63
Preliminary Physics Notes Second Edition
Preliminary Physics Notes
Second Edition
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Contents
Second Edition, First Revision
1 History 7
2 Units and Dimensions 9
2.1 SI Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
2.2 Related units and constants . . . . . . . . . . . . . . . . . . . . . . . . . . 10
2.3 Applications to physical problems . . . . . . . . . . . . . . . . . . . . . . . 11
2.4 Example: Dimensional analysis . . . . . . . . . . . . . . . . . . . . . . . . 12
2.5 Example: Checking a solution . . . . . . . . . . . . . . . . . . . . . . . . . 13
2.6 Example: Conversion of units . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.7 Units on graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
2.8 Problem Set: Dimensions and Units . . . . . . . . . . . . . . . . . . . . . . 15
3 Solving Problems 17
3.1 Steps in problem solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.1.1 Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17
3.1.2 Directions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.1.3 Definitions or Data . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.1.4 Diagnosis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.1.5 Derivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.1.6 Determination . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
3.1.7 Dimensions (and limiting cases) . . . . . . . . . . . . . . . . . . . . 19
3.1.8 Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3.2 Important considerations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3.2.1 Scalars and vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3.2.2 Choice of origin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.2.3 Selection of axes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.2.4 Superposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
3.2.5 Approximations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
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3.2.6 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
3.2.7 This is physics, not mathematics . . . . . . . . . . . . . . . . . . . 22
4 Measurement 23
4.1 Types of experimental errors . . . . . . . . . . . . . . . . . . . . . . . . . . 23
4.2 Significant figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
4.3 Example: Significant figures vs percentage errors . . . . . . . . . . . . . . . 25
4.4 Error estimation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
4.4.1 Addition and subtraction . . . . . . . . . . . . . . . . . . . . . . . . 27
4.4.2 Multiplication and Division . . . . . . . . . . . . . . . . . . . . . . 27
4.4.3 Raising to a power . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
4.5 Example: Estimating error in resistivity . . . . . . . . . . . . . . . . . . . 28
4.6 Graphs and errors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
4.7 Problem Set: Measurement . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
5 Kinematics 31
5.1 Linear Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
5.1.1 Equations describing linear motion under constant acceleration . . . 31
5.2 Rotational Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
5.3 Rotational Quantities as Vectors . . . . . . . . . . . . . . . . . . . . . . . . 34
5.4 Motion in a circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
5.5 Problem Set: Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
6 Energy 39
6.1 Potential and kinetic energy . . . . . . . . . . . . . . . . . . . . . . . . . . 39
6.2 Internal energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
6.3 Law of conservation of energy . . . . . . . . . . . . . . . . . . . . . . . . . 40
6.4 Potential energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
6.5 Potential energy diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
6.6 The zero of potential energy . . . . . . . . . . . . . . . . . . . . . . . . . . 42
6.7 Potential energy as a function of separation . . . . . . . . . . . . . . . . . 43
6.8 Algebraic expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
6.9 Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
6.10 Potentials in two dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . 44
6.11 Potentials in three dimensions . . . . . . . . . . . . . . . . . . . . . . . . . 46
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7 Forces 47
7.1 Agents of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
7.2 Newton’s First Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
7.3 Forces in two or more dimensions . . . . . . . . . . . . . . . . . . . . . . . 48
7.4 Algebraic definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
7.5 Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
7.6 Gravitational and electrostatic fields . . . . . . . . . . . . . . . . . . . . . 51
7.7 Reaction forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
7.8 Contact forces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53
7.9 Free body diagrams . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
7.10 Tension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
7.11 Gravitation at the surface of the Earth . . . . . . . . . . . . . . . . . . . . 57
7.12 Hooke’s Law . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58
7.13 Momentum and Newton’s Second Law . . . . . . . . . . . . . . . . . . . . 59
7.14 Kinetic energy and work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
7.15 Example: Loop the loop . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
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Chapter 1
History
Physics began when mankind first started observing the world around him and trying toexplain why things happened the way they did. Early civilisations noted patterns andcycles in the movements of the moon, stars and planets. It is impossible to say exactlywhen physics began, but it is likely that it began as soon as mankind started thinkingabout his environment.
In the European or western tradition, the ancient Greeks are often credited with thebeginning of physics as a systematic and experimental science. Aristotle (ca. 384 - 322BCE), who was a philosopher and a student of Plato, was a powerful man in his owntime, and a teacher and mentor of Alexander the Great. His influence was strong inscience for over a thousand years after he died. Aristotle advocated the use of the sensesand experiments to learn about the natural world, rather than the use of pure logic andmathematics as many of his colleagues did. However, he was still inclined to use logic evenwhen it contradicted observations. For example, in spite of experimental investigations,he concluded that objects with different masses fell at different rates under gravity.
Unfortunately for the development of physics, Aristotelian natural philosophy becameunquestioned and unquestionable dogma in the 12th century. In fact, many students stillhave what are often termed Aristotelian (mis)conceptions about physics. This is not to saythat Aristotle did not make great contributions; those who later turned his conceptionsinto dogma which denied experimental evidence were in opposition to Aristotle’s ownphilosophical stance that we learn through the use of our senses.
In the same period, Archimedes (ca. 287–212 BCE) discovered that objects displace theirown volume in fluids. He also made contributions to mathematics and mechanics. OtherGreek physicists, or natural philosophers as they were known, deduced that the Earthwas spherical and estimated its diameter using trigonometry (another Greek invention).
Shortly afterwards, Ptolemy (ca. 85–165 CE) came up with the geocentric model of thesolar system, which was to be the dominant theory in the western world until the scientificrevolution, which started around the 16th to 17th century (historians debate exactly when).
The principal natural philosophers of the scientific revolution were Copernicus (1473–1543), Kepler (1571–1630), Galileo (1564–1642), and Newton (1643–1727). Copernicuscame up with a heliocentric (sun centred) model of the universe, but it was not publisheduntil his death in 1543. He did, however, give public lectures on his model and commu-nicate with scientists across Europe about the heliocentric model, and was not, unlikeGalileo, reprimanded by the Catholic church. It is generally believed, however, that hedelayed publishing his theories out of fear of the response of the church.
Kepler refined Copernicus’ model by making the planetary orbits elliptical rather thancircular, improving the match between astronomical observations and predictions of thetheory. In physics, the test of any theory is not only how well it explains earlier ob-servations, but its predictive power. A theory which has no predictive power would be
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considered unscientific by most physicists. Galileo provided the new observations, of themovements of Jupiter’s moons and the phases of Venus, which agreed with the Coperni-can model of a heliocentric solar system with elliptical orbits. Galileo was tried by theinquisition in 1633 for his public support and teaching of the heliocentric model.
Newton was the first to demonstrate that the same natural laws govern both motion onEarth and celestial motion. Much of the mechanics which fills the first few chapters ofalmost every physics textbook was discovered by Newton. Newton’s laws of forces replacedthe earlier Aristotelian notions of impetus, and his theory of gravitation allowed a moregeneral understanding of Kepler’s laws of planetary motion as being due to the same forcethat makes objects fall to the ground on Earth. Newton also made great contributionsto the study of light and thermal physics. You will learn much of the physics of Newtonfrom these volumes and at the scholar school in January. Apart from his contributionsto physics, Newton also invented the gold standard in currency as Master of the RoyalMint, and has been credited with inventing the cat-door.
The other big development in science during the scientific revolution was the introductionof the scientific method - in particular the use of experiment beyond simple observation.Francis Bacon (1561 - 1626) (who some believe was actually the author of Shakespeare’splays) was a great advocate of empiricism, which is the systematic study by experiment,starting with observation rather than known (or assumed) facts.
Over the next few hundred years after the scientific revolution great progress was madein the understanding of electromagnetism, the wave nature of light, thermodynamicsand statistical mechanics. You will learn about these aspects of physics in your studiesat school, and at the scholar school in January. Important contributions were made byBoyle (1627–1691) and Young (1773–1829), who did much work on thermodynamics. Thesteam engine was developed at around the same time, and the industrial revolution alsobegan.
In the 19th century Maxwell (1831–1879), Faraday (1791–1867), Ohm (1787–1854) andothers developed electromagnetism. Faraday was a keen experimentalist, making manyobservations which led to the development of the electromagnetic theory. Maxwell unifiedmuch of electromagnetism, and also provided the link between optics and electromag-netism in his formulation of what are now known as Maxwell’s equations.
At the start of the 20th century there was a second revolution in physics. Newton’s lawswere found to be an approximation to what really happens, accurate in the limits oflow velocities, large masses and weak gravitational fields. Two new theories emergedin the early 20th century, largely as a result of the inability of classical physics (alsoknown as Newtonian physics) to explain various experimental results. These two newtheories were relativity and quantum mechanics. Einstein played an important role inquantum physics and was largely responsible for both special and general relativity. Thedevelopment of quatum theory led to a technological revolution as condensed matterphysics, which is largely applied quantum physics, was developed and allowed us to starttailor-making materials with desirable properties, such as the semiconductors on whichmodern electronics is based.
Relativity has changed the way we view the universe, and how we describe it; we nowconsider space-time to be a single four dimensional “fabric” rather than space and timeas separate, independent dimensions.
Physics continues to develop as we make new discoveries and invent new ways of under-standing them. It is by no means finished, with much left to be done for those who havethe will and are prepared to do the work!
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Chapter 2
Units and Dimensions
An appreciation of units and dimensions is very helpful when analysing problems. Unitsand dimensions are not the same. An angle is a ratio of two lengths and thus has nodimensions. Angle is measured in the unit radians, where one radian is defined to be theangle subtended at the centre of a circle by a length of arc equal to the radius of thatcircle. θ radians thus has no dimensions. The unit of angular velocity is radians/s, thedimension is time−1 or T−1. The basic units and dimensions are given in Table 2.1.
All measurements in physics are ratios, for example the ratio of the length of an object toa standard length or the ratio of the density of an object to the density of water. In somecases, such as relative density, this ratio is a pure number, but in most cases the numbermust be followed by the denominator of the ratio, e.g. if an object is 5m long, the ratioof its length to the standard length of 1 metre is 5.
The history of how various units such as metres, seconds, amperes etc. came to be definedis fascinating. For example, in the 18th century there were two competing approachesto the definition of a standard unit of length. Some suggested defining the metre as thelength of a pendulum having a half-period of one second; others suggested defining themetre as one ten-millionth of the length of the Earth’s meridian along a quadrant (onefourth of the circumference of the Earth). In 1791 the French Academy of Sciences chosethe meridian definition over the pendulum definition because the force of gravity variesslightly over the surface of the Earth, affecting the period of the pendulum. So the metrewas intended to equal 10−7 (one ten-millionth) of the length of the meridian through Parisfrom the north pole to the equator. In 1889, a new international prototype was made of analloy of platinum with 10 percent iridium, that was to be measured at the melting pointof ice. In 1927, the conditions under which the bar was to be measured were specifiedmore thoroughly. In 1960 the metre was redefined based upon a wavelength of krypton-86radiation. In 1983 it was redefined again as the distance traveled by light in vacuum in1/(299,792,458) second. While this removes the need for a standard metre-long objectto be kept as a standard, it depends on the definition of the second. The second is nowdefined relative to the time of a particular atomic transition. The system of units definedby the French Academy is the Systeme Internationale, and it is now the system of unitsadopted by scientists and engineers through most of the world, and the system which youwill be using.
2.1 SI Units
Table 2.1 shows the seven base units of the SI. There is large collection of fundamentalconstants of nature associated with these units.
