||Fgrav |

)= tan−1

(2× 70 N

98 N

)= 55 .

Art! example 10. The abstract sculpture shown in figure v contains a cube ofmass m and sides of length b. The cube rests on top of a cylinder,which is off-center by a distance a. Find the tension in the cable.

. There are four forces on the cube: a gravitational force mg, theforce FT from the cable, the upward normal force from the cylin-der, FN , and the horizontal static frictional force from the cylinder,Fs.

The total force on the cube in the vertical direction is zero:

FN −mg = 0 .

As our axis for defining torques, it’s convenient to choose the pointof contact between the cube and the cylinder, because then nei-ther Fs nor FN makes any torque. The cable’s torque is counter-clockwise, the torque due to gravity is clockwise. Letting counter-clockwise torques be positive, and using the convenient equationτ = r⊥F , we find the equation for the total torque:

bFT −mga = 0 .

We could also write down the equation saying that the total hori-zontal force is zero, but that would bring in the cylinder’s frictionalforce on the cube, which we don’t know and don’t need to find. Wealready have two equations in the two unknowns FT and FN , sothere’s no need to make it into three equations in three unknowns.Solving the first equation for FN = mg, we then substitute into thesecond equation to eliminate FN , and solve for FT = (a/b)mg.

As a check, our result makes sense when a = 0; the cube isbalanced on the cylinder, so the cable goes slack.

Section 15.5 Statics 401

w / Stable and unstable equi-libria.

x / The dancer’s equilibriumis unstable. If she didn’t con-stantly make tiny adjustments,she would tip over.

y / Example 11.

Stable and unstable equilibria

A pencil balanced upright on its tip could theoretically be inequilibrium, but even if it was initially perfectly balanced, it wouldtopple in response to the first air current or vibration from a pass-ing truck. The pencil can be put in equilibrium, but not in stableequilibrium. The things around us that we really do see staying stillare all in stable equilibrium.

Why is one equilibrium stable and another unstable? Try push-ing your own nose to the left or the right. If you push it a millimeterto the left, your head responds with a gentle force to the right, whichkeeps your nose from flying off of your face. If you push your nose acentimeter to the left, your face’s force on your nose becomes muchstronger. The defining characteristic of a stable equilibrium is thatthe farther the object is moved away from equilibrium, the strongerthe force is that tries to bring it back.

The opposite is true for an unstable equilibrium. In the topfigure, the ball resting on the round hill theoretically has zero totalforce on it when it is exactly at the top. But in reality the totalforce will not be exactly zero, and the ball will begin to move off toone side. Once it has moved, the net force on the ball is greater thanit was, and it accelerates more rapidly. In an unstable equilibrium,the farther the object gets from equilibrium, the stronger the forcethat pushes it farther from equilibrium.

This idea can be rephrased in terms of energy. The differencebetween the stable and unstable equilibria shown in figure w is thatin the stable equilibrium, the potential energy is at a minimum, andmoving to either side of equilibrium will increase it, whereas theunstable equilibrium represents a maximum.

Note that we are using the term “stable” in a weaker sense thanin ordinary speech. A domino standing upright is stable in the sensewe are using, since it will not spontaneously fall over in response toa sneeze from across the room or the vibration from a passing truck.We would only call it unstable in the technical sense if it could betoppled by any force, no matter how small. In everyday usage, ofcourse, it would be considered unstable, since the force required totopple it is so small.

An application of calculus example 11. Nancy Neutron is living in a uranium nucleus that is undergoingfission. Nancy’s potential energy as a function of position can beapproximated by PE = x4 − x2, where all the units and numeri-cal constants have been suppressed for simplicity. Use calculusto locate the equilibrium points, and determine whether they arestable or unstable.

. The equilibrium points occur where the PE is at a minimum ormaximum, and minima and maxima occur where the derivative


z / The biceps muscle flexesthe arm.

aa / The triceps extends thearm.

(which equals minus the force on Nancy) is zero. This deriva-tive is dPE/dx = 4x3 − 2x , and setting it equal to zero, we havex = 0,±1/

√2. Minima occur where the second derivative is pos-

itive, and maxima where it is negative. The second derivativeis 12x2 − 2, which is negative at x = 0 (unstable) and positive atx = ±1/

√2 (stable). Interpretation: the graph of the PE is shaped

like a rounded letter ‘W,’ with the two troughs representing the twohalves of the splitting nucleus. Nancy is going to have to decidewhich half she wants to go with.

15.6 Simple machines: the leverAlthough we have discussed some simple machines such as the pul-ley, without the concept of torque we were not yet ready to ad-dress the lever, which is the machine nature used in designing livingthings, almost to the exclusion of all others. (We can speculate whatlife on our planet might have been like if living things had evolvedwheels, gears, pulleys, and screws.) The figures show two examplesof levers within your arm. Different muscles are used to flex andextend the arm, because muscles work only by contraction.

Analyzing example z physically, there are two forces that dowork. When we lift a load with our biceps muscle, the muscle doespositive work, because it brings the bone in the forearm in the direc-tion it is moving. The load’s force on the arm does negative work,because the arm moves in the direction opposite to the load’s force.This makes sense, because we expect our arm to do positive work onthe load, so the load must do an equal amount of negative work onthe arm. (If the biceps was lowering a load, the signs of the workswould be reversed. Any muscle is capable of doing either positiveor negative work.)

There is also a third force on the forearm: the force of the upperarm’s bone exerted on the forearm at the elbow joint (not shownwith an arrow in the figure). This force does no work, because theelbow joint is not moving.

Because the elbow joint is motionless, it is natural to define ourtorques using the joint as the axis. The situation now becomesquite simple, because the upper arm bone’s force exerted at theelbow neither does work nor creates a torque. We can ignore itcompletely. In any lever there is such a point, called the fulcrum.

If we restrict ourselves to the case in which the forearm rotateswith constant angular momentum, then we know that the totaltorque on the forearm is zero,

τmuscle + τload = 0 .

If we choose to represent counterclockwise torques as positive, thenthe muscle’s torque is positive, and the load’s is negative. In terms

Section 15.6 Simple machines: the lever 403

of their absolute values,

|τmuscle| = |τload| .

Assuming for simplicity that both forces act at angles of 90 withrespect to the lines connecting the axis to the points at which theyact, the absolute values of the torques are

rmuscleFmuscle = rloadFarm ,

where rmuscle, the distance from the elbow joint to the biceps’ pointof insertion on the forearm, is only a few cm, while rload might be 30cm or so. The force exerted by the muscle must therefore be aboutten times the force exerted by the load. We thus see that this leveris a force reducer. In general, a lever may be used either to increaseor to reduce a force.

Why did our arms evolve so as to reduce force? In general,your body is built for compactness and maximum speed of motionrather than maximum force. This is the main anatomical differencebetween us and the Neanderthals (their brains covered the samerange of sizes as those of modern humans), and it seems to haveworked for us.

As with all machines, the lever is incapable of changing theamount of mechanical work we can do. A lever that increases forcewill always reduce motion, and vice versa, leaving the amount ofwork unchanged.

It is worth noting how simple and yet how powerful this analysiswas. It was simple because we were well prepared with the conceptsof torque and mechanical work. In anatomy textbooks, whose read-ers are assumed not to know physics, there is usually a long andcomplicated discussion of the different types of levers. For example,the biceps lever, z, would be classified as a class III lever, since ithas the fulcrum and load on the ends and the muscle’s force actingin the middle. The triceps, aa, is called a class I lever, because theload and muscle’s force are on the ends and the fulcrum is in themiddle. How tiresome! With a firm grasp of the concept of torque,we realize that all such examples can be analyzed in much the sameway. Physics is at its best when it lets us understand many appar-ently complicated phenomena in terms of a few simple yet powerfulconcepts.


ab / The two atoms cover thesame angle in a given timeinterval.

ac / Their velocity vectors,however, differ in both magnitudeand direction.

15.7 Rigid-body rotationKinematics

When a rigid object rotates, every part of it (every atom) movesin a circle, covering the same angle in the same amount of time,ab. Every atom has a different velocity vector, ac. Since all thevelocities are different, we can’t measure the speed of rotation ofthe top by giving a single velocity. We can, however, specify itsspeed of rotation consistently in terms of angle per unit time. Letthe position of some reference point on the top be denoted by itsangle θ, measured in a circle around the axis. For reasons that willbecome more apparent shortly, we measure all our angles in radians.Then the change in the angular position of any point on the top canbe written as dθ, and all parts of the top have the same value of dθover a certain time interval dt. We define the angular velocity, ω(Greek omega),

ω =dθ

dt,

[definition of angular velocity; θ in units of radians]

which is similar to, but not the same as, the quantity ω we definedearlier to describe vibrations. The relationship between ω and tis exactly analogous to that between x and t for the motion of aparticle through space.

self-check BIf two different people chose two different reference points on the topin order to define θ=0, how would their θ-t graphs differ? What effectwould this have on the angular velocities? . Answer, p. 513

The angular velocity has units of radians per second, rad/s.However, radians are not really units at all. The radian measureof an angle is defined, as the length of the circular arc it makes,divided by the radius of the circle. Dividing one length by anothergives a unitless quantity, so anything with units of radians is reallyunitless. We can therefore simplify the units of angular velocity, andcall them inverse seconds, s−1.

A 78-rpm record example 12. In the early 20th century, the standard format for music record-ings was a plastic disk that held a single song and rotated at 78rpm (revolutions per minute). What was the angular velocity ofsuch a disk?

. If we measure angles in units of revolutions and time in unitsof minutes, then 78 rpm is the angular velocity. Using standardphysics units of radians/second, however, we have

78 revolutions1 minute

× 2π radians1 revolution

× 1 minute60 seconds

= 8.2 s−1 .

Section 15.7 Rigid-body rotation 405

ad / Analogies between rota-tional and linear quantities.

ae / We construct a coordi-nate system that coincides withthe location and motion of themoving point of interest at acertain moment.

In the absence of any torque, a rigid body will rotate indefinitelywith the same angular velocity. If the angular velocity is changingbecause of a torque, we define an angular acceleration,

α =dω

dt, [definition of angular acceleration]

The symbol is the Greek letter alpha. The units of this quantity arerad/s2, or simply s−2.

The mathematical relationship between ω and θ is the same asthe one between v and x, and similarly for α and a. We can thusmake a system of analogies, ad, and recycle all the familiar kinematicequations for constant-acceleration motion.

The synodic period example 13Mars takes nearly twice as long as the Earth to complete an orbit.If the two planets are alongside one another on a certain day,then one year later, Earth will be back at the same place, butMars will have moved on, and it will take more time for Earth tofinish catching up. Angular velocities add and subtract, just asvelocity vectors do. If the two planets’ angular velocities are ω1and ω2, then the angular velocity of one relative to the other isω1−ω2. The corresponding period, 1/(1/T1−1/T2) is known asthe synodic period.

A neutron star example 14. A neutron star is initially observed to be rotating with an angularvelocity of 2.0 s−1, determined via the radio pulses it emits. If itsangular acceleration is a constant −1.0 × 10−8 s−2, how manyrotations will it complete before it stops? (In reality, the angularacceleration is not always constant; sudden changes often occur,and are referred to as “starquakes!”)

. The equation v2f −v2

i =2a∆x can be translated intoω2f −ω

2i =2α∆θ,

giving

∆θ = (ω2f −ω

2i )/2α

= 2.0× 108 radians

= 3.2× 107 rotations .

Relations between angular quantities and motion of a point

It is often necessary to be able to relate the angular quantitiesto the motion of a particular point on the rotating object. As wedevelop these, we will encounter the first example where the advan-tages of radians over degrees become apparent.

The speed at which a point on the object moves depends on boththe object’s angular velocity ω and the point’s distance r from theaxis. We adopt a coordinate system, ae, with an inward (radial)


af / Even if the rotating object haszero angular acceleration, everypoint on it has an accelerationtowards the center.

axis and a tangential axis. The length of the infinitesimal circulararc ds traveled by the point in a time interval dt is related to dθby the definition of radian measure, dθ = ds/r, where positive andnegative values of ds represent the two possible directions of motionalong the tangential axis. We then have vt = ds/dt = rdθ/dt = ωr,or

vt = ωr . [tangential velocity of a point at a

distance r from the axis of rotation]

The radial component is zero, since the point is not moving inwardor outward,

vr = 0 . [radial velocity of a point at a

distance r from the axis of rotation]

Note that we had to use the definition of radian measure in thisderivation. Suppose instead we had used units of degrees for our an-gles and degrees per second for angular velocities. The relationshipbetween dθdegrees and ds is dθdegrees = (360/2π)s/r, where the extraconversion factor of (360/2π) comes from that fact that there are 360degrees in a full circle, which is equivalent to 2π radians. The equa-tion for vt would then have been vt = (2π/360)(ωdegrees per second)(r),which would have been much messier. Simplicity, then, is the rea-son for using radians rather than degrees; by using radians we avoidinfecting all our equations with annoying conversion factors.

Since the velocity of a point on the object is directly proportionalto the angular velocity, you might expect that its acceleration wouldbe directly proportional to the angular acceleration. This is not true,however. Even if the angular acceleration is zero, i.e., if the objectis rotating at constant angular velocity, every point on it will havean acceleration vector directed toward the axis, af. As derived onpage 229, the magnitude of this acceleration is

ar = ω2r . [radial acceleration of a point

at a distance r from the axis]

For the tangential component, any change in the angular velocitydω will lead to a change dω ·r in the tangential velocity, so it is easilyshown that

at = αr . [tangential acceleration of a point


self-check CPositive and negative signs of ω represent rotation in opposite direc-tions. Why does it therefore make sense physically that ω is raised tothe first power in the equation for vt and to the second power in the onefor ar ? . Answer, p. 513


Radial acceleration at the surface of the Earth example 15. What is your radial acceleration due to the rotation of the earthif you are at the equator?

. At the equator, your distance from the Earth’s rotation axis isthe same as the radius of the spherical Earth, 6.4 × 106 m. Yourangular velocity is

ω =2π radians

1 day= 7.3× 10−5 s−1 ,

which gives an acceleration of

ar = ω2r

= 0.034 m/s2 .

The angular velocity was a very small number, but the radius wasa very big number. Squaring a very small number, however, givesa very very small number, so the ω2 factor “wins,” and the finalresult is small.

If you’re standing on a bathroom scale, this small acceleration isprovided by the imbalance between the downward force of gravityand the slightly weaker upward normal force of the scale on yourfoot. The scale reading is therefore a little lower than it shouldbe.

Dynamics

If we want to connect all this kinematics to anything dynamical,we need to see how it relates to torque and angular momentum.Our strategy will be to tackle angular momentum first, since angu-lar momentum relates to motion, and to use the additive propertyof angular momentum: the angular momentum of a system of par-ticles equals the sum of the angular momenta of all the individualparticles. The angular momentum of one particle within our rigidlyrotating object, L = mv⊥r, can be rewritten as L = r p sin θ,where r and p are the magnitudes of the particle’s r and momen-tum vectors, and θ is the angle between these two vectors. (The rvector points outward perpendicularly from the axis to the parti-cle’s position in space.) In rigid-body rotation the angle θ is 90 ,so we have simply L = rp. Relating this to angular velocity, wehave L = rp = (r)(mv) = (r)(mωr) = mr2ω. The particle’s con-tribution to the total angular momentum is proportional to ω, witha proportionality constant mr2. We refer to mr2 as the particle’scontribution to the object’s total moment of inertia, I, where “mo-ment” is used in the sense of “important,” as in “momentous” — abigger value of I tells us the particle is more important for deter-mining the total angular momentum. The total moment of inertia


ag / Analogies between rota-tional and linear quantities.

ah / Example 16

is

I =∑

mir2i , [definition of the moment of inertia;

for rigid-body rotation in a plane; r is the distance

from the axis, measured perpendicular to the axis]

The angular momentum of a rigidly rotating body is then

L = Iω . [angular momentum of

rigid-body rotation in a plane]

Since torque is defined as dL/dt, and a rigid body has a constantmoment of inertia, we have τ = dL/dt = Idω/dt = Iα,

τ = Iα , [relationship between torque and

angular acceleration for rigid-body rotation in a plane]

which is analogous to F = ma.

The complete system of analogies between linear motion andrigid-body rotation is given in figure ag.

A barbell example 16. The barbell shown in figure ah consists of two small, dense,massive balls at the ends of a very light rod. The balls havemasses of 2.0 kg and 1.0 kg, and the length of the rod is 3.0m. Find the moment of inertia of the rod (1) for rotation about itscenter of mass, and (2) for rotation about the center of the moremassive ball.

. (1) The ball’s center of mass lies 1/3 of the way from the greatermass to the lesser mass, i.e., 1.0 m from one and 2.0 m from theother. Since the balls are small, we approximate them as if theywere two pointlike particles. The moment of inertia is

I = (2.0 kg)(1.0 m)2 + (1.0 kg)(2.0 m)2

= 2.0 kg·m2 + 4.0 kg·m2

= 6.0 kg·m2

Perhaps counterintuitively, the less massive ball contributes farmore to the moment of inertia.

(2) The big ball theoretically contributes a little bit to the momentof inertia, since essentially none of its atoms are exactly at r=0.However, since the balls are said to be small and dense, we as-sume all the big ball’s atoms are so close to the axis that we canignore their small contributions to the total moment of inertia:

I = (1.0 kg)(3.0 m)2

= 9.0 kg·m2

This example shows that the moment of inertia depends on thechoice of axis. For example, it is easier to wiggle a pen about itscenter than about one end.


The parallel axis theorem example 17. Generalizing the previous example, suppose we pick any axisparallel to axis 1, but offset from it by a distance h. Part (2) ofthe previous example then corresponds to the special case ofh = −1.0 m (negative being to the left). What is the moment ofinertia about this new axis?

. The big ball’s distance from the new axis is (1.0 m)+h, and thesmall one’s is (2.0 m)-h. The new moment of inertia is

I = (2.0 kg)[(1.0 m)+h]2 + (1.0 kg)[(2.0 m)− h]2

= 6.0 kg·m2 + (4.0 kg·m)h − (4.0 kg·m)h + (3.0 kg)h2 .

The constant term is the same as the moment of inertia about thecenter-of-mass axis, the first-order terms cancel out, and the thirdterm is just the total mass multiplied by h2. The interested readerwill have no difficulty in generalizing this to any set of particles,resulting in the parallel axis theorem: If an object of total mass Mrotates about a line at a distance h from its center of mass, thenits moment of inertia equals Icm + Mh2, where Icm is the momentof inertia for rotation about a parallel line through the center ofmass.

Scaling of the moment of inertia example 18. (1) Suppose two objects have the same mass and the sameshape, but one is less dense, and larger by a factor k . How dotheir moments of inertia compare?(2) What if the densities are equal rather than the masses?

. (1) This is like increasing all the distances between atoms by afactor k . All the r ’s become greater by this factor, so the momentof inertia is increased by a factor of k2.(2) This introduces an increase in mass by a factor of k3, so themoment of inertia of the bigger object is greater by a factor ofk5.

Iterated integrals

In various places in this book, starting with subsection 15.7.5,we’ll come across integrals stuck inside other integrals. These areknown as iterated integrals, or double integrals, triple integrals, etc.Similar concepts crop up all the time even when you’re not doingcalculus, so let’s start by imagining such an example. Suppose youwant to count how many squares there are on a chess board, and youdon’t know how to multiply eight times eight. You could start fromthe upper left, count eight squares across, then continue with thesecond row, and so on, until you how counted every square, givingthe result of 64. In slightly more formal mathematical language,we could write the following recipe: for each row, r, from 1 to 8,consider the columns, c, from 1 to 8, and add one to the count for


each one of them. Using the sigma notation, this becomes

8∑r=1

8∑c=1

1 .

If you’re familiar with computer programming, then you can thinkof this as a sum that could be calculated using a loop nested insideanother loop. To evaluate the result (again, assuming we don’t knowhow to multiply, so we have to use brute force), we can first evaluatethe inside sum, which equals 8, giving

8∑r=1

8 .

Notice how the “dummy” variable c has disappeared. Finally we dothe outside sum, over r, and find the result of 64.

Now imagine doing the same thing with the pixels on a TVscreen. The electron beam sweeps across the screen, painting thepixels in each row, one at a time. This is really no different than theexample of the chess board, but because the pixels are so small, younormally think of the image on a TV screen as continuous ratherthan discrete. This is the idea of an integral in calculus. Supposewe want to find the area of a rectangle of width a and height b, andwe don’t know that we can just multiply to get the area ab. Thebrute force way to do this is to break up the rectangle into a grid ofinfinitesimally small squares, each having width dx and height dy,and therefore the infinitesimal area dA = dxdy. For convenience,we’ll imagine that the rectangle’s lower left corner is at the origin.Then the area is given by this integral:

area =

∫ b

y=0

∫ a

x=0dA

=

∫ b

y=0

∫ a

x=0dx dy

Notice how the leftmost integral sign, over y, and the rightmost dif-ferential, dy, act like bookends, or the pieces of bread on a sandwich.Inside them, we have the integral sign that runs over x, and the dif-ferential dx that matches it on the right. Finally, on the innermostlayer, we’d normally have the thing we’re integrating, but here’s it’s1, so I’ve omitted it. Writing the lower limits of the integrals withx = and y = helps to keep it straight which integral goes with with


differential. The result is

area =

∫ b

y=0

∫ a

x=0dA

=

∫ b

y=0

∫ a

x=0dx dy

=

∫ b

y=0

(∫ a

x=0dx

)dy

=

∫ b

y=0a dy

= a

∫ b

y=0dy

= ab .

Area of a triangle example 19. Find the area of a 45-45-90 right triangle having legs a.

. Let the triangle’s hypotenuse run from the origin to the point(a, a), and let its legs run from the origin to (0, a), and then to(a, a). In other words, the triangle sits on top of its hypotenuse.Then the integral can be set up the same way as the one before,but for a particular value of y , values of x only run from 0 (on they axis) to y (on the hypotenuse). We then have

area =∫ a

y=0

∫ y

x=0dA

=∫ a

y=0

∫ y

x=0dx dy

=∫ a

y=0

(∫ y

x=0dx)

dy

=∫ a

y=0y dy

=12

a2

Note that in this example, because the upper end of the x valuesdepends on the value of y , it makes a difference which order wedo the integrals in. The x integral has to be on the inside, and wehave to do it first.

Volume of a cube example 20. Find the volume of a cube with sides of length a.

. This is a three-dimensional example, so we’ll have integralsnested three deep, and the thing we’re integrating is the volumedV = dxdydz.


volume =∫ a

z=0

∫ a

y=0

∫ a

x=0dx dy dz

=∫ a

z=0

∫ a

y=0a dy dz

= a∫ a

z=0

∫ a

y=0dy dz

= a∫ a

z=0a dz

= a3

Area of a circle example 21. Find the area of a circle.

. To make it easy, let’s find the area of a semicircle and thendouble it. Let the circle’s radius be r , and let it be centered on theorigin and bounded below by the x axis. Then the curved edgeis given by the equation r2 = x2 + y2, or y =

√r2 − x2. Since

the y integral’s limit depends on x , the x integral has to be on theoutside. The area is

area =∫ r

x=−r

∫ √r2−x2

y=0dy dx

=∫ r

x=−r

√r2 − x2dx

= r∫ r

x=−r

√1− (x/r )2 dx .

Substituting u = x/r ,

area = r2∫ 1

u=−1

√1− u2 du

The definite integal equals π, as you can find using a trig substi-tution or simply by looking it up in a table, and the result is, asexpected, πr2/2 for the area of the semicircle.

Finding moments of inertia by integration

When calculating the moment of inertia of an ordinary-sized ob-ject with perhaps 1026 atoms, it would be impossible to do an actualsum over atoms, even with the world’s fastest supercomputer. Cal-culus, however, offers a tool, the integral, for breaking a sum downto infinitely many small parts. If we don’t worry about the exis-tence of atoms, then we can use an integral to compute a moment


of inertia as if the object was smooth and continuous throughout,rather than granular at the atomic level. Of course this granular-ity typically has a negligible effect on the result unless the objectis itself an individual molecule. This subsection consists of threeexamples of how to do such a computation, at three distinct levelsof mathematical complication.

Moment of inertia of a thin rod

What is the moment of inertia of a thin rod of mass M andlength L about a line perpendicular to the rod and passing throughits center? We generalize the discrete sum

I =∑

mir2i

to a continuous one,

I =

∫r2dm

=

∫ L/2

−L/2x2 M

Ldx [r = |x|, so r2 = x2]

=1

12ML2

In this example the object was one-dimensional, which madethe math simple. The next example shows a strategy that can beused to simplify the math for objects that are three-dimensional,but possess some kind of symmetry.

Moment of inertia of a disk

What is the moment of inertia of a disk of radius b, thickness t,and mass M , for rotation about its central axis?

We break the disk down into concentric circular rings of thick-ness dr. Since all the mass in a given circular slice has essentiallythe same value of r (ranging only from r to r + dr), the slice’s con-tribution to the total moment of inertia is simply r2dm. We thenhave

I =

∫r2dm

=

∫r2ρ dV ,

where V = πb2t is the total volume, ρ = M/V = M/πb2t is thedensity, and the volume of one slice can be calculated as the volumeenclosed by its outer surface minus the volume enclosed by its innersurface, dV = π(r + dr)2t− πr2t = 2πtrdr.

I =

∫ b

0r2 M

πb2t2πt r dr

=1

2Mb2 .


In the most general case where there is no symmetry about therotation axis, we must use iterated integrals, as discussed in subsec-tion 15.7.4. The example of the disk possessed two types of symme-try with respect to the rotation axis: (1) the disk is the same whenrotated through any angle about the axis, and (2) all slices perpen-dicular to the axis are the same. These two symmetries reduced thenumber of layers of integrals from three to one. The following ex-ample possesses only one symmetry, of type (2), and we simply setit up as a triple integral. You may not have seen multiple integralsyet in a math course. If so, just skim this example.

Moment of inertia of a cube

What is the moment of inertia of a cube of side b, for rotationabout an axis that passes through its center and is parallel to fourof its faces? Let the origin be at the center of the cube, and let xbe the rotation axis.

I =

∫r2dm

= ρ

∫r2dV

= ρ

∫ b/2

−b/2

∫ b/2

−b/2

∫ b/2

−b/2

(y2 + z2

)dx dy dz

= ρb

∫ b/2

−b/2

∫ b/2

−b/2

(y2 + z2

)dy dz

The fact that the last step is a trivial integral results from the sym-metry of the problem. The integrand of the remaining double in-tegral breaks down into two terms, each of which depends on onlyone of the variables, so we break it into two integrals,

I = ρb

∫ b/2

−b/2

∫ b/2

−b/2y2dy dz + ρb

∫ b/2

−b/2

∫ b/2

−b/2z2dy dz

which we know have identical results. We therefore only need toevaluate one of them and double the result:

I = 2ρb

∫ b/2

−b/2

∫ b/2

−b/2z2dy d z

= 2ρb2∫ b/2

−b/2z2dz

=1

6ρb5

=1

6Mb2

Figure ai shows the moments of inertia of some shapes, whichwere evaluated with techniques like these.


aj / Example 23.

ai / Moments of inertia of somegeometric shapes.

The hammer throw example 22. In the men’s Olympic hammer throw, a steel ball of radius 6.1 cmis swung on the end of a wire of length 1.22 m. What fraction ofthe ball’s angular momentum comes from its rotation, as opposedto its motion through space?