G is called the gravitational constant, c is the speed of light in a vacuum, h is Planck’sconstant, k is Boltzmann’s constant, µ0 is the permeability of free space and NA is Avo-
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Dimension SI Unit Related Constants
Mass (M) kilogram (kg) G = 6.67 × 10−11 N kg−2 m2 [≡ kg−1m3s−2]
Length (L) metre (m) c = 3.00 × 108 m s−1
Time (T ) second (s) h = 6.63 × 10−34 J s [≡ kg m2s−1]
Temperature (K) kelvin (K) k = 1.38 × 10−23 J K−1 [≡ kg m2s−2K−1]
Current (I) ampere (A) µ0 = 4π × 10−7 H m−1 [≡ kg m2s−2A−2]
Amount of substance mole (mol) NA = 6.02 × 1023 mol−1
Luminous intensity candela (cd)
Table 2.1: Basic units and some related fundamental constants
gadro’s number. In hindsight physicists should have given a simple value to as many ofthe most common fundamental constants as possible, as with µ0, and used them to definethe SI. Standards were in general set before the significance of the constants was realised.It is not necessary to memorise these constants but you may end up doing so as they areused very frequently. Note that in the SI, units written in full are never capitalized, withthe exception of the degree Celsius.
2.2 Related units and constants
coulomb: The coulomb is the unit of charge. One coulomb is equal to 6.25× 1018 timesthe magnitude of the charge of one electron. The coulomb is related to the ampere,which is the SI unit of electric current. An ampere is defined as “that constantcurrent which, if maintained in two straight parallel conductors of infinite length, ofnegligible circular cross section, and placed 1 meter apart in vacuum, would producebetween these conductors a force equal to 2 × 10−7 newton per meter of length”. Itis related to the coulomb by 1 A = 1 C s−1.
degree Celsius: In the SI, temperature is measured in kelvin (K). The word “degree” isnot used. To convert from degrees Celsius (C) to kelvin, add 273.15. If a temperatureis to be multiplied it must be in K. When dealing with temperature differences, thenumeric values in C and K are equal.
Permittivity of free space ǫ0: The permittivity of free space is a fundamental constantrelated to electrostatics and electromagnetism. This constant was shown by Maxwellto be equal to 1/µ0c
2. Since µ0 has the value 4π × 10−7 H m−1 by definition and c isvery close to 3.00 × 108 m s−1, the value of ǫ0 is approximately 8.85 × 10−12 F m−1.Note that this constant has an exact value in the SI as both µ0 and c are fixed exactly.The SI unit for ǫ0 is F m−1 where F represents the farad, the unit of capacitance. Notethe similarity between the units of µ0 and ǫ0.
Gas constant: R = NAk = 8.314 J K−1 mole−1. k is Boltzmann’s constant, and NA isAvogadro’s number which is the number of particles in a mole.
Other fundamental quantities: There are many other constants in nature. Most aredetermined experimentally. For example:
Mass of an electron: me ≃ 9.1 × 10−31 kg
Charge of an electron: e ≃ 1.6 × 10−19 C
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2.3 Applications to physical problems
All answers in physics problems must be followed by the appropriate unit.This is easy to understand when you appreciate that the number is merely the ratio ofyour measurement to the standard unit. If you do not include the unit then all work donein substituting numbers will have been in vain because your answer cannot be correct.The only exception to this is where the answer is a pure ratio or number, such as relativedensity.
Some textbooks and teachers prefer to use the basic SI unit as an indication of thedimensions of a quantity; for example energy is in kg m2s−2. Others use the capitalletters M , L, T to indicate the dimensions, so energy becomes ML2T−2. Either methodis acceptable as long as it is consistent. What must be remembered when using the firstis that an algebraic equation has dimensions but it does not have units. It is not correctto say that
v = u + at ms−1 .
One could say
(v) ms−1 = (u + at) ms−1 ,
but such a statement contains no new information and is thus pointless.
The dimensions of a quantity can be determined by building it up from more basic quan-tities. For example, the following chain of ideas will lead to energy:
Length L, velocity LT−1, acceleration LT−2, force MLT−2, work ML2T−2.
In any formula all terms must have the same dimensions. If you are uncertainabout the structure of a formula check the dimensions. In the expression
s = ut +1
2at2
all terms are lengths (and thus have SI units of metres). On the other hand
v2 − u2 = 2at
is incorrect because the dimensions of the first two terms are L2T−2 and of the last areLT−2T or LT−1. Note, however, that this can tell you nothing about the dimensionlessconstants like 2 in the equation.
In any problem it pays to stop occasionally and check that all terms in the expressionsor equations have the same dimensions. If they have you may proceed. If they have notthen you are most definitely wrong.
Substituting a numerical value for any physical constant will mean that it is no longerpossible to determine whether a problem is dimensionally consistent (see Example 2.5).Furthermore it is dimensionally incorrect to substitute a number without the units for adimensioned quantity. Many students cannot resist the temptation of writing 9.8 in placeof g in a problem, which takes away this valuable asset, is dimensionally incorrect andwill be marked as such.
Never make any numerical substitution when solving a problem until the fi-nal algebraic expression has been derived, written out in its simplest form and had itsdimensions checked.
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It must be emphasised that while a solution is necessarily wrong if the dimensionsare wrong, the converse is not true. Dimensionless quantities are not checked by thismethod and no information about them can be obtained. Numerical factors, trigonometricratios, etc. will never be detected by dimensional analysis.
There are quantities which have the same dimensions but are not equivalent. One suchpair is torque and work. Torque is the vector product of force and radial distance, workis the scalar product of force and a parallel distance. Both have units of newton metre,N m, but the first cannot be expressed in joules as the second can, and they cannot beequated. Torque is a vector quantity, work is a scalar quantity.
In all mathematical functions such as sin(x), cos(x), log(x), ln(x), tan(x), ex, and so on,the argument of the function, in this case x, must be a dimensionless quantity. Thereis no expression which allows one to take, for instance, log(10kg). Thus one of the firstchecks on a derived formula can always be to check that any such argument of a math-ematical function is dimensionless. Of course, the function itself is also dimensionless soany quantity which is equal to that function must itself be dimensionless. For example,the equation (from simple harmonic motion) Y = A sin(ωt) is saying that Y/A = sin(ωt)where both sides of the equation are dimensionless, since Y/A is the ratio of the displace-ment to the maximum displacement. Similarly the argument of the sine function, ωt, isalso dimensionless.
There are many times when an answer is best given in units which are not strictly SI. Forinstance if a problem is set in kilometres per hour it would not be incorrect to give ananswer in the same units. The calculations throughout must, however, be givenin SI units.
Any competent examiner in physics will, if the solution to a problem is incorrect, en-deavour to locate the mistake and then work logically through the problem to the givenanswer. If there are no further mistakes and the problem has not been trivialized theexaminer will then deduct the mark for the error and the student will benefit.
This means that if a student purely transcribes a number incorrectly or makes an arith-metic mistake the error can be identified. On the other hand if the student makes atranscription error or an algebraic error which renders the equation dimensionally incor-rect, everything which follows is dimensionally incorrect, and it is known that the studentcould have checked this, so examiners will be less lenient.
It is thus in your interests to ensure that all equations are dimensionally soundthroughout.
2.4 Example: Dimensional analysis
Dimensional analysis is a useful technique in which the dimensions or units are used toestablish a relationship between different parameters. The basic method is describedbelow. This method will allow you to determine the relationship between parameters towithin a numerical constant.
The frequency of a vibrating string depends on its length, its linear density and the ten-
sion. Use dimensional analysis to determine this relationship, given that the dimensionless
constant is 1
2in SI units.
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Set f = laµbT c where: l is the length in metresµ is the linear density in kilogram per metreT is the tension in newtons
The dimensions of each side must be equal, so
T−1 = (L)a (ML−1)b (MLT−2)c ,
and equating indices one has
M : 0 = b + c (2.1)
L : 0 = a − b + c (2.2)
T : −1 = −2c . (2.3)
From Eq. 2.3 we get
c =1
2,
from Eq. 2.1
b = −c = −1
2
and from Eq. 2.2
a = b − c = −1 .
Hence
f = kl−1µ−1/2T 1/2 =k
l
(
T
µ
)1/2
where k is a dimensionless constant.
From the given data k = 1
2, so
f =1
2l
(
T
µ
)1/2
which is the desired expression.
2.5 Example: Checking a solution
If in a force problem an intermediate step using the normal notation read
m1g sin(θ) − µm1g cos(θ) − m2g = (m1 + m2 +2
5m3)a ,
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you would note that all terms have units of mass × acceleration (MLT−2) and wouldproceed. If however it read
m1g sin(θ) − µm1 cos(θ) − m2g = (m1 + m2 +2
5m3)a ,
there is clearly something wrong here since the second term has dimensions M and notMLT−2. You therefore know not to proceed but to go back and find the error.
If you substituted for g and µ and had written
10m1 sin(θ) − m1 cos(θ) − 10m2 = (m1 + m2 +2
5m3)a ,
you would have lost the use of this powerful tool, and been marked incorrect, as such anequation implies that a is dimensionless.
2.6 Example: Conversion of units
Students often have difficulty converting between units.
Question: Convert the density of water from lb ft−3 to kg m−3 given that the density is62.516 lb ft−3.
Solution:
1 lb
ft3=
1 lb
ft3×
(
1 ft
12 in
)3
×
(
1 in
2.54 cm
)3
×
(
100 cm
1 m
)3
×1 kg
2.2 lb
=106
123 × 2.543 × 2.2kg m−3
= 16.0 kg m−3
The density of water is 62.5 lb ft−3, so
ρwater = 62.5 lb ft−3 = 62.5 × 16.0 kg m−3
= 1000 kg m−3
as expected.
2.7 Units on graphs
Many physics problems, either in the laboratory or in theory, involve drawing graphs.Each graph involves a pair of physical quantities, each of which has units. In general thedifferential (dx, dy, etc.) can be considered to be a very small portion of x or y and hencehas the same dimensions and units. The slope, dY/dX, will have units of the quotientY/X, and the area under the graph,
∫
Y dX, will have units of the product Y X.
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In the special case of a closed loop the area inside the loop is the change in the productY X on going once around the loop. If the product Y X has the units of energy, the areaunder the curve gives the energy change in moving from the initial to the final state andthe area inside a loop gives the energy added or subtracted in proceeding around the loop.
2.8 Problem Set: Dimensions and Units
Use the method of dimensional analysis to determine these relationships. All constantsare for formulae in SI units.
1. The period T of a simple pendulum can depend on the length, the mass of the boband the value of g. Determine this relationship. If the dimensionless constant in thiscase is 2π, what is the relationship?
2. Viscosity is measured in pascal second (Pa s). Determine the viscous drag force F onan object of mass m and radius r moving with velocity v through a fluid of viscosityη. You will need to assume that F does not depend on m to complete the problem.If the value of the dimensionless constant is 6π, what is the relationship? Draw adiagram of the situation.
3. The rate of emission of radiant energy per unit area from a perfectly radiating bodyis given by the Stefan-Boltzmann formula P = σT 4 where P is in watts per squaremetre, T is the temperature in kelvin and σ is the Stefan-Boltzmann constant. Ex-press σ in terms of the fundamental constants of nature given in Table 2.1. (You willneed to make the assumption that as radiation does not have mass or charge, G andthe µ0 are not relevant). If there is a constant 2π5/15, determine the value of σ.
4. For each of the following units, give
• the quantity being measured (e.g. coulomb/gram would be charge per unitmass)
• the dimensions in terms of M , L, T , K and I.
• the value in SI units
Show all working.
(a) kilometre per hour
(b) litre
(c) atmosphere
(d) gram per cubic centimetre
(e) year
(f) light year
(g) kilowatt hour
(h) horsepower-fortnight per cubic furlong
15
16
Chapter 3
Solving Problems
Solving text book and other simple problems usually means that you are taking an ex-isting theory and applying it to an unknown situation to see what the theory predicts.Sometimes these problems are the other way around, finding what happened earlier toproduce a given situation. In the laboratory you will be measuring and experimenting totest a theory. If you pursue your studies in physics far enough you will have the opportu-nity to engage in the entire process, developing and testing your own theories. Describedhere is a method for tackling problems. Although sometimes physicists seem to solveproblems with a sudden burst of inspiration, more usually a process like this takes place,and even when it doesn’t, it happens afterwards just to check the answer!
3.1 Steps in problem solving
Often the most difficult step in solving a problem is getting started. The first thing youmust always do is read the question carefully, then read it again. Then you are readyto move on to the steps below — the 7 D’s, which provide guidance in tackling problems:
1. Diagram
2. Directions
3. Definitions or Data
4. Diagnosis
5. Derivation
6. Determination
7. Dimensions
and then, if it is appropriate, Substitution.
3.1.1 Diagram
You need a big diagram, no, a REALLY big diagram, which extends across the width ofone page. The thickness of your lines should be minimal. You may need to draw a set
17
of diagrams to depict a sequence of events and you may need to draw specific diagramsof subsystems. In force problems you need to draw a free body diagram for each of thesubsystems considered. In some cases you may need to sketch a graph as well, such as ap − V diagram.
3.1.2 Directions
For all problems, vector or scalar, you need to indicate the positive direction. In a probleminvolving vectors you need to clearly show the co-ordinate axes and directions, in two orthree dimensions, whether these be Cartesian or polar. It is often helpful to resolve vectorquantities along your axes.
3.1.3 Definitions or Data
Define all symbols which you are using on the diagram and make their meanings clear,even if they follow a convention with which you are familiar. Show any subscriptedvariables very clearly. However, do not give definitions of variables given in the question,and do use these given variables in your solution. Never use any algebraic symbol withoutdefining it. “v = 10.6 ms−1” is hardly illuminating if you did not define v.
Show expressions for all parameters like moments of inertia, momentum, angular velocity,torque, frictional forces etc. on the diagram and write the relationships you need to obtainthem, if known. For example, Ff = µmg cos(θ), I = 2/5mr2.
3.1.4 Diagnosis
Write in English words the principle(s) you will be following and if possible follow with amore specific statement in the same form, e.g.
energy is conservedenergy before = energy after + energy lost
PE at the top of the hill = KE at the bottom + energy lost to friction .
Nearly all problems in mechanics can be solved using one or more of the 5 principles listedbelow:
• Conservation of energy
• Sum of the forces = mass × acceleration
• Sum of the torques = moment of inertia × angular acceleration
• Momentum is conserved
• Angular momentum is conserved
If you are unable to proceed for any reason, it may help you to write down these fiveprinciples on a spare page. It might remind you of the principle you have forgotten,which is very often conservation of angular momentum.
18
3.1.5 Derivation
Now, for each of the identified principles, write an equation representing the situationusing the variables which you have defined. If new parameters are required, then addthem to the diagram. Don’t introduce new symbols before defining them.
Now count your unknowns and your equations. If you have more unknowns than equa-tions then you must find additional relationships between your variables. No amount ofrearrangement will solve the problem. If you have more equations than unknowns youhave either repeated yourself or the problem is overspecified. Find out which. You nowhave N simultaneous equations in N unknown variables. Put a small box around eachone of them and number them. You should at this point do a dimension check on each ofthese equations (some notes on doing this are given below).
You have not yet tackled any algebra, but have almost completed the physics.
3.1.6 Determination
This step could also be called d’algebra. Now you must solve a system of N equationsin N unknown variables, eliminating N − 1 of these variables to yield a single equationin one unknown variable — the answer you want. Write this algebraic equation with thesingle unknown on the left hand side. Remember that such an expression does NOT haveunits.
This is your answer. Draw a double line box around it. Next you must to check itsdimensions.
3.1.7 Dimensions (and limiting cases)
Check that the dimensions of the LHS of any algebraic equation are the same as those ofeach term on the RHS. If they are not then work backwards until the error is found.
This process should be applied at intermediate steps. Aim to become a methodical prob-lem solver, and devise some symbol to put against each expression when you have checkedthe dimensions and found them to be correct. You might, for instance, like to draw asmiley face after each equation once you have checked the dimensions.
Before proceeding, you should check limiting cases.
• Try setting one of the parameters to zero. Does it give a simple solution which youcan easily interpret?
• Could a denominator become zero?
• Is there a square root of something which could become negative?
• Could you be asking for a cosine or sine to be greater than one?
Once you have satisfactorily answered these and similar questions you have finished theproblem and should draw an even more definite box around your answer. If applicablewrite something in words to qualify your answer which might help to explain the directionof flow, direction of force or something similar. Make sure that if you were asked for avector quantity, your answer is given as a vector.
19
3.1.8 Substitution
This is the least important step. Many students believe that the number is what counts.This is not true. It is the understanding of the problem that counts. However, if theproblem gives data to produce a numeric solution, you must take care and get it right.
There always is an enormous temptation to substitute numbers before this point. Resistit! If the problem requires a numeric solution you now substitute but never until theabove steps have been completed.
First, do a rough calculation before you begin, without your calculator. It is quitelikely that this will give you an answer which is good enough for the occasion and mayeven be your final answer. Sometimes many factors cancel. With this answer you cancheck your solution to make sure that it is physically reasonable.
If required, use a calculator to produce a slightly more precise answer. What sometimeshappens here is that a completely different answer will appear. You should repeat bothcalculations until they agree. Students love to argue that the “rough answer” step isunnecessary. Far from it. It is the only check that you inserted numbers into yourcalculator correctly.
There are plenty of traps even when you are satisfied with your numerical answer.
• You must include the correct units with your answer.
• You should not express your answer to a greater precision than the least precise dataused. Your calculator will encourage a much more precise answer than is justified.
• Check to see that your answer makes physical sense. For instance, the depth of amineshaft should probably not be greater than the radius of the Earth.
3.2 Important considerations
Here are a few more points which are important in solving a physics problem. Manyof them you will find repeated elsewhere; the more times they are repeated the moreimportant they are!
ALWAYS START WITH FIRST PRINCIPLES
DON’T EVER JUST “FORMULA FIT”
Collecting parameters, finding a formula, and substituting them in is NOT problem solvingand teaches no physics at all. To understand a problem it is necessary to see each problemin its physical context and go back to first principles. This is what is meant by theDIAGNOSIS of the problem.
There are very few formulae that must be remembered. Most can be easily derived whenneeded.
3.2.1 Scalars and vectors
If it is possible to do a problem by either a vector or a scalar method then the scalarmethod is often less complicated. For example, many problems can be done using eitherforces or energies, and the energy method is often simpler.
20
3.2.2 Choice of origin
In simpler problems the choice of origin may be irrelevant but as complication grows theposition of the origin becomes more important. An absolute zero is sometimes necessary,such as when determining energies from temperatures. It is not necessary to have anythingbut an arbitrary zero when considering potential differences or energy differences.
3.2.3 Selection of axes
For most problems a set of 3D Cartesian axes is adequate, and selecting the direction ofthese axes carefully can help. Often, problems with spherical or cylindrical symmetry aremuch easier to solve using polar coordinates. It is always essential to specify the axesused.
3.2.4 Superposition
Often in physics relationships are linear, and so quantities involved in the relationshipsadd when two problems are superimposed on each other. Sometimes, an apparentlycomplex problem is just the superposition of two simpler problems which can be solvedseparately, giving the solution to the complicated problem as the sum of their solutions.This is not always true, but works in many situations and checking to see if there is aneasy decomposition of the problem is worthwhile. An example of this is drawing fieldlines. The pattern produced by two charges is the sum of the two patterns produced byeach charge considered separately.
3.2.5 Approximations
Almost no real physical problems have exact solutions. Many physical problems begin,and end, with the assumption that certain effects like friction, variation in length, variationin g, etc. can be ignored.
When solving a physics problem you have the same options available. You must use thedata available to find the best solution and realise that it will not be exact. If you cannotsolve the problem as it stands then first try to eliminate the effect of certain variables.Ignore friction, ignore heat flow etc. (provided that it is not a critical parameter!). Ifyou are then able to solve the problem, go back and seek a better solution without theapproximation.
Once you have arrived at a set of physical equations, the first step is to work out whatis valuable and what is not. In the equations, if one parameter is small, there are zerothorder terms and a number of first order terms which might be of the order of one percentof the zeroth order terms. Unless you cannot go further these should be retained. Theseare the “buggerall” terms. You might ignore them but you would prefer not to do so.Then you may find second order terms which consist of products of the first order terms.These are the “buggerall squared” terms and certainly can, and should, be eliminatedin most cases. This will make your algebra much simpler and you will lose very little(buggerall) in accuracy.
Often there is a simple algebraic method of making this approximation, particularly in
21
the form of the binomial expansion:
(1 + x)n = 1 + nx + higher order terms in x
The higher order terms in x are less important if x is small (much less than one). Thatis, if x is “buggerall”, then x squared is “buggerall squared” and may be neglected ifnecessary.
The equations that come out of real life problems are often difficult or impossible to solvewithout making some simplifying approximations. For example,
y2 = y − sin(y) + x
is impossible to solve analytically for y(x) . If, however, it were given that y is smallthen sin y could be approximated by y leaving y2 = x, a much simpler equation to solve.Remember that the accuracy of the answer is limited by the accuracy of the data. Someuseful approximations are given in Table 3.1.
Quantity Approximation Validity Notes
sin(x) x (radians) small x Error of 2% at x = 0.17
tan(x) x (radians) small x Error of 2% at x = 0.17
(1 + x)n 1 + nx nx ≪ 1 Error depends on n
ln(1 + x) x x ≪ 1 Error of 1% at x = 0.02Cx
C + x x x ≪ C Error of 1% at x = 0.01C
Quantity Value (3 sf) Approximation Error
π 3.14 22/7 (25/8) 0.04% (0.5%)
π2 9.87 10 1.3%
g 9.80 m s−2 10 m s−2 2%
seconds per year 3.15 × 107 π × 107 0.4%
Table 3.1: Useful approximations
3.2.6 Symmetry
Symmetry is one of the greatest tools of a physicist, and can reduce a seemingly impossibleproblem to a simple one. It can simplify equations, reduce the number of dimensions orcancel terms. With practice, you will learn to recognise symmetry more and more easily.
3.2.7 This is physics, not mathematics
To be able to do physics a knowledge of mathematics is necessary but not sufficient.Mathematics is the language and tool kit of physics, but it is not in itself physics. Justas it is possible to have an excellent set of tools, but be unable to dismantle an engine, itis likewise possible to have an excellent ability with mathematics but little ability to gainphysical insight into how the world works. Keep in mind that ultimately your answershould have physical meaning.
22
Chapter 4
Measurement
This section refers mainly to laboratory work and the significance of much of the detailmay not, at this stage, be apparent. Read through and absorb as much as you can butdon’t be distressed if the full relevance does not manifest itself until your first laboratoryexperiment.
A measurement of a physical quantity inherently involves some uncertainty; the best thatcan be done in any measurement is to place a constraint on the possible values of thequantity. This is usually done by including an estimate of the uncertainty or error with thenumerical result of the measurement. If, for example, you measure the length of an objectwith an ordinary ruler you might get a value of 16 mm; with a pair of callipers 16.3 mm;with a micrometer 16.28 mm; with a micrometer microscope 16.276 mm. Note thateach of these has an “error”; namely, ±1 mm, ±0.05 mm, ±0.005 mm and ±0.0005 mmrespectively. But none of them is “wrong”. Each is as good a measurement as could bemade with the measuring device available.
The results of measurements can be influenced in many ways including:
• instrumental errors — zero errors and calibration errors in instrumentation
• human errors — e.g. parallax errors in reading scales, reaction time in usingstopwatches
• extraneous influences — e.g. the effect of temperature changes during a measure-ment
• statistical fluctuations — e.g. radioactive decay, variations from the “norm” insamples of biological populations
• disturbance of the system by observation — e.g. immersing a cold thermometerin a hot liquid
4.1 Types of experimental errors
Experimental errors can usually be classified as systematic errors or random errors.