. It’s always important to solve problems symbolically first, andplug in numbers only at the end, so let the radius of the ball be b,and the length of the wire `. If the time the ball takes to go oncearound the circle is T , then this is also the time it takes to revolveonce around its own axis. Its speed is v = 2π`/T , so its angularmomentum due to its motion through space is mv` = 2πm`2/T .Its angular momentum due to its rotation around its own cen-ter is (4π/5)mb2/T . The ratio of these two angular momenta is(2/5)(b/`)2 = 1.0×10−3. The angular momentum due to the ball’sspin is extremely small.

Toppling a rod example 23. A rod of length b and mass m stands upright. We want to strikethe rod at the bottom, causing it to fall and land flat. Find themomentum, p, that should be delivered, in terms of m, b, andg. Can this really be done without having the rod scrape on thefloor?

. This is a nice example of a question that can very nearly beanswered based only on units. Since the three variables, m, b,and g, all have different units, they can’t be added or subtracted.The only way to combine them mathematically is by multiplicationor division. Multiplying one of them by itself is exponentiation, soin general we expect that the answer must be of the form

p = Amjbkg l ,

where A, j , k , and l are unitless constants. The result has to haveunits of kg·m/s. To get kilograms to the first power, we need

j = 1 ,


meters to the first power requires

k + l = 1 ,

and seconds to the power −1 implies

l = 1/2 .

We find j = 1, k = 1/2, and l = 1/2, so the solution must be of theform

p = Am√

bg .

Note that no physics was required!

Consideration of units, however, won’t help us to find the unit-less constant A. Let t be the time the rod takes to fall, so that(1/2)gt2 = b/2. If the rod is going to land exactly on its side,then the number of revolutions it completes while in the air mustbe 1/4, or 3/4, or 5/4, . . . , but all the possibilities greater than 1/4would cause the head of the rod to collide with the floor prema-turely. The rod must therefore rotate at a rate that would causeit to complete a full rotation in a time T = 4t , and it has angularmomentum L = (π/6)mb2/T .

The momentum lost by the object striking the rod is p, and byconservation of momentum, this is the amount of momentum, inthe horizontal direction, that the rod acquires. In other words,the rod will fly forward a little. However, this has no effect onthe solution to the problem. More importantly, the object strikingthe rod loses angular momentum bp/2, which is also transferredto the rod. Equating this to the expression above for L, we findp = (π/12)m

√bg.

Finally, we need to know whether this can really be done withouthaving the foot of the rod scrape on the floor. The figure showsthat the answer is no for this rod of finite width, but it appearsthat the answer would be yes for a sufficiently thin rod. This isanalyzed further in homework problem 46 on page 443.


15.8 Angular momentum in three dimensionsConservation of angular momentum produces some surprising phe-nomena when extended to three dimensions. Try the following ex-periment, for example. Take off your shoe, and toss it in to the air,making it spin along its long (toe-to-heel) axis. You should observea nice steady pattern of rotation. The same happens when you spinthe shoe about its shortest (top-to-bottom) axis. But somethingunexpected happens when you spin it about its third (left-to-right)axis, which is intermediate in length between the other two. Insteadof a steady pattern of rotation, you will observe something morecomplicated, with the shoe changing its orientation with respect tothe rotation axis.

Rigid-body kinematics in three dimensions

How do we generalize rigid-body kinematics to three dimensions?When we wanted to generalize the kinematics of a moving particleto three dimensions, we made the numbers r, v, and a into vectorsr, v, and a. This worked because these quantities all obeyed thesame laws of vector addition. For instance, one of the laws of vectoraddition is that, just like addition of numbers, vector addition givesthe same result regardless of the order of the two quantities beingadded. Thus you can step sideways 1 m to the right and thenstep forward 1 m, and the end result is the same as if you steppedforward first and then to the side. In order words, it didn’t matterwhether you took ∆r1 + ∆r2 or ∆r2 + ∆r1. In math this is calledthe commutative property of addition.

ak / Performing the rotations inone order gives one result, 3,while reversing the order gives adifferent result, 5.

Angular motion, unfortunately doesn’t have this property, asshown in figure ak. Doing a rotation about the x axis and then


al / The right-hand rule forassociating a vector with adirection of rotation.

about y gives one result, while doing them in the opposite ordergives a different result. These operations don’t “commute,” i.e., itmakes a difference what order you do them in.

This means that there is in general no possible way to constructa ∆θ vector. However, if you try doing the operations shown infigure ak using small rotation, say about 10 degrees instead of 90,you’ll find that the result is nearly the same regardless of whatorder you use; small rotations are very nearly commutative. Notonly that, but the result of the two 10-degree rotations is about thesame as a single, somewhat larger, rotation about an axis that liessymmetrically at between the x and y axes at 45 degree angles toeach one. This is exactly what we would expect if the two smallrotations did act like vectors whose directions were along the axisof rotation. We therefore define a dθ vector whose magnitude isthe amount of rotation in units of radians, and whose direction isalong the axis of rotation. Actually this definition is ambiguous,because there it could point in either direction along the axis. Wetherefore use a right-hand rule as shown in figure al to define thedirection of the dθ vector, and the ω vector, ω = dθ/dt, based on it.Aliens on planet Tammyfaye may decide to define it using their lefthands rather than their right, but as long as they keep their scientificliterature separate from ours, there is no problem. When entering aphysics exam, always be sure to write a large warning note on yourleft hand in magic marker so that you won’t be tempted to use itfor the right-hand rule while keeping your pen in your right.

self-check DUse the right-hand rule to determine the directions of the ω vectors ineach rotation shown in figures ak/1 through ak/5. . Answer, p. 513

Because the vector relationships among dθ, ω, and α are strictlyanalogous to the ones involving dr, v, and a (with the proviso thatwe avoid describing large rotations using ∆θ vectors), any operationin r-v-a vector kinematics has an exact analog in θ-ω-α kinematics.

Result of successive 10-degree rotations example 24. What is the result of two successive (positive) 10-degree rota-tions about the x and y axes? That is, what single rotation about asingle axis would be equivalent to executing these in succession?

. The result is only going to be approximate, since 10 degreesis not an infinitesimally small angle, and we are not told in whatorder the rotations occur. To some approximation, however, wecan add the ∆θ vectors in exactly the same way we would add ∆rvectors, so we have

∆θ ≈ ∆θ1 + ∆θ2

≈ (10 degrees)x + (10 degrees)y .

This is a vector with a magnitude of√

(10 deg)2 + (10 deg)2 =

Section 15.8 Angular momentum in three dimensions 419

14 deg, and it points along an axis midway between the x and yaxes.

Angular momentum in three dimensions

The vector cross product

In order to expand our system of three-dimensional kinematics toinclude dynamics, we will have to generalize equations like vt = ωr,τ = rF sin θrF , and L = rp sin θrp, each of which involves threequantities that we have either already defined as vectors or that wewant to redefine as vectors. Although the first one appears to differfrom the others in its form, it could just as well be rewritten asvt = ωr sin θωr, since θωr = 90 , and sin θωr = 1.

It thus appears that we have discovered something general aboutthe physically useful way to relate three vectors in a multiplicativeway: the magnitude of the result always seems to be proportional tothe product of the magnitudes of the two vectors being “multiplied,”and also to the sine of the angle between them.

Is this pattern just an accident? Actually the sine factor hasa very important physical property: it goes to zero when the twovectors are parallel. This is a Good Thing. The generalization ofangular momentum into a three-dimensional vector, for example, ispresumably going to describe not just the clockwise or counterclock-wise nature of the motion but also from which direction we wouldhave to view the motion so that it was clockwise or counterclock-wise. (A clock’s hands go counterclockwise as seen from behind theclock, and don’t rotate at all as seen from above or to the side.) Nowsuppose a particle is moving directly away from the origin, so thatits r and p vectors are parallel. It is not going around the originfrom any point of view, so its angular momentum vector had betterbe zero.

Thinking in a slightly more abstract way, we would expect theangular momentum vector to point perpendicular to the plane ofmotion, just as the angular velocity vector points perpendicular tothe plane of motion. The plane of motion is the plane containingboth r and p, if we place the two vectors tail-to-tail. But if r andp are parallel and are placed tail-to-tail, then there are infinitelymany planes containing them both. To pick one of these planes inpreference to the others would violate the symmetry of space, sincethey should all be equally good. Thus the zero-if-parallel propertyis a necessary consequence of the underlying symmetry of the lawsof physics.

The following definition of a kind of vector multiplication is con-sistent with everything we’ve seen so far, and later we’ll prove thatthe definition is unique, i.e., if we believe in the symmetry of space,it is essentially the only way of defining the multiplication of two


am / The right-hand rule forthe direction of the vector crossproduct.

an / The magnitude of thecross product is the area of theshaded parallelogram.

ao / A cyclic change in the x ,y , and z subscripts.

vectors to produce a third vector:

Definition of the vector cross product:The cross product A × B of two vectors A and B is defined asfollows:(1) Its magnitude is defined by |A×B| = |A||B| sin θAB, where θABis the angle between A and B when they are placed tail-to-tail.(2) Its direction is along the line perpendicular to both A and B.Of the two such directions, it is the one that obeys the right-handrule shown in figure am.

The name “cross product” refers to the symbol, and distin-guishes it from the dot product, which acts on two vectors butproduces a scalar.

Although the vector cross-product has nearly all the propertiesof numerical multiplication, e.g., A × (B + C) = A × B + A ×C, it lacks the usual property of commutativity. Try applying theright-hand rule to find the direction of the vector cross productB × A using the two vectors shown in the figure. This requiresstarting with a flattened hand with the four fingers pointing alongB, and then curling the hand so that the fingers point along A. Theonly possible way to do this is to point your thumb toward the floor,in the opposite direction. Thus for the vector cross product we have

A×B = −B×A ,

a property known as anticommutativity. The vector cross productis also not associative, i.e., A× (B×C) is usually not the same as(A×B)×C.

A geometric interpretation of the cross product, an, is that ifboth A and B are vectors with units of distance, then the mag-nitude of their cross product can be interpreted as the area of theparallelogram they form when placed tail-to-tail.

A useful expression for the components of the vector cross prod-uct in terms of the components of the two vectors being multipliedis as follows:

(A×B)x = AyBz −ByAz(A×B)y = AzBx −BzAx(A×B)z = AxBy −BxAy

I’ll prove later that these expressions are equivalent to the previ-ous definition of the cross product. Although they may appearformidable, they have a simple structure: the subscripts on the rightare the other two besides the one on the left, and each equation isrelated to the preceding one by a cyclic change in the subscripts,


ap / The position and momentumvectors of an atom in the spinningtop.

aq / The right-hand rule forthe atom’s contribution to theangular momentum.

ao. If the subscripts were not treated in some completely symmet-ric manner like this, then the definition would provide some way todistinguish one axis from another, which would violate the symme-try of space.

self-check EShow that the component equations are consistent with the rule A×B =−B× A. . Answer, p. 513

Angular momentum in three dimensions

In terms of the vector cross product, we have:

v = ω × r

L = r× p

τ = r× F

But wait, how do we know these equations are even correct?For instance, how do we know that the quantity defined by r × pis in fact conserved? Well, just as we saw on page 332 that thedot product is unique (i.e., can only be defined in one way whileobserving rotational invariance), the cross product is also unique,as proved on page 430. If r×p was not conserved, then there couldnot be any generally conserved quantity that would reduce to ourold definition of angular momentum in the special case of planerotation. This doesn’t prove conservation of angular momentum— only experiments can prove that — but it does prove that ifangular momentum is conserved in three dimensions, there is onlyone possible way to generalize from two dimensions to three.

Angular momentum of a spinning top example 25As an illustration, we consider the angular momentum of a spin-ning top. Figures ap and aq show the use of the vector crossproduct to determine the contribution of a representative atom tothe total angular momentum. Since every other atom’s angularmomentum vector will be in the same direction, this will also bethe direction of the total angular momentum of the top. This hap-pens to be rigid-body rotation, and perhaps not surprisingly, theangular momentum vector is along the same direction as the an-gular velocity vector.

Three important points are illustrated by this example: (1)When we do the full three-dimensional treatment of angular mo-mentum, the “axis” from which we measure the position vectors isjust an arbitrarily chosen point. If this had not been rigid-bodyrotation, we would not even have been able to identify a single lineabout which every atom circled. (2) Starting from figure ap, we hadto rearrange the vectors to get them tail-to-tail before applying theright-hand rule. If we had attempted to apply the right-hand ruleto figure ap, the direction of the result would have been exactly theopposite of the correct answer. (3) The equation L = r× p cannot


ar / A top is supported at itstip by a pinhead. (More practicaldevices to demonstrate thiswould use a double bearing.)

as / The torque made by gravityis in the horizontal plane.

at / The ∆L vector is in thesame direction as the torque, outof the page.

be applied all at once to an entire system of particles. The totalmomentum of the top is zero, which would give an erroneous resultof zero angular momentum (never mind the fact that r is not welldefined for the top as a whole).

Doing the right-hand rule like this requires some practice. Iurge you to make models like aq out of rolled up pieces of paper andto practice with the model in various orientations until it becomesnatural.

Precession example 26Figure ar shows a counterintuitive example of the concepts we’vebeen discussing. One expects the torque due to gravity to causethe top to flop down. Instead, the top remains spinning in the hor-izontal plane, but its axis of rotation starts moving in the directionshown by the shaded arrow. This phenomenon is called preces-sion. Figure as shows that the torque due to gravity is out of thepage. (Actually we should add up all the torques on all the atomsin the top, but the qualitative result is the same.) Since torqueis the rate of change of angular momentum, τ = dL/dt , the ∆Lvector must be in the same direction as the torque (division bya positive scalar doesn’t change the direction of the vector). Asshown in at, this causes the angular momentum vector to twist inspace without changing its magnitude.

For similar reasons, the Earth’s axis precesses once every 26,000years (although not through a great circle, since the angle betweenthe axis and the force isn’t 90 degrees as in figure ar). This pre-cession is due to a torque exerted by the moon. If the Earth wasa perfect sphere, there could be no precession effect due to sym-metry. However, the Earth’s own rotation causes it to be slightlyflattened (oblate) relative to a perfect sphere, giving it “love han-dles” on which the moon’s gravity can act. The moon’s gravity onthe nearer side of the equatorial bulge is stronger, so the torques donot cancel out perfectly. Presently the earth’s axis very nearly linesup with the star Polaris, but in 12,000 years, the pole star will beVega instead.

The frisbee example 27The flow of the air over a flying frisbee generates lift, and the liftat the front and back of the frisbee isn’t necessarily balanced. Ifyou throw a frisbee without rotating it, as if you were shooting abasketball with two hands, you’ll find that it pitches, i.e., its nosegoes either up or down. When I do this with my frisbee, it goesnose down, which apparently means that the lift at the back ofthe disc is greater than the lift at the front. The two torques areunbalanced, resulting in a total torque that points to the left.

The way you actually throw a frisbee is with one hand, putting alot of spin on it. If you throw backhand, which is how most peo-ple first learn to do it, the angular momentum vector points down


au / Example 28.

(assuming you’re right-handed). On my frisbee, the aerodynamictorque to the left would therefore tend to make the angular mo-mentum vector precess in the clockwise direction as seen by thethrower. This would cause the disc to roll to the right, and there-fore follow a curved trajectory. Some specialized discs, used inthe sport of disc golf, are actually designed intentionally to showthis behavior; they’re known as “understable” discs. However, thetypical frisbee that most people play with is designed to be stable:as the disc rolls to one side, the airflow around it is altered in waythat tends to bring the disc back into level flight. Such a disc willtherefore tend to fly in a straight line, provided that it is thrownwith enough angular momentum.

Finding a cross product by components example 28. What is the torque produced by a force given by x + 2y + 3z (inunits of Newtons) acting on a point whose radius vector is 4x + 2y(in meters)?

. It’s helpful to make a table of the components as shown in thefigure. The results are

τx = ryFz − Fy rz = 15 N·mτy = rzFx − Fzrx =−12 N·mτz = rxFy − Fx ry = 3 N·m

Torque and angular momentum example 29In this example, we prove explicitly the consistency of the equa-

tions involving torque and angular momentum that we provedabove based purely on symmetry. Starting from the definition oftorque, we have

τ =dLdt

=ddt

∑i

ri × pi

=∑

i

ddt

(ri × pi ) .

The derivative of a cross product can be evaluated in the sameway as the derivative of an ordinary scalar product:

τ =∑

i

[(dri

dt× pi

)+(

ri ×dpi

dt

)]The first term is zero for each particle, since the velocity vector isparallel to the momentum vector. The derivative appearing in thesecond term is the force acting on the particle, so

τ =∑

i

ri × Fi ,

which is the relationship we set out to prove.


Rigid-body dynamics in three dimensions

The student who is not madly in love with mathematics maywish to skip the rest of this section after absorbing the statementthat, for a typical, asymmetric object, the angular momentum vectorand the angular velocity vector need not be parallel. That is, onlyfor a body that possesses symmetry about the rotation axis is it truethat L = Iω (the rotational equivalent of p = mv) for some scalarI.

Let’s evaluate the angular momentum of a rigidly rotating sys-tem of particles:

L =∑i

ri × pi

=∑i

miri × vi

=∑i

miri × (ω × ri)

An important mathematical skill is to know when to give up andback off. This is a complicated expression, and there is no reasonto expect it to simplify and, for example, take the form of a scalarmultiplied by ω. Instead we examine its general characteristics. Ifwe expanded it using the equation that gives the components of avector cross product, every term would have one of the ω compo-nents raised to the first power, multiplied by a bunch of other stuff.The most general possible form for the result is

Lx = Ixxωx + Ixyωy + Ixzωz

Ly = Iyxωx + Iyyωy + Iyzωz

Lz = Izxωx + Izyωy + Izzωz ,

which you may recognize as a case of matrix multiplication. Themoment of inertia is not a scalar, and not a three-component vector.It is a matrix specified by nine numbers, called its matrix elements.

The elements of the moment of inertia matrix will depend on ourchoice of a coordinate system. In general, there will be some specialcoordinate system, in which the matrix has a simple diagonal form:

Lx = Ixxωx

Ly = Iyyωy

Lz = Izzωz .

The three special axes that cause this simplification are calledthe principal axes of the object, and the corresponding coordinatesystem is the principal axis system. For symmetric shapes such asa rectangular box or an ellipsoid, the principal axes lie along theintersections of the three symmetry planes, but even an asymmetricbody has principal axes.


We can also generalize the plane-rotation equationK = (1/2)Iω2

to three dimensions as follows:

K =∑i

1

2miv

2i

=1

2

∑i

mi(ω × ri) · (ω × ri)

We want an equation involving the moment of inertia, and this hassome evident similarities to the sum we originally wrote down forthe moment of inertia. To massage it into the right shape, we needthe vector identity (A×B)·C = (B×C)·A, which we state withoutproof. We then write

K =1

2

∑i

mi [ri × (ω × ri)] · ω

=1

2ω ·∑i

miri × (ω × ri)

=1

2L · ω

As a reward for all this hard work, let’s analyze the problem ofthe spinning shoe that I posed at the beginning of the chapter. Thethree rotation axes referred to there are approximately the principalaxes of the shoe. While the shoe is in the air, no external torques areacting on it, so its angular momentum vector must be constant inmagnitude and direction. Its kinetic energy is also constant. That’sin the room’s frame of reference, however. The principal axis frameis attached to the shoe, and tumbles madly along with it. In theprincipal axis frame, the kinetic energy and the magnitude of theangular momentum stay constant, but the actual direction of theangular momentum need not stay fixed (as you saw in the caseof rotation that was initially about the intermediate-length axis).Constant |L| gives

L2x + L2

y + L2z = constant .

In the principal axis frame, it’s easy to solve for the componentsof ω in terms of the components of L, so we eliminate ω from theexpression 2K = L · ω, giving

1

IxxL2x +

1

IyyL2y +

1

IzzL2z = constant #2.

The first equation is the equation of a sphere in the three di-mensional space occupied by the angular momentum vector, whilethe second one is the equation of an ellipsoid. The top figure cor-responds to the case of rotation about the shortest axis, which hasthe greatest moment of inertia element. The intersection of the two


av / Visualizing surfaces ofconstant energy and angularmomentum in Lx -Ly -Lz space.

aw / The Explorer I satellite.

surfaces consists only of the two points at the front and back of thesphere. The angular momentum is confined to one of these points,and can’t change its direction, i.e., its orientation with respect to theprincipal axis system, which is another way of saying that the shoecan’t change its orientation with respect to the angular momentumvector. In the bottom figure, the shoe is rotating about the longestaxis. Now the angular momentum vector is trapped at one of thetwo points on the right or left. In the case of rotation about theaxis with the intermediate moment of inertia element, however, theintersection of the sphere and the ellipsoid is not just a pair of iso-lated points but the curve shown with the dashed line. The relativeorientation of the shoe and the angular momentum vector can andwill change.

One application of the moment of inertia tensor is to video gamesthat simulate car racing or flying airplanes.

One more exotic example has to do with nuclear physics. Al-though you have probably visualized atomic nuclei as nothing morethan featureless points, or perhaps tiny spheres, they are often el-lipsoids with one long axis and two shorter, equal ones. Althougha spinning nucleus normally gets rid of its angular momentum viagamma ray emission within a period of time on the order of picosec-onds, it may happen that a deformed nucleus gets into a state inwhich has a large angular momentum is along its long axis, whichis a very stable mode of rotation. Such states can live for secondsor even years! (There is more to the story — this is the topic onwhich I wrote my Ph.D. thesis — but the basic insight applies eventhough the full treatment requires fancy quantum mechanics.)

Our analysis has so far assumed that the kinetic energy of ro-tation energy can’t be converted into other forms of energy such asheat, sound, or vibration. When this assumption fails, then rota-tion about the axis of least moment of inertia becomes unstable,and will eventually convert itself into rotation about the axis whosemoment of inertia is greatest. This happened to the U.S.’s first ar-tificial satellite, Explorer I, launched in 1958. Note the long, floppyantennas, which tended to dissipate kinetic energy into vibration. Ithad been designed to spin about its minimimum-moment-of-inertiaaxis, but almost immediately, as soon as it was in space, it beganspinning end over end. It was nevertheless able to carry out itsscience mission, which didn’t depend on being able to maintain astable orientation, and it discovered the Van Allen radiation belts.

15.9 ? Proof of Kepler’s elliptical orbit lawKepler determined purely empirically that the planets’ orbits wereellipses, without understanding the underlying reason in terms ofphysical law. Newton’s proof of this fact based on his laws of motion

Section 15.9 ? Proof of Kepler’s elliptical orbit law 427

ax / The r −φ representation of acurve.

ay / Proof that the two an-gles labeled φ are in fact equal:The definition of an ellipse is thatthe sum of the distances fromthe two foci stays constant. If wemove a small distance ` along theellipse, then one distance shrinksby an amount ` cosφ1, while theother grows by ` cosφ2. Theseare equal, so φ1 = φ2.

and law of gravity was considered his crowning achievement bothby him and by his contemporaries, because it showed that the samephysical laws could be used to analyze both the heavens and theearth. Newton’s proof was very lengthy, but by applying the morerecent concepts of conservation of energy and angular momentumwe can carry out the proof quite simply and succinctly, and withoutcalculus.

The basic idea of the proof is that we want to describe the shapeof the planet’s orbit with an equation, and then show that this equa-tion is exactly the one that represents an ellipse. Newton’s originalproof had to be very complicated because it was based directly onhis laws of motion, which include time as a variable. To make anystatement about the shape of the orbit, he had to eliminate timefrom his equations, leaving only space variables. But conservationlaws tell us that certain things don’t change over time, so they havealready had time eliminated from them.

There are many ways of representing a curve by an equation, ofwhich the most familiar is y = ax + b for a line in two dimensions.It would be perfectly possible to describe a planet’s orbit using anx − y equation like this, but remember that we are applying con-servation of angular momentum, and the space variables that occurin the equation for angular momentum are the distance from theaxis, r, and the angle between the velocity vector and the r vector,which we will call φ. The planet will have φ=90 when it is movingperpendicular to the r vector, i.e., at the moments when it is at itssmallest or greatest distances from the sun. When φ is less than90 the planet is approaching the sun, and when it is greater than90 it is receding from it. Describing a curve with an r−φ equationis like telling a driver in a parking lot a certain rule for what direc-tion to steer based on the distance from a certain streetlight in themiddle of the lot.

The proof is broken into the three parts for easier digestion.The first part is a simple and intuitively reasonable geometrical factabout ellipses, whose proof we relegate to the caption of figure ay;you will not be missing much if you merely absorb the result withoutreading the proof.

(1) If we use one of the two foci of an ellipse as an axis fordefining the variables r and φ, then the angle between the tangentline and the line drawn to the other focus is the same as φ, i.e., thetwo angles labeled φ in figure ay are in fact equal.

The other two parts form the meat of our proof. We state theresults first and then prove them.

(2) A planet, moving under the influence of the sun’s gravitywith less then the energy required to escape, obeys an equation of


az / Proof of part (3).

the form

sinφ =1√

−pr2 + qr,

where p and q are positive constants that depend on the planet’senergy and angular momentum.

(3) A curve is an ellipse if and only if its r−φ equation is of theform

sinφ =1√

−pr2 + qr,

where p and q are constants that depend on the size and shape ofthe ellipse and p is greater than zero.

Proof of part (2)

The component of the planet’s velocity vector that is perpen-dicular to the r vector is v⊥ = v sinφ, so conservation of angularmomentum tells us that L = mrv sinφ is a constant. Since theplanet’s mass is a constant, this is the same as the condition

rv sinφ = constant .

Conservation of energy gives

1

2mv2 − GMm

r= constant .

We solve the first equation for v and plug into the second equationto eliminate v. Straightforward algebra then leads to the equationclaimed above, with the constant p being positive because of ourassumption that the planet’s energy is insufficient to escape fromthe sun, i.e., its total energy is negative.

Proof of part (3)

We define the quantities α, d, and s as shown in the figure. Thelaw of cosines gives

d2 = r2 + s2 − 2rs cosα .

Using α = 180 −2φ and the trigonometric identities cos(180 −x) =− cosx and cos 2x = 1− 2 sin2 x, we can rewrite this as

d2 = r2 + s2 − 2rs(2 sin2 φ− 1

).

Straightforward algebra transforms this into

sin φ =

√(r + s)2 − d2

4rs.

Since r + s is constant, the top of the fraction is constant, and thedenominator can be rewritten as 4rs = 4r(constant − r), which isequivalent to the desired form.

Section 15.9 ? Proof of Kepler’s elliptical orbit law 429

15.10 Some theorems and proofsIn this theorem I prove three theorems stated earlier, and state afourth theorem whose proof is left as an exercise.

Uniqueness of the cross product

The vector cross product as we have defined it has the followingproperties:(1) It does not violate rotational invariance.(2) It has the property A× (B + C) = A×B + A×C.(3) It has the property A× (kB) = k(A×B), where k is a scalar.

Theorem: The definition we have given is the only possible methodof multiplying two vectors to make a third vector which has theseproperties, with the exception of trivial redefinitions which just in-volve multiplying all the results by the same constant or swappingthe names of the axes. (Specifically, using a left-hand rule ratherthan a right-hand rule corresponds to multiplying all the results by−1.)

Proof: We prove only the uniqueness of the definition, withoutexplicitly proving that it has properties (1) through (3).

Using properties (2) and (3), we can break down any vectormultiplication (Axx +Ayy +Azz)× (Bxx +Byy +Bzz) into termsinvolving cross products of unit vectors.

A “self-term” like x × x must either be zero or lie along the xaxis, since any other direction would violate property (1). If was notzero, then (−x) × (−x) would have to lie in the opposite directionto avoid breaking rotational invariance, but property (3) says that(−x)×(−x) is the same as x×x, which is a contradiction. Thereforethe self-terms must be zero.