A systematic error influences all of a set of measurements in the same way. Systematicerrors may arise from the instrument or from the observer, or they may be due to someenvironmental condition. Such errors may be constant in magnitude or they may vary insome regular way. Repeated observations may not reveal them. Attention should be paidto this type of error when designing experiments. However, the exclusion of disturbingfactors which give rise to systematic errors is quite difficult. An example of a source ofsystematic error is a bad calibration of an instrument, which causes the instrument toread consistently high or low, either by some absolute or relative amount.
23
A random error is one which varies in magnitude and which is equally likely to bepositive or negative. Random errors arise from various sources, including actual variationsthat exist in the required physical magnitude. The occurrence of this type of error isusually revealed by repeated observations, which show a spread of results about the truevalue, the difference varying in both magnitude and sign. Random errors are alwayspresent in an experiment and, in the absence of systematic errors, they cause successivereadings to spread symmetrically about the true value of the quantity. This can berepresented diagrammatically as follows:
value of quantity
readings
true value
Figure 4.1: Random errors only
If, in addition, a systematic error is present, the readings will spread about some displacedvalue, i.e.:
value ofquantity
randomerrors
readings
true value average reading
errorsystematic
Figure 4.2: Both random and systematic error
The term precise indicates that random errors are small; the term accurate indicatesthat systematic errors are small.
4.2 Significant figures
Any measurement in physics has an uncertainty, or error, which must be stated with themeasurement. A number which is expressed as a set of digits without a specified error,for example a number which is written as 123.45 rather than 123.45 ± 0.02, implies thatthe last figure has an error as large as the limit of reading which is, without furtherinformation, assumed to be one half of the last digit. Thus 123.45 must, without anyfurther information, be read as 123.45± 0.005. This is because 123.45 means larger than123.445 (otherwise you would have written 123.44) and less than 123.455 (otherwise youwould have written 123.46). The error which follows the ± in this case is known as theabsolute error and has the same units as the principal measurement.
The number of meaningful digits in a number is called the number of significant figures.For instance the number 123.45 has five significant figures. The way in which digits arewritten usually makes the number of significant figures clear. Any trailing zero occurringto the right of a decimal point is considered significant; for example 123.00 is seen a
24
five significant figure number as the zeros would not have been included if they were notsignificant.
The possible error in 12300 is ambiguous since the last two trailing zeros may be significantor may purely be included to denote the correct magnitude. To avoid this ambiguityscientific notation is used and the number above would be written as 1.23×105, 1.230×105
or 1.2300 × 105 depending on whether 3, 4 or 5 figures were significant.
The number of significant figures is thus an indication of the precision of a number. Thenumber 123.45, a number of five significant figures, read as 123.45±0.005, has an precisionof 5 parts in 123450, say 5 parts in 100 000 or 5 parts in 105. 123, a number of 3 significantfigures, which is 123 ± 0.5, has an precision of 5 parts in 103. Note, however, that 999, anumber of 3 significant figures, 999± 0.5, has an precision of 5 parts in 104 while 1001, anumber of 4 significant figures, 1001 ± 0.5, has the same precision of 5 parts in 104. Theexpression 5 parts in 104 is known as the relative error which is a ratio and has no units.The relative error could be expressed as a fraction, 5/10000 or 1/2000, or as a percentage,0.05%.
The next important thing to note is that the answer to a calculation can never be moreprecise than the least precise piece of data. When quantities are added or subtracted, theabsolute errors add; when quantities are multiplied or divided, the relative errors add;errors are never subtracted.
A rule of thumb is that since no answer can be any more precise than the least precisepiece of input data, no answer may be specified with more significant figures than thepiece of data with the smallest number of significant figures. This is only a rule of thumb,and if the relative error is sufficiently small, an extra figure may be used.
4.3 Example: Significant figures vs percentage errors
The speed of a jet plane is to be measured over a distance of 2000.0 m (carefully measuredto five significant figures). A watch with half-second markings is used to measure the timethe plane took to travel this distance, and it is found to be 1.5 seconds. The average speedof the plane is
v = 2000/1.5 m s−1
= 1.3333 × 103 m s−1 .
The answer is mathematically correct but what of the precision of the result? The meaningof 1.5 in this case is that the time was less than 1.75 (otherwise you would have said 2.0)and more than 1.25 (otherwise you would have said 1.0). The average speed must thenbe greater than
v = 2000/1.75 m s−1
= 1.14 × 103 m s−1
and less than
v = 2000/1.25 m s−1
= 1.60 × 103m s−1 .
The best answer we can possibly give is that it lies somewhere between these two valuesso the average speed is given by
v = (1.3 ± 0.3) × 103 m s−1 .
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Note that 1.3 has the same number of significant figures, two, as the original 1.5. Buteven the number 1.3 by itself is incorrect because this implies greater than 1.25 and lessthan 1.35 while the answer really can lie between 1.14 and 1.60. This implies that evengiving an answer to the same number of significant figures might be more precise thancan be allowed.
Note that the 2000.0 is precise to five significant figures but this will have no effect onthe precision of the result. It may just mean that someone wasted their time measuringto this precision.
If a more precise stopwatch is used and the measured time is 1.52 seconds then
v = 2000/1.52 m s−1
= 1.32 × 103 m s−1 .
This time a more convenient method of determining and expressing errors is used. Sincethe time is given effectively as 1.52 ± 0.005, the time is only precise to 5 parts in 1520or 1 part in 300, which is slightly more than 0.3%. Percentage errors are not given moreprecisely than this. The error in the distance is 0.05 in 2000.0 which is 0.003% andnegligible.
The maximum error in the final result will thus be slightly more than 0.3%. 0.3% of 1.32is around 0.004 so the result is 1.32 ± 0.005 and should read
v = (1.316 ± 0.005) × 103 m s−1 .
Note that an extra significant figure has been given to match the precision implied by thecalculated error. The precision of the value should always match the error. Also noticehow this number is written to incorporate the error before the multiplier and the unit.This is much tidier than writing
v = 1.316 × 103 ms−1 ± 0.005 × 103 m s−1 .
If a number of equally reliable measurements are made of a quantity, then the best estimatewe can make of the true value of the quantity is the average of all of these measurements.Once an average has been determined, the error in this value can be regarded as thelargest difference between the measured values and the average. For example, for the setof measurements
10.7, 11.0, 9.9, 10.5, 10.1, 10.2 ,
the best value (the average) is 10.4 and the maximum error 11.0 − 10.4 = 0.6. Thus thequoted value should be 10.4 ± 0.6.
4.4 Error estimation
In many cases the magnitude of a physical quantity depends upon a number of quantitiesseparately determined. Each of these quantities will have an associated error and theerror in the calculated magnitude must be determined.
Consider two experimental quantities, A±∆A and B±∆B, and a resultant Z±∆Z where∆A, ∆B and ∆Z are the uncertainties or absolute errors in A, B and Z respectively.
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4.4.1 Addition and subtraction
When a quantity is equal to the sum or difference of two measured quantities, themaximum possible error in the resultant is equal to the sum of the absoluteerrors in the separate measurements,
∆Z = ∆A + ∆B .
Notice that the new error is always the sum of the original errors, even if the two principlevalues are subtracted. This can generate some very large errors from individually precisemeasurements.
For example, the mass of an object was determined before and after an experiment to be69.34 ± 0.01 kg and 69.86 ± 0.01 kg respectively. Each measurement is precise to 1 partin 700 or 0.15%. The change in mass is 0.52 ± 0.02 (since errors must always add) andthis is precise to 2 parts in 50 or 4%. This emphasises that when small differences arerequired, the measurements must initially be very precise.
4.4.2 Multiplication and Division
When multiplying the two quantities,
Z ± ∆Z = (A ± ∆A) (B ± ∆B)
= A
(
1 ±∆A
A
)
B
(
1 ±∆B
B
)
= AB
(
1 ±∆A
A±
∆B
B±
∆A∆B
AB
)
.
Since ∆A and ∆B are small (“buggerall” terms), ∆A∆BAB
can be ignored (“buggerallsquared”) so
Z ± ∆Z = AB
(
1 ±∆A
A±
∆B
B
)
,
and since Z ± ∆Z = Z(
1 ± ∆ZZ
)
∆Z
Z=
∆A
A+
∆B
B.
The maximum error always comes from adding the errors as was found with addition andsubtraction. So the relative error of a product is the sum of the relative errors of thecomponents. The same argument can be set up for a quotient. Hence when a quantity isequal to the product or quotient of two measured quantities, the maximum possiblerelative error in the result is equal to the sum of the relative errors in theseparate measurements.
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4.4.3 Raising to a power
If the principal measurement is raised to a power then the error can be found by consid-ering the exponentiation to be consecutive multiplication, and adding the relative errorsfor each multiplication. Hence if Z equals A raised to the nth power, An,
∆Z
Z= n
∆A
A;
the relative error in the result is the power times the relative error in theoriginal measured quantity. If a measurement is to be raised to a large power then itis important that this measurement must be made as precise as possible.
4.5 Example: Estimating error in resistivity
The resistivity ρ of a wire of length L, diameter D and resistance R is given by
ρ =RA
L=
RπD2
4L.
For a certain specimen of wire:
R = (15.5 ± 0.1) Ω relative error 1 part in 155 = 0.6%
D = (0.0050 ± 0.0002) m relative error 2 parts in 50 = 4%
L = (1.2345 ± 0.0002) m relative error 2 parts in 12000 < 0.1%
The relative error in ρ is thus 0.6%+2(4%)+0.1% = 8.7%. Note that numbers such as 4and π are known exactly and therefore do not contribute to the uncertainty in the result.
The numerical value of ρ is 2.46 × 10−4; 9% of 2.46 is 0.22 so the final answer is
ρ = (2.5 ± 0.2) × 10−4 Ω m .
Only one significant figure should be given for the error, or sometimes two if the first digitis 1 and the second digit is below 5.
4.6 Graphs and errors
Quite often it is possible to reduce errors by plotting a graph, such as the load-extensiongraph used in the determination of Young’s Modulus. Here, the effects of random errors ineither load or extension are minimised by drawing the line of best fit through the plottedpoints. In such cases, the errors in the quantities involved are represented by error bars:
The elimination of systematic errors is one of the best reasons to use graphs. If there isa systematic error in a plotted quantity, then it appears as an intercept, and does not
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Lo
ad
, y
Extension, x
Upper limitof load
Lower limitof extension
Lower limitof load
Upper limitof extension
+∆y
−∆y
−∆x +∆x
Error bars
Figure 4.3: Error bars
affect the gradient of the graph. If the gradient is used to obtain a physical parameter,the systematic error does not cause inaccuracy. If a single point were used, then thesystematic error would affect the result. Note that this is the most important reason thatone must NEVER force a line of best fit through the origin.
The best estimates of load and extension (the mean values) are used as co-ordinates toplot a point on the graph in the usual way. The errors involved are incorporated in thegraph as horizontal and vertical lines whose lengths are equal to the size of the errors (seeinsert in Figure 4.3).
The dotted line is the maximum slope which passes through the error flags, and thereforerepresents the likely upper limit of the slope. A similar line may be drawn to obtain thelower limit of error in the slope. The solid line is the line of best fit and represents themean slope, i.e. the mean value of load per unit extension in this case.
4.7 Problem Set: Measurement
1. If a power supply provides a DC voltage of 6.0± 0.2V , and the DC load is known tohave a resistance of 56Ω ± 2Ω, what is the uncertainty in the current?Hint: the relationship between current and voltage for a resistive load is
I =V
R
where V is the voltage, R the resistance, and I the current.
29
2. During their laboratory sessions, two students both complete an experiment designedto measure the constant γ in the equation A = γB2. Charlotte and Ferdinand obtainthe results in the accompanying tables.
(a) Plot graphs for both students on graph paper. Don’t forget to include errorbars.
(b) Calculate, using lines of best and worst fit, the constant γ and its uncertaintyfor both students’ results.
(c) The results obtained by Charlotte and Ferdinand differ both in value, and pre-cision. Which types of error are most likely to be the result of each difference?Given that the true value of the constant is close to 0.1 ms−2, what can beconcluded about the relative accuracy and precision of the students’ results?
A m(± 0.05 m) B2 s2(± 1 s2)
1 9
3 28
5 40
7 73
9 82
11 110
Charlotte Ferdinand
A m(± 0.05 m) B2 s2(± 1 s2)
2 21
4 44
6 67
8 89
10 112
12 135
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Chapter 5
Kinematics
Kinematics is the study of motion of bodies, without reference to the cause of the motion.Some people do not consider this physics, but just an application of simple mathematicalrelationships, although physical quantities are involved.
5.1 Linear Kinematics
We define:
• t as the time taken from a prescribed origin (it may be negative)
• s(t) as the displacement of an object from the prescribed origin at time t
• v(t) as the velocity of the object at time t
• a(t) as the acceleration of the object at time t
where s = xi+ yj+ zk, v = vxi+ vyj + vzk and a = axi + ayj + azk. The velocity v andacceleration a are given by
v =ds
dt=
dx
dti +
dy
dtj +
dz
dtk
and
a =dv
dt=
d2s
dt2=
d2x
dt2i +
d2y
dt2j +
d2z
dt2k .
It is important to note that acceleration, velocity and displacement in any direction areunaffected by those quantities in either of the other two directions. They depend uponeach other only through the time variable. Note that velocity is a vector. The magnitudeof the velocity is the speed.
5.1.1 Equations describing linear motion under constant acceleration
For one-dimensional motion and constant acceleration, we can generate five equations.
We shall set the origin at s(0) = 0 and call the initial velocity v(0) = u.
Start with a = dv/dt. Then v =∫
a dt = at + constant, and since we have set v(0) = u
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v = u + at . (5.1)
Now v = ds/dt , so s =∫
v dt =∫
(u + at) dt = ut + 1
2at2 + constant and if s = 0 when
t = 0 then
s = ut +1
2at2 . (5.2)
Substituting t from 5.1 into 5.2 gives
v2 = u2 + 2as . (5.3)
Exercise: derive two more equations relating the quantites mentioned here.
If a = 0, the problem is trivial. This case often appears as a transverse component inother problems. One example of an applied force causing a constant acceleration is whenay = g, the acceleration due to gravity. Since the directions in all problems with constanta can be defined by two vectors — an initial velocity u and an acceleration a, each ofthese problems can be done in two dimensions by appropriate selection of the x and yaxis. The acceleration is usually chosen along one of these axes, for instance in a fallingbody problem the y axis is usually chosen as vertical to give g a zero component in the xdirection. In other problems the acceleration would be chosen to be in the most convenientdirection.
5.2 Rotational Kinematics
In this section we shall consider the motion of a rigid body about a single unchangingaxis with a constant angular acceleration. A constant force gives a constant torque, whichgives a constant angular acceleration. This restriction means that the rotation is limitedto one plane, so vectors are not needed in this section.
Consider a general rotating object as in Fig. 5.1, rotating around O.
O
r
r
ds
P’
P
ωdθ
v
Figure 5.1: Rotating Body
32
By the definition of radian measure dθ =ds
r, so the distance PP ′ travelled by the point
P when the body rotates through dθ radians is given by ds = rdθ (where r is the lengthOP = OP ′). Note that rotating a rigid body by an angle dθ rotates all points on thebody by the same angle dθ about O.
Just as we defined velocity v = dsdt
, so we have angular velocity ω = dθdt
.
Angular velocity is usually given the symbol ω, the Greek letter omega, which is variouslypronounced ‘o-mi-ga’ or ‘oh-me-ga’ depending on your physics teacher, watchmaker orclergyman, but under no circumstances is it pronounced double-you.
The angular velocity may change with an angular acceleration, α = dωdt
.
Note that
• θ is measured in radians (a dimensionless unit).
• ω is measured in radians s−1 (dimensions: time−1).
• α is measured in radians s−2 (dimensions: time−2).
By strict analogy with one dimensional linear motion:
Since α =dω
dt, the angular velocity is given by ω =
∫
α dt = αt + ω0 .
Since ω =dθ
dt, the angle is given by θ =
∫
ω dt =
∫
(ω0 + αt) dt = ω0t +1
2αt2 .
We can derive the full set of equations (for an unchanging axis and constant angularacceleration) corresponding to those for linear motion:
ω = ω0 + αt [v = u + at]
θ = ω0t + 1
2αt2
[
s = ut + 1
2at2
]
ω2 = ω20
+ 2αθ [v2 = u2 + 2as]
Exercise: find the other two possible equations.
The similarity is evident, and these rotational relationships need not be memorised, justhow to change to the linear equations. In fact we can easily transform between rotationaland linear motion,
s =
∫
ds =
∫
r dθ = rθ , v = rω , a = rα .
Note that these equations only hold if r is constant.
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5.3 Rotational Quantities as Vectors
Just as it is useful to define vectors describing linear motion (e.g. v), it is also convenientto define vector quantities describing rotational motion.
Consider the case for the vector quantity ω. The only possible unique direction for theangular velocity vector is along the axis perpendicular to the plane of the diagram, as inFigure 5.1. However whether ω should point out of the page or into it remains ambiguous.
Figure 5.2: Right-Hand rule giving the direction of ω
To resolve this ambiguity we define a right-hand rule: curl the fingers of your right handin the sense of the motion and your thumb will point in the direction of ω (see Figure5.2). In the case of Figure 5.1, ω points out of the page. Note that for left-handed co-ordinate systems, the direction of ω (and the direction of most rotational quantities) willbe reversed.
Technically, ω isn’t a vector as its direction depends on the co-ordinate system used.Quantities like these that reverse direction when the handedness of the co-ordinate systemis changed are called ‘pseudo-vectors’, as otherwise they behave as vectors (ω1+ω2 = ω2+ω1, etc). Similarly it is useful to define a pseudo-vector angular acceleration α = dω/dtwith the direction of α being given by the right hand rule.
Note, however that the quantity θ is neither a vector nor a pseudo-vector as additiondoes not necessarily commute: ie θ1 + θ2 6= θ2 + θ1. An example of such a situation isillustrated in Figure 5.3. In (a) a 90 degree rotation is firstly made about the negative xaxis, followed by a 90 degree rotation about the negative y axis. The reverse is done in(b). Clearly the end results are not the same, ie θ1 + θ2 6= θ2 + θ1.
However dθ is a pseudo-vector. The reason for this is that as the magnitude of the θi’sare reduced, the difference between θ1 + θ2 and θ2 + θ1 reduces as well.
The advantage of defining ω and α is that relationships can then be obtained betweenthese rotational quantities and their linear counterparts. For example: v = ω × r, wherev is the velocity due to the rotation of a particle at r relative to a point on the axis ofrotation.
5.4 Motion in a circle
Consider a particle moving in a circle about a fixed point O.
34
x
y
z
(b)
(a)
Figure 5.3: Demonstration that θ1 + θ2 6= θ2 + θ1
a
a
r
Figure 5.4: General circular motion
At any given time the particle has a tangential acceleration a‖ (parallel to the directionof travel) due to the changes in the angular velocity about O (a = rα).
Now consider a solution for which α = 0 (ω constant): there is no longer any tangentialacceleration. Consider two points along the path of the motion, with associated velocitiesv, v′. Clearly although |v| = |v′|, v 6= v′. Thus there must be an acceleration, and wedraw a diagram (Figure 5.5) to find the direction and magnitude of this acceleration.
From the diagram,
a =v2
r.
The acceleration is in the direction of dv, towards the centre of the circle; it is thus “centreseeking” or “centripetal”. Note that v is perpendicular to a.
Since v = rω , we can write
a = ω2r ,
or a = −ω2r
to indicate that a is in the opposite direction to r.
35
dθ O
r vds
r
Note: |v’| = |v|
From the diagram:
dv ~ v.dθds = r.dθ
dv dθdt dt
a = = = =vr
vωv2
dθ
vv’
dv
~dv
Direction of a
Figure 5.5: Deriving the centripetal acceleration
Representing the angular velocity as a vector, the above equation can also be written as
a = ω × (ω × r) .
In the general case where there is also a tangential acceleration a‖, the net accelerationis the vector sum a⊥ + a‖.
5.5 Problem Set: Kinematics
1. A batter hits a ball at a height 1.2 m above ground so that its angle of projection is45 and it would hit the ground at a distance of 120 m from the batter. There is an8 m fence located 110 m from the batter. Will the ball clear the fence? If so, by howmuch?
2. An antitank gun is located on the edge of a plateau that is 60 m above the surroundingplain. The gun crew sights an enemy tank stationary on the plain at a horizontaldistance of 2.2 km from the gun. At the same moment, the tank crew sees the gunand starts to move directly away from it with an acceleration of 0.90 ms−2. If theantitank gun fires a shell with a muzzle speed of 240 ms−1 at an elevation angle of10 above the horizontal, how long should the gun crew wait before firing if they areto hit the tank?
3. A uniform disc rotates about a fixed axis starting from rest and accelerates withconstant angular acceleration. At one time it is rotating at 10 rev/s. After undergoing60 more complete revolutions its angular speed is 15 rev/s. Calculate:
(a) The angular acceleration,
(b) the time required to complete the 60 revolutions mentioned,
(c) the time required to attain the 10 rev/s angular speed, and,
(d) the number of revolutions from rest until the time the disc attained the 10 rev/sangular speed.
36
4. A pulsar is a rapidly rotating neutron star that emits radio pulses with precise syn-chronisation, there being one such pulse for each rotation of the star. The period Tof rotation is found by measuring the time between pulses. At present, the pulsar inthe central region of the Crab nebula has a period of rotation of T = 0.033s, and thisis observed to be increasing at the rate of 1.26 × 10−5 s/y.
(a) Show that the star’s angular velocity is related to the period of rotation by
ω =2π
T
(b) What is the value of the angular acceleration in rad s−2?
(c) If its angular acceleration is constant, when will the pulsar stop rotating?
(d) The pulsar is the remnant of a super-nova explosion in the year 1054 A.D. Whatwas the period of rotation of the pulsar when it was born? (Assume constantangular acceleration).
5. An astronaut is being tested in a centrifuge. The centrifuge has a radius of 10 mand, in starting, rotates according to θ = 0.3t2, where t in seconds gives θ in radians.When t = 5.0 s, what are the astronaut’s
(a) angular velocity,
(b) tangential velocity,
(c) tangential acceleration,
(d) radial acceleration, and
(e) total acceleration?
6. Four pulleys (A, B, B′, C) are connected by two belts (a, b). Pulley A (radius 15 cm)is the drive pulley and it rotates at 10 rad/s. Pulley B (radius 10 cm) is connectedby belt a to pulley A. Pulley B′ (radius 5 cm) is concentric with pulley B and isrigidly attached to it. Pulley C (radius 25 cm) is connected by belt b to pulley B′.Calculate
(a) the linear speed of a point on belt a,
(b) the angular speed of pulley B,
(c) the angular speed of pulley B′,
(d) the linear speed of a point on belt b, and
(e) the angular speed of pulley C.
7. Consider an object on the surface of the Earth:
(a) What is the centripetal acceleration required to maintain the rotation of anobject on the surface of the earth if the object is:
(i) At the equator?
(ii) 21S of the equator?
(b) What would the period of rotation of the earth have to be in order that theobject have a centripetal acceleration equal to 9.8 m s−2 if the object is:
(i) At the equator?
(ii) 21S of the equator?
(c) Why doesn’t part (b) make sense if the object is at one of the poles?