An “other-term” like x× y could conceivably have componentsin the x-y plane and along the z axis. If there was a nonzero compo-nent in the x-y plane, symmetry would require that it lie along thediagonal between the x and y axes, and similarly the in-the-planecomponent of (−x)× y would have to be along the other diagonal inthe x-y plane. Property (3), however, requires that (−x)× y equal−(x× y), which would be along the original diagonal. The only wayit can lie along both diagonals is if it is zero.

We now know that x × y must lie along the z axis. Since weare not interested in trivial differences in definitions, we can fixx × y = z, ignoring peurile possibilities such as x × y = 7z or theleft-handed definition x× y = −z. Given x× y = z, the symmetryof space requires that similar relations hold for y× z and z× x, withat most a difference in sign. A difference in sign could always beeliminated by swapping the names of some of the axes, so ignoringpossible trivial differences in definitions we can assume that thecyclically related set of relations x× y = z, y× z = x, and z× x = y


holds. Since the arbitrary cross-product with which we started canbe broken down into these simpler ones, the cross product is uniquelydefined.

Choice of axis theorem

Theorem: Suppose a closed system of material particles conservesangular momentum in one frame of reference, with the axis takento be at the origin. Then conservation of angular momentum isunaffected if the origin is relocated or if we change to a frame ofreference that is in constant-velocity motion with respect to thefirst one. The theorem also holds in the case where the system isnot closed, but the total external force is zero.

Proof: In the original frame of reference, angular momentum isconserved, so we have dL/dt=0. From example 29 on page 424, thisderivative can be rewritten as

dL

dt=∑i

ri × Fi ,

where Fi is the total force acting on particle i. In other words, we’readding up all the torques on all the particles.

By changing to the new frame of reference, we have changedthe position vector of each particle according to ri → ri + k − ut,where k is a constant vector that indicates the relative position ofthe new origin at t = 0, and u is the velocity of the new frame withrespect to the old one. The forces are all the same in the new frameof reference, however. In the new frame, the rate of change of theangular momentum is

dL

dt=∑i

(ri + k− ut)× Fi

=∑i

ri × Fi + (k− ut)×∑i

Fi .

The first term is the expression for the rate of change of the angu-lar momentum in the original frame of reference, which is zero byassumption. The second term vanishes by Newton’s third law; sincethe system is closed, every force Fi cancels with some force Fj . (Ifexternal forces act, but they add up to zero, then the sum can bebroken up into a sum of internal forces and a sum of external forces,each of which is zero.) The rate of change of the angular momentumis therefore zero in the new frame of reference.

Spin theorem

Theorem: An object’s angular momentum with respect to someoutside axis A can be found by adding up two parts:(1) The first part is the object’s angular momentum found by usingits own center of mass as the axis, i.e., the angular momentum the

Section 15.10 Some theorems and proofs 431

object has because it is spinning.(2) The other part equals the angular momentum that the objectwould have with respect to the axis A if it had all its mass concen-trated at and moving with its center of mass.

Proof: Let the system’s center of mass be at rcm, and let particlei lie at position rcm + di. Then the total angular momentum is

L =∑i

(rcm + di)× pi

= rcm ×∑i

pi +∑i

di × pi ,

which establishes the result claimed, since we can identify the firstterm with (2) and the second with (1).

Parallel axis theorem

Suppose an object has mass m, and moment of inertia Io for ro-tation about some axis A passing through its center of mass. Givena new axis B, parallel to A and lying at a distance h from it, theobject’s moment of inertia is given by Io +mh2.

The proof of this theorem is left as an exercise (problem 27,p. 440).


SummarySelected vocabularyangular momen-tum . . . . . . . .

a measure of rotational motion; a conservedquantity for a closed system

axis . . . . . . . . An arbitrarily chosen point used in the defini-tion of angular momentum. Any object whosedirection changes relative to the axis is consid-ered to have angular momentum. No matterwhat axis is chosen, the angular momentum ofa closed system is conserved.

torque . . . . . . the rate of change of angular momentum; anumerical measure of a force’s ability to twiston an object

equilibrium . . . a state in which an object’s momentum andangular momentum are constant

stable equilibrium one in which a force always acts to bring theobject back to a certain point

unstable equilib-rium . . . . . . . .

one in which any deviation of the object fromits equilibrium position results in a force push-ing it even farther away

NotationL . . . . . . . . . . angular momentumt . . . . . . . . . . torqueT . . . . . . . . . the period the time required for a rigidly ro-

tating body to complete one rotationω . . . . . . . . . . the angular velocity, dθ/dtmoment of iner-tia, I . . . . . . .

the proportionality constant in the equationL = Iω

Summary

Angular momentum is a measure of rotational motion which isconserved for a closed system. This book only discusses angularmomentum for rotation of material objects in two dimensions. Notall rotation is rigid like that of a wheel or a spinning top. An exampleof nonrigid rotation is a cyclone, in which the inner parts take lesstime to complete a revolution than the outer parts. In order to definea measure of rotational motion general enough to include nonrigidrotation, we define the angular momentum of a system by dividingit up into small parts, and adding up all the angular momenta ofthe small parts, which we think of as tiny particles. We arbitrarilychoose some point in space, the axis, and we say that anythingthat changes its direction relative to that point possesses angularmomentum. The angular momentum of a single particle is

L = mv⊥r ,

where v⊥ is the component of its velocity perpendicular to the linejoining it to the axis, and r is its distance from the axis. Positive and

Summary 433

negative signs of angular momentum are used to indicate clockwiseand counterclockwise rotation.

The choice of axis theorem states that any axis may be used fordefining angular momentum. If a system’s angular momentum isconstant for one choice of axis, then it is also constant for any otherchoice of axis.

The spin theorem states that an object’s angular momentumwith respect to some outside axis A can be found by adding up twoparts:

(1) The first part is the object’s angular momentum found byusing its own center of mass as the axis, i.e., the angular momentumthe object has because it is spinning.

(2) The other part equals the angular momentum that the ob-ject would have with respect to the axis A if it had all its massconcentrated at and moving with its center of mass.

Torque is the rate of change of angular momentum. The torquea force can produce is a measure of its ability to twist on an object.The relationship between force and torque is

|τ | = r|F⊥| ,

where r is the distance from the axis to the point where the force isapplied, and F⊥ is the component of the force perpendicular to theline connecting the axis to the point of application. Statics problemscan be solved by setting the total force and total torque on an objectequal to zero and solving for the unknowns.

In the special case of a rigid body rotating in a single plane, wedefine

ω =dθ

dt[angular velocity]

and

α =dω

dt, [angular acceleration]

in terms of which we have

L = Iω

and

τ = Iα ,

where the moment of inertia, I, is defined as

I =∑

mir2i ,


summing over all the atoms in the object (or using calculus to per-form a continuous sum, i.e. an integral). The relationship betweenthe angular quantities and the linear ones is

vt = ωr [tangential velocity of a point]

vr = 0 [radial velocity of a point]

at = αr . [radial acceleration of a point]


ar = ω2r [radial acceleration of a point]


In three dimensions, torque and angular momentum are vectors,and are expressed in terms of the vector cross product, which is theonly rotationally invariant way of defining a multiplication of twovectors that produces a third vector:

L = r× p

τ = r× F

In general, the cross product of vectors b and c has magnitude

|b× c| = |b| |c| sin θbc ,

which can be interpreted geometrically as the area of the parallel-ogram formed by the two vectors when they are placed tail-to-tail.The direction of the cross product lies along the line which is per-pendicular to both vectors; of the two such directions, we choose theone that is right-handed, in the sense that if we point the fingers ofthe flattened right hand along b, then bend the knuckles to pointthe fingers along c, the thumb gives the direction of b× c. In termsof components, the cross product is

(b× c)x = bycz − cybz(b× c)y = bzcx − czbx(b× c)z = bxcy − cxby

The cross product has the disconcerting properties

a× b = −b× a [noncommutative]

and

a× (b× c) 6= (a× b)× c [nonassociative] ,

and there is no “cross-division.”

For rigid-body rotation in three dimensions, we define an angularvelocity vector ω, which lies along the axis of rotation and bears aright-hand relationship to it. Except in special cases, there is noscalar moment of inertia for which L = Iω; the moment of inertiamust be expressed as a matrix.

Summary 435

Problem 4.

ProblemsKey√



1 A skilled motorcyclist can ride up a ramp, fly through the air,and land on another ramp. Why would it be useful for the rider tospeed up or slow down the back wheel while in the air?

2 An object thrown straight up in the air is momentarily at restwhen it reaches the top of its motion. Does that mean that it is inequilibrium at that point? Explain.

3 An object is observed to have constant angular momentum.Can you conclude that no torques are acting on it? Explain. [Basedon a problem by Serway and Faughn.]

4 The sun turns on its axis once every 26.0 days. Its mass is2.0 × 1030 kg and its radius is 7.0 × 108 m. Assume it is a rigidsphere of uniform density.(a) What is the sun’s angular momentum?

√

In a few billion years, astrophysicists predict that the sun will useup all its sources of nuclear energy, and will collapse into a ball ofexotic, dense matter known as a white dwarf. Assume that its radiusbecomes 5.8 × 106 m (similar to the size of the Earth.) Assume itdoes not lose any mass between now and then. (Don’t be fooledby the photo, which makes it look like nearly all of the star wasthrown off by the explosion. The visually prominent gas cloud isactually thinner than the best laboratory vacuum ever produced onearth. Certainly a little bit of mass is actually lost, but it is not atall unreasonable to make an approximation of zero loss of mass aswe are doing.)(b) What will its angular momentum be?(c) How long will it take to turn once on its axis?

√

5 (a) Alice says Cathy’s body has zero momentum, but Bobsays Cathy’s momentum is nonzero. Nobody is lying or making amistake. How is this possible? Give a concrete example.(b) Alice and Bob agree that Dong’s body has nonzero momentum,but disagree about Dong’s angular momentum, which Alice says iszero, and Bob says is nonzero. Explain.

6 Two objects have the same momentum vector. Assume thatthey are not spinning; they only have angular momentum due totheir motion through space. Can you conclude that their angularmomenta are the same? Explain. [Based on a problem by Serwayand Faughn.]


Problem 8.

Problem 9.

Problem 13.

7 You are trying to loosen a stuck bolt on your RV using a bigwrench that is 50 cm long. If you hang from the wrench, and yourmass is 55 kg, what is the maximum torque you can exert on thebolt?

√

8 The figure shows scale drawing of a pair of pliers beingused to crack a nut, with an appropriately reduced centimeter grid.Warning: do not attempt this at home; it is bad manners. If theforce required to crack the nut is 300 N, estimate the force requiredof the person’s hand. . Solution, p. 507

9 Make a rough estimate of the mechanical advantage of thelever shown in the figure. In other words, for a given amount offorce applied on the handle, how many times greater is the resultingforce on the cork?

10 A physical therapist wants her patient to rehabilitate hisinjured elbow by laying his arm flat on a table, and then lifting a2.1 kg mass by bending his elbow. In this situation, the weight is33 cm from his elbow. He calls her back, complaining that it hurtshim to grasp the weight. He asks if he can strap a bigger weightonto his arm, only 17 cm from his elbow. How much mass shouldshe tell him to use so that he will be exerting the same torque? (Heis raising his forearm itself, as well as the weight.)

√

11 Two horizontal tree branches on the same tree have equaldiameters, but one branch is twice as long as the other. Give aquantitative comparison of the torques where the branches join thetrunk. [Thanks to Bong Kang.]

12 A ball is connected by a string to a vertical post. The ball isset in horizontal motion so that it starts winding the string aroundthe post. Assume that the motion is confined to a horizontal plane,i.e., ignore gravity. Michelle and Astrid are trying to predict thefinal velocity of the ball when it reaches the post. Michelle saysthat according to conservation of angular momentum, the ball hasto speed up as it approaches the post. Astrid says that according toconservation of energy, the ball has to keep a constant speed. Whois right? [Hint: How is this different from the case where you whirla rock in a circle on a string and gradually reel in the string?]

13 A person of weight W stands on the ball of one foot. Findthe tension in the calf muscle and the force exerted by the shinboneson the bones of the foot, in terms of W , a, and b. For simplicity,assume that all the forces are at 90-degree angles to the foot, i.e.,neglect the angle between the foot and the floor.

Problems 437

Problem 14.

Problems 15 and 16.

Problem 17.

14 The rod in the figure is supported by the finger and thestring.(a) Find the tension, T , in the string, and the force, F , from thefinger, in terms of m, b,L, and g.

√

(b) Comment on the cases b = L and b = L/2.(c) Are any values of b unphysical?

15 A uniform ladder of mass m and length L leans against asmooth wall, making an angle q with respect to the ground. The dirtexerts a normal force and a frictional force on the ladder, producinga force vector with magnitude F1 at an angle φ with respect to theground. Since the wall is smooth, it exerts only a normal force onthe ladder; let its magnitude be F2.(a) Explain why φ must be greater than θ. No math is needed.(b) Choose any numerical values you like for m and L, and showthat the ladder can be in equilibrium (zero torque and zero totalforce vector) for θ = 45.00 and φ = 63.43 .

16 Continuing the previous problem, find an equation for φ interms of θ, and show that m and L do not enter into the equation.Do not assume any numerical values for any of the variables. Youwill need the trig identity sin(a − b) = sin a cos b − sin b cos a. (Asa numerical check on your result, you may wish to check that theangles given in part b of the previous problem satisfy your equation.)√

?

17 (a) Find the minimum horizontal force which, applied atthe axle, will pull a wheel over a step. Invent algebra symbols forwhatever quantities you find to be relevant, and give your answerin symbolic form. [Hints: There are four forces on the wheel atfirst, but only three when it lifts off. Normal forces are alwaysperpendicular to the surface of contact. Note that the corner of thestep cannot be perfectly sharp, so the surface of contact for thisforce really coincides with the surface of the wheel.](b) Under what circumstances does your result become infinite?Give a physical interpretation.

18 In the 1950’s, serious articles began appearing in magazineslike Life predicting that world domination would be achieved by thenation that could put nuclear bombs in orbiting space stations, fromwhich they could be dropped at will. In fact it can be quite difficultto get an orbiting object to come down. Let the object have energyE = KE +PE and angular momentum L. Assume that the energyis negative, i.e., the object is moving at less than escape velocity.Show that it can never reach a radius less than

rmin =GMm

2E

(−1 +

√1 +

2EL2

G2M2m3

).

[Note that both factors are negative, giving a positive result.]


Problem 19.

19 You wish to determine the mass of a ship in a bottle withouttaking it out. Show that this can be done with the setup shown inthe figure, with a scale supporting the bottle at one end, providedthat it is possible to take readings with the ship slid to severaldifferent locations. Note that you can’t determine the position ofthe ship’s center of mass just by looking at it, and likewise for thebottle. In particular, you can’t just say, “position the ship right ontop of the fulcrum” or “position it right on top of the balance.” ?

20 Two atoms will interact via electrical forces between theirprotons and electrons. One fairly good approximation to the poten-tial energy is the Lennard-Jones potential,

PE(r) = k

[(ar

)12− 2

(ar

)6]

,

where r is the center-to-center distance between the atoms.

Show that (a) there is an equilibrium point at r = a, (b) the equi-librium is stable, and (c) the energy required to bring the atomsfrom their equilibrium separation to infinity is k. [Hints: The firsttwo parts can be done more easily by setting a = 1, since the valueof a only changes the distance scale. One way to do part b is bygraphing.]

21 Suppose that we lived in a universe in which Newton’s lawof gravity gave forces proportional to r−7 rather than r−2. Which,if any, of Kepler’s laws would still be true? Which would be com-pletely false? Which would be different, but in a way that could becalculated with straightforward algebra?

22 Show that a sphere of radius R that is rolling without slippinghas angular momentum and momentum in the ratio L/p = (2/5)R.

23 Suppose a bowling ball is initially thrown so that it has noangular momentum at all, i.e., it is initially just sliding down thelane. Eventually kinetic friction will get it spinning fast enough sothat it is rolling without slipping. Show that the final velocity of theball equals 5/7 of its initial velocity. [Hint: You’ll need the result ofproblem 22.]

24 Penguins are playful animals. Tux the Penguin invents a newgame using a natural circular depression in the ice. He waddles attop speed toward the crater, aiming off to the side, and then hopsinto the air and lands on his belly just inside its lip. He then belly-surfs, moving in a circle around the rim. The ice is frictionless, sohis speed is constant. Is Tux’s angular momentum zero, or nonzero?What about the total torque acting on him? Take the center of thecrater to be the axis. Explain your answers.

Problems 439

Problem 28.

Problem 29.

Problem 30.

Problem 32

25 A massless rod of length ` has weights, each of mass m, at-tached to its ends. The rod is initially put in a horizontal position,and laid on an off-center fulcrum located at a distance b from therod’s center. The rod will topple. (a) Calculate the total gravita-tional torque on the rod directly, by adding the two torques. (b)Verify that this gives the same result as would have been obtainedby taking the entire gravitational force as acting at the center ofmass.

26 Use analogies to find the equivalents of the following equationsfor rotation in a plane:

K = p2/2m

∆x = vo∆t+ (1/2)a∆t2

W = F∆x

Example: v = ∆x/∆t→ ω = ∆θ/∆t

27 Prove the parallel axis theorem stated on page 432.

28 The box shown in the figure is being accelerated by pullingon it with the rope.(a) Assume the floor is frictionless. What is the maximum forcethat can be applied without causing the box to tip over?

. Hint, p. 494√

(b) Repeat part a, but now let the coefficient of friction be µ.√

(c) What happens to your answer to part b when the box is suffi-ciently tall? How do you interpret this?

29 (a) The bar of mass m is attached at the wall with a hinge,and is supported on the right by a massless cable. Find the tension,T , in the cable in terms of the angle θ.

√

(b) Interpreting your answer to part a, what would be the best angleto use if we wanted to minimize the strain on the cable?(c) Again interpreting your answer to part a, for what angles doesthe result misbehave mathematically? Interpet this physically.

30 (a) The two identical rods are attached to one another witha hinge, and are supported by the two massless cables. Find theangle α in terms of the angle β, and show that the result is a purelygeometric one, independent of the other variables involved.

√

(b) Using your answer to part a, sketch the configurations for β → 0,β = 45 , and β = 90 . Do your results make sense intuitively?

31 (a) Find the angular velocities of the earth’s rotation and ofthe earth’s motion around the sun.

√

(b) Which motion involves the greater acceleration?

32 Give a numerical comparison of the two molecules’ momentsof inertia for rotation in the plane of the page about their centersof mass.


Problem 39

33 Find the angular momentum of a particle whose position isr = 3x− y+ z (in meters) and whose momentum is p = −2x+ y+ z(in kg·m/s).

√

34 Find a vector that is perpendicular to both of the followingtwo vectors:

x + 2y + 3z

4x + 5y + 6z

√

35 Prove property (3) of the vector cross product from thetheorem on page 430.

36 Prove the anticommutative property of the vector cross prod-uct, A×B = −B×A, using the expressions for the components ofthe cross product.

37 Find three vectors with which you can demonstrate that thevector cross product need not be associative, i.e., that A× (B×C)need not be the same as (A×B)×C.

38 Which of the following expressions make sense, and which arenonsense? For those that make sense, indicate whether the result isa vector or a scalar.(a) (A×B)×C(b) (A×B) ·C(c) (A ·B)×C

39 (a) As suggested in the figure, find the area of the infinites-imal region expressed in polar coordinates as lying between r andr + dr and between θ and θ + dθ.

√

(b) Generalize this to find the infinitesimal element of volume incylindrical coordinates (r, θ, z), where the Cartesian z axis is per-pendicular to the directions measured by r and θ.

√

(c) Find the moment of inertia for rotation about its axis of a conewhose mass is M , whose height is h, and whose base has a radiusb.

√

40 Find the moment of inertia of a solid rectangular box of massM and uniform density, whose sides are of length a, b, and c, forrotation about an axis through its center parallel to the edges oflength a.

√

Problems 441

41 The nucleus 168Er (erbium-168) contains 68 protons (whichis what makes it a nucleus of the element erbium) and 100 neutrons.It has an ellipsoidal shape like an American football, with one longaxis and two short axes that are of equal diameter. Because thisis a subatomic system, consisting of only 168 particles, its behaviorshows some clear quantum-mechanical properties. It can only havecertain energy levels, and it makes quantum leaps between theselevels. Also, its angular momentum can only have certain values,which are all multiples of 2.109× 10−34 kg ·m2/s. The table showssome of the observed angular momenta and energies of 168Er, in SIunits (kg ·m2/s and joules).L× 1034 E × 1014

0 02.109 1.27864.218 4.23116.327 8.79198.437 14.873110.546 22.379812.655 31.13514.764 41.20616.873 52.223

(a) These data can be described to a good approximation as a rigidend-over-end rotation. Estimate a single best-fit value for the mo-ment of inertia from the data, and check how well the data agreewith the assumption of rigid-body rotation. . Hint, p. 494

√

(b) Check whether this moment of inertia is on the right order ofmagnitude. The moment of inertia depends on both the size andthe shape of the nucleus. For the sake of this rough check, ignorethe fact that the nucleus is not quite spherical. To estimate its size,use the fact that a neutron or proton has a volume of about 1 fm3

(one cubic femtometer, where 1 fm = 10−15 m), and assume theyare closely packed in the nucleus.

42 (a) Prove the identity a × (b × c) = b(a · c) − c(a · b)by expanding the product in terms of its components. Note thatbecause the x, y, and z components are treated symmetrically inthe definitions of the vector cross product, it is only necessary tocarry out the proof for the x component of the result.(b) Applying this to the angular momentum of a rigidly rotatingbody, L =

∫r× (ω× r) dm, show that the diagonal elements of the

moment of inertia tensor can be expressed as, e.g., Ixx =∫

(y2 +z2)dm.(c) Find the diagonal elements of the moment of inertia matrix ofan ellipsoid with axes of lengths a, b, and c, in the principal-axisframe, and with the axis at the center.

√?


Problem 44.

43 When we talk about rigid-body rotation, the concept ofa perfectly rigid body can only be an idealization. In reality, anyobject will compress, expand, or deform to some extent when sub-jected to the strain of rotation. However, if we let it settle down fora while, perhaps it will reach a new equilibrium. As an example,suppose we fill a centrifuge tube with some compressible substancelike shaving cream or Wonder Bread. We can model the contents ofthe tube as a one-dimensional line of mass, extending from r = 0 tor = `. Once the rotation starts, we expect that the contents will bemost compressed near the “floor” of the tube at r = `; this is bothbecause the inward force required for circular motion increases withr for a fixed ω, and because the part at the floor has the greatestamount of material pressing “down” (actually outward) on it. Thelinear density dm/dr, in units of kg/m, should therefore increase asa function of r. Suppose that we have dm/dr = µer/`, where µ is aconstant. Find the moment of inertia.

√

44 Two bars of length L are connected with a hinge and placedon a frictionless cylinder of radius r. (a) Show that the angle θ shownin the figure is related to the unitless ratio r/L by the equation

r

L=

cos2 θ

2 tan θ.

(b) Discuss the physical behavior of this equation for very large andvery small values of r/L. ?

45 Let two sides of a triangle be given by the vectors A andB, with their tails at the origin, and let mass m be uniformly dis-tributed on the interior of the triangle. (a) Show that the distanceof the triangle’s center of mass from the intersection of sides A andB is given by 1

3 |A + B|.(b) Consider the quadrilateral with mass 2m, and vertices at theorigin, A, B, and A + B. Show that its moment of inertia, forrotation about an axis perpendicular to it and passing through itscenter of mass, is m

6 (A2 +B2).(c) Show that the moment of inertia for rotation about an axis per-pendicular to the plane of the original triangle, and passing throughits center of mass, is m

18(A2 +B2−A ·B). Hint: Combine the resultsof parts a and b with the result of problem 27. ?

46 In example 23 on page 416, prove that if the rod is sufficientlythin, it can be toppled without scraping on the floor.

. Solution, p. 508 ?

Problems 443

47 A yo-yo of total mass m consists of two solid cylindersof radius R, connected by a small spindle of negligible mass andradius r. The top of the string is held motionless while the stringunrolls from the spindle. Show that the acceleration of the yo-yois g/(1 + R2/2r2). [Hint: The acceleration and the tension in thestring are unknown. Use τ = ∆L/∆t and F = ma to determinethese two unknowns.] ?

48 We have n identical books of width w, and we wish to stackthem at the edge of a table so that they extend the maximum possi-ble distance Ln beyond the edge. Surprisingly, it is possible to havevalues of Ln that are greater than w, even with fairly small n. Forlarge n, however, Ln begins to grow very slowly. Our goal is to findLn for a given n. (We assume here, as everyone seems to do in pos-ing this classic problem, that only one book is ever used at a givenheight. I do not know of any rigorous proof that applies when thisrestriction is relaxed, but I suspect that the result is unaffected.) (a)Use proof by induction to find Ln, expressing your result as a sum.(b) Find a sufficiently tight lower bound on this sum, as a closed-form expression, to prove that 1,202,604 books suffice for L > 7w.

?

49 A certain function f takes two vectors as inputs and gives anoutput that is also a vector. The function can be defined in such away that it is rotationally invariant. It takes on the following valuesfor the following inputs:

f(x, y) = −z

f(2x, y) = −8z

f(x, 2y) = −2z

Prove that the given information uniquely determines f , and givean explicit expression for it. ?

50 (a) Find the moment of inertia of a uniform square of massm and with sides of length b, for rotation in its own plane, aboutone of its corners.(b) The square is balanced on one corner on a frictionless surface.An infinitesimal perturbation causes it to topple. Find its angularvelocity at the moment when its side slaps the surface. ?


Exercise 15: TorqueEquipment:

• rulers with holes in them

• spring scales (two per group)

While one person holds the pencil which forms the axle for the ruler, the other members of thegroup pull on the scale and take readings. In each case, calculate the total torque on the ruler,and find out whether it equals zero to roughly within the accuracy of the experiment. Finishthe calculations for each part before moving on to the next one.

Exercise 15: Torque 445

Vibrations and resonance

446

448

The vibrations of this electric bassstring are converted to electricalvibrations, then to sound vibra-tions, and finally to vibrations ofour eardrums.

Chapter 16

Vibrations

Dandelion. Cello. Read those two words, and your brain instantlyconjures a stream of associations, the most prominent of which haveto do with vibrations. Our mental category of “dandelion-ness” isstrongly linked to the color of light waves that vibrate about half amillion billion times a second: yellow. The velvety throb of a cellohas as its most obvious characteristic a relatively low musical pitch— the note you are spontaneously imagining right now might beone whose sound vibrations repeat at a rate of a hundred times asecond.

Evolution has designed our two most important senses aroundthe assumption that not only will our environment be drenched withinformation-bearing vibrations, but in addition those vibrations willoften be repetitive, so that we can judge colors and pitches by therate of repetition. Granting that we do sometimes encounter non-repeating waves such as the consonant “sh,” which has no recogniz-able pitch, why was Nature’s assumption of repetition neverthelessso right in general?

Repeating phenomena occur throughout nature, from the orbitsof electrons in atoms to the reappearance of Halley’s Comet every 75years. Ancient cultures tended to attribute repetitious phenomena

449

a / If we try to draw a non-repeating orbit for Halley’sComet, it will inevitably end upcrossing itself.

b / A spring has an equilib-rium length, 1, and can bestretched, 2, or compressed, 3. Amass attached to the spring canbe set into motion initially, 4, andwill then vibrate, 4-13.

like the seasons to the cyclical nature of time itself, but we nowhave a less mystical explanation. Suppose that instead of Halley’sComet’s true, repeating elliptical orbit that closes seamlessly uponitself with each revolution, we decide to take a pen and draw awhimsical alternative path that never repeats. We will not be able todraw for very long without having the path cross itself. But at sucha crossing point, the comet has returned to a place it visited oncebefore, and since its potential energy is the same as it was on thelast visit, conservation of energy proves that it must again have thesame kinetic energy and therefore the same speed. Not only that,but the comet’s direction of motion cannot be randomly chosen,because angular momentum must be conserved as well. Althoughthis falls short of being an ironclad proof that the comet’s orbit mustrepeat, it no longer seems surprising that it does.

Conservation laws, then, provide us with a good reason whyrepetitive motion is so prevalent in the universe. But it goes deeperthan that. Up to this point in your study of physics, I have beenindoctrinating you with a mechanistic vision of the universe as agiant piece of clockwork. Breaking the clockwork down into smallerand smaller bits, we end up at the atomic level, where the electronscircling the nucleus resemble — well, little clocks! From this pointof view, particles of matter are the fundamental building blocksof everything, and vibrations and waves are just a couple of thetricks that groups of particles can do. But at the beginning ofthe 20th century, the tables were turned. A chain of discoveriesinitiated by Albert Einstein led to the realization that the so-calledsubatomic “particles” were in fact waves. In this new world-view,it is vibrations and waves that are fundamental, and the formationof matter is just one of the tricks that waves can do.

16.1 Period, frequency, and amplitude

Figure b shows our most basic example of a vibration. With noforces on it, the spring assumes its equilibrium length, b/1. It canbe stretched, 2, or compressed, 3. We attach the spring to a wallon the left and to a mass on the right. If we now hit the mass witha hammer, 4, it oscillates as shown in the series of snapshots, 4-13.If we assume that the mass slides back and forth without frictionand that the motion is one-dimensional, then conservation of energyproves that the motion must be repetitive. When the block comesback to its initial position again, 7, its potential energy is the sameagain, so it must have the same kinetic energy again. The motionis in the opposite direction, however. Finally, at 10, it returns to itsinitial position with the same kinetic energy and the same directionof motion. The motion has gone through one complete cycle, andwill now repeat forever in the absence of friction.

The usual physics terminology for motion that repeats itself over

450 Chapter 16 Vibrations

c / The locomotive’s wheelsspin at a frequency of f cyclesper second, which can alsobe described as ω radians persecond. The mechanical link-ages allow the linear vibration ofthe steam engine’s pistons, atfrequency f , to drive the wheels.

d / Example 1.

and over is periodic motion, and the time required for one repetitionis called the period, T . (The symbol P is not used because of thepossible confusion with momentum.) One complete repetition of themotion is called a cycle.

We are used to referring to short-period sound vibrations as“high” in pitch, and it sounds odd to have to say that high pitcheshave low periods. It is therefore more common to discuss the rapid-ity of a vibration in terms of the number of vibrations per second,a quantity called the frequency, f . Since the period is the numberof seconds per cycle and the frequency is the number of cycles persecond, they are reciprocals of each other,

f = 1/T .

The forms of various equations turn out to be simpler whenthey are expressed not in terms of f but in terms of ω = 2πf .It’s not a coincidence that this relationship looks the same as theone relating angular velocity and frequency in circular motion. Inmachines, mechanical linkages are used to convert back and forthbetween vibrational motion and circular motion. For example, a carengine’s pistons oscillate in their cylinders at a frequency f , drivingthe crankshaft at the same frequency f . Either of these motions canbe described using ω instead of f , even though only in the case ofthe crankshaft’s rotational motion does it make sense to interpretω as the number of radians per second. When the motion is notrotational, we usually refer to ω as the angular frequency, and weoften use the word “frequency” to mean either f or ω, relying oncontext to make the meaning clear.

A carnival game example 1In the carnival game shown in figure d, the rube is supposed topush the bowling ball on the track just hard enough so that it goesover the hump and into the valley, but does not come back outagain. If the only types of energy involved are kinetic and poten-tial, this is impossible. Suppose you expect the ball to come backto a point such as the one shown with the dashed outline, thenstop and turn around. It would already have passed through thispoint once before, going to the left on its way into the valley. Itwas moving then, so conservation of energy tells us that it can-not be at rest when it comes back to the same point. The motionthat the customer hopes for is physically impossible. There isa physically possible periodic motion in which the ball rolls backand forth, staying confined within the valley, but there is no wayto get the ball into that motion beginning from the place where westart. There is a way to beat the game, though. If you put enoughspin on the ball, you can create enough kinetic friction so that asignificant amount of heat is generated. Conservation of energythen allows the ball to be at rest when it comes back to a point

Section 16.1 Period, frequency, and amplitude 451

e / 1. The amplitude of thevibrations of the mass on a springcould be defined in two differentways. It would have units ofdistance. 2. The amplitude of aswinging pendulum would morenaturally be defined as an angle.

like the outlined one, because kinetic energy has been convertedinto heat.

Period and frequency of a fly’s wing-beats example 2A Victorian parlor trick was to listen to the pitch of a fly’s buzz, re-produce the musical note on the piano, and announce how manytimes the fly’s wings had flapped in one second. If the fly’s wingsflap, say, 200 times in one second, then the frequency of theirmotion is f = 200/1 s = 200 s−1. The period is one 200th of asecond, T = 1/f = (1/200) s = 0.005 s.

Units of inverse second, s−1, are awkward in speech, so an abbre-viation has been created. One Hertz, named in honor of a pioneerof radio technology, is one cycle per second. In abbreviated form,1 Hz = 1 s−1. This is the familiar unit used for the frequencies onthe radio dial.

Frequency of a radio station example 3. KKJZ’s frequency is 88.1 MHz. What does this mean, and whatperiod does this correspond to?

. The metric prefix M- is mega-, i.e., millions. The radio wavesemitted by KKJZ’s transmitting antenna vibrate 88.1 million timesper second. This corresponds to a period of

T = 1/f = 1.14× 10−8 s .

This example shows a second reason why we normally speak interms of frequency rather than period: it would be painful to haveto refer to such small time intervals routinely. I could abbreviateby telling people that KKJZ’s period was 11.4 nanoseconds, butmost people are more familiar with the big metric prefixes thanwith the small ones.

Units of frequency are also commonly used to specify the speedsof computers. The idea is that all the little circuits on a computerchip are synchronized by the very fast ticks of an electronic clock, sothat the circuits can all cooperate on a task without getting aheador behind. Adding two numbers might require, say, 30 clock cycles.Microcomputers these days operate at clock frequencies of about agigahertz.

We have discussed how to measure how fast something vibrates,but not how big the vibrations are. The general term for this isamplitude, A. The definition of amplitude depends on the systembeing discussed, and two people discussing the same system maynot even use the same definition. In the example of the block on theend of the spring, e/1, the amplitude will be measured in distanceunits such as cm. One could work in terms of the distance traveledby the block from the extreme left to the extreme right, but itwould be somewhat more common in physics to use the distancefrom the center to one extreme. The former is usually referred to as


f / Sinusoidal and non-sinusoidalvibrations.

the peak-to-peak amplitude, since the extremes of the motion looklike mountain peaks or upside-down mountain peaks on a graph ofposition versus time.

In other situations we would not even use the same units for am-plitude. The amplitude of a child on a swing, or a pendulum, e/2,would most conveniently be measured as an angle, not a distance,since her feet will move a greater distance than her head. The elec-trical vibrations in a radio receiver would be measured in electricalunits such as volts or amperes.

16.2 Simple harmonic motionWhy are sine-wave vibrations so common?

If we actually construct the mass-on-a-spring system discussedin the previous section and measure its motion accurately, we willfind that its x−t graph is nearly a perfect sine-wave shape, as shownin figure f/1. (We call it a “sine wave” or “sinusoidal” even if it isa cosine, or a sine or cosine shifted by some arbitrary horizontalamount.) It may not be surprising that it is a wiggle of this generalsort, but why is it a specific mathematically perfect shape? Why isit not a sawtooth shape like 2 or some other shape like 3? The mys-tery deepens as we find that a vast number of apparently unrelatedvibrating systems show the same mathematical feature. A tuningfork, a sapling pulled to one side and released, a car bouncing onits shock absorbers, all these systems will exhibit sine-wave motionunder one condition: the amplitude of the motion must be small.

It is not hard to see intuitively why extremes of amplitude wouldact differently. For example, a car that is bouncing lightly on itsshock absorbers may behave smoothly, but if we try to double theamplitude of the vibrations the bottom of the car may begin hittingthe ground, f/4. (Although we are assuming for simplicity in thischapter that energy is never dissipated, this is clearly not a veryrealistic assumption in this example. Each time the car hits theground it will convert quite a bit of its potential and kinetic en-ergy into heat and sound, so the vibrations would actually die outquite quickly, rather than repeating for many cycles as shown in thefigure.)

The key to understanding how an object vibrates is to know howthe force on the object depends on the object’s position. If an objectis vibrating to the right and left, then it must have a leftward forceon it when it is on the right side, and a rightward force when it is onthe left side. In one dimension, we can represent the direction of theforce using a positive or negative sign, and since the force changesfrom positive to negative there must be a point in the middle wherethe force is zero. This is the equilibrium point, where the objectwould stay at rest if it was released at rest. For convenience of

Section 16.2 Simple harmonic motion 453

g / The force exerted by anideal spring, which behavesexactly according to Hooke’s law.

notation throughout this chapter, we will define the origin of ourcoordinate system so that x equals zero at equilibrium.

The simplest example is the mass on a spring, for which the forceon the mass is given by Hooke’s law,

F = −kx .

We can visualize the behavior of this force using a graph of F versusx, as shown in figure g. The graph is a line, and the spring constant,k, is equal to minus its slope. A stiffer spring has a larger value ofk and a steeper slope. Hooke’s law is only an approximation, butit works very well for most springs in real life, as long as the springisn’t compressed or stretched so much that it is permanently bentor damaged.

The following important theorem relates the motion graph tothe force graph.

Theorem: A linear force graph makes a sinusoidal motiongraph.

If the total force on a vibrating object depends only on theobject’s position, and is related to the objects displacementfrom equilibrium by an equation of the form F = −kx, thenthe object’s motion displays a sinusoidal graph with frequencyω =

√k/m.

Proof: By Newton’s second law, −kx = ma, so we need a functionx(t) that satisfies the equation d2x/dt2 = −cx, where for conve-nience we write c for k/m. This type of equation is called a differ-ential equation, because it relates a function to its own derivative(in this case the second derivative).

Just to make things easier to think about, suppose that we hap-pen to have an oscillator with c = 1. Then our goal is to finda function whose second derivative is equal to minus the originalfunction. We know of two such functions, the sine and the cosine.These two solutions can be combined to make anything of the formP cos t + Q cos t, where P and Q are constants, and the result willstill be a solution. Using trig identities, such an expression canalways be rewritten as A cos(t+ δ).

Now what about the more general case where c need not equal1? The role of c in d2x/dt2 = −cx is to set the time scale. Forexample, suppose we produce a fake video of an object oscillatingaccording to A cos(t + δ), which violates Newton’s second law be-cause c doesn’t equal 1, so the acceleration is too small. We canalways make the video physically accurate by speeding it up. Thissuggests generalizing the solution to A cos(ωt + δ). Plugging in tothe differential equation, we find that ω =

√k/m, and T = 2π/ω

brings us to the claimed result.

We’ve proved that anything of this form is a solution, but we


i / Seen from close up, anyF − x curve looks like a line.

h / Because simple harmonic mo-tion involves sinusoidal functions,it is equivalent to circular motionthat has been projected into onedimension. This figure shows asimulated view of Jupiter and itsfour largest moons at intervals ofthree hours. Seen from the sidefrom within the plane of the solarsystem, the circular orbits appearlinear. In coordinates with the ori-gin at Jupiter, a moon has coordi-nates x = r cos θ and y = r sin θ,where θ = ωt . If the view isalong the y axis, then we see onlythe x motion, which is of the formA cos(ωt).

haven’t shown that any solution must be of this form. Physically,this must be true because the motion is fully determined by theoscillator’s initial position and initial velocity, which can always bematched by choosing A and δ appropriately. Mathematically, theuniqueness result is a standard one about second-order differentialequations.

This may seem like only an obscure theorem about the mass-on-a-spring system, but figure i shows it to be far more general thanthat. Figure i/1 depicts a force curve that is not a straight line. Asystem with this F −x curve would have large-amplitude vibrationsthat were complex and not sinusoidal. But the same system wouldexhibit sinusoidal small-amplitude vibrations. This is because anycurve looks linear from very close up. If we magnify the F −x graphas shown in figure i/2, it becomes very difficult to tell that the graphis not a straight line. If the vibrations were confined to the regionshown in i/2, they would be very nearly sinusoidal. This is the rea-son why sinusoidal vibrations are a universal feature of all vibratingsystems, if we restrict ourselves to small amplitudes. The theoremis therefore of great general significance. It applies throughout theuniverse, to objects ranging from vibrating stars to vibrating nuclei.A sinusoidal vibration is known as simple harmonic motion.

This relates to the fundamental idea behind differential calcu-lus, which is that up close, any smooth function looks linear. Tocharacterize small oscillations about the equilibrium at x = 0 in fig-ure g, all we need to know is the derivative dF/dx|0, which equals−k. That is, a force function F (x) has no “individuality” except asdefined by k.

Spring constant related to potential energy example 4


j / Three functions with thesame curvature at x = 0.

k / Example 5. The rod piv-ots on the hinge at the bottom.

The same idea about lack of individuality can be expressed interms of energy.

On a graph of PE versus x , an equilibrium is a local minimum.We can imagine an oscillation about this equilibrium point as if amarble was rolling back and forth in the depression of the graph.Let’s choose a coordinate system in which x = 0 is the equilib-rium, and since the potential energy is only well defined up to anadditive constant, we’ll simply define it to be zero at equilibrium:

PE(0) = 0

Since x = 0 is a local minimum,

dPEdx

(0) = 0 .

There are still infinitely many functions that could satisfy thesecriteria, including the three shown in figure j, which are x2/2,x2/2(1+x2), and (e3x +e−3x−2)/18. Note, however, how all threefunctions are virtually identical right near the minimum. That’s be-cause they all have the same curvature. More specifically, eachfunction has its second derivative equal to 1 at x = 0, and thesecond derivative is a measure of curvature. Since the F =−dPE/dx and k = −dPE/dx , k equals the second derivativeof the PE,

d2PEdx2 (0) = k .

As shown in figure j, any two functions that have PE(0) = 0,dPE/dx = 0, and d2PE/dx2 = k , with the same value of k , arevirtually indistinguishable for small values of x , so if we want toanalyze small oscillations, it doesn’t even matter which functionwe assume. For simplicity, we can always use PE(x) = (1/2)kx2,which is the form that gives a constant second derivative.

A spring and a lever example 5. What is the period of small oscillations of the system shown inthe figure? Neglect the mass of the lever and the spring. Assumethat the spring is so stiff that gravity is not an important effect.The spring is relaxed when the lever is vertical.

. This is a little tricky, because the spring constant k , although it isrelevant, is not the k we should be putting into the equation ω =√

k/m. I find this easier to understand by working with energyrather than force. (Another method would be to use torque, as inproblem 15.) The k that goes into

√k/m has to be the second

derivative of PE with respect to the position, x , of the mass that’smoving. The energy PE stored in the spring depends on how farthe tip of the lever is from the center. This distance equals (L/b)x ,


l / Example 6.

so the energy in the spring is

PE =12

k(

Lb

x)2

=kL2

2b2 x2 ,

and the k we have to put in T = 2π√

m/k is

d2PEdx2 =

kL2

b2 .

The result is

ω =

√kL2

mb2

=Lb

√km

The leverage of the lever makes it as if the spring was stronger,decreasing the period of the oscillations by a factor of b/L.

Water in a U-shaped tube example 6. The U-shaped tube in figure l has cross-sectional area A, andthe density of the water inside is ρ. Find the gravitational poten-tial energy as a function of the quantity y shown in the figure,show that there is an equilibrium at y=0, and find the frequencyof oscillation of the water.

. Potential energy is only well defined up to an additive constant.To fix this constant, let’s define PE to be zero when y=0. Thedifference between PE(y ) and PE(0) is the energy that would berequired to lift a water column of height y out of the right side, andplace it above the dashed line, on the left side, raising it througha height y . This water column has height y and cross-sectionalarea A, so its volume is Ay , its mass is ρAy , and the energyrequired is mgy=(ρAy )gy=ρgAy2. We then have PE(y ) = PE(0)+ρgAy2 = ρgAy2.

The “spring constant” is

k =d2PEdy2

= 2ρgA .

This is an interesting example, because k can be calculated with-out any approximations, but the kinetic energy requires an ap-proximation, because we don’t know the details of the pattern


of flow of the water. It could be very complicated. There willbe a tendency for the water near the walls to flow more slowlydue to friction, and there may also be swirling, turbulent motion.However, if we make the approximation that all the water moveswith the same velocity as the surface, dy/dt , then the mass-on-a-spring analysis applies. Letting L be the total length of the filledpart of the tube, the mass is ρLA, and we have

ω =√

k/m

=

√2ρgAρLA

=

√2gL

.

Period is approximately independent of amplitude, if theamplitude is small.

Until now we have not even mentioned the most counterintuitiveaspect of the equation ω =

√k/m: it does not depend on ampli-

tude at all. Intuitively, most people would expect the mass-on-a-spring system to take longer to complete a cycle if the amplitudewas larger. (We are comparing amplitudes that are different fromeach other, but both small enough that the theorem applies.) Infact the larger-amplitude vibrations take the same amount of timeas the small-amplitude ones. This is because at large amplitudes,the force is greater, and therefore accelerates the object to higherspeeds.

Legend has it that this fact was first noticed by Galileo dur-ing what was apparently a less than enthralling church service. Agust of wind would now and then start one of the chandeliers in thecathedral swaying back and forth, and he noticed that regardlessof the amplitude of the vibrations, the period of oscillation seemedto be the same. Up until that time, he had been carrying out hisphysics experiments with such crude time-measuring techniques asfeeling his own pulse or singing a tune to keep a musical beat. Butafter going home and testing a pendulum, he convinced himself thathe had found a superior method of measuring time. Even withouta fancy system of pulleys to keep the pendulum’s vibrations fromdying down, he could get very accurate time measurements, becausethe gradual decrease in amplitude due to friction would have no ef-fect on the pendulum’s period. (Galileo never produced a modern-style pendulum clock with pulleys, a minute hand, and a secondhand, but within a generation the device had taken on the formthat persisted for hundreds of years after.)


The pendulum example 7. Compare the frequencies of pendula having bobs with differentmasses.

. From the equation ω =√

k/m, we might expect that a largermass would lead to a lower frequency. However, increasing themass also increases the forces that act on the pendulum: gravityand the tension in the string. This increases k as well as m, sothe frequency of a pendulum is independent of m.


SummarySelected vocabularyperiodic motion . motion that repeats itself over and overperiod . . . . . . . the time required for one cycle of a periodic

motionfrequency . . . . . the number of cycles per second, the inverse of

the periodamplitude . . . . the amount of vibration, often measured from

the center to one side; may have different unitsdepending on the nature of the vibration

simple harmonicmotion . . . . . .

motion whose x− t graph is a sine wave

NotationT . . . . . . . . . periodf . . . . . . . . . . frequencyA . . . . . . . . . amplitudek . . . . . . . . . . the slope of the graph of F versus x, where

F is the total force acting on an object andx is the object’s position; for a spring, this isknown as the spring constant.

ω (Greek letter“omega”) . . . . .

2πf

Other terminology and notationν . . . . . . . . . . The Greek letter ν, nu, is used in many books

for frequency.

Summary

Periodic motion is common in the world around us because ofconservation laws. An important example is one-dimensional motionin which the only two forms of energy involved are potential andkinetic; in such a situation, conservation of energy requires that anobject repeat its motion, because otherwise when it came back tothe same point, it would have to have a different kinetic energy andtherefore a different total energy.

Not only are periodic vibrations very common, but small-amplitudevibrations are always sinusoidal as well. That is, the x− t graph is asine wave. This is because the graph of force versus position will al-ways look like a straight line on a sufficiently small scale. This typeof vibration is called simple harmonic motion. In simple harmonicmotion, the frequency is independent of the amplitude, and is givenby

ω =√k/m .


Problem 3.

Problem 5.

ProblemsKey√



1 Find an equation for the frequency of simple harmonic motionin terms of k and m.

√

2 Many single-celled organisms propel themselves through waterwith long tails, which they wiggle back and forth. (The most obviousexample is the sperm cell.) The frequency of the tail’s vibration istypically about 10-15 Hz. To what range of periods does this rangeof frequencies correspond?

3 The figure shows the oscillation of a microphone in responseto the author whistling the musical note “A.” The horizontal axis,representing time, has a scale of 1.0 ms per square. Find the periodT , the frequency f , and the angular frequency ω.

√

4 (a) Pendulum 2 has a string twice as long as pendulum 1. Ifwe define x as the distance traveled by the bob along a circle awayfrom the bottom, how does the k of pendulum 2 compare with thek of pendulum 1? Give a numerical ratio. [Hint: the total forceon the bob is the same if the angles away from the bottom are thesame, but equal angles do not correspond to equal values of x.]

(b) Based on your answer from part (a), how does the period of pen-dulum 2 compare with the period of pendulum 1? Give a numericalratio.

5 A pneumatic spring consists of a piston riding on top of theair in a cylinder. The upward force of the air on the piston is givenby Fair = ax−β, where β = 1.4 and a is a constant with funnyunits of N ·m1.4. For simplicity, assume the air only supports theweight, FW , of the piston itself, although in practice this deviceis used to support some other object. The equilibrium position,x0, is where FW equals −Fair. (Note that in the main text I haveassumed the equilibrium position to be at x = 0, but that is not thenatural choice here.) Assume friction is negligible, and consider acase where the amplitude of the vibrations is very small. Find theangular frequency of oscillation.

√

Problems 461

Problem 9.

6 Verify that energy is conserved in simple harmonic motion.

7 Consider the same pneumatic piston described in problem5, but now imagine that the oscillations are not small. Sketch agraph of the total force on the piston as it would appear over thiswider range of motion. For a wider range of motion, explain whythe vibration of the piston about equilibrium is not simple harmonicmotion, and sketch a graph of x vs t, showing roughly how the curveis different from a sine wave. [Hint: Acceleration corresponds to thecurvature of the x − t graph, so if the force is greater, the graphshould curve around more quickly.]

8 Archimedes’ principle states that an object partly or whollyimmersed in fluid experiences a buoyant force equal to the weightof the fluid it displaces. For instance, if a boat is floating in water,the upward pressure of the water (vector sum of all the forces ofthe water pressing inward and upward on every square inch of itshull) must be equal to the weight of the water displaced, becauseif the boat was instantly removed and the hole in the water filledback in, the force of the surrounding water would be just the rightamount to hold up this new “chunk” of water. (a) Show that a cubeof mass m with edges of length b floating upright (not tilted) in afluid of density ρ will have a draft (depth to which it sinks belowthe waterline) h given at equilibrium by h0 = m/b2ρ. (b) Find thetotal force on the cube when its draft is h, and verify that pluggingin h − h0 gives a total force of zero. (c) Find the cube’s period ofoscillation as it bobs up and down in the water, and show that canbe expressed in terms of and g only.

√

9 The figure shows a see-saw with two springs at Codornices Parkin Berkeley, California. Each spring has spring constant k, and akid of mass m sits on each seat. (a) Find the period of vibration interms of the variables k, m, a, and b. (b) Discuss the special casewhere a = b, rather than a > b as in the real see-saw. (c) Show thatyour answer to part a also makes sense in the case of b = 0.

√?

10 Show that the equation T = 2π√m/k has units that make

sense.


Problem 13.

Problem 14.

11 A hot scientific question of the 18th century was the shapeof the earth: whether its radius was greater at the equator than atthe poles, or the other way around. One method used to attack thisquestion was to measure gravity accurately in different locationson the earth using pendula. If the highest and lowest latitudesaccessible to explorers were 0 and 70 degrees, then the the strengthof gravity would in reality be observed to vary over a range fromabout 9.780 to 9.826 m/s2. This change, amounting to 0.046 m/s2,is greater than the 0.022 m/s2 effect to be expected if the earthhad been spherical. The greater effect occurs because the equatorfeels a reduction due not just to the acceleration of the spinningearth out from under it, but also to the greater radius of the earthat the equator. What is the accuracy with which the period of aone-second pendulum would have to be measured in order to provethat the earth was not a sphere, and that it bulged at the equator?

12 A certain mass, when hung from a certain spring, causesthe spring to stretch by an amount h compared to its equilibriumlength. If the mass is displaced vertically from this equilibrium, itwill oscillate up and down with a period Tosc. Give a numericalcomparison between Tosc and Tfall, the time required for the massto fall from rest through a height h, when it isn’t attached to thespring. .

√

13 Find the period of vertical oscillations of the mass m. Thespring, pulley, and ropes have negligible mass.

. Hint, p. 494√

14 The equilibrium length of each spring in the figure is b, sowhen the mass m is at the center, neither spring exerts any forceon it. When the mass is displaced to the side, the springs stretch;their spring constants are both k.(a) Find the energy, U , stored in the springs, as a function of y, thedistance of the mass up or down from the center.

√

(b) Show that the period of small up-down oscillations is infinite.

Problems 463

15 For a one-dimensional harmonic oscillator, the solution tothe energy conservation equation,

U +K =1

2kx2 +

1

2mv2 = constant ,

is an oscillation with frequency ω =√k/m.

Now consider an analogous system consisting of a bar magnet hungfrom a thread, which acts like a magnetic compass. A normal com-pass is full of water, so its oscillations are strongly damped, butthe magnet-on-a-thread compass has very little friction, and will os-cillate repeatedly around its equilibrium direction. The magneticenergy of the bar magnet is

U = −Bm cos θ ,

where B is a constant that measures the strength of the earth’smagnetic field, m is a constant that parametrizes the strength ofthe magnet, and θ is the angle, measured in radians, between thebar magnet and magnetic north. The equilibrium occurs at θ = 0,which is the minimum of U .

(a) Problem 26 on p. 440 gave some examples of how to constructanalogies between rotational and linear motion. Using the sametechnique, translate the equation defining the linear quantity k toone that defines an analogous angular one κ (Greek letter kappa).Applying this to the present example, find an expression for κ. (As-sume the thread is so thin that its stiffness does not have any sig-nificant effect compared to earth’s magnetic field.)(b) Find the frequency of the compass’s vibrations.

16 A mass m on a spring oscillates around an equilibrium atx = 0. Any function F (x) with an equilibrium at x = 0, F (0) = 0,can be approximated as F (x) = −kx, and if the spring’s behavioris symmetric with respect to positive and negative values of x, sothat F (−x) = −F (x), then the next level of improvement in suchan approximation would be F (x) = −kx + bx3. The general ideahere is that any smooth function can be approximated locally by apolynomial, and if you want a better approximation, you can use apolynomial with more terms in it. When you ask your calculator tocalculate a function like sin or ex, it’s using a polynomial approxima-tion with 10 or 12 terms. Physically, a spring with a positive valueof b gets stiffer when stretched strongly than an “ideal” spring withb = 0. A spring with a negative b is like a person who cracks understress — when you stretch it too much, it becomes more elastic thanan ideal spring would. We should not expect any spring to give to-tally ideal behavior no matter no matter how much it is stretched.For example, there has to be some point at which it breaks.