37
38
Chapter 6
Energy
Energy is a scalar quantity which is conserved under all physical interactions. The precisenature of energy is mostly a matter of definition and will not be discussed here.
6.1 Potential and kinetic energy
Energy can exist in numerous forms. Two are
• kinetic energy, often designated by the letter K, which is the energy possessed byan object by virtue of its motion relative to other objects, and
• potential energy, sometimes assigned the letter U , which is the energy due to theposition of an object relative to other objects or the configuration of a system.
Although a physical system may possess many different forms of energy, when solving aproblem about an interaction involving energy transfer, only the energy forms which areaffected by the interaction need be considered.
Other common manifestations of energy are included in these two forms. Some of theseforms are listed below.
• Mechanical energy of an object or a macroscopic system of particles is the energywhich can be measured macroscopically. The total mechanical enrgy is the sum ofthe macroscopic kinetic and potential energies.
• Chemical energy is the potential energy of the electrostatic bonds between atoms orions.
• Elastic energy is the potential energy arising from deformation of these bonds
• Thermal energy in gases is the kinetic energy of the molecules.
• Thermal energy in solids is energy due to the vibration of molecules and movementof electrons. There is a constant interchange of energy between the kinetic energy ofthe particles and the electrostatic potential energy of the bonds between them.
6.2 Internal energy
A system may contain stored energy which is not immediately visible; it is not present aseither macroscopic kinetic energy or macroscopic potential energy. This is known as theinternal energy of an object.
39
All objects contain internal energy, due in part to the motion of molecules within them.Under certain conditions this internal energy can interact with the other energies of thesystem; this is normally associated with a temperature gradient.
Mass is also a form of internal energy. Unless relativistic effects are important, which isthe case when particles have extreme velocities or during nuclear reactions and similarprocesses, the mass energy of a system need not be considered when solving problemsusing energy conservation.
6.3 Law of conservation of energy
For any isolated system (where energy is not allowed to move between the system andthe surroundings), the total energy of the system remains constant. This is often calledthe law of conservation of energy:
energy of the system before interaction = energy of the system after interaction ,
where the interaction is some rearrangement of the system which does not involve energybeing given to or taken from the surroundings.
If, however, energy is lost to the surroundings, the energy balance equation becomes
energy of the system before interaction = energy of the system after interaction
+ energy lost to the surroundings .
The energy loss may be negative indicating energy gained from the surroundings.
The energy of the system might be in the form of macroscopic kinetic or potential energy,but it may also include internal energy.
6.4 Potential energy
Potential energy is the energy of a system due to its configuration, and its position relativeto the surroundings. In general it is the relative positions of charges or masses that giverise to potential energy.
The word ‘potential’ in this instance is related to the Latin ‘potens’ meaning ability orpower. As this suggests, potential energy is energy associated with a system which enablesit to do work.
Placing one object with mass adjacent to another object with mass will, in general, involvedoing work; this work is stored in the system as potential energy. A similar argumentapplies to placing one object with a charge adjacent to another charged object. This workmay be positive or negative; it may be done on the object or by the object. Thus, anobject on the Earth’s surface moved away from the Earth will have work done on it andits potential energy will increase. A negatively charged object moved further away fromanother negatively charged object will do work on the surroundings and thus its potentialenergy will decrease. The potential energy stored in a spring is electrostatic. However, itis not practical to calculate the total potential energy from the relative positions of all theatoms in the spring. In such cases an empirical macroscopic description of the potentialenergy is more appropriate.
40
Personal experience indicates that objects will spontaneously decrease their potentialenergy in the absence of any restraining force. The everyday expression for this is thatan object will “fall” unless restrained.
6.5 Potential energy diagrams
It is often useful to plot potential energy versus position, in one or more dimensions. Hereonly one dimension is considered.
A simple way to interpret one of these plots is to imagine the line as the height of theground and to place an object on it. The object slides downhill to a point of lowerpotential energy.
Figure 6.1(a) shows a potential well, a potential field that attains a minimum. An hypo-thetical object would move (along the x-axis) to the point of minimum potential energy,where the particle is in equilibrium since the slope of the U vs x graph is zero. This posi-tion is one of stable equilibrium since if the object is displaced a little from equilibrium,it will return to the equilibrium position.
U
x
U
x
Figure 6.1: (a) Potential well, (b) Potential hill
Figure 6.1(b) shows a potential hill, a plot which attains a maximum. Again, an objectwould move in the direction of decreasing potential energy, which could be to either sideof the hill. If, however, the particle happens to be on top of the hill where the slope is zerothe potential energy will not decrease spontaneously as the particle is in equilibrium. If itis slightly perturbed either way, the slope is not zero and the particle’s potential energydecreases; that is the top of the hill is a position of unstable equilibrium.
Figure 6.2 (a) shows a point of inflection where the slope is zero. Again the situationis unstable because any displacement to the right will result in the particle’s potentialenergy decreasing and x increasing spontaneously.
In summary a system is in equilibrium at a point if the potential energy of the systemdoes not change with position at that point, i.e. the slope of the U vs x curve
(
dUdx
)
iszero at that point.
An equilibrium point is stable if the potential energy there is less than that at all adjacentpoints, and unstable if the potential energy there is greater than that at an adjacentpoint.
41
U
x
U
x
Figure 6.2: (a) Point of inflection, (b) Position of equilibrium
If no dissipative forces (like friction) are acting, mechanical energy will be conserved, soif a particle is perturbed from stable equilibrium, it will oscillate in the potential well.Like a swing at the bottom of its arc, it has maximum kinetic energy as it passes throughthe equilibrium point, converting this to potential energy as it climbs the well. When allkinetic energy is converted to potential, the particle stops, and begins to fall back downthe well. The motion is hence periodic. If instead energy is dissipated by friction, theamplitude of the oscillation will decrease and the particle will eventually come to rest atequilibrium.
Figure 6.2 shows two equilibria, one stable, the other unstable. For sufficiently smallperturbations about the stable equilibrium, a particle will remain in the potential well. If,however, the perturbation is large enough to push the particle over the adjacent potentialhill, it will escape the well and continue moving to the right. This is an example of anenergy barrier, where the particle is confined unless it can overcome an energy threshold.
Note that potential energy can be discrete rather than continuous in some cases, partic-ularly in quantum mechanics.
6.6 The zero of potential energy
Consider two masses. As they are brought closer together the gravitational potentialenergy of each decreases. If the zero of potential energy is chosen to be that of one masswhen the masses have any finite separation, this separation will appear as a constant inthe expressions for the potential energy. The limiting value of the potential energy as theseparation becomes large with this choice of zero is finite and positive, and the limitingvalue as the separation goes to zero is negative infinity. So it is common to choose thezero for potential energy to be that when the masses have infinite separation. This choiceof zero is independent of any other body in the universe. The same argument appliesto electrostatic interactions. Note that in a laboratory experiment, a few metres from acharge is a good approximation to infinity.
42
6.7 Potential energy as a function of separation
Consider a mass very far from the Earth which is moving very slowly, so its kinetic energyis negligible. Since we choose the zero of potential energy to be infinitely far from theEarth, its potential energy is also negligible. This mass is attracted towards the Earth.As the separation between the mass and the Earth decreases its potential energy alsodecreases. Since energy is conserved, its kinetic energy must increase so that the totalenergy is constant. In this case the initial total energy is zero, so
K + U = 0 , and so K = −U .
Figure 6.3 (a) shows a plot of potential energy as a function of distance from the centreof the Earth. Note that the mass will stop and reach its minimum potential energy whenthe mass is a distance r0, the radius of the Earth, from the Earth’s centre, as it collideswith the surface of the Earth.
U
r
(Attraction) U
r
(Repulsion)
r0
r0
Figure 6.3: (a) Potential energy of attraction, (b) Potential energy of repulsion
For the electrostatic problem, the plot of potential energy against position will be identicalfor opposite charges, since the field is attractive. For like charges there is a repulsion andwork must be done on the system to make the charges come together. The potentialenergy plot in this case is shown in Figure 6.3(b).
6.8 Algebraic expressions
The formula for the gravitational potential energy of one mass due to another must
• be zero when the distance between the two masses is infinite,
• be symmetric in the two masses since interchanging the masses does not change thepotential energy,
• be linear in the masses, and
• give negative values for all finite distances between the two masses and increase withincreasing r.
The formula below for gravitational potential energy was first given by Newton. r is thedistance between the two masses, M and m. The dependence on r was determined byNewton using observations made by Kepler and others.
43
U = −GMm
r(6.1)
The gravitational constant G = 6.67 × 10−11 J m kg−2 (N m2 kg−2) is determined empiri-cally.
The formula for the electrostatic potential energy of a charge Q separated by a distancer from a second charge q is
U =1
4πε0
r. (6.2)
Note the similarities to equation 6.1.
The constant ε0 = 8.85× 10−12 F m−1 is determined through the relationship c2µ0ε0 = 1,where c is the speed of light in a vacuum and µ0 = 4π×10−7 H m−1 is set as a fundamentalS.I. value. Note that this implies that 1/4πε0 = c2µ0/4π, a quick way to approximatelyevaluate this common value.
6.9 Potential
Often physical problems consider the interaction (one at a time) of a variety of masses(or charges) with one system. The potential energy depends on the mass (or charge) ofboth objects. In general these objects may be any collection of masses (charges). Thepotential energy per unit mass and the potential energy per unit charge both dependonly on the properties of the system and so are useful in such cases. These quantitiesare known as the gravitational potential, measured in J kg−1, and the electrostaticpotential (or electric potential or just potential) which is measured in J C−1. This latterunit has a special name, volts (V), and the electric potential is commonly called voltage.The symbols V and Φ are often used for potential. The expression for the gravitationalpotential is
V = −GM
r,
and that for electrostatic potential is
V =1
4πε0
Q
r.
6.10 Potentials in two dimensions
In this section the potential as a function of position in two dimensions is considered.
44
Figure 6.4: Two-dimensional potential hill
x
y
Figure 6.5: An equipotential diagram
Consider a pair of Cartesian axes x and y. Let there be a potential field of the formV (x, y). A plot of V (x, y) by against both x and y for a potential hill is shown in Figure6.4.
As in one dimension, an equilibrium point is one where the slope of V (x, y) is zero inall directions. In two dimensions if at some point the slope is zero in two perpendiculardirections, e.g. the x and y directions, then it is zero in all directions and that point is apoint of equilibrium. The slope in the x direction at the point (x0, y0) is
dV (x, y0)
dx, i.e.
∂V
∂x,
evaluated at x = x0. The definitions of stable and unstable equilibrium are the same asin one dimension.
Considering all points (x, y) at which V (x, y) = V0 is equivalent to taking the curve ofintersection between V (x, y) and the horizontal plane V = V0. This curve is called anequipotential, since all points on it have the same potential V0. In all cases, equipotentialcurves are continuous, closed (have no beginning or end) and do not cross. Twodifferent equipotentials cannot intersect as this would imply that a single point was attwo different potentials. It is, however, possible to have a loop which intersects itself, e.g.a figure-eight, or two or more completely disconnected closed loops. This latter case couldbe produced by two hills side by side.
A diagram showing equipotentials for a range of values of V is an equipotential diagram.Figure 6.5 shows an equipotential diagram for the potential in figure 6.4. A topographiccontour map is an equipotential diagram, as it has lines which link points of equal heightand gravitational potential.
Choose any point A on an equipotential and place a mass there. Move along that equipo-tential to a point B. From B, move to another equipotential, and call this point C. Thisrequires that work be done on (by) the mass, as its potential energy is raised (lowered).Now move it along the second equipotential to another point D, a process in which noenergy change occurs because CD is an equipotential. Now move the mass back to the
45
original equipotential at point A. In this last step exactly the same amount of work is doneby (on) the mass as was done on (by) it in step BC, because the difference in potentialenergy between A and D is exactly the same as between B and C. The mass has thusmoved around the circuit ABCDA with no net consumption of energy and has returnedto the same point with the same potential energy as before. Any field for which the totalwork done in moving any particle along any path back to its starting position is zero is aconservative field. Both gravitational and electrostatic fields are conservative fields.