Do a numerical simulation of the oscillation of a mass on a springwhose force has a nonvanishing b. Is the period still independent of


amplitude? Is the amplitude-independent equation for the periodstill approximately valid for small enough amplitudes? Does theaddition of a positive x3 term tend to increase the period, or decreaseit? Include a printout of your program and its output with yourhomework paper.

17 An idealized pendulum consists of a pointlike mass m on theend of a massless, rigid rod of length L. Its amplitude, θ, is the anglethe rod makes with the vertical when the pendulum is at the endof its swing. Write a numerical simulation to determine the periodof the pendulum for any combination of m, L, and θ. Examine theeffect of changing each variable while manipulating the others. ?

Problems 465

Exercise 16: VibrationsEquipment:

• air track and carts of two different masses

• springs

• spring scales

Place the cart on the air track and attach springs so that it can vibrate.

1. Test whether the period of vibration depends on amplitude. Try at least one moderateamplitude, for which the springs do not go slack, at least one amplitude that is large enough sothat they do go slack, and one amplitude that’s the very smallest you can possibly observe.

2. Try a cart with a different mass. Does the period change by the expected factor, based onthe equation T = 2π

√m/k?

3. Use a spring scale to pull the cart away from equilibrium, and make a graph of force versusposition. Is it linear? If so, what is its slope?

4. Test the equation T = 2π√m/k numerically.


Top: A series of images froma film of the Tacoma NarrowsBridge vibrating on the day it wasto collapse. Middle: The bridgeimmediately before the collapse,with the sides vibrating 8.5 me-ters (28 feet) up and down. Notethat the bridge is over a mile long.Bottom: During and after the fi-nal collapse. The right-hand pic-ture gives a sense of the massivescale of the construction.

Chapter 17

Resonance

Soon after the mile-long Tacoma Narrows Bridge opened in July1940, motorists began to notice its tendency to vibrate frighteninglyin even a moderate wind. Nicknamed “Galloping Gertie,” the bridgecollapsed in a steady 42-mile-per-hour wind on November 7 of thesame year. The following is an eyewitness report from a newspapereditor who found himself on the bridge as the vibrations approachedthe breaking point.

“Just as I drove past the towers, the bridge began to sway vi-olently from side to side. Before I realized it, the tilt became soviolent that I lost control of the car... I jammed on the brakes and

467

got out, only to be thrown onto my face against the curb.

“Around me I could hear concrete cracking. I started to get mydog Tubby, but was thrown again before I could reach the car. Thecar itself began to slide from side to side of the roadway.

“On hands and knees most of the time, I crawled 500 yards ormore to the towers... My breath was coming in gasps; my kneeswere raw and bleeding, my hands bruised and swollen from grippingthe concrete curb... Toward the last, I risked rising to my feet andrunning a few yards at a time... Safely back at the toll plaza, Isaw the bridge in its final collapse and saw my car plunge into theNarrows.”

The ruins of the bridge formed an artificial reef, one of theworld’s largest. It was not replaced for ten years. The reason forits collapse was not substandard materials or construction, nor wasthe bridge under-designed: the piers were hundred-foot blocks ofconcrete, the girders massive and made of carbon steel. The bridgewas destroyed because of the physical phenomenon of resonance,the same effect that allows an opera singer to break a wine glasswith her voice and that lets you tune in the radio station you want.The replacement bridge, which has lasted half a century so far, wasbuilt smarter, not stronger. The engineers learned their lesson andsimply included some slight modifications to avoid the resonancephenomenon that spelled the doom of the first one.

17.1 Energy in vibrationsOne way of describing the collapse of the bridge is that the bridgekept taking energy from the steadily blowing wind and building upmore and more energetic vibrations. In this section, we discuss theenergy contained in a vibration, and in the subsequent sections wewill move on to the loss of energy and the adding of energy to avibrating system, all with the goal of understanding the importantphenomenon of resonance.

Going back to our standard example of a mass on a spring, wefind that there are two forms of energy involved: the potential energystored in the spring and the kinetic energy of the moving mass. Wemay start the system in motion either by hitting the mass to put inkinetic energy by pulling it to one side to put in potential energy.Either way, the subsequent behavior of the system is identical. Ittrades energy back and forth between kinetic and potential energy.(We are still assuming there is no friction, so that no energy isconverted to heat, and the system never runs down.)

The most important thing to understand about the energy con-tent of vibrations is that the total energy is proportional to the

468 Chapter 17 Resonance

a / Example 1.

square of the amplitude. Although the total energy is constant, itis instructive to consider two specific moments in the motion of themass on a spring as examples. When the mass is all the way toone side, at rest and ready to reverse directions, all its energy ispotential. We have already seen that the potential energy storedin a spring equals (1/2)kx2, so the energy is proportional to thesquare of the amplitude. Now consider the moment when the massis passing through the equilibrium point at x = 0. At this point ithas no potential energy, but it does have kinetic energy. The veloc-ity is proportional to the amplitude of the motion, and the kineticenergy, (1/2)mv2, is proportional to the square of the velocity, soagain we find that the energy is proportional to the square of theamplitude. The reason for singling out these two points is merelyinstructive; proving that energy is proportional to A2 at any pointwould suffice to prove that energy is proportional to A2 in general,since the energy is constant.

Are these conclusions restricted to the mass-on-a-spring exam-ple? No. We have already seen that F = −kx is a valid approxima-tion for any vibrating object, as long as the amplitude is small. Weare thus left with a very general conclusion: the energy of any vibra-tion is approximately proportional to the square of the amplitude,provided that the amplitude is small.

Water in a U-tube example 1If water is poured into a U-shaped tube as shown in the figure, itcan undergo vibrations about equilibrium. The energy of such avibration is most easily calculated by considering the “turnaroundpoint” when the water has stopped and is about to reverse direc-tions. At this point, it has only potential energy and no kineticenergy, so by calculating its potential energy we can find the en-ergy of the vibration. This potential energy is the same as thework that would have to be done to take the water out of the right-hand side down to a depth A below the equilibrium level, raise itthrough a height A, and place it in the left-hand side. The weightof this chunk of water is proportional to A, and so is the heightthrough which it must be lifted, so the energy is proportional toA2.

The range of energies of sound waves example 2. The amplitude of vibration of your eardrum at the threshold ofpain is about 106 times greater than the amplitude with whichit vibrates in response to the softest sound you can hear. Howmany times greater is the energy with which your ear has to copefor the painfully loud sound, compared to the soft sound?

. The amplitude is 106 times greater, and energy is proportionalto the square of the amplitude, so the energy is greater by a factor

Section 17.1 Energy in vibrations 469

of 1012 . This is a phenomenally large factor!

We are only studying vibrations right now, not waves, so we arenot yet concerned with how a sound wave works, or how the energygets to us through the air. Note that because of the huge range ofenergies that our ear can sense, it would not be reasonable to havea sense of loudness that was additive. Consider, for instance, thefollowing three levels of sound:

barely audible windquiet conversation . . . . 105 times more energy than the

windheavy metal concert . . 1012 times more energy than the

wind

In terms of addition and subtraction, the difference between thewind and the quiet conversation is nothing compared to the differ-ence between the quiet conversation and the heavy metal concert.Evolution wanted our sense of hearing to be able to encompass allthese sounds without collapsing the bottom of the scale so that any-thing softer than the crack of doom would sound the same. So ratherthan making our sense of loudness additive, mother nature made itmultiplicative. We sense the difference between the wind and thequiet conversation as spanning a range of about 5/12 as much as thewhole range from the wind to the heavy metal concert. Althougha detailed discussion of the decibel scale is not relevant here, thebasic point to note about the decibel scale is that it is logarithmic.The zero of the decibel scale is close to the lower limit of humanhearing, and adding 1 unit to the decibel measurement correspondsto multiplying the energy level (or actually the power per unit area)by a certain factor.

17.2 Energy lost from vibrationsNumerical treatment

An oscillator that has friction is referred to as damped. Let’suse numerical techniques to find the motion of a damped oscillatorthat is released away from equilibrium, but experiences no drivingforce after that. We can expect that the motion will consist ofoscillations that gradually die out. Newton’s second law, a = F/m,gives a = (−kx− bv)/m. This becomes a little prettier if we rewriteit in the form

ma+ bv + kx = 0 ,

which gives symmetric treatment to three terms involving x and itsfirst and second derivatives, v and a.

1 import math

2 k=39.4784 # chosen to give a period of 1 second

3 m=1.


4 b=0.211 # chosen to make the results simple

5 x=1.

6 v=0.

7 t=0.

8 dt=.01

9 n=1000

10 for j in range(n):

11 x=x+v*dt

12 a=(-k*x-b*v)/m

13 if (v>0) and (v+a*dt<0) :

14 print("turnaround at t=",t,", x=",x)

15 v=v+a*dt

16 t=t+dt

turnaround at t= 0.99 , x= 0.899919262445

turnaround at t= 1.99 , x= 0.809844934046

turnaround at t= 2.99 , x= 0.728777519477

turnaround at t= 3.99 , x= 0.655817260033

turnaround at t= 4.99 , x= 0.590154191135

turnaround at t= 5.99 , x= 0.531059189965

turnaround at t= 6.99 , x= 0.477875914756

turnaround at t= 7.99 , x= 0.430013546991

turnaround at t= 8.99 , x= 0.386940256644

turnaround at t= 9.99 , x= 0.348177318484

The spring constant, k = 4π = 39.4784 N/m, is designed sothat if the undamped equation f = (1/2π)

√k/m was still true, the

frequency would be 1 Hz. We start by noting that the addition of asmall amount of damping doesn’t seem to have changed the periodat all, or at least not to within the accuracy of the calculation. Youcan check for yourself, however, that a large value of b, say 5 N·s/m,does change the period significantly.

We release the mass from x = 1 m, and after one cycle, it onlycomes back to about x = 0.9 m. I chose b = 0.211 N·s/m by fiddlingaround until I got this result, since a decrease of exactly 10% iseasy to discuss. Notice how the amplitude after two cycles is about0.81 m, i.e., 1 m times 0.92: the amplitude has again dropped byexactly 10%. This pattern continues for as long as the simulationruns, e.g., for the last two cycles, we have 0.34818/0.38694=0.89982,or almost exactly 0.9 again. It might have seemed capricious when Ichose to use the unrealistic equation F = −bv, but this is the payoff.Only with −bv friction do we get this kind of mathematically simpleexponential decay.

Analytic treatment

Taking advantage of this unexpectedly simple result, let’s findan analytic solution for the motion. The numerical output suggests

Section 17.2 Energy lost from vibrations 471

b / A damped sine wave, ofthe form x = Ae−ctsin(ωf t + δ).

c / Self-check A.

d / A damped sine wave iscompared with an undampedone, with m and k kept the sameand only b changed.

that we assume a solution of the form

x = Ae−ct sin(ωf t+ δ) ,

where the unknown constants ωf and c will presumably be related tom, b, and k. The constant c indicates how quickly the oscillations dieout. The constant ωf is, as before, defined as 2π times the frequency,with the subscript f to indicate a free (undriven) solution. Allour equations will come out much simpler if we use ωs everywhereinstead of fs from now on, and, as physicists often do, I’ll generallyuse the word “frequency” to refer to ω when the context makes itclear what I’m talking about. The phase angle δ has no real physicalsignificance, since we can define t = 0 to be any moment in time welike.

self-check AIn figure c, which graph has the greater value of c? . Answer, p. 513

The factor A for the initial amplitude can also be omitted with-out loss of generality, since the equation we’re trying to solve, ma+bv + kx = 0, is linear. That is, v and a are the first and secondderivatives of x, and the derivative of Ax is simply A times thederivative of x. Thus, if x(t) is a solution of the equation, thenmultiplying it by a constant gives an equally valid solution. This isanother place where we see that a damping force proportional to v isthe easiest to handle mathematically. For a damping force propor-tional to v2, for example, we would have had to solve the equationma+ bv2 + kx = 0, which is nonlinear.

For the purpose of determining ωf and c, the most general formwe need to consider is therefore x = e−ct sinωf t , whose first andsecond derivatives are v = e−ct (−c sinωf t+ ω cosωf t) and a =

e−ct(c2 sinωf t− 2ωfc cosωf t− ω2

f sinωf t)

. Plugging these into the

equation ma + bv + kx = 0 and setting the sine and cosine partsequal to zero gives, after some tedious algebra,

c =b

2m

and

ωf =

√k

m− b2

4m2.

Intuitively, we expect friction to “slow down” the motion, as whenwe ride a bike into a big patch of mud. “Slow down,” however, couldhave more than one meaning here. It could mean that the oscillatorwould take more time to complete each cycle, or it could mean thatas time went on, the oscillations would die out, thus giving smallervelocities.

Our mathematical results show that both of these things hap-pen. The first equation says that c, which indicates how quickly the


e / 1. Pushing a child on aswing gradually puts more andmore energy into her vibrations.2. A fairly realistic graph of thedriving force acting on the child.3. A less realistic, but moremathematically simple, drivingforce.

f / The amplitude approachesa maximum.

oscillations damp out, is directly related to b, the strength of thedamping.

The second equation, for the frequency, can be compared withthe result from page 453 of

√k/m for the undamped system. Let’s

refer to this now as ωo, to distinguish it from the actual frequencyωf of the free oscillations when damping is present. The result forωf will be less than ωo, due to the presence of the b2/4m2 term. Thistells us that the addition of friction to the system does increase thetime required for each cycle. However, it is very common for theb2/4m2 term to be negligible, so that ωf ≈ ωo.

Figure d shows an example. The damping here is quite strong:after only one cycle of oscillation, the amplitude has already beenreduced by a factor of 2, corresponding to a factor of 4 in energy.However, the frequency of the damped oscillator is only about 1%lower than that of the undamped one; after five periods, the ac-cumulated lag is just barely visible in the offsetting of the arrows.We can see that extremely strong damping — even stronger thanthis — would have been necessary in order to make ωf ≈ ωo a poorapproximation.

17.3 Putting energy into vibrations

When pushing a child on a swing, you cannot just apply a con-stant force. A constant force will move the swing out to a certainangle, but will not allow the swing to start swinging. Nor can yougive short pushes at randomly chosen times. That type of ran-dom pushing would increase the child’s kinetic energy whenever youhappened to be pushing in the same direction as her motion, but itwould reduce her energy when your pushing happened to be in theopposite direction compared to her motion. To make her build upher energy, you need to make your pushes rhythmic, pushing at thesame point in each cycle. In other words, your force needs to form arepeating pattern with the same frequency as the normal frequencyof vibration of the swing. Graph e/1 shows what the child’s x − tgraph would look like as you gradually put more and more energyinto her vibrations. A graph of your force versus time would prob-ably look something like graph 2. It turns out, however, that it ismuch simpler mathematically to consider a vibration with energybeing pumped into it by a driving force that is itself a sine-wave, 3.A good example of this is your eardrum being driven by the forceof a sound wave.

Now we know realistically that the child on the swing will notkeep increasing her energy forever, nor does your eardrum end upexploding because a continuing sound wave keeps pumping more andmore energy into it. In any realistic system, there is energy goingout as well as in. As the vibrations increase in amplitude, there is an

Section 17.3 Putting energy into vibrations 473

increase in the amount of energy taken away by damping with eachcycle. This occurs for two reasons. Work equals force times distance(or, more accurately, the area under the force-distance curve). Asthe amplitude of the vibrations increases, the damping force is beingapplied over a longer distance. Furthermore, the damping forceusually increases with velocity (we usually assume for simplicitythat it is proportional to velocity), and this also serves to increasethe rate at which damping forces remove energy as the amplitudeincreases. Eventually (and small children and our eardrums arethankful for this!), the amplitude approaches a maximum value, f,at which energy is removed by the damping force just as quickly asit is being put in by the driving force.

This process of approaching a maximum amplitude happens ex-tremely quickly in many cases, e.g., the ear or a radio receiver, andwe don’t even notice that it took a millisecond or a microsecondfor the vibrations to “build up steam.” We are therefore mainlyinterested in predicting the behavior of the system once it has hadenough time to reach essentially its maximum amplitude. This isknown as the steady-state behavior of a vibrating system.

Now comes the interesting part: what happens if the frequencyof the driving force is mismatched to the frequency at which thesystem would naturally vibrate on its own? We all know that aradio station doesn’t have to be tuned in exactly, although there isonly a small range over which a given station can be received. Thedesigners of the radio had to make the range fairly small to makeit possible eliminate unwanted stations that happened to be nearbyin frequency, but it couldn’t be too small or you wouldn’t be ableto adjust the knob accurately enough. (Even a digital radio canbe tuned to 88.0 MHz and still bring in a station at 88.1 MHz.)The ear also has some natural frequency of vibration, but in thiscase the range of frequencies to which it can respond is quite broad.Evolution has made the ear’s frequency response as broad as pos-sible because it was to our ancestors’ advantage to be able to heareverything from a low roars to a high-pitched shriek.

The remainder of this section develops four important facts aboutthe response of a system to a driving force whose frequency is notnecessarily the same as the system’s natural frequency of vibration.The style is approximate and intuitive, but proofs are given in sec-tion 17.4.

First, although we know the ear has a frequency — about 4000Hz — at which it would vibrate naturally, it does not vibrate at4000 Hz in response to a low-pitched 200 Hz tone. It always re-sponds at the frequency at which it is driven. Otherwise all pitcheswould sound like 4000 Hz to us. This is a general fact about drivenvibrations:


g / The collapsed section ofthe Nimitz Freeway.

(1) The steady-state response to a sinusoidal driving force oc-curs at the frequency of the force, not at the system’s own naturalfrequency of vibration.

Now let’s think about the amplitude of the steady-state response.Imagine that a child on a swing has a natural frequency of vibrationof 1 Hz, but we are going to try to make her swing back and forth at3 Hz. We intuitively realize that quite a large force would be neededto achieve an amplitude of even 30 cm, i.e., the amplitude is less inproportion to the force. When we push at the natural frequency of1 Hz, we are essentially just pumping energy back into the systemto compensate for the loss of energy due to the damping (friction)force. At 3 Hz, however, we are not just counteracting friction. Weare also providing an extra force to make the child’s momentumreverse itself more rapidly than it would if gravity and the tensionin the chain were the only forces acting. It is as if we are artificiallyincreasing the k of the swing, but this is wasted effort because wespend just as much time decelerating the child (taking energy outof the system) as accelerating her (putting energy in).

Now imagine the case in which we drive the child at a verylow frequency, say 0.02 Hz or about one vibration per minute. Weare essentially just holding the child in position while very slowlywalking back and forth. Again we intuitively recognize that theamplitude will be very small in proportion to our driving force.Imagine how hard it would be to hold the child at our own head-levelwhen she is at the end of her swing! As in the too-fast 3 Hz case, weare spending most of our effort in artificially changing the k of theswing, but now rather than reinforcing the gravity and tension forceswe are working against them, effectively reducing k. Only a verysmall part of our force goes into counteracting friction, and the restis used in repetitively putting potential energy in on the upswingand taking it back out on the downswing, without any long-termgain.

We can now generalize to make the following statement, whichis true for all driven vibrations:

(2) A vibrating system resonates at its own natural frequency.1

That is, the amplitude of the steady-state response is greatest inproportion to the amount of driving force when the driving forcematches the natural frequency of vibration.

An opera singer breaking a wine glass example 3In order to break a wineglass by singing, an opera singer mustfirst tap the glass to find its natural frequency of vibration, andthen sing the same note back.


Collapse of the Nimitz Freeway in an earthquake example 4I led off the chapter with the dramatic collapse of the TacomaNarrows Bridge, mainly because a it was well documented by alocal physics professor, and an unknown person made a movieof the collapse. The collapse of a section of the Nimitz Freewayin Oakland, CA, during a 1989 earthquake is however a simplerexample to analyze.

An earthquake consists of many low-frequency vibrations that oc-cur simultaneously, which is why it sounds like a rumble of inde-terminate pitch rather than a low hum. The frequencies that wecan hear are not even the strongest ones; most of the energy isin the form of vibrations in the range of frequencies from about 1Hz to 10 Hz.

Now all the structures we build are resting on geological layersof dirt, mud, sand, or rock. When an earthquake wave comesalong, the topmost layer acts like a system with a certain natu-ral frequency of vibration, sort of like a cube of jello on a platebeing shaken from side to side. The resonant frequency of thelayer depends on how stiff it is and also on how deep it is. Theill-fated section of the Nimitz freeway was built on a layer of mud,and analysis by geologist Susan E. Hough of the U.S. GeologicalSurvey shows that the mud layer’s resonance was centered onabout 2.5 Hz, and had a width covering a range from about 1 Hzto 4 Hz.

When the earthquake wave came along with its mixture of fre-quencies, the mud responded strongly to those that were close toits own natural 2.5 Hz frequency. Unfortunately, an engineeringanalysis after the quake showed that the overpass itself had a res-onant frequency of 2.5 Hz as well! The mud responded strongly tothe earthquake waves with frequencies close to 2.5 Hz, and thebridge responded strongly to the 2.5 Hz vibrations of the mud,causing sections of it to collapse.

Collapse of the Tacoma Narrows Bridge example 5Let’s now examine the more conceptually difficult case of theTacoma Narrows Bridge. The surprise here is that the wind wassteady. If the wind was blowing at constant velocity, why did itshake the bridge back and forth? The answer is a little compli-cated. Based on film footage and after-the-fact wind tunnel exper-iments, it appears that two different mechanisms were involved.

The first mechanism was the one responsible for the initial, rel-atively weak vibrations, and it involved resonance. As the windmoved over the bridge, it began acting like a kite or an airplanewing. As shown in the figure, it established swirling patterns of airflow around itself, of the kind that you can see in a moving cloudof smoke. As one of these swirls moved off of the bridge, therewas an abrupt change in air pressure, which resulted in an up or


down force on the bridge. We see something similar when a flagflaps in the wind, except that the flag’s surface is usually verti-cal. This back-and-forth sequence of forces is exactly the kind ofperiodic driving force that would excite a resonance. The fasterthe wind, the more quickly the swirls would get across the bridge,and the higher the frequency of the driving force would be. At justthe right velocity, the frequency would be the right one to excitethe resonance. The wind-tunnel models, however, show that thepattern of vibration of the bridge excited by this mechanism wouldhave been a different one than the one that finally destroyed thebridge.

The bridge was probably destroyed by a different mechanism, inwhich its vibrations at its own natural frequency of 0.2 Hz set upan alternating pattern of wind gusts in the air immediately aroundit, which then increased the amplitude of the bridge’s vibrations.This vicious cycle fed upon itself, increasing the amplitude of thevibrations until the bridge finally collapsed.

As long as we’re on the subject of collapsing bridges, it is worthbringing up the reports of bridges falling down when soldiers march-ing over them happened to step in rhythm with the bridge’s naturalfrequency of oscillation. This is supposed to have happened in 1831in Manchester, England, and again in 1849 in Anjou, France. Manymodern engineers and scientists, however, are suspicious of the anal-ysis of these reports. It is possible that the collapses had more to dowith poor construction and overloading than with resonance. TheNimitz Freeway and Tacoma Narrows Bridge are far better docu-mented, and occurred in an era when engineers’ abilities to analyzethe vibrations of a complex structure were much more advanced.

Emission and absorption of light waves by atoms example 6In a very thin gas, the atoms are sufficiently far apart that they canact as individual vibrating systems. Although the vibrations are ofa very strange and abstract type described by the theory of quan-tum mechanics, they nevertheless obey the same basic rules asordinary mechanical vibrations. When a thin gas made of a cer-tain element is heated, it emits light waves with certain specificfrequencies, which are like a fingerprint of that element. As withall other vibrations, these atomic vibrations respond most stronglyto a driving force that matches their own natural frequency. Thusif we have a relatively cold gas with light waves of various fre-quencies passing through it, the gas will absorb light at preciselythose frequencies at which it would emit light if heated.

(3) When a system is driven at resonance, the steady-state vi-brations have an amplitude that is proportional to Q.

This is fairly intuitive. The steady-state behavior is an equilib-


h / The definition of the fullwidth at half maximum.

rium between energy input from the driving force and energy lossdue to damping. A low-Q oscillator, i.e., one with strong damping,dumps its energy faster, resulting in lower-amplitude steady-statemotion.

self-check BIf an opera singer is shopping for a wine glass that she can impress herfriends by breaking, what should she look for? . Answer, p. 513

Piano strings ringing in sympathy with a sung note example 7. A sufficiently loud musical note sung near a piano with the lidraised can cause the corresponding strings in the piano to vibrate.(A piano has a set of three strings for each note, all struck by thesame hammer.) Why would this trick be unlikely to work with aviolin?

. If you have heard the sound of a violin being plucked (the pizzi-cato effect), you know that the note dies away very quickly. Inother words, a violin’s Q is much lower than a piano’s. This meansthat its resonances are much weaker in amplitude.

Our fourth and final fact about resonance is perhaps the mostsurprising. It gives us a way to determine numerically how widea range of driving frequencies will produce a strong response. Asshown in the graph, resonances do not suddenly fall off to zero out-side a certain frequency range. It is usual to describe the width of aresonance by its full width at half-maximum (FWHM) as illustratedin figure h.

(4) The FWHM of a resonance is related to its Q and its resonantfrequency fres by the equation

FWHM =fresQ

.

(This equation is only a good approximation when Q is large.)

Why? It is not immediately obvious that there should be anylogical relationship between Q and the FWHM. Here’s the idea. Aswe have seen already, the reason why the response of an oscillatoris smaller away from resonance is that much of the driving force isbeing used to make the system act as if it had a different k. Roughlyspeaking, the half-maximum points on the graph correspond to theplaces where the amount of the driving force being wasted in thisway is the same as the amount of driving force being used pro-ductively to replace the energy being dumped out by the dampingforce. If the damping force is strong, then a large amount of forceis needed to counteract it, and we can waste quite a bit of drivingforce on changing k before it becomes comparable to the dampingforce. If, on the other hand, the damping force is weak, then even a


small amount of force being wasted on changing k will become sig-nificant in proportion, and we cannot get very far from the resonantfrequency before the two are comparable.

The response is in general out of phase with the driving force byan angle δ.

i / Dependence of the amplitudeand phase angle on the drivingfrequency. The undamped caseis Q = ∞, and the other curvesrepresent Q=1, 3, and 10. Fm, m,and ωo are all set to 1.

Changing the pitch of a wind instrument example 8. A saxophone player normally selects which note to play bychoosing a certain fingering, which gives the saxophone a cer-tain resonant frequency. The musician can also, however, changethe pitch significantly by altering the tightness of her lips. Thiscorresponds to driving the horn slightly off of resonance. If thepitch can be altered by about 5% up or down (about one musi-cal half-step) without too much effort, roughly what is the Q of asaxophone?


j / Example 12. 1. A com-pass needle vibrates about theequilibrium position under theinfluence of the earth’s magneticforces. 2. The orientation of aproton’s spin vibrates around itsequilibrium direction under theinfluence of the magnetic forcescoming from the surroundingelectrons and nuclei.

k / A member of the author’sfamily, who turned out to behealthy.

l / A three-dimensional com-puter reconstruction of the shapeof a human brain, based onmagnetic resonance data.

. Five percent is the width on one side of the resonance, so thefull width is about 10%, FWHM / fres = 0.1. This implies a Qof about 10, i.e., once the musician stops blowing, the horn willcontinue sounding for about 10 cycles before its energy falls off bya factor of 535. (Blues and jazz saxophone players will typicallychoose a mouthpiece that has a low Q, so that they can producethe bluesy pitch-slides typical of their style. “Legit,” i.e., classicallyoriented players, use a higher-Q setup because their style onlycalls for enough pitch variation to produce a vibrato.)

Decay of a saxophone tone example 9. If a typical saxophone setup has a Q of about 10, how long willit take for a 100-Hz tone played on a baritone saxophone to diedown by a factor of 535 in energy, after the player suddenly stopsblowing?

. A Q of 10 means that it takes 10 cycles for the vibrations to diedown in energy by a factor of 535. Ten cycles at a frequency of100 Hz would correspond to a time of 0.1 seconds, which is notvery long. This is why a saxophone note doesn’t “ring” like a noteplayed on a piano or an electric guitar.

Q of a radio receiver example 10. A radio receiver used in the FM band needs to be tuned in towithin about 0.1 MHz for signals at about 100 MHz. What is itsQ?

. Q = fres/FWHM = 1000. This is an extremely high Q comparedto most mechanical systems.

Q of a stereo speaker example 11We have already given one reason why a stereo speaker shouldhave a low Q: otherwise it would continue ringing after the end ofthe musical note on the recording. The second reason is that wewant it to be able to respond to a large range of frequencies.

Nuclear magnetic resonance example 12If you have ever played with a magnetic compass, you have un-doubtedly noticed that if you shake it, it takes some time to settledown, j/1. As it settles down, it acts like a damped oscillator of thetype we have been discussing. The compass needle is simply asmall magnet, and the planet earth is a big magnet. The magneticforces between them tend to bring the needle to an equilibriumposition in which it lines up with the planet-earth-magnet.

Essentially the same physics lies behind the technique called Nu-clear Magnetic Resonance (NMR). NMR is a technique used todeduce the molecular structure of unknown chemical substances,and it is also used for making medical images of the inside of peo-ple’s bodies. If you ever have an NMR scan, they will actually tellyou you are undergoing “magnetic resonance imaging” or “MRI,”because people are scared of the word “nuclear.” In fact, the


nuclei being referred to are simply the non-radioactive nuclei ofatoms found naturally in your body.

Here’s how NMR works. Your body contains large numbers ofhydrogen atoms, each consisting of a small, lightweight electronorbiting around a large, heavy proton. That is, the nucleus of ahydrogen atom is just one proton. A proton is always spinningon its own axis, and the combination of its spin and its electricalcharge cause it to behave like a tiny magnet. The principle iden-tical to that of an electromagnet, which consists of a coil of wirethrough which electrical charges pass; the circling motion of thecharges in the coil of wire makes it magnetic, and in the sameway, the circling motion of the proton’s charge makes it magnetic.

Now a proton in one of your body’s hydrogen atoms finds itselfsurrounded by many other whirling, electrically charged particles:its own electron, plus the electrons and nuclei of the other nearbyatoms. These neighbors act like magnets, and exert magneticforces on the proton, j/2. The k of the vibrating proton is simply ameasure of the total strength of these magnetic forces. Depend-ing on the structure of the molecule in which the hydrogen atomfinds itself, there will be a particular set of magnetic forces actingon the proton and a particular value of k . The NMR apparatusbombards the sample with radio waves, and if the frequency ofthe radio waves matches the resonant frequency of the proton,the proton will absorb radio-wave energy strongly and oscillatewildly. Its vibrations are damped not by friction, because there isno friction inside an atom, but by the reemission of radio waves.

By working backward through this chain of reasoning, one can de-termine the geometric arrangement of the hydrogen atom’s neigh-boring atoms. It is also possible to locate atoms in space, allowingmedical images to be made.

Finally, it should be noted that the behavior of the proton cannotbe described entirely correctly by Newtonian physics. Its vibra-tions are of the strange and spooky kind described by the laws ofquantum mechanics. It is impressive, however, that the few sim-ple ideas we have learned about resonance can still be appliedsuccessfully to describe many aspects of this exotic system.

Discussion question

A Nikola Tesla, one of the inventors of radio and an archetypical madscientist, told a credulous reporter the following story about an applica-tion of resonance. He built an electric vibrator that fit in his pocket, andattached it to one of the steel beams of a building that was under construc-tion in New York. Although the article in which he was quoted didn’t sayso, he presumably claimed to have tuned it to the resonant frequency ofthe building. “In a few minutes, I could feel the beam trembling. Graduallythe trembling increased in intensity and extended throughout the wholegreat mass of steel. Finally, the structure began to creak and weave, and


the steelworkers came to the ground panic-stricken, believing that therehad been an earthquake. ... [If] I had kept on ten minutes more, I couldhave laid that building flat in the street.” Is this physically plausible?

17.4 ? ProofsOur first goal is to predict the amplitude of the steady-state vibra-tions as a function of the frequency of the driving force and theamplitude of the driving force. With that equation in hand, we willthen prove statements 2, 3, and 4 from section 17.3.

We have an external driving force F = Fm sinωt, where theconstant Fm indicates the maximum strength of the force in eitherdirection. The equation of motion is

[1] ma+ bv + kx = Fm sinωt .

For the steady-state motion, we’re going to look for a solution ofthe form

x = A sin(ωt+ δ) .

The left-hand side of the equation of motion will clearly be a sinu-soidal function with frequency ω, so it can only equal the right-handside if, as we have already implicitly assumed, the frequency of themotion matches the frequency of the driving force. This provesstatement (1).

In contrast to the undriven case, here it’s not possible to sweep Aand δ under the rug. The amplitude of the steady-state motion, A,is actually the most interesting thing to know about the steady-statemotion, and it’s not true that we still have a solution no matter howwe fiddle with A; if we have a solution for a certain value of A, thenmultiplying A by some constant would break the equality betweenthe two sides of the equation of motion. It’s also no longer true thatwe can get rid of δ simply be redefining when we start the clock;here δ represents a difference in time between the start of one cycleof the driving force and the start of the corresponding cycle of themotion.

The velocity and acceleration are v = ωA cos(ωt + δ) and a =−ω2A sin(ωt+ δ), and if we plug these into the equation of motion,[1], and simplify a little, we find

[2] (k −mω2) sin(ωt+ δ) + ωb cos(ωt+ δ) =FmA

sinωt .

The sum of any two sinusoidal functions with the same frequencyis also a sinusoidal, so the whole left side adds up to a sinusoidal.By fiddling with A and δ we can make the amplitudes and phasesof the two sides of the equation match up.

Using the trig identities for the sine of a sum and cosine of a


sum, we can change equation [2] into the form[(−mω2 + k) cos δ − bω sin δ − Fm/A

]sinωt

+[(−mω2 + k) sin δ + bω cos δ

]cosωt = 0 .

Both the quantities in square brackets must equal zero, which givesus two equations we can use to determine the unknowns A and δ.The results are

[3] δ = tan−1 ωωo

Q(ω2o − ω2)

and

[4] A =Fm

m

√(ω2 − ω2

o)2 + ω2oω

2Q−2

.

Statement 2: maximum amplitude at resonance

Equation [4] makes it plausible that the amplitude is maximizedwhen the system is driven at close to its resonant frequency. Atf = fo, the first term inside the square root vanishes, and thismakes the denominator as small as possible, causing the amplitudeto be as big as possible. (Actually this is only approximately true,because it is possible to make A a little bigger by decreasing f a littlebelow fo, which makes the second term smaller. This technical issueis addressed in homework problem 3 on page 487.)

Statement 3: amplitude at resonance proportional to Q

Equation [4] shows that the amplitude at resonance is propor-tional to 1/b, and the Q of the system is inversely proportional tob, so the amplitude at resonance is proportional to Q.

Statement 4: FWHM related to Q

We will satisfy ourselves by proving only the proportionalityFWHM ∝ fo/Q, not the actual equation FWHM = fo/Q. Theenergy is proportional to A2, i.e., to the inverse of the quantityinside the square root in equation [4]. At resonance, the first terminside the square root vanishes, and the half-maximum points occurat frequencies for which the whole quantity inside the square rootis double its value at resonance, i.e., when the two terms are equal.At the half-maximum points, we have

f2 − f2o =

(fo ±

FWHM

2

)2

− f2o

= ±fo · FWHM +1

4FWHM2

If we assume that the width of the resonance is small compared tothe resonant frequency, then the FWHM2 term is negligible com-pared to the fo · FWHM term, and setting the terms in equation 4equal to each other gives

4π2m2 (foFWHM)2 = b2f2 .

Section 17.4 ? Proofs 483

We are assuming that the width of the resonance is small comparedto the resonant frequency, so f and fo can be taken as synonyms.Thus,

FWHM =b

2πm.

We wish to connect this to Q, which can be interpreted as the en-ergy of the free (undriven) vibrations divided by the work done bydamping in one cycle. The former equals kA2/2, and the latter isproportional to the force, bv ∝ bAfo, multiplied by the distancetraveled, A. (This is only a proportionality, not an equation, sincethe force is not constant.) We therefore find that Q is proportionalto k/bfo. The equation for the FWHM can then be restated as aproportionality FWHM ∝ k/Qfom ∝ fo/Q.


SummarySelected vocabularydamping . . . . . the dissipation of a vibration’s energy into

heat energy, or the frictional force that causesthe loss of energy

quality factor . . the number of oscillations required for a sys-tem’s energy to fall off by a factor of 535 dueto damping

driving force . . . an external force that pumps energy into a vi-brating system

resonance . . . . the tendency of a vibrating system to respondmost strongly to a driving force whose fre-quency is close to its own natural frequencyof vibration

steady state . . . the behavior of a vibrating system after it hashad plenty of time to settle into a steady re-sponse to a driving force

NotationQ . . . . . . . . . the quality factorfo . . . . . . . . . the natural (resonant) frequency of a vibrating

system, i.e., the frequency at which it wouldvibrate if it was simply kicked and left alone

f . . . . . . . . . . the frequency at which the system actually vi-brates, which in the case of a driven system isequal to the frequency of the driving force, notthe natural frequency

Summary

The energy of a vibration is always proportional to the square ofthe amplitude, assuming the amplitude is small. Energy is lost froma vibrating system for various reasons such as the conversion to heatvia friction or the emission of sound. This effect, called damping,will cause the vibrations to decay exponentially unless energy ispumped into the system to replace the loss. A driving force thatpumps energy into the system may drive the system at its ownnatural frequency or at some other frequency. When a vibratingsystem is driven by an external force, we are usually interested inits steady-state behavior, i.e., its behavior after it has had time tosettle into a steady response to a driving force. In the steady state,the same amount of energy is pumped into the system during eachcycle as is lost to damping during the same period.

The following are four important facts about a vibrating systembeing driven by an external force:

(1) The steady-state response to a sinusoidal driving force oc-curs at the frequency of the force, not at the system’s own naturalfrequency of vibration.

Summary 485

(2) A vibrating system resonates at its own natural frequency.That is, the amplitude of the steady-state response is greatest inproportion to the amount of driving force when the driving forcematches the natural frequency of vibration.

(3) When a system is driven at resonance, the steady-state vi-brations have an amplitude that is proportional to Q.

(4) The FWHM of a resonance is related to its Q and its resonantfrequency fo by the equation

FWHM =fo

Q.

(This equation is only a good approximation when Q is large.)


ProblemsKey√



1 If one stereo system is capable of producing 20 watts of soundpower and another can put out 50 watts, how many times greateris the amplitude of the sound wave that can be created by the morepowerful system? (Assume they are playing the same music.)

2 Many fish have an organ known as a swim bladder, an air-filledcavity whose main purpose is to control the fish’s buoyancy an allowit to keep from rising or sinking without having to use its muscles.In some fish, however, the swim bladder (or a small extension of it)is linked to the ear and serves the additional purpose of amplifyingsound waves. For a typical fish having such an anatomy, the bladderhas a resonant frequency of 300 Hz, the bladder’s Q is 3, and themaximum amplification is about a factor of 100 in energy. Over whatrange of frequencies would the amplification be at least a factor of50?

3 As noted in section 17.4, it is only approximately true that theamplitude has its maximum at f = (1/2π)

√k/m. Being more care-

ful, we should actually define two different symbols, f0 = (1/2π)√k/m

and fo for the slightly different frequency at which the amplitudeis a maximum, i.e., the actual resonant frequency. In this notation,the amplitude as a function of frequency is

A =F

2π√

4π2m2(f2 − f2

0

)2+ b2f2

.

Show that the maximum occurs not at fo but rather at the frequency

fo =

√f2

0 −b2

8π2m2=

√f2

0 −1

2FWHM2

Hint: Finding the frequency that minimizes the quantity inside thesquare root is equivalent to, but much easier than, finding the fre-quency that maximizes the amplitude.

Problems 487

4 (a) Let W be the amount of work done by friction in the firstcycle of oscillation, i.e., the amount of energy lost to heat. Findthe fraction of the original energy E that remains in the oscillationsafter n cycles of motion.

(b) From this, prove the equation(1− W

E

)Q= e−2π

(recalling that the number 535 in the definition of Q is e2π).

(c) Use this to prove the approximation 1/Q ≈ (1/2π)W/E. (Hint:Use the approximation ln(1 +x) ≈ x, which is valid for small valuesof x.)

5 (a) We observe that the amplitude of a certain free oscillationdecreases from Ao to Ao/Z after n oscillations. Find its Q.

√

(b) The figure is from Shape memory in Spider draglines, Emile,Le Floch, and Vollrath, Nature 440:621 (2006). Panel 1 shows anelectron microscope’s image of a thread of spider silk. In 2, a spi-der is hanging from such a thread. From an evolutionary point ofview, it’s probably a bad thing for the spider if it twists back andforth while hanging like this. (We’re referring to a back-and-forthrotation about the axis of the thread, not a swinging motion like apendulum.) The authors speculate that such a vibration could makethe spider easier for predators to see, and it also seems to me thatit would be a bad thing just because the spider wouldn’t be ableto control its orientation and do what it was trying to do. Panel 3shows a graph of such an oscillation, which the authors measuredusing a video camera and a computer, with a 0.1 g mass hung from itin place of a spider. Compared to human-made fibers such as kevlaror copper wire, the spider thread has an unusual set of properties:

1. It has a low Q, so the vibrations damp out quickly.

2. It doesn’t become brittle with repeated twisting as a copperwire would.

3. When twisted, it tends to settle in to a new equilibrium angle,rather than insisting on returning to its original angle. Youcan see this in panel 2, because although the experimentersinitially twisted the wire by 35 degrees, the thread only per-formed oscillations with an amplitude much smaller than ±35degrees, settling down to a new equilibrium at 27 degrees.

4. Over much longer time scales (hours), the thread eventuallyresets itself to its original equilbrium angle (shown as zero de-grees on the graph). (The graph reproduced here only showsthe motion over a much shorter time scale.) Some human-made materials have this “memory” property as well, but they


typically need to be heated in order to make them go back totheir original shapes.

Focusing on property number 1, estimate the Q of spider silk fromthe graph.

√

Problem 5.

6 An oscillator with sufficiently strong damping has its maximumresponse at ω = 0. Using equation [4] on p. 483 , find the value ofQ at which this behavior sets in.

. Hint, p. 494 . Answer, p. 514

Problems 489

7 The goal of this problem is to refine the proportionalityFWHM ∝ fres/Q into the equation FWHM = fres/Q, i.e., to provethat the constant of proportionality equals 1.

(a) Show that the work done by a damping force F = −bv over onecycle of steady-state motion equals Wdamp = −2π2bfA2. Hint: Itis less confusing to calculate the work done over half a cycle, fromx = −A to x = +A, and then double it.

(b) Show that the fraction of the undriven oscillator’s energy lost todamping over one cycle is |Wdamp|/E = 4π2bf/k.

(c) Use the previous result, combined with the result of problem 4,to prove that Q equals k/2πbf .

(d) Combine the preceding result for Q with the equation FWHM =b/2πm from section 17.4 to prove the equation FWHM = fres/Q.

?

8 An oscillator has Q=6.00, and, for convenience, let’s assumeFm = 1.00, ωo = 1.00, and m = 1.00. The usual approximationswould give

ωres = ωo ,

Ares = 6.00 , and

∆ω = 1/6.00 .

Determine these three quantities numerically using equation [4] onp. 483 , and compare with the approximations.


Exercise 17: Resonance1. Compare the oscillator’s energies at A, B, C, and D.

2. Compare the Q values of the two oscillators.

3. Match the x-t graphs in #2 with the amplitude-frequency graphs below.

Exercise 17: Resonance 491

Chapter AAppendix 1: Programming withPythonThe purpose of this tutorial is to help you getfamiliar with a computer programming lan-guage called Python, which I’ve chosen be-cause (a) it’s free, and (b) it’s easy to use inter-actively. I won’t assume you have any previousexperience with computer programming; youwon’t need to learn very much Python, andwhat little you do need to learn I’ll explainexplicitly. If you really want to learn Pythonmore thoroughly, there are a couple of excel-lent books that you can download for free onthe Web:

How to Think Like a Computer Sci-entist (Python Version), Allen B. Dow-ney, Jeffrey Elkner, Moshe Zadka,http://www.ibiblio.org/obp/

Dive Into Python, Mark Pilgrim,http://diveintopython.org/

The first book is meant for people who havenever programmed before, while the second isa more complete introduction aimed at vet-eran programmers who know a different lan-guage already.

Using Python as a calculator

The easiest way to get Python going is to goto the web site ideone.com. Under “choosea language,” select Python. Inside the win-dow where it says “paste your source code orinsert template or sample,” type print(2+2).Click on the “submit” button. The result, 4,is shown under “output.” In other words, youcan use Python just like a calculator.

For compactness, I’ll show examples in the fol-lowing style:

>>> print(2+2)

4

Here the >>> is not something you would typeyourself; it’s just a marker to distinguish yourinput from the program’s output. (In someother versions of Python, the computer willactually print out >>> as a prompt to tell youit’s ready to type something.)

There are only a couple of things to watch outfor. First, Python distinguishes between inte-gers and real numbers, so the following givesan unexpected result:

>>> print(2/3)

0

To get it to treat these values as real numbers,you have to use decimal points:

>>> print(2./3.)

0.6666666666666666666663

Multiplication is represented by “*”:

>>> print(2.*3.)

6.0

Also, Python doesn’t know about its own li-brary of math functions unless you tell it ex-plicitly to load them in:

>>> print (sqrt(2.))

Traceback (most recent call last):

File ‘‘<stdin>’’, line 1, in ?

NameError: There is no variable named ‘sqrt’

Here are the steps you have to go through tocalculate the square root of 2 successfully:

492

>>> import math

>>> print(math.sqrt(2.))

1.4142135623730951

The first line is just something you can make ahabit of doing every time you start up Python.In the second line, the name of the squareroot function had to be prefixed with “math.”to tell Python where you wanted to get this“sqrt” function from. (All of this may seemlike a nuisance if you’re just using Python as acalculator, but it’s a good way to design a pro-gramming language so that names of functionsnever conflict.)

Try it. Experiment and figure out whetherPython’s trig functions assume radians ordegrees.

Variables

Python lets you define variables and assignvalues to them using an equals sign:

>>> dwarfs=7

>>> print(dwarfs)

>>> print(dwarfs+3)

7

10

Note that a variable in computer programmingisn’t quite like a variable in algebra. In alge-bra, if a=7 then a=7 always, throughout aparticular calculation. But in a programminglanguage, the variable name really representsa place in memory where a number can bestored, so you can change its value:

>>> dwarfs=7

>>> dwarfs=37

>>> print(dwarfs)

37

You can even do stuff like this,

>>> dwarfs=37

>>> dwarfs=dwarfs+1

>>> print(dwarfs)

38

In algebra it would be nonsense to have a vari-able equal to itself plus one, but in a com-puter program, it’s not an assertion that thetwo things are equal, its a command to calcu-late the value of the expression on the rightside of the equals, and then put that numberinto the memory location referred to by thevariable name on the left.

Try it. What happens if you do dwarfs+1 =dwarfs? Do you understand why?

Functions

Somebody had to teach Python how to dofunctions like sqrt, and it’s handy to be ableto define your own functions in the same way.Here’s how to do it:

>>> def double(x):

>>> return 2.*x

>>> print(double(5.))

10.0

Note that the indentation is mandatory. Thefirst and second lines define a function calleddouble. The final line evaluates that functionwith an input of 5.

LoopsSuppose we want to add up all the numbersfrom 0 to 99.

Automating this kind of thing is exactly whatcomputers are best at, and Python provides amechanism for this called a loop:

>>> sum=0

>>> for j in range(100):

>>> sum=sum+j

>>> print(sum)

4950

The stuff that gets repeated — the inside ofthe loop — has to be indented, just like ina function definition. Python always countsloops starting from 0, so for j in range(100)

actually causes j to range from 0 to 99, notfrom 1 to 100.

493

Hints

Page 218, problem 18: The easiest way to do this problem is to use two different coordinatesystems: one that’s tilted to coincide with the upper slope, and one that’s tilted to coincidewith the lower one.

Page 218, problem 19: Consider a section of the rope subtending a very small angle,and find an approximate equation relating the normal force to the tension. Apply small-angleapproximations to any trig functions occurring in your result. Eliminate all variables except forthe tension and the angle, and separate these variables.

Page ??, problem ??: What does the total energy have to be if the projectile’s velocity isexactly escape velocity? Write down conservation of energy, change v to dr/dt, separate thevariables, and integrate.

Page 345, problem 25: The analytic approach is a little cumbersome, although it can bedone by using approximations like 1/

√1 + ε ≈ 1− (1/2)ε. A more straightforward, brute-force

method is simply to write a computer program that calculates U/m for a given point in sphericalcoordinates. By trial and error, you can fairly rapidly find the r that gives a desired value ofU/m.

Page 344, problem 20: You can use the geometric interpretation of the dot product.

Page 440, problem 28: The choice of axis theorem only applies to a closed system, or to asystem acted on by a total force of zero. Even if the box is not going to rotate, its center ofmass is going to accelerate, and this can still cause a change in its angular momentum, unlessthe right axis is chosen. For example, if the axis is chosen at the bottom right corner, then thebox will start accumulating clockwise angular momentum, even if it is just accelerating to theright without rotating. Only by choosing the axis at the center of mass (or at some other pointon the same horizontal line) do we get a constant, zero angular momentum.

Page 442, problem 41: You’ll need the result of problem 26 in order to relate the energyand angular momentum of a rigidly rotating body. Since this relationship involves a variableraised to a power, you can’t just graph the data and get the moment of inertia directly. Oneway to get around this is to manipulate one of the variables to make the graph linear. Hereis an example of this technique from another context. Suppose you were given a table of themasses, m, of cubical pieces of wood, whose sides had various lengths, b. You want to find abest-fit value for the density of the wood. The relationship is m = ρb3. The graph of m versusb would be a curve, and you would not have any easy way to get the density from such a graph.But by graphing m versus b3, you can produce a graph that is linear, and whose slope equalsthe density.

Page 463, problem 13: The spring constant of this spring, k, is not the quantity you needin the equation for the period. What you need in that equation is the second derivative of thespring’s energy with respect to the position of the thing that’s oscillating. You need to startby finding the energy stored in the spring as a function of the vertical position, y, of the mass.This is similar to example 5 on page 456.

Page 489, problem 6: The whole expression for the amplitude has maxima where the stuffinside the square root is at a minimum, and vice versa, so you can save yourself a lot of workby just working on the stuff inside the square root. For normal, large values of Q, the thereare two extrema, one at ω = 0 and one at resonance; one of these is a maximum and one is a

494

minimum. You want to find out at what value of Q the zero-frequency extremum switches overfrom being a maximum to being a minimum.

Solutions

Solutions to selected problems

Solutions for chapter 0Page 33, problem 1:

134 mg× 10−3 g

1 mg× 10−3 kg

1 g= 1.34× 10−4 kg

Page 33, problem 7:(a) Let’s do 10.0 g and 1000 g. The arithmetic mean is 505 grams. It comes out to be 0.505 kg,which is consistent. (b) The geometric mean comes out to be 100 g or 0.1 kg, which is consistent.(c) If we multiply meters by meters, we get square meters. Multiplying grams by grams shouldgive square grams! This sounds strange, but it makes sense. Taking the square root of squaregrams (g2) gives grams again. (d) No. The superduper mean of two quantities with units ofgrams wouldn’t even be something with units of grams! Related to this shortcoming is the factthat the superduper mean would fail the kind of consistency test carried out in the first twoparts of the problem.

Page 34, problem 10:(a) They’re all defined in terms of the ratio of side of a triangle to another. For instance, thetangent is the length of the opposite side over the length of the adjacent side. Dividing metersby meters gives a unitless result, so the tangent, as well as the other trig functions, is unitless.(b) The tangent function gives a unitless result, so the units on the right-hand side had bettercancel out. They do, because the top of the fraction has units of meters squared, and so doesthe bottom.

Solutions for chapter 1Page 51, problem 1:Scaling down the linear dimensions by a factor of 1/10 reduces the volume by a factor of(1/10)3 = 1/1000, so if the whole cube is a liter, each small one is one milliliter.

Page 51, problem 2:

1 mm2 ×(

1 cm

10 mm

)2

= 10−2 cm2

Page 51, problem 3:The bigger scope has a diameter that’s ten times greater. Area scales as the square of the lineardimensions, so its light-gathering power is a hundred times greater (10× 10).

Page 51, problem 4:The cone of mixed gin and vermouth is the same shape as the cone of vermouth, but its lineardimensions are doubled, so its volume is 8 times greater. The ratio of gin to vermouth is 7 to 1.

495

Page 51, problem 8:Since they differ by two steps on the Richter scale, the energy of the bigger quake is 10000 timesgreater. The wave forms a hemisphere, and the surface area of the hemisphere over which theenergy is spread is proportional to the square of its radius. If the amount of vibration was thesame, then the surface areas much be in the ratio of 10000:1, which means that the ratio of theradii is 100:1.

Page 54, problem 22:Let’s estimate the Great Wall’s volume, and then figure out how many bricks that would repre-sent. The wall is famous because it covers pretty much all of China’s northern border, so let’ssay it’s 1000 km long. From pictures, it looks like it’s about 10 m high and 10 m wide, so thetotal volume would be 106 m× 10 m× 10 m = 108 m3. If a single brick has a volume of 1 liter,or 10−3 m3, then this represents about 1011 bricks. If one person can lay 10 bricks in an hour(taking into account all the preparation, etc.), then this would be 1010 man-hours.

Solutions for chapter 2Page 92, problem 1:Since the lines are at intervals of one m/s and one second, each box represents one meter. Fromt = 0 to t = 2 s, the area under the curve represents a positive ∆x of 6 m. (The triangle has halfthe area of the 2× 6 rectangle it fits inside.) After t = 2 s, the area above the curve representsnegative ∆x. To get −6 m worth of area, we need to go out to t = 6 s, at which point thetriangle under the axis has a width of 4 s and a height of 3 m/s, for an area of 6 m (half of3× 4).

Page 93, problem 8:(a) Let f and g be functions. Then the chain rule states that if we construct the functionf(g(x)), its derivative is

df

dx=

df

dg· dg

dx.

On the right-hand side, the units of dg on the top cancel with the units of dg on the bottom,so the units do match up with those of df/dx on the left.(b) The cosine function requires a unitless input and produces a unitless output. Therefore Amust have units of meters, and b must have units of s−1 (inverse seconds, or “per second”). Ais the distance the object moves on either side of the origin, and b is a measure of how fast itvibrates back and forth (how many radians it passes through per second).(b) The derivative is v = dx/dt = −Ab cos(bt), where the factor of b in front comes from thechain rule. The product Ab does have units of m/s. If we hadn’t put in the factor of b asrequired by the chain rule, the units would have been wrong. Physically, it also makes sensethat a larger b, indicating a more rapid vibration, produces a greater v.

Page 93, problem 10:In one second, the ship moves v meters to the east, and the person moves v meters north relativeto the deck. Relative to the water, he traces the diagonal of a triangle whose length is givenby the Pythagorean theorem, (v2 + v2)1/2 =

√2v. Relative to the water, he is moving at a

45-degree angle between north and east.