6.11 Potentials in three dimensions
Although it is not possible to graph a function of three variables, such as a potentialin three dimensions, it is possible to draw equipotential surfaces, which are the three-dimensional equivalent of equipotential curves.
By considering the symmetries of a point mass (or charge) the shape of the equipotentialabout it may be determined. If the mass or charge is at a single point, then no specific di-rection is any different from any other. The equipotentials, therefore, must be completelysymmetric under rotations in all directions about the point in question. The only shapefor which this is the case is the sphere, and so the equipotentials are spherical surfaces.
The same result is obtained from the formula for V given earlier. Since V = kr
for someconstant k, V constant implies that r is constant. Hence the equipotentials are surfacesof constant r, i.e. spheres. Since V is proportional to the reciprocal of r, the distancebetween spheres increases with r for equal increments of V with the sphere of V = 0 beingat infinity. For an attractive force the values of V will increase with r from −∞ to zerowhile for a repulsive force they will decrease with r from ∞ to zero.
A special case of this is the gravitational field at the surface of the Earth. Because heightsat the surface of the Earth are very small compared with the radius of the Earth (metrescompared with thousands of kilometres) the equipotential spheres are seen as a series ofparallel planes and the incremental distances for equal changes in V will be the same (theyare not exactly the same but we do not normally observe the tiny difference). A personon the surface of the Earth sees the gravitational field as a series of equally spaced parallelhorizontal equipotential planes and concludes that the gravitational force (see later) isboth uniform and vertical.
46
Chapter 7
Forces
Forces are agents of motion, that is, an object will be accelerated if acted upon by a singleforce. There are four physical origins of forces according to our current understanding:
• The gravitational force, which holds together aggregations of matter.
• The electromagnetic force, which gives rise to electric and magnetic fields and theproduction of photons (electromagnetic radiation).
• The strong nuclear force, which holds together quarks and hence the particles inthe nucleus.
• The weak nuclear force, which involves quarks and leptons (e.g. electrons).
Here discussion will be limited to the gravitational force and the electromagnetic force.Note that throughout this section relativistic effects are ignored.
7.1 Agents of motion
U
x
Figure 7.1: Potential hill
Refer again to the diagram of the one dimensional potential hill, repeated as Figure 7.1.Recall that a particle at a position where the gradient is zero is in equilibrium, and inother places the particle would move so as to reduce its potential energy. The agent forthis is called a force. A force is a vector and acts on a particle, object or system.
Consider Figure 7.1. If the slope of the line is negative (it slopes down to the right), theforce on the object is to the right and therefore positive. If the slope of the line is positive(it slopes down to the left), the force on the object is to the left and therefore negative.
47
7.2 Newton’s First Law
The above discussion implies that if an object is at rest at an equilibrium point it willremain at rest as the force is zero at this point. To change the state of motion of anobject, for example start an object at rest moving, a force is required. A force is alsorequired to change the velocity of an object. This leads us to Newton’s first law:
“Every body at rest will remain at rest and every body moving in a straight line will
continue to move in that straight line unless acted upon by a force.”
Although this statement may seem obvious, it was not recognized until the 17th century.
7.3 Forces in two or more dimensions
The same conclusions can be drawn from a two dimensional potential hill, such as theone in Figure 6.4. Forces can have any direction on the x − y plane consistent with thepotential energy diagram. For a potential hill, the force at points close to the top isdirected away from the top of the hill which is an unstable equilibrium point.
It is observed that particles spontaneously decrease their potential energy if possibleby moving to an adjacent point with lower potential. So at any point, the force mustbe in the direction of the greatest potential energy decrease. This direction must beperpendicular to the equipotential passing through the point, because moving parallel tothe equipotential does not change the energy, and so adding a component of motion inthis direction cannot result in a bigger decrease of potential energy. This means that theforce has no component along the equipotential.
Moving an infinitesimal amount in the direction of the force at each point traces out acurve always perpendicular to the equipotentials. This is called a line of force, and thetangent to the curve at any point is in the direction of the force at that point. Since acurve can only have one tangent at any point and there is only one direction of a force atany point, it follows that lines of force cannot intersect except at a point of equilibrium.
O
y
x
Figure 7.2: Lines of force
Lines of force have the following properties:
48
• Lines of force are always at right angles to equipotentials.
• Lines of force cannot intersect except at a point of equilibrium.
• Lines of force may either be of infinite extent or start or finish at an equilibriumpoint.
• The tangent to a line of force gives the direction of the force at that point.
Using this picture of lines of force a picture of a force field can be built up, as shown inFigure 7.3. Shown on the figure is the field about an irregular potential hill. The closedlines are the equipotentials and the lines with directions (they are vectors) are the linesof force.
O
y
x
Figure 7.3: Equipotentials and lines of force
7.4 Algebraic definition
Matching the properties described above, a one-dimensional force is defined by
F = −dU
dx,
i.e. the force is the derivative of the potential energy. If the energy is in joules then theforce is in joules per metre, or newtons (N).
Refer once more to the potential energy diagram of the two dimensional hill, Figure 7.2.The direction of the force at any point is perpendicular to the equipotential at that point.In the x − y plane this force can be decomposed into two components, one in the xdirection and one in the y direction. The force vector F at any arbitrary point can beresolved into these two components. A graphical representation of this is in Figure 7.4.
The force in each of these directions is again the derivative of the potential energy, butconsidering only the direction of interest to be changing. In other words, the componentsof the force in the x and y directions are the partial derivatives of the potential energywith respect to x and y,
49
O
y
x
F
x force component
y forcecomponent
Figure 7.4: Components of the force vector
Fx = −∂U
∂x,
Fy = −∂U
∂y.
The force is the vector sum of these components given the appropriate directions.
Extension to the z direction gives
Fz = −∂U
∂z.
Thus the total force vector is
F = −∂U
∂xi −
∂U
∂yj −
∂U
∂zk ,
where i, j and k are the unit Cartesian vectors. We could rewrite this as
F = −
(
∂
∂xi +
∂
∂yj +
∂
∂zk
)
U
= −∇U .
This equation shows the definition of the gradient operator ∇. More importantly, itstates that the force is the gradient of the potential energy, whether in one, two or threedimensions.
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7.5 Fields
The lines of force discussed in the last section depend on the mass or charge of thebody placed in the field. As for potential, it is useful to know what force an object willexperience due to an existing set of other objects. The field strength is the force perunit mass, or force per unit charge.
The force of gravity on a particle is proportional to m, the mass of the particle undergravitational attraction. So for gravitational fields the field strength is the force per unitmass, which is given the symbol g and is measured in newtons per kilogram. The forceis then F = mg and g does not depend on the mass of the object you place in thegravitational field.
In an electric field the force is proportional to q, the charge of the particle placed inthe field. The electrostatic field strength is thus F/q and is given the symbol E. E hasunits of newtons per coulomb (N C−1). Since a newton is a joule per metre, the unit ofelectric field strength is joule per metre per coulomb, which is also volt per metre. Thuselectrostatic field strength can be measured in either newton per coulomb or volt permetre; they are identical units. This is useful when different information is wanted - tofind the force on a particle it is most convenient to use the former, for the field strengthdue to a known potential difference it is more convenient to use the latter.
7.6 Gravitational and electrostatic fields
The simplest gravitational field geometry is that due to a uniform spherical mass. Sincethe field must be spherically symmetric the equipotentials consist, as discussed before, ofa series of concentric spheres becoming further apart with the sphere of V = 0 at infinity.The equipotentials get closer together on approaching the surface of the mass. The valueof the potential at each successive equipotential decreases, i.e. becomes more negative, asthe potential is attractive.
Field lines are always at right angles to equipotentials, so in this case they point radiallyinward, corresponding to a decrease in potential with a decrease in r. Lines of force startat infinity (which has V = 0 everywhere and is therefore in equilibrium, but unstableequilibrium) and finish on the surface of the sphere. The surface of the sphere is anequilibrium surface, because the balancing forces make any further approach to the centreof the sphere impossible.
Note that this is a conservative field.
Mathematically the potential is
U = −GMm
r.
Note that the situation is spherically symmetric, and so equipotentials have constantvalues of r, and U is a function of r alone. Spherical polar coordinates are hence the mostconvenient here.
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Fr = −∂U
∂r
= −∂
∂r
(
−GMm
r
)
= GMm∂
∂r
(
1
r
)
= −GMm
r2.
This is the only component of the force, and is in the direction of r (the unit vector alongthe direction of the increasing axis of r). So the force is
F = −GMm
r2r .
This is Newton’s Law of Gravitation.
This defines the gravitational field strength,
g =F
m= −
GM
r2r .
It is possible to obtain the potential energy from the force by integrating.
Electrostatic fields behave in the same way as gravitational fields, except that the potentialof a charge may be positive or negative. The formula for the electrostatic field of a pointcharge is derived in the same way as above. The formulae are summarised below.
Potential Energy Force Potential Field Strength
Gravitational U = −GMm
rF = −
GMm
r2r V = −
GM
rg = −
GM
r2r
Electrostatic U =1
4πε0
rF =
1
4πε0
r2r V =
1
4πε0
Q
rE =
1
4πε0
Q
r2r
7.7 Reaction forces
In the table it is given that the force on a charge q due to charge Q is given by
F =1
4πε0
r2r ,
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where r is the distance from Q to q and r is the radial unit vector measured away fromQ. This is Coulomb’s law. This expression is symmetric in Q and q, which means thatif Q and q are interchanged the formula remains the same. Thus if we consider the forceon Q, we obtain exactly the same expression. The only difference is that r is now theradial unit vector measured away from q and is in the opposite direction.
So the force on q due to Q is exactly equal to and in the opposite direction to the forceon Q due to q. Note that these two equal and opposite forces are acting on differentbodies. This is an example of an action-reaction pair in the sense of Newton’s ThirdLaw which states that
“For every action there is an equal and opposite reaction.”
Whenever one body applies a force to another, the second body applies a force equal inmagnitude but opposite in direction to the first. The important thing to remember isthat action and reaction apply to two different bodies. This law can also be written asFAB = −FBA. The force on A due to B is equal in magnitude and opposite in directionto the force on B due to A.
7.8 Contact forces
Contact forces arise due to the interaction of surface atoms on two surfaces close together.They occur either in response to other forces, or to relative motion of the surfaces.
If two objects come very close together, and a force attempts to push them into eachother, then the surfaces of the objects interact to oppose this. If an object is placed on atable, say, then the gravitational force on the object would accelerate it towards the table.The fact that this does not happen means that the table exerts a force on the object,opposing gravity. This is a contact force.
If there is a component of a force pushing an object directly against another, i.e. a forcein the normal direction, then the surfaces exert a force against this. The component offorce in the normal direction exerted by a surface is called the normal force. This is theforce that keeps an object on a table.
If there is a component of a force on an object parallel to the surface, rather than normalto it, then if the surfaces are very smooth, or not pressed together very much, the surfacesare not forced to interact. If the surfaces are rougher, or there is more force pushing themtogether, then there is more adhesion between the surfaces and more resistance to themotion. This force is the force of static friction.
Both the normal force and static friction take the value necessary to balance an appliedforce. However, if the applied force is sufficiently large, the contact forces cannot supplyenough force and the object will move. In the case of the normal force, this means thatthe object is permanently deformed or broken, e.g. sitting on a very flimsy chair.
Static friction is described by the coefficient of static friction, µs, and obeys
Ff ≤ µsN ,
where N is the normal force pushing the objects together. Static friction will oppose anyforce pushing an object along the surface, up to its maximum value. After this point theobject will start to slide.