Page 93, problem 11:Velocity is relative, so having to lean tells you nothing about the train’s velocity. Fullerton ismoving at a huge speed relative to Beijing, but that doesn’t produce any noticeable effect ineither city. The fact that you have to lean tells you that the train is changing its speed, but it

496

doesn’t tell you what the train’s current speed is.

Page 93, problem 13:To the person riding the moving bike, bug A is simply going in circles. The only differencebetween the motions of the two wheels is that one is traveling through space, but motion isrelative, so this doesn’t have any effect on the bugs. It’s equally hard for each of them.

Solutions for chapter 3Page 111, problem 1:Taking g to be 10 m/s2, the bullet loses 10 m/s of speed every second, so it will take 10 s tocome to a stop, and then another 10 s to come back down, for a total of 20 s.

Page 111, problem 4:

v =dx

dt= 10− 3t2

a =dv

dt= −6t

= −18 m/s2


Page 111, problem 7:(a) We choose a coordinate system with positive pointing to the right. Some people mightexpect that the ball would slow down once it was on the more gentle ramp. This may be trueif there is significant friction, but Galileo’s experiments with inclined planes showed that whenfriction is negligible, a ball rolling on a ramp has constant acceleration, not constant speed. Thespeed stops increasing as quickly once the ball is on the more gentle slope, but it still keeps onincreasing. The a-t graph can be drawn by inspecting the slope of the v-t graph.

(b) The ball will roll back down, so the second half of the motion is the same as in part a. In

497

the first (rising) half of the motion, the velocity is negative, since the motion is in the oppositedirection compared to the positive x axis. The acceleration is again found by inspecting theslope of the v-t graph.

Page 111, problem 8:This is a case where it’s probably easiest to draw the acceleration graph first. While the ballis in the air (bc, de, etc.), the only force acting on it is gravity, so it must have the same,constant acceleration during each hop. Choosing a coordinate system where the positive x axispoints up, this becomes a negative acceleration (force in the opposite direction compared to theaxis). During the short times between hops when the ball is in contact with the ground (cd,ef, etc.), it experiences a large acceleration, which turns around its velocity very rapidly. Theseshort positive accelerations probably aren’t constant, but it’s hard to know how they’d reallylook. We just idealize them as constant accelerations. Similarly, the hand’s force on the ballduring the time ab is probably not constant, but we can draw it that way, since we don’t knowhow to draw it more realistically. Since our acceleration graph consists of constant-accelerationsegments, the velocity graph must consist of line segments, and the position graph must consistof parabolas. On the x graph, I chose zero to be the height of the center of the ball above thefloor when the ball is just lying on the floor. When the ball is touching the floor and compressed,as in interval cd, its center is below this level, so its x is negative.

Page 112, problem 11:(a) Solving for ∆x = 1

2at2 for a, we find a = 2∆x/t2 = 5.51 m/s2. (b) v =

√2a∆x = 66.6 m/s.

(c) The actual car’s final velocity is less than that of the idealized constant-acceleration car. Ifthe real car and the idealized car covered the quarter mile in the same time but the real carwas moving more slowly at the end than the idealized one, the real car must have been going

498

faster than the idealized car at the beginning of the race. The real car apparently has a greateracceleration at the beginning, and less acceleration at the end. This make sense, because everycar has some maximum speed, which is the speed beyond which it cannot accelerate.

Page 112, problem 13:∆x = 1

2at2, so for a fixed value of ∆x, we have t ∝ 1/

√a. Decreasing a by a factor of 3 means

that t will increase by a factor of√

3 = 1.7. (The given piece of data, 3, only has one sig fig,but rounding the final result off to one sig fig, giving 2 rather then 1.7, would be a little toosevere. As discussed in section 0.10, sig figs are only a rule of thumb, and when in doubt, youcan change the input data to see how much the output would have changed. The ratio of thegravitational fields on Earth and Mars must be in the range from 2.5 to 3.5, since otherwisethe given data would not have been rounded off to 3. Using this range of inputs, the possiblerange of values for the final result becomes 1.6 to 1.9. The final digit in the 1.7 is therefore alittle uncertain, but it’s not complete garbage. It carries useful information, and should not bethrown out.)

Page 112, problem 15:We have v2

f = 2a∆x, so the distance is proportional to the square of the velocity. To get up tohalf the speed, the ball needs 1/4 the distance, i.e., L/4.

Solutions for chapter 4Page 141, problem 1:a = ∆v/∆t, and also a = F/m, so

∆t =∆v

a

=m∆v

F

=(1000 kg)(50 m/s− 20 m/s)

3000 N= 10 s

Page 141, problem 4:(a) This is a measure of the box’s resistance to a change in its state of motion, so it measuresthe box’s mass. The experiment would come out the same in lunar gravity.(b) This is a measure of how much gravitational force it feels, so it’s a measure of weight. Inlunar gravity, the box would make a softer sound when it hit.(c) As in part a, this is a measure of its resistance to a change in its state of motion: its mass.Gravity isn’t involved at all.

Solutions for chapter 5Page 168, problem 1:(a) The swimmer’s acceleration is caused by the water’s force on the swimmer, and the swimmermakes a backward force on the water, which accelerates the water backward. (b) The club’snormal force on the ball accelerates the ball, and the ball makes a backward normal force on theclub, which decelerates the club. (c) The bowstring’s normal force accelerates the arrow, andthe arrow also makes a backward normal force on the string. This force on the string causes thestring to accelerate less rapidly than it would if the bow’s force was the only one acting on it.(d) The tracks’ backward frictional force slows the locomotive down. The locomotive’s forwardfrictional force causes the whole planet earth to accelerate by a tiny amount, which is too small

499

to measure because the earth’s mass is so great.

Page 168, problem 2:The person’s normal force on the box is paired with the box’s normal force on the person. Thedirt’s frictional force on the box pairs with the box’s frictional force on the dirt. The earth’sgravitational force on the box matches the box’s gravitational force on the earth.

Page 168, problem 3:(a) A liter of water has a mass of 1.0 kg. The mass is the same in all three locations. Massindicates how much an object resists a change in its motion. It has nothing to do with gravity.(b) The term “weight” refers to the force of gravity on an object. The bottle’s weight on earthis FW = mg = 9.8 N. Its weight on the moon is about one sixth that value, and its weight ininterstellar space is zero.


(a)

top spring’s rightward force on connector...connector’s leftward force on top springbottom spring’s rightward force on connector...connector’s leftward force on bottom springhand’s leftward force on connector...connector’s rightward force on hand

Looking at the three forces on the connector, we see that the hand’s force must be double theforce of either spring. The value of x− xo is the same for both springs and for the arrangementas a whole, so the spring constant must be 2k. This corresponds to a stiffer spring (more forceto produce the same extension).

(b) Forces in which the left spring participates:

hand’s leftward force on left spring...left spring’s rightward force on handright spring’s rightward force on left spring...left spring’s leftward force on right spring

Forces in which the right spring participates:

left spring’s leftward force on right spring...right spring’s rightward force on left springwall’s rightward force on right spring...right spring’s leftward force on wall

Since the left spring isn’t accelerating, the total force on it must be zero, so the two forces actingon it must be equal in magnitude. The same applies to the two forces acting on the right spring.The forces between the two springs are connected by Newton’s third law, so all eight of theseforces must be equal in magnitude. Since the value of x−xo for the whole setup is double whatit is for either spring individually, the spring constant of the whole setup must be k/2, whichcorresponds to a less stiff spring.

Page 173, problem 28:(a) Spring constants in parallel add, so the spring constant has to be proportional to the cross-sectional area. Two springs in series give half the spring constant, three springs in series give

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1/3, and so on, so the spring constant has to be inversely proportional to the length. Summa-rizing, we have k ∝ A/L. (b) With the Young’s modulus, we have k = (A/L)E.The springconstant has units of N/m, so the units of E would have to be N/m2.

Solutions for chapter 8Page 214, problem 3:We want to find out about the velocity vector vBG of the bullet relative to the ground, so we needto add Annie’s velocity relative to the ground vAG to the bullet’s velocity vector vBA relativeto her. Letting the positive x axis be east and y north, we have

vBA,x = (140 mi/hr) cos 45

= 100 mi/hr

vBA,y = (140 mi/hr) sin 45

= 100 mi/hr

and

vAG,x = 0

vAG,y = 30 mi/hr .

The bullet’s velocity relative to the ground therefore has components

vBG,x = 100 mi/hrand

vBG,y = 130 mi/hr .

Its speed on impact with the animal is the magnitude of this vector

|vBG| =√

(100 mi/hr)2 + (130 mi/hr)2

= 160 mi/hr

(rounded off to 2 significant figures).

Page 216, problem 9:Since its velocity vector is constant, it has zero acceleration, and the sum of the force vectorsacting on it must be zero. There are three forces acting on the plane: thrust, lift, and gravity.We are given the first two, and if we can find the third we can infer the plane’s mass. The sumof the y components of the forces is zero, so

0 = Fthrust,y + Flift,y + Fg,y

= |Fthrust| sin θ + |Flift| cos θ −mg .

The mass is

m = (|Fthrust| sin θ + |Flift| cos θ)/g

= 6.9× 104 kg .

Page 217, problem 13:(a) If there was no friction, the angle of repose would be zero, so the coefficient of static friction,µs, will definitely matter. We also make up symbols θ, m and g for the angle of the slope, the

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mass of the object, and the acceleration of gravity. The forces form a triangle just like the onein section 8.3, but instead of a force applied by an external object, we have static friction, whichis less than µs|FN |. As in that example, |Fs| = mg sin θ, and |Fs| < µs|FN |, so

mg sin θ < µs|FN | .

From the same triangle, we have |FN | = mg cos θ, so

mg sin θ < µsmg cos θ .

Rearranging,

θ < tan−1 µs .

(b) Both m and g canceled out, so the angle of repose would be the same on an asteroid.

Page 217, problem 14:(a) Since the wagon has no acceleration, the total forces in both the x and y directions mustbe zero. There are three forces acting on the wagon: FT , FW , and the normal force from theground, FN . If we pick a coordinate system with x being horizontal and y vertical, then theangles of these forces measured counterclockwise from the x axis are 90 −φ, 270 , and 90 +θ,respectively. We have

Fx,total = |FT | cos(90 − φ) + |FW | cos(270 ) + |FN | cos(90 + θ)

Fy,total = |FT | sin(90 − φ) + |FW | sin(270 ) + |FN | sin(90 + θ) ,

which simplifies to

0 = |FT | sinφ− |FN | sin θ0 = |FT | cosφ− |FW |+ |FN | cos θ.

The normal force is a quantity that we are not given and do not with to find, so we shouldchoose it to eliminate. Solving the first equation for |FN | = (sinφ/ sin θ)|FT |, we eliminate |FN |from the second equation,

0 = |FT | cosφ− |FW |+ |FT | sinφ cos θ/ sin θ

and solve for |FT |, finding

|FT | =|FW |

cosφ+ sinφ cos θ/ sin θ.

Multiplying both the top and the bottom of the fraction by sin θ, and using the trig identity forsin(θ + φ) gives the desired result,

|FT | =sin θ

sin(θ + φ)|FW | .

(b) The case of φ = 0, i.e., pulling straight up on the wagon, results in |FT | = |FW |: we simplysupport the wagon and it glides up the slope like a chair-lift on a ski slope. In the case ofφ = 180 − θ, |FT | becomes infinite. Physically this is because we are pulling directly into theground, so no amount of force will suffice.

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Solutions for chapter 9Page 235, problem 3:(a) The inward normal force must be sufficient to produce circular motion, so

|FN | = mv2/r .

We are searching for the minimum speed, which is the speed at which the static friction force isjust barely able to cancel out the downward gravitational force. The maximum force of staticfriction is

|Fs| = µs|FN | ,

and this cancels the gravitational force, so

|Fs| = mg .

Solving these three equations for v gives

v =

√gr

µs.

(b) Greater by a factor of√

3.

Page 236, problem 5:The inward force must be supplied by the inward component of the normal force,

|FN | sin θ = mv2/r .

The upward component of the normal force must cancel the downward force of gravity,

|FN | cos θ = mg.

Eliminating |FN | and solving for θ, we find

θ = tan−1

(v2

gr

).

Page 237, problem 10:Each cyclist has a radial acceleration of v2/r = 5 m/s2. The tangential accelerations of cyclistsA and B are 375 N/75 kg = 5 m/s2.

Solutions for chapter 10Page 261, problem 1:Newton’s law of gravity tells us that her weight will be 6000 times smaller because of theasteroid’s smaller mass, but 132 = 169 times greater because of its smaller radius. Putting these

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two factors together gives a reduction in weight by a factor of 6000/169, so her weight will be(400 N)(169)/(6000) = 11 N.

Page 261, problem 3:(a) The asteroid’s mass depends on the cube of its radius, and for a given mass the surfacegravity depends on r−2. The result is that surface gravity is directly proportional to radius.Half the gravity means half the radius, or one eighth the mass. (b) To agree with a, Earth’smass would have to be 1/8 Jupiter’s. We assumed spherical shapes and equal density. Bothplanets are at least roughly spherical, so the only way out of the contradiction is if Jupiter’sdensity is significantly less than Earth’s.

Page 262, problem 6:Newton’s law of gravity depends on the inverse square of the distance, so if the two planets’masses had been equal, then the factor of 0.83/0.059 = 14 in distance would have caused theforce on planet c to be 142 = 2.0 × 102 times weaker. However, planet c’s mass is 3.0 timesgreater, so the force on it is only smaller by a factor of 2.0× 102/3.0 = 65.

Page 265, problem 20:Newton’s law of gravity says F = Gm1m2/r

2, and Newton’s second law says F = m2a, soGm1m2/r

2 = m2a. Since m2 cancels, a is independent of m2.

Page 265, problem 21:Newton’s second law gives

F = mDaD ,

where F is Ida’s force on Dactyl. Using Newton’s universal law of gravity, F= GmImD/r2,and

the equation a = v2/r for circular motion, we find

GmImD/r2 = mDv

2/r.

Dactyl’s mass cancels out, giving

GmI/r2 = v2/r.

Dactyl’s velocity equals the circumference of its orbit divided by the time for one orbit: v =2πr/T . Inserting this in the above equation and solving for mI , we find

mI =4π2r3

GT 2,

so Ida’s density is

ρ = mI/V

=4π2r3

GV T 2.

Page 265, problem 22:Any fractional change in r results in double that amount of fractional change in 1/r2. Forexample, raising r by 1% causes 1/r2 to go down by very nearly 2%. The fractional change in1/r2 is actually

2× (1/27) cm

3.84× 105 km× 1 km

105 cm= 2× 10−12

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Solutions for chapter 11Page 293, problem 6:A force is an interaction between two objects, so while the bullet is in the air, there is no force.There is only a force while the bullet is in contact with the book. There is energy the wholetime, and the total amount doesn’t change. The bullet has some kinetic energy, and transferssome of it to the book as heat, sound, and the energy required to tear a hole through the book.

Page 293, problem 7:(a) The energy stored in the gasoline is being changed into heat via frictional heating, and alsoprobably into sound and into energy of water waves. Note that the kinetic energy of the propellerand the boat are not changing, so they are not involved in the energy transformation. (b) Thecrusing speed would be greater by a factor of the cube root of 2, or about a 26% increase.

Page 293, problem 8:We don’t have actual masses and velocities to plug in to the equation, but that’s OK. We justhave to reason in terms of ratios and proportionalities. Kinetic energy is proportional to massand to the square of velocity, so B’s kinetic energy equals

(13.4 J)(3.77)/(2.34)2 = 9.23 J

Page 293, problem 11:Room temperature is about 20 C. The fraction of the energy that actually goes into heatingthe water is

(250 g)/(0.24 g· C/J)× (100 C− 20 C)

(1.25× 103 J/s) (126 s)= 0.53

So roughly half of the energy is wasted. The wasted energy might be in several forms: heatingof the cup, heating of the oven itself, or leakage of microwaves from the oven.

Solutions for chapter 12Page 311, problem 6:

Etotal,i = Etotal,f

PEi + heati = PEf +KEf + heatf1

2mv2 = PEi − PEf + heati − heatf

= −∆PE −∆heat

v =

√2

(−∆PE −∆heat

m

)= 6.4 m/s

Page 312, problem 10:(a) Example: As one child goes up on one side of a see-saw, another child on the other sidecomes down. (b) Example: A pool ball hits another pool ball, and transfers some KE.

Page 312, problem 12:Suppose the river is 1 m deep, 100 m wide, and flows at a speed of 10 m/s, and that the fallsare 100 m tall. In 1 second, the volume of water flowing over the falls is 103 m3, with a mass of106 kg. The potential energy released in one second is (106 kg)(g)(100 m) = 109 J, so the power

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is 109 W. A typical household might have 10 hundred-watt applicances turned on at any giventime, so it consumes about 103 watts on the average. The plant could supply a about millionhouseholds with electricity.

Page 313, problem 16:Let θ be the angle by which he has progressed around the pipe. Conservation of energy gives

Etotal,i = Etotal,f

PEi = PEf +KEf

Let’s make PE = 0 at the top, so

0 = mgr(cos θ − 1) +1

2mv2 .

While he is still in contact with the pipe, the radial component of his acceleration is

ar =v2

r,

and making use of the previous equation we find

ar = 2g(1− cos θ) .

There are two forces on him, a normal force from the pipe and a downward gravitation forcefrom the earth. At the moment when he loses contact with the pipe, the normal force is zero,so the radial component, mg cos θ, of the gravitational force must equal mar,

mg cos θ = 2mg(1− cos θ) ,

which gives

cos θ =2

3.

The amount by which he has dropped is r(1− cos θ), which equals r/3 at this moment.

Solutions for chapter 13Page 339, problem 4:No. Work describes how energy was transferred by some process. It isn’t a measurable propertyof a system.

Solutions for chapter 14Page 376, problem 3:By conservation of momentum, the total momenta of the pieces after the explosion is the sameas the momentum of the firework before the explosion. However, there is no law of conservationof kinetic energy, only a law of conservation of energy. The chemical energy in the gunpowderis converted into heat and kinetic energy when it explodes. All we can say about the kineticenergy of the pieces is that their total is greater than the kinetic energy before the explosion.

Page 377, problem 8:Let m be the mass of the little puck and M = 2.3m be the mass of the big one. All we needto do is find the direction of the total momentum vector before the collision, because the totalmomentum vector is the same after the collision. Given the two components of the momentum

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vector px = mv and py = Mv, the direction of the vector is tan−1(py/px) = 23 counterclockwisefrom the big puck’s original direction of motion.

Page 378, problem 12:Momentum is a vector. The total momentum of the molecules is always zero, since the momentain different directions cancal out on the average. Cooling changes individual molecular momenta,but not the total.

Page 378, problem 15:(a) Particle i had velocity vi in the center-of-mass frame, and has velocity vi + u in the newframe. The total kinetic energy is

1

2m1 (v1 + u)2 + . . . ,

where “. . . ” indicates that the sum continues for all the particles. Rewriting this in terms ofthe vector dot product, we have

1

2m1 (v1 + u) · (v1 + u) + . . . =

1

2m1 (v1 · v1 + 2u · v1 + u · u) + . . . .

When we add up all the terms like the first one, we get Kcm. Adding up all the terms like thethird one, we get M |u|2/2. The terms like the second term cancel out:

m1u · v1 + . . . = u · (m1v1 + . . .) ,

where the sum in brackets equals the total momentum in the center-of-mass frame, which iszero by definition.(b) Changing frames of reference doesn’t change the distances between the particles, so thepotential energies are all unaffected by the change of frames of reference. Suppose that in agiven frame of reference, frame 1, energy is conserved in some process: the initial and finalenergies add up to be the same. First let’s transform to the center-of-mass frame. The potentialenergies are unaffected by the transformation, and the total kinetic energy is simply reducedby the quantity M |u1|2/2, where u1 is the velocity of frame 1 relative to the center of mass.Subtracting the same constant from the initial and final energies still leaves them equal. Nowwe transform to frame 2. Again, the effect is simply to change the initial and final energies byadding the same constant.

Page 378, problem 16:A conservation law is about addition: it says that when you add up a certain thing, the totalalways stays the same. Funkosity would violate the additive nature of conservation laws, becausea two-kilogram mass would have twice as much funkosity as a pair of one-kilogram masses movingat the same speed.

Solutions for chapter 15Page 437, problem 8:The pliers are not moving, so their angular momentum remains constant at zero, and the totaltorque on them must be zero. Not only that, but each half of the pliers must have zero totaltorque on it. This tells us that the magnitude of the torque at one end must be the same asthat at the other end. The distance from the axis to the nut is about 2.5 cm, and the distancefrom the axis to the centers of the palm and fingers are about 8 cm. The angles are closeenough to 90 that we can pretend they’re 90 degrees, considering the rough nature of the otherassumptions and measurements. The result is (300 N)(2.5 cm) = (F )(8 cm), or F = 90 N.

507

Page 443, problem 46:The foot of the rod is moving in a circle relative to the center of the rod, with speed v = πb/T ,and acceleration v2/(b/2) = (π2/8)g. This acceleration is initially upward, and is greater inmagnitude than g, so the foot of the rod will lift off without dragging. We could also worryabout whether the foot of the rod would make contact with the floor again before the rodfinishes up flat on its back. This is a question that can be settled by graphing, or simply byinspection of figure aj on page 416. The key here is that the two parts of the acceleration areboth independent of m and b, so the result is univeral, and it does suffice to check a graph ina single example. In practical terms, this tells us something about how difficult the trick is todo. Because π2/8 = 1.23 isn’t much greater than unity, a hit that is just a little too weak (bya factor of 1.231/2 = 1.11) will cause a fairly obvious qualitative change in the results. This iseasily observed if you try it a few times with a pencil.

Answers to Self-Checks

Answers to self-checks for chapter 0

Page 13, self-check A:If only he has the special powers, then his results can never be reproduced.

Page 15, self-check B:They would have had to weigh the rays, or check for a loss of weight in the object from whichthey were have emitted. (For technical reasons, this was not a measurement they could actuallydo, hence the opportunity for disagreement.)

Page 21, self-check C:A dictionary might define “strong” as “possessing powerful muscles,” but that’s not an oper-ational definition, because it doesn’t say how to measure strength numerically. One possibleoperational definition would be the number of pounds a person can bench press.

Page 25, self-check D:A microsecond is 1000 times longer than a nanosecond, so it would seem like 1000 seconds, orabout 20 minutes.

Page 26, self-check E:Exponents have to do with multiplication, not addition. The first line should be 100 timeslonger than the second, not just twice as long.

Page 29, self-check F:The various estimates differ by 5 to 10 million. The CIA’s estimate includes a ridiculous numberof gratuitous significant figures. Does the CIA understand that every day, people in are bornin, die in, immigrate to, and emigrate from Nigeria?

Page 30, self-check G:(1) 4; (2) 2; (3) 2


Page 38, self-check A:1 yd2 × (3 ft/1 yd)2 = 9 ft2

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1 yd3 × (3 ft/1 yd)3 = 27 ft3


Page 63, self-check A:Coasting on a bike and coasting on skates give one-dimensional center-of-mass motion, butrunning and pedaling require moving body parts up and down, which makes the center of massmove up and down. The only example of rigid-body motion is coasting on skates. (Coasting ona bike is not rigid-body motion, because the wheels twist.)

Page 63, self-check B:By shifting his weight around, he can cause the center of mass not to coincide with the geometriccenter of the wheel.

Page 64, self-check C:(1) a point in time; (2) time in the abstract sense; (3) a time interval

Page 66, self-check D:Zero, because the “after” and “before” values of x are the same.

Page 71, self-check E:(1) The effect only occurs during blastoff, when their velocity is changing. Once the rocketengines stop firing, their velocity stops changing, and they no longer feel any effect. (2) It isonly an observable effect of your motion relative to the air.

Page 83, self-check F:At v = 0, we get γ = 1, so t = T . There is no time distortion unless the two frames of referenceare in relative motion.


Page 99, self-check A:Its speed increases at a steady rate, so in the next second it will travel 19 cm.


Page 129, self-check A:(1) The case of ρ = 0 represents an object falling in a vacuum, i.e., there is no density of air.The terminal velocity would be infinite. Physically, we know that an object falling in a vacuumwould never stop speeding up, since there would be no force of air friction to cancel the force ofgravity. (2) The 4-cm ball would have a mass that was greater by a factor of 4× 4× 4, but itscross-sectional area would be greater by a factor of 4× 4. Its terminal velocity would be greaterby a factor of

√43/42 = 2. (3) It isn’t of any general importance. It’s just an example of one

physical situation. You should not memorize it.

Page 132, self-check B:(1) This is motion, not force. (2) This is a description of how the sub is able to get the waterto produce a forward force on it. (3) The sub runs out of energy, not force.


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Page 147, self-check A:The sprinter pushes backward against the ground, and by Newton’s third law, the ground pushesforward on her. (Later in the race, she is no longer accelerating, but the ground’s forward forceis needed in order to cancel out the backward forces, such as air friction.)

Page 154, self-check B:(1) It’s kinetic friction, because her uniform is sliding over the dirt. (2) It’s static friction,because even though the two surfaces are moving relative to the landscape, they’re not slippingover each other. (3) Only kinetic friction creates heat, as when you rub your hands together. Ifyou move your hands up and down together without sliding them across each other, no heat isproduced by the static friction.

Page 155, self-check C:By the POFOSTITO mnemonic, we know that each of the bird’s forces on the trunk will be ofthe same type as the corresponding force of the tree on the bird, but in the opposite direction.The bird’s feet make a normal force on the tree that is to the right and a static frictional forcethat is downward.

Page 155, self-check D:Frictionless ice can certainly make a normal force, since otherwise a hockey puck would sinkinto the ice. Friction is not possible without a normal force, however: we can see this from theequation, or from common sense, e.g., while sliding down a rope you do not get any frictionunless you grip the rope.

Page 156, self-check E:(1) Normal forces are always perpendicular to the surface of contact, which means right or leftin this figure. Normal forces are repulsive, so the cliff’s force on the feet is to the right, i.e., awayfrom the cliff. (2) Frictional forces are always parallel to the surface of contact, which meansright or left in this figure. Static frictional forces are in the direction that would tend to keepthe surfaces from slipping over each other. If the wheel was going to slip, its surface would bemoving to the left, so the static frictional force on the wheel must be in the direction that wouldprevent this, i.e., to the right. This makes sense, because it is the static frictional force thataccelerates the dragster. (3) Normal forces are always perpendicular to the surface of contact.In this diagram, that means either up and to the left or down and to the right. Normal forcesare repulsive, so the ball is pushing the bat away from itself. Therefore the ball’s force is downand to the right on this diagram.


Page 179, self-check A:The wind increases the ball’s overall speed. If you think about it in terms of overall speed,it’s not so obvious that the increased speed is exactly sufficient to compensate for the greaterdistance. However, it becomes much simpler if you think about the forward motion and thesideways motion as two separate things. Suppose the ball is initially moving at one meter persecond. Even if it picks up some sideways motion from the wind, it’s still getting closer to thewall by one meter every second.


Page 191, self-check A:

510

v = ∆r/∆t

Page 191, self-check B:

Page 195, self-check C:A −B is equivalent to A + (−B), which can be calculated graphically by reversing B to form−B, and then adding it to A.


Page 206, self-check A:(1) It is speeding up, because the final velocity vector has the greater magnitude. (2) The resultwould be zero, which would make sense. (3) Speeding up produced a ∆v vector in the samedirection as the motion. Slowing down would have given a ∆v that pointed backward.

Page 207, self-check B:As we have already seen, the projectile has ax = 0 and ay = −g, so the acceleration vector ispointing straight down.


Page 225, self-check A:(1) Uniform. They have the same motion as the drum itself, which is rotating as one solid piece.No part of the drum can be rotating at a different speed from any other part. (2) Nonuniform.Gravity speeds it up on the way down and slows it down on the way up.


Page 244, self-check A:It would just stay where it was. Plugging v = 0 into eq. [1] would give F = 0, so it would notaccelerate from rest, and would never fall into the sun. No astronomer had ever observed anobject that did that!

Page 245, self-check B:

F ∝ mr/T 2 ∝ mr/(r3/2)2 ∝ mr/r3 = m/r2

Page 248, self-check C:The equal-area law makes equally good sense in the case of a hyperbolic orbit (and observationsverify it). The elliptical orbit law had to be generalized by Newton to include hyperbolas. Thelaw of periods doesn’t make sense in the case of a hyperbolic orbit, because a hyperbola nevercloses back on itself, so the motion never repeats.

Page 253, self-check D:Above you there is a small part of the shell, comprising only a tiny fraction of the earth’s mass.This part pulls you up, while the whole remainder of the shell pulls you down. However, the

511

part above you is extremely close, so it makes sense that its force on you would be far out ofproportion to its small mass.


Page 280, self-check A:(1) A spring-loaded toy gun can cause a bullet to move, so the spring is capable of storing energyand then converting it into kinetic energy. (2) The amount of energy stored in the spring relatesto the amount of compression, which can be measured with a ruler.


Page 304, self-check A:Both balls start from the same height and end at the same height, so they have the same ∆y.This implies that their losses in potential energy are the same, so they must both have gainedthe same amount of kinetic energy.


Page 316, self-check A:Work is defined as the transfer of energy, so like energy it is a scalar with units of joules.

Page 319, self-check B:Whenever energy is transferred out of the spring, the same amount has to be transferred intothe ball, and vice versa. As the spring compresses, the ball is doing positive work on the spring(giving up its KE and transferring energy into the spring as PE), and as it decompresses theball is doing negative work (extracting energy).

Page 322, self-check C:(a) No. The pack is moving at constant velocity, so its kinetic energy is staying the same. Itis only moving horizontally, so its gravitational potential energy is also staying the same. Noenergy transfer is occurring. (b) No. The horse’s upward force on the pack forms a 90-degreeangle with the direction of motion, so cos θ = 0, and no work is done.

Page 325, self-check D:Only in (a) can we use Fd to calculate work. In (b) and (c), the force is changing as the distancechanges.


Page 372, self-check A:When m = 0, we have E = p (or E = pc, in units with c 6= 1), which is what we expect.


Page 393, self-check A:1, 2, and 4 all have the same sigm, because they are trying to twist the wrench clockwise. Thesign of torque 3 is opposite to the signs of the others. The magnitude of torque 3 is the greatest,

512

since it has a large r, and the force is nearly all perpendicular to the wrench. Torques 1 and 2are the same because they have the same values of r and F⊥. Torque 4 is the smallest, due toits small r.

Page 405, self-check B:One person’s θ-t graph would simply be shifted up or down relative to the others. The derivativeequals the slope of the tangent line, and this slope isn’t changed when you shift the graph, soboth people would agree on the angular velocity.

Page 407, self-check C:Reversing the direction of ω also reverses the direction of motion, and this is reflected by therelationship between the plus and minus signs. In the equation for the radial acceleration, ω issquared, so even if ω is negative, the result is positive. This makes sense because the accelerationis always inward in circular motion, never outward.

Page 419, self-check D:All the rotations around the x axis give ω vectors along the positive x axis (thumb pointing alongpositive x), and all the rotations about the y axis have ω vectors with positive y components.

Page 422, self-check E:For example, if we take (A×B)x = AyBz−ByAz and reverse the A’s and B’s, we get (B×A)x =ByAz −AyBz, which is just the negative of the original expression.


Page 472, self-check A:The two graphs start off with the same amplitude, but the solid curve loses amplitude morerapidly. For a given time, t, the quantity e−ct is apparently smaller for the solid curve, meaningthat ct is greater. The solid curve has the higher value of c.

Page 478, self-check B:She should tap the wine glasses she finds in the store and look for one with a high Q, i.e., onewhose vibrations die out very slowly. The one with the highest Q will have the highest-amplituderesponse to her driving force, making it more likely to break.

Answers to Selected Homework Problems

Answers for chapter 1

Page 52, problem 10:Check: The actual number of species of lupine occurring in the San Gabriels is 22. You shouldfind that your answer comes out in the same ballpark as this figure, but not exactly the same,of course, because the scaling rule is only a generalization.


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Page 187, problem 5:(a) R = (2v2/g) sin θ cos θ (c) 45


Page 201, problem 6:(a) The optimal angle is about 40 , and the resulting range is about 124 meters, which is aboutthe length of a home run. (b) It goes about 9 meters farther. For comparison with reality, thestadium’s web site claims a home run goes about 11 meters farther there than in a sea-levelstadium.


Page 489, problem 6:Q = 1/

√2

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Photo creditsExcept as specifically noted below or in a parenthetical credit in the caption of a figure, all the illustrations inthis book are under my own copyright, and are copyleft licensed under the same license as the rest of the book.

In some cases it’s clear from the date that the figure is public domain, but I don’t know the name of the artistor photographer; I would be grateful to anyone who could help me to give proper credit. I have assumed thatimages that come from U.S. government web pages are copyright-free, since products of federal agencies fall intothe public domain. I’ve included some public-domain paintings; photographic reproductions of them are notcopyrightable in the U.S. (Bridgeman Art Library, Ltd. v. Corel Corp., 36 F. Supp. 2d 191, S.D.N.Y. 1999).

When “PSSC Physics” is given as a credit, it indicates that the figure is from the first edition of the textbookentitled Physics, by the Physical Science Study Committee. The early editions of these books never had theircopyrights renewed, and are now therefore in the public domain. There is also a blanket permission given inthe later PSSC College Physics edition, which states on the copyright page that “The materials taken from theoriginal and second editions and the Advanced Topics of PSSC PHYSICS included in this text will be availableto all publishers for use in English after December 31, 1970, and in translations after December 31, 1975.”

Credits to Millikan and Gale refer to the textbooks Practical Physics (1920) and Elements of Physics (1927).Both are public domain. (The 1927 version did not have its copyright renewed.) Since it is possible that some ofthe illustrations in the 1927 version had their copyrights renewed and are still under copyright, I have only usedthem when it was clear that they were originally taken from public domain sources.

In a few cases, I have made use of images under the fair use doctrine. However, I am not a lawyer, and the lawson fair use are vague, so you should not assume that it’s legal for you to use these images. In particular, fair uselaw may give you less leeway than it gives me, because I’m using the images for educational purposes, and givingthe book away for free. Likewise, if the photo credit says “courtesy of ...,” that means the copyright owner gaveme permission to use it, but that doesn’t mean you have permission to use it.

Cover Shukhov Tower: Wikipedia user Arssenev, CC-BY-SA licensed.

Contents See photo credits below, referring to the places where the images appear in the main text.

11 Mars Climate Orbiter: NASA/JPL/CIT. 22 Standard kilogram: Bo Bengtsen, GFDL licensed. Furtherretouching by Wikipedia user Greg L and by B. Crowell. 37 Bee: Wikipedia user Fir0002, CC-BY-SA licensed.52 Galaxy: ESO, CC-BY license. 53 E. Coli bacteria: Eric Erbe, digital colorization by Christopher Pooley,both of USDA, ARS, EMU. A public-domain product of the Agricultural Research Service.. 60 Trapeze: CalvertLitho. Co., Detroit, ca. 1890. 63 High jumper: Dunia Young. 63 Gymnastics wheel: Copyright Hans Genten,Aachen, Germany. “The copyright holder of this file allows anyone to use it for any purpose, provided that thisremark is referenced or copied.”. 70 Aristotle: Francesco Hayez, 1811. 70 Angel Stadium: U.S. MarineCorps, Staff Sgt. Chad McMeen, public domain work of the U.S. Government. 70 Shanghai: AgnieszkaBojczuk, CC-BY-SA. 70 Jets over New York: U.S. Air Force, Tech. Sgt. Sean Mateo White, public domainwork of the U.S. Government. 70 Rocket sled: U.S. Air Force, public domain work of the U.S. Government.97 Galileo’s trial: Cristiano Banti (1857). 103 Gravity map: US Navy, European Space Agency, D. Sandwell,and W. Smith. 112 Astronaut jumping: NASA. 114 Flea jumping: Burrows and Sutton, “Biomechanics ofjumping in the flea,” J. Exp. Biology 214 (2011) 836. Used under the U.S. fair use exception to copyright. 117Newton: Godfrey Kneller, 1702. 146 Space shuttle launch: NASA. 147 Swimmer: Karen Blaha, CC-BY-SAlicensed. 154 Partridge: Redrawn from K.P. Dial, “Wing-Assisted Incline Running and the Evolution ofFlight,” Science 299 (2003) 402. 157 Wind tunnel: Jeff Caplan/NASA Langley, public domain. 157 Cropduster: NASA Langley Research Center, public domain. 157 Locomotive: Locomotive Cyclopedia of AmericanPractice, 1922, public domain. 157 Series of vortices: Wikimedia Commons user Onera, CC-BY license.157 Turbulence: C. Fukushima and J. Westerweel, Technical University of Delft, The Netherlands, CC-BYlicense. 158 Golf ball: Wikimedia Commons user Paolo Neo, CC-BY-SA license. 158 Shark: WikimediaCommons user Pterantula, CC-BY-SA license. 158 Hummer: Wikimedia commons user Bull-Doser, publicdomain. 158 Prius: Wikimedia commons user IFCAR, public domain. 162 Golden Gate Bridge: Wikipediauser Dschwen, CC-BY-SA licensed. 168 Football player and old lady: Hazel Abaya. 177 Ring toss: ClarenceWhite, 1899. 189 Aerial photo of Mondavi vineyards: NASA. 203 Galloping horse: Eadweard Muybridge,1878. 209 Sled: Modified from Millikan and Gale, 1920. 214 Hurricane track: Public domain, NASA andWikipedia user Nilfanion. 217 Hanging boy: Millikan and Gale, 1927. 224 Crane fly: Wikipedia user Pinzo,public domain. 227 Motorcyclist: Wikipedia user Fir0002, CC-BY-SA licensed. 232 Space colony: NASA.241 Saturn: Voyager 2 team, NASA. 242 Tycho Brahe: public domain. 247 Pluto and Charon: Hubble

515

Space Telescope, STSCi. 254 Simplified Cavendish experiment: Wikimedia commons user Chris Burks, publicdomain.. 258 WMAP: NASA. 261 Earth: Apollo 11, NASA. 261 Uranus: Voyager 2 team, NASA.263 New Horizons spacecraft image: Wikipedia user NFRANGA, CC-BY-SA. 263 New Horizons trajectory:Wikimedia commons user Martinw89, CC-BY-SA. 271 Jupiter: Images from the Hubble Space Telescope,NASA, not copyrighted. 276 Hoover Dam: U.S. Department of the Interior, Bureau of Reclamation, LowerColorado Region, not copyrighted. 295 Hydraulic ram: Millikan and Gale, 1920. 297 Bonfire, grapes:CC-BY-SA licensed, by Wikipedia user Fir0002. 300 Skater in pool: Courtesy of J.D. Rogge. 305 Plutoniumpellet: U.S. Department of Energy, public domain.. 308 PET body scan: Jens Langner, public domain. 308Photo of PET scanner: Wikipedia user Hg6996, public domain. 308 Ring of detectors in PET scanner:Wikipedia user Damato, public domain. 309 Eclipse: 1919, public domain. 309 Newspaper headline: 1919,public domain. 313 Skateboarder on top of pipe: Oula Lehtinen, Wikimedia Commons, CC-BY-SA. 317Baseball pitch: Wikipedia user Rick Dikeman, CC-BY-SA. 322 Breaking Trail: Art by Walter E. Bohl. Imagecourtesy of the University of Michigan Museum of Art/School of Information and Library Studies. 351 DeepSpace 1 engine: NASA. 352 Nucleus of Halley’s comet: NASA, not copyrighted. 358 Chadwick’s apparatus:Redrawn from the public-domain figure in Chadwick’s original paper. 359 Wrench: PSSC Physics. 361Jupiter: Uncopyrighted image from the Voyager probe. Line art by the author. 363 Air bag: DaimlerChryslerAG, CC-BY-SA licensed.. 381 Tornado: NOAA Photo Library, NOAA Central Library; OAR/ERL/NationalSevere Storms Laboratory (NSSL); public-domain product of the U.S. government. 382 Longjump: ThomasEakins, public domain. 389 Pendulum: PSSC Physics. 390 Diver: PSSC Physics. 396 Cow: Drawnby the author, from a CC-BY-SA-licensed photo on commons.wikimedia.org by user B.navez.. 399 Modernwindmill farm, Tehachapi, CA: U.S. Department of Energy, not copyrighted. 399 Old-fashioned windmill:Photo by the author. 402 Ballerina: Rick Dikeman, 1981, CC-BY-SA license, www.gnu.org/copyleft/fdl.html,from the Wikipedia article on ballet (retouched by B. Crowell). 436 White dwarf: Image of NGC 2440 fromthe Hubble Space Telescope, H. Bond and R. Ciardullo. 451 Locomotive linkages: David Ingham, CC-BY-SAlicensed.. 455 Jupiter: Uncopyrighted image from the Voyager probe. Line art by the author. 467 TacomaNarrows Bridge: Public domain, from Stillman Fires Collection: Tacoma Fire Dept, www.archive.org. 475Nimitz Freeway: Unknown photographer, courtesy of the UC Berkeley Earth Sciences and Map Library. 480Three-dimensional brain: R. Malladi, LBNL. 480 Two-dimensional MRI: Image of the author’s wife. 489Spider oscillations: Emile, Le Floch, and Vollrath, Nature 440 (2006) 621.

516

Indexacceleration, 101

as a vector, 206

constant, 108

definition, 108

negative, 104

alchemists, 271

alchemy, 13

amplitude

defined, 452

peak-to-peak, 453

related to energy, 469

angular acceleration, 406

angular frequency, 451

angular momentum

choice of axis theorem, 390, 431

defined, 383

definition, 384

introduction to, 381

related to area swept out, 387

spin theorem, 390, 431

angular velocity, 405

area

operational definition, 37

scaling of, 39

astrology, 13

Bacon, Francis, 17

Big Bang, 257

black hole, 309

Brahe, Tycho, 242

calculus

with vectors, 210

cathode rays, 15

causality, 75

center of mass, 60

frame of reference, 361

motion of, 61

related to momentum, 359

center-of-mass motion, 61

centi- (metric prefix), 20

Chadwick, James

discovery of neutron, 357

choice of axis theorem, 390

proof, 431

circular motion, 223

inward force, 229

no forward force, 229

no outward force, 229

nonuniform, 225

uniform, 225

CMB, 258

coefficient of kinetic friction, 155

coefficient of static friction, 154

collision

defined, 354

comet, 449

component

defined, 181

conduction of heat

distinguished from work, 316

conversions of units, 26

coordinate system

defined, 66

Copernicus, 69

correspondence principle, 275

defined, 75

for time dilation, 75

cosmic microwave background, 258

cosmological constant, 257

cross product, 420

damped oscillations, 470

dark energy, 257

Darwin, 16

decibel scale, 470

delta notation, 64

Dialogues Concerning the Two New Sciences,40

dot product of two vectors, 323, 330

driving force, 473

eardrum, 473

Einstein, Albert, 450

electrical force

in atoms, 357

electron, 357

element, chemical, 272

517

elliptical orbit law, 427energy

distinguished from force, 131gravitational potential energy, 302potential, 300related to amplitude, 469

equilibriumdefined, 399

ether, 87

falling objects, 97Feynman, 100Feynman, Richard, 100force

analysis of forces, 158Aristotelian versus Newtonian, 118as a vector, 209contact, 120distinguished from energy, 131frictional, 153gravitational, 153net, 122noncontact, 120normal, 152positive and negative signs of, 121transmission, 161

forcesclassification of, 149

four-vector, 334frame of reference

defined, 66inertial

in Newtonian mechanics, 134rotating, 224

French Revolution, 20frequency

angular, 451defined, 451

frictionfluid, 157kinetic, 153, 154static, 153, 154

fulcrum, 403full width at half-maximum, 478FWHM, 478

Galileo, 458Galileo Galilei, 39gamma ray, 357

gamma rays, 15garage paradox, 85grand jete, 61graphs

of position versus time, 67velocity versus time, 74

Halley’s Comet, 449heat

as a fluid, 298as a form of kinetic energy, 298

heat conductiondistinguished from work, 316

high jump, 63homogeneity of spacetime, 80Hooke’s law, 163, 454hypothesis, 12

inertiaprinciple of, 69

infinitesimal number, 65Ives-Stilwell experiment, 289

joule (unit), 276Joyce, James, 298

Keplerelliptical orbit law, 427law of equal areas, 387

Kepler’s laws, 242, 243elliptical orbit law, 243equal-area law, 243law of periods, 243, 245

Kepler, Johannes, 242kilo- (metric prefix), 20kilogram, 22kinetic energy, 281

compared to momentum, 352

Laplace, 14Leibniz, Gottfried, 65lever, 403light, 14Lorentz transformation, 81Lorentz, Hendrik, 81

magnitude of a vectordefined, 190

matter, 14mega- (metric prefix), 20

518 Index

meter (metric unit), 22metric system, 20

prefixes, 20Michelson-Morley experiment, 87micro- (metric prefix), 20microwaves, 15milli- (metric prefix), 20mks units, 23model

scientific, 153models, 61moment of inertia, 408

tabulated for various shapes, 415momentum

compared to kinetic energy, 352defined, 349examples in three dimensions, 365of light, 352rate of change of, 362related to center of mass, 359transfer of, 362

motionperiodic, 451rigid-body, 59types of, 59

Muybridge, Eadweard, 203

nano- (metric prefix), 20Neanderthals, 404neutron

discovery of, 357Newton

first law of motion, 121second law of motion, 125

Newton’s laws of motionin three dimensions, 183

Newton’s third law, 146Newton, Isaac, 20

definition of time, 23nucleus, 357

operational definitionacceleration, 105energy, 285power, 285

operational definitions, 21order-of-magnitude estimates, 47oscillations

damped, 470

parabolamotion of projectile on, 182

parallel axis theorem, 410, 432, 440particle zoo, 297Pauli exclusion principle, 16period

defined, 451of uniform circular motion, 230

perpetual motion machine, 272physics, 14pitch, 449POFOSTITO, 148Pope, 40positron, 308potential energy

electrical, 304gravitational, 302, 327nuclear, 305of a spring, 326related to work, 326

power, 283projectiles, 182proton, 357pulley, 164

quarks, 298

radial componentdefined, 232

radio waves, 15reductionism, 17Renaissance, 11resonance

defined, 475RHIC accelerator, 85rigid rotation

defined, 383Robinson, Abraham, 65rotation, 59

scalardefined, 190

scalar (dot) product, 323, 330scale height of atmosphere, 138scaling, 39scientific method, 12second (unit), 21shell theorem, 252SI units, 23

Index 519

significant figures, 28simple harmonic motion

defined, 455simple machine

defined, 164slam dunk, 61slingshot effect, 361spin theorem, 390

proof, 431spring

potential energy of, 326work done by, 326

spring constant, 163Stanford, Leland, 203statics, 399steady-state behavior, 474Stevin, Simon, 275strain, 163Swift, Jonathan, 39swing, 473

temperatureas a measure of energy per atom, 299

tension, 162theory, 12thermodynamics, 299time

duration, 64point in, 64

torquedefined, 391due to gravity, 395related to force, 424relationship to force, 392

transmission of forces, 161triangle inequality, 335tuning fork, 453twin paradox, 335

unit vectors, 196units, conversion of, 26

vectoracceleration, 206addition, 190defined, 190force, 209four-vector, 334magnitude of, 190

velocity, 204vector cross product, 420vector product, cross, 420velocity

addition of, 72as a vector, 204negative, 72

volumeoperational definition, 37scaling of, 39

Voyager space probe, 94

watt (unit), 284weight force

defined, 120relationship to mass, 128

workdefined, 316distinguished from heat conduction, 316done by a spring, 326done by a varying force, 325, 450, 453in three dimensions, 321positive and negative, 319related to potential energy, 326

work-kinetic energy theorem, 329

x-rays, 15

Young’s modulus, 173

520 Index

Trig Tableθ sin θ cos θ tan θ θ sin θ cos θ tan θ θ sin θ cos θ tan θ

0 0.000 1.000 0.000 30 0.500 0.866 0.577 60 0.866 0.500 1.7321 0.017 1.000 0.017 31 0.515 0.857 0.601 61 0.875 0.485 1.8042 0.035 0.999 0.035 32 0.530 0.848 0.625 62 0.883 0.469 1.8813 0.052 0.999 0.052 33 0.545 0.839 0.649 63 0.891 0.454 1.9634 0.070 0.998 0.070 34 0.559 0.829 0.675 64 0.899 0.438 2.0505 0.087 0.996 0.087 35 0.574 0.819 0.700 65 0.906 0.423 2.1456 0.105 0.995 0.105 36 0.588 0.809 0.727 66 0.914 0.407 2.2467 0.122 0.993 0.123 37 0.602 0.799 0.754 67 0.921 0.391 2.3568 0.139 0.990 0.141 38 0.616 0.788 0.781 68 0.927 0.375 2.4759 0.156 0.988 0.158 39 0.629 0.777 0.810 69 0.934 0.358 2.605

10 0.174 0.985 0.176 40 0.643 0.766 0.839 70 0.940 0.342 2.74711 0.191 0.982 0.194 41 0.656 0.755 0.869 71 0.946 0.326 2.90412 0.208 0.978 0.213 42 0.669 0.743 0.900 72 0.951 0.309 3.07813 0.225 0.974 0.231 43 0.682 0.731 0.933 73 0.956 0.292 3.27114 0.242 0.970 0.249 44 0.695 0.719 0.966 74 0.961 0.276 3.48715 0.259 0.966 0.268 45 0.707 0.707 1.000 75 0.966 0.259 3.73216 0.276 0.961 0.287 46 0.719 0.695 1.036 76 0.970 0.242 4.01117 0.292 0.956 0.306 47 0.731 0.682 1.072 77 0.974 0.225 4.33118 0.309 0.951 0.325 48 0.743 0.669 1.111 78 0.978 0.208 4.70519 0.326 0.946 0.344 49 0.755 0.656 1.150 79 0.982 0.191 5.14520 0.342 0.940 0.364 50 0.766 0.643 1.192 80 0.985 0.174 5.67121 0.358 0.934 0.384 51 0.777 0.629 1.235 81 0.988 0.156 6.31422 0.375 0.927 0.404 52 0.788 0.616 1.280 82 0.990 0.139 7.11523 0.391 0.921 0.424 53 0.799 0.602 1.327 83 0.993 0.122 8.14424 0.407 0.914 0.445 54 0.809 0.588 1.376 84 0.995 0.105 9.51425 0.423 0.906 0.466 55 0.819 0.574 1.428 85 0.996 0.087 11.43026 0.438 0.899 0.488 56 0.829 0.559 1.483 86 0.998 0.070 14.30127 0.454 0.891 0.510 57 0.839 0.545 1.540 87 0.999 0.052 19.08128 0.469 0.883 0.532 58 0.848 0.530 1.600 88 0.999 0.035 28.63629 0.485 0.875 0.554 59 0.857 0.515 1.664 89 1.000 0.017 57.290

90 1.000 0.000 ∞

Index 521

Mathematical ReviewAlgebra

Quadratic equation:

The solutions of ax2 + bx+ c = 0are x = −b±

√b2−4ac

2a .

Logarithms and exponentials:

ln(ab) = ln a+ ln b

ea+b = eaeb

ln ex = eln x = x

ln(ab) = b ln a

Geometry, area, and volume

area of a triangle of base b and height h = 12bh

circumference of a circle of radius r = 2πrarea of a circle of radius r = πr2

surface area of a sphere of radius r = 4πr2

volume of a sphere of radius r = 43πr

3

Trigonometry with a right triangle

sin θ = o/h cos θ = a/h tan θ = o/a

Pythagorean theorem: h2 = a2 + o2

Trigonometry with any triangle

Law of Sines:

sinα

A=

sinβ

B=

sin γ

C

Law of Cosines:

C2 = A2 +B2 − 2AB cos γ

Properties of the derivative and integral (forstudents in calculus-based courses)

Let f and g be functions of x, and let c be a con-stant.

Linearity of the derivative:

d

dx(cf) = c

df

dx

d

dx(f + g) =

df

dx+

dg

dx

The chain rule:

d

dxf(g(x)) = f ′(g(x))g′(x)

Derivatives of products and quotients:

d

dx(fg) =

df

dxg +

dg

dxf

d

dx

(f

g

)=f ′

g− fg′

g2

Some derivatives:d

dxxm = mxm−1, except for m = 0

ddx sinx = cosx d

dx cosx = − sinxd

dxex = ex d

dx lnx = 1x

The fundamental theorem of calculus:∫df

dxdx = f

Linearity of the integral:∫cf(x)dx = c

∫f(x)dx

∫[f(x) + g(x)] =

∫f(x)dx+

∫g(x)dx

Integration by parts:∫fdg = fg −

∫gdf

522 Index

Useful Data

Metric Prefixes

M- mega- 106

k- kilo- 103

m- milli- 10−3

µ- (Greek mu) micro- 10−6

n- nano- 10−9

p- pico- 10−12

f- femto- 10−15

(Centi-, 10−2, is used only in the centimeter.)

The Greek Alphabet

α A alpha ν N nuβ B beta ξ Ξ xiγ Γ gamma o O omicronδ ∆ delta π Π piε E epsilon ρ P rhoζ Z zeta σ Σ sigmaη H eta τ T tauθ Θ theta υ Y upsilonι I iota φ Φ phiκ K kappa χ X chiλ Λ lambda ψ Ψ psiµ M mu ω Ω omega

Subatomic Particles

particle mass (kg) radius (fm)electron 9.109× 10−31 . 0.01proton 1.673× 10−27 ∼ 1.1neutron 1.675× 10−27 ∼ 1.1

The radii of protons and neutrons can only be given approx-

imately, since they have fuzzy surfaces. For comparison, a

typical atom is about a million fm in radius.

Notation and Units

quantity unit symboldistance meter, m x, ∆xtime second, s t, ∆tmass kilogram, kg mdensity kg/m3 ρvelocity m/s vacceleration m/s2 aforce N = kg·m/s2 Fpressure Pa=1 N/m2 Penergy J = kg·m2/s2 Epower W = 1 J/s Pmomentum kg·m/s pangular momentum kg·m2/s or J·s Lperiod s Twavelength m λfrequency s−1 or Hz fgamma factor unitless γprobability unitless Pprob. distribution various D

electron wavefunction m−3/2 Ψ

Earth, Moon, and Sun

body mass (kg) radius (km) radius of orbit (km)earth 5.97× 1024 6.4× 103 1.49× 108

moon 7.35× 1022 1.7× 103 3.84× 105

sun 1.99× 1030 7.0× 105 —

Fundamental Constants

gravitational constant G = 6.67× 10−11 N·m2/kg2

Coulomb constant k = 8.99× 109 N·m2/C2

quantum of charge e = 1.60× 10−19 Cspeed of light c = 3.00× 108 m/sPlanck’s constant h = 6.63× 10−34 J·s

Index 523

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