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Once an object starts sliding, a different contact force acts. The surfaces are still in closecontact, and momentarily adhere to one another as the object slides. These adhesionscause a resistance to the relative motion of the surfaces, called kinetic friction.
It is important to recognise that kinetic friction always opposes relative motion of two
surfaces. As such, it is possible for kinetic friction to act either in the direction of thevelocity or opposite to it, if both objects are considered to be moving. Kinetic frictiondoes not vary with applied force, and is described by
Ff = µkN ,
where µk is the coefficient of kinetic friction. For a given pair of materials, µk is generallyless than µs.
7.9 Free body diagrams
It is important to know how to use diagrams to solve force problems.
For example, consider a mass m being attracted by the Earth, mass M . Each is attractedto the other by gravitational forces; by Newton’s Third Law these are equal and opposite.These forces can be depicted as in Figure 7.5.
mass M
FMm
mass m
FmM
Figure 7.5: Gravitational attraction
Considering the problem from the Earth’s perspective, the mass m is falling towards theEarth which, because of its size, is relatively unperturbed (although it does fall toward theobject to some extent). Since the PE is proportional to r−1 and negative, the potentialenergy of the mass m will become very large and negative as it approaches the surface ofthe Earth, its target.
Let the two bodies now touch one another. A contact force (normal force) is exerted byboth m and M , each acting on the other. These contact forces must, by the third law,be equal and opposite. This situation is depicted in Figure 7.6, where the contact forcesare shown as thin lines. By convention, the gravitational forces are shown as acting atthe centre of mass (thick lines) and the contact forces at the point of contact. Since theseforces are collinear and acting on solid objects, the difference is of no consequence in thiscase.
This depicts a system of two particles with 4 forces, which may be considered as eithertwo pairs of third law forces, or one pair of forces acting on one object and one pair offorces acting on the other. If, however, the object is much smaller than the Earth, thediagram becomes Figure 7.7, and the surface of the Earth is an equipotential plane.
To analyse the motion, it is convenient to consider one object only as the system inquestion, existing in the gravitational field of the other. The Earth still exists, but only
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massm
massM
Figure 7.6: Contact force
mass m
Earth, mass M
Figure 7.7: Gravitation force at the surface
to provide the geometry and the gravitational field. In this case the situation is depictedby a free body diagram, as shown in Figure 7.8.
Free body diagrams are essential when solving force problems as they depict the forceson a system due to its surroundings, including any fields present.
Figure 7.8 shows two forces, the gravitational force, or weight, W = mg, and the normalforce N, acting on a mass m. Since velocity of the system is not changing the net force iszero, by Newton’s second law (see section 7.13). Note that forces are vectors and must bedrawn as such. Note that it is NOT the third law which causes these forces to be equaland opposite; the third law refers to forces acting on different bodies.
The situation becomes more complicated if the surface of the Earth is not horizontalbut inclined at θ above the horizontal as shown in Figure 7.9. The dotted lines showthe horizontal equipotentials. Remember that the normal force acts at right angles tothe plane. The mass m, from experience, rolls or slides down the incline with increasingvelocity, so there must be a net force.
To calculate the force, choose a set of Cartesian axes x and y parallel to the plane and itsnormal respectively. The weight W is resolved into components along the x and y axes,W cos θ and W sin θ respectively, as in Figure 7.10. Since the mass does not accelerate inthe y direction the net force in the y direction is zero, i.e. N = W cos θ, and the resultantforce has magnitude W sin θ and is directed down the plane.
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mass m
Weight
Normal
Figure 7.8: Free body diagram
θ
Weight, W=mg
Normal, N
mass m
x
y
Figure 7.9: Inclined plane problem
7.10 Tension
In the above discussion two bodies have been kept apart by contact forces. Tension arisesin a situation in which two bodies are attempting to separate but are constrained.
Consider an object of mass m hanging from a rope. The other end of the rope is fixed toa rigid support, as in Figure 7.11. Assume the rope to be massless and inextensible.
Consider first the free body diagram for the mass, Figure 7.12 (a). There are two forcesacting — the weight, W, and the tension, T1. Since the mass is in equilibrium these mustbe equal, by Newton’s second law.
The free-body diagram for the rope, Figure 7.12 (b), is similar. The force T2 is caused bythe mass pulling down on the rope, and the force T3 is caused by the support pulling upon the rope. Since the rope is not moving these forces must be equal and opposite. Wherethe rope is in contact with the mass, the mass and the rope exert equal and opposite forceson one another by Newton’s third law, and hence T1 and T2 are equal. These forces arecalled tension.
A massless, frictionless pulley can be used to change the direction but not the magnitudeof the tension in a rope.
Note that the tension in the rope is not constant if the rope has mass, is extensible or isconnected to an object of non-negligible mass at any other point.
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θ
y component of weight= mg cos θ
x component of weight= mg sin θ
Normal, Nmass m x
y
Figure 7.10: Resolution of weight force
mass m
rope
Figure 7.11: The result of tension
7.11 Gravitation at the surface of the Earth
It was stated earlier that the gravitational potential energy is proportional to the inverseof the separation, which does not at first sight appear consistent with the expressionP.E. = mgh which you may have encountered.
Recall that the gravitational force on any object is the product of its mass and the field
strength g, where g is equal toGM
R2(where, for g at the Earth’s surface, R is the radius
of the Earth). Since variations in height above the Earth’s surface are small comparedwith R, this value of g is approximately constant for any point on the Earth’s surface.This means that the gravitational force on any object, or its weight, is simply given bymg and directed downwards, i.e. towards the surface of the Earth, where g is roughlyconstant at 9.80 N kg−1.
To determine the potential energy difference consider a mass moving from a height hA
to a height hB above the surface of the Earth, where hA, hB ≪ R. The potential energy
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tension, T1 tension, T3
tension, T2
rope
weight, W
mass m
Figure 7.12: (a) Free body diagram for mass, (b) Free body diagram for rope
difference is
UB − UA =
(
−GMm
R + hB
)
−
(
−GMm
R + hA
)
UB − UA = GMm
(
−(R + hA) + (R + hB)
(R + hA)(R + hB)
)
≃GMm(hB − hA)
R2since hA, hB << R
= mg(hB − hA) .
Note that if the approximation had been made in the first line the answer would havebeen zero, because this difference in potential energy is very small compared with thetotal potential energy. This is a good example of the need to exercise care when makingapproximations.
7.12 Hooke’s Law
Hooke’s law is an empirical relationship between the force applied to a material and theextension produced. This theory applies to the extension by force of most (but certainlynot all) materials, provided that the force is below what is called the elastic limit of thatmaterial.
Take some material and try to stretch or extend the object along some line, like a spring.Hooke’s law states that the restoring force is proportional to the extension. Since therestoring force F is in a direction opposite to that of the extension, (see figure 7.13),
F = −kx .
The potential energy, assuming that the force is in the x direction and that k is constant,
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x
F
equilibrium
Figure 7.13: The restoring force of a spring
is
U = −
∫
F dx
= −
∫
−kx dx
=1
2kx2 .
The constant k is the force required to produce unit extension and has units of newtonsper metre. If k is constant no mechanical energy is lost in the process, i.e. perfect elasticforces are conservative forces.
7.13 Momentum and Newton’s Second Law
Momentum, defined byp = mv , (7.1)
is proportional to both mass and velocity; a large mass with a small velocity can havesimilar momentum to a small mass with a large velocity. The unit for momentum iskg m s−1. Newton’s Second Law of motion defines force as the time rate of change ofmomentum (p),
F =dp
dt. (7.2)
Substituting equation 7.1 into equation 7.2,
F =d
dt(mv)
= vdm
dt+ m
dv
dt. (7.3)
The first term cannot be ignored in general, as there are many forces which depend ona changing mass — such as the loading of a moving train or the thrust on a rocket. If,
59
however, the mass is constant then equation 7.3 reduces to
F = mdv
dt
= ma . (7.4)
Remember that equation 7.4, F = ma, applies only if m is constant. Both F and a arevectors and are in the same direction when only one particle is involved. When more thanone force is acting, the sum of the forces (nett force) determines the acceleration,
∑
F = ma . (7.5)
This is the most commonly used form of Newton’s Second Law of motion.
7.14 Kinetic energy and work
The kinetic energy of a mass m with speed v is defined by
K =1
2mv2 . (7.6)
The work done on the mass is defined here as the change in kinetic energy of the mass,that is
W = ∆K . (7.7)
Note that the work done by the mass is the negative of the work done on the mass.
Consider the spatial derivative of the kinetic energy of a mass moving in one dimension,
dK
dx=
1
2m
d
dxv2 . (7.8)
Applying the chain rule,
dK
dx=
1
2m
(
d
dvv2
) (
dv
dx
)
=1
2m (2v)
dv
dx. (7.9)
Rewriting v as dx/dt and applying the chain rule in reverse leads to
dK
dx= m
dx
dt
dv
dx
= mdv
dt= max
= Fx , (7.10)
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using Newton’s second law with constant mass.
Rearranging 7.10 and then integrating from the initial position xi and kinetic energy Ki
to the final position xf and kinetic energy Kf gives
dK = Fxdx
∫ Kf
Ki
dK =
∫ xf
xi
Fx dx
∴ ∆K = Kf − Ki =
∫ xf
xi
Fx dx . (7.11)
Using the definition of work given in equation 7.7
W =
∫ xf
xi
Fx dx . (7.12)
If the mass is free to move in three dimensions then equation 7.12 becomes
W =
∫
xf
xi
F · dx . (7.13)
In differential form 7.13 is written δW = F·dx, where the delta symbol δ is used to denotean infinitesimal amount, in this case of work done, of a quantity that is not a function.
If the force doing work on the mass is conservative, i.e. it is the gradient of some potentialenergy, then
W =
∫
xf
xi
F · dx
=
∫
xf
xi
−∇U · dx
= −∆U . (7.14)
This says that the work done on an object by a conservative force is the negative of thechange in potential energy of the object.
If a constant force acts on a mass at constant angle θ to the direction of motion, thenW = F · s = |F||s| cos θ, where s is the displacement. Specifically, if a constant force isalways in the direction of motion, then W = Fs.
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7.15 Example: Loop the loop
h
R
v
Figure 7.14: A block performing a loop-the-loop
Consider a block released from height h sliding without friction down a slope, and thenproceeding to complete a loop-the-loop of radius R. Find the minimum height h suchthat the block makes it around the loop.
Solution
Assume the block has mass m, and the acceleration due to gravity is g. We draw theforces on the mass at the top of the loop as this will be the critical point: if the blockdoes not fall off at the top of the loop (and makes it there) then it will make it aroundthe remainder of the loop. (See Figure 7.15: N is the normal reaction force exerted bythe track on the block).
v
+
mg
N
Figure 7.15: Diagram showing the forces on the block at the top of the loop
‘Conservation of Energy’ and ‘ΣF = ma’ problem.
The net force on the block must provide the centripetal acceleration required for theblock to progress around the loop. Considering the forces in the vertical direction withthe positive direction defined as down as indicated on Figure 7.15,
∑
F = N + mg = ma = mac = mv2
R.
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Applying conservation of energy,
Ebefore = Eafter
mgh = mg(2R) +1
2mv2
g(h − 2R) =1
2v2
v2 = 2g(h − 2R) ,
so
N + mg = mv2
R= m
2g(h − 2R)
R.
As the block is released from greater heights, the speed of the block at the top of the loopincreases, and the normal force increases to provide the greater centripetal accelerationrequired to keep the block moving in a circle. Thus the minimum height from which theblock can be released will occur when the normal force takes its minimum, i.e. zero asthe normal force cannot be exerted upwards. Thus the minimum value of h is given by
mg = 2mghmin − 2R
R,
1
2R = hmin − 2R ⇒ hmin =
5
2R .
Clearly the dimensions are correct, and this answer makes sense as hmin > 2R.
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