7 FE1 MOTION - The University of · PDF file7 FE1 MOTION OBJECTIVES Aims From ... Introduction...

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FE1 MOTION

OBJECTIVES

AimsFrom this chapter you should develop your understanding of the ways that physicists describemotion. You will appreciate the differences between the everyday meanings of common words likevelocity and acceleration and their scientific meanings - which can be quite different! As well aslearning this new word-language you should also learn how graphs can be used to describe thequantitative details of motion. You will come to appreciate, we hope, that such graphs are much moreuseful and comprehensive than memorised formulas and equations - which have limited validity inany case.Minimum learning goalsWhen you have finished studying this chapter you should be able to do all of the following.1. Explain, interpret and use the terms

position, displacement, vector, scalar, component, time, time interval, average velocity,velocity [instantaneous velocity], speed, acceleration, centripetal acceleration.

2. Plot, interpret and find values of kinematic quantities from graphs of displacement component,velocity component, acceleration component, distance travelled and speed, all plotted againsttime interval. Solve kinematic problems using graphical techniques.

3. State, apply and explain the relation among centripetal acceleration, radius and speed.4. State the relation between acceleration and total force and apply the relation qualitatively to

some simple examples.

PRE-LECTURE

IntroductionThis chapter deals with ways of describing the motion of objects. The basic concepts are introducedusing examples involving motion in only one dimension (i.e. motion along a straight line). Theseideas are then extended to treat motion in two and three dimensions.

1-1 POSITION AND DISPLACEMENTA basic description of the motion of an object can be developed using the ideas of position andtime. As time goes on a moving object changes its position relative to some reference point. As theobject moves it also traces out a path, which is usually curved and whose length, measured along thepath, is the distance travelled by the object.

To illustrate the idea of position, consider a car journey from Lane Cove to the University viathe Gladesville Bridge (figure 1.1). The car is continually changing its speed and direction of travel.(Remember that going up and down hills also involves changing direction.)

FE1: Motion 8

Lane Cove

Gladesville

Town Hall

University

Initial position

Final position

Position

Figure 1.1 Position vectors

At any instant, we can specify the position of the car from some reference point, the SydneyTown Hall for instance. The position could be described by the length and direction of an imaginaryarrow, the car's position vector, drawn from the Town Hall to the car.

However it is usually much more convenient to describe position using a set of rectangularcoordinates, x ,y, z. These three quantities are also called components. In this example x could bethe easterly component of position, y the northerly component, and z the altitude component. Infigure 1.2 x has a negative value (-6.3 km, for example) because the car is to the west, rather than theeast, of the origin (Town Hall).

2.1 km

Lane Cove

Town Hall

University

6.3 km

x,y( )Position

Figure 1.2 Coordinates or components of positionThe components of position are scalar quantities, which can have positive or negative values.

A change in position is called a displacement. It can be specified by quoting the changes ineach of the three position components. Notice that the magnitude of the displacement and distancetravelled are quite different quantities.

Lane Cove

Town Hall

University

Distancetravelled

Displacement

Figure 1.3 Displacement and distance travelledDisplacement, or change in position, is a vector quantity. Distance travelled is a scalar quantity which

cannot be negative.

FE1: Motion 9

1-2 VELOCITY IN ONE DIMENSIONIn order to develop the other concepts used to describe motion - velocity and acceleration - we canuse one-dimensional examples, with a single position component. The treatment of the other twocomponents in more general cases is exactly the same.Example 1.1: One-dimensional motion of a car

Consider a car rolling downhill - out of gear. Let's call its position component x and let the time intervalsince it started to roll be t. Suppose that its motion is described by the following table:

t ime t /s

position componentx /m

0.0 01.0 12.0 43.0 94.0 165.0 256.0 36

Q1.1 It is often more revealing to display this information pictorially on a graph. Plot the position componentas a function of time elapsed on the graph paper below (figure 1.4).

0 1 2 3 4 5 60

10

20

30

40

x/m

t/s

Positioncoordinate

Elapsed time

Figure 1.4 A position-time graph.You should plot the graph yourself using the data tabulated above.

We define average velocity over some time interval:

average velocity = change in position

time taken .

In the general case we would need three components to describe the position, so we wouldneed three components to describe the velocity. In this example we have only one positioncomponent and one component of average velocity:

average velocity component = change in xchange in t ;

or vx— =

∆x∆t ... (1.1)

(The bar on top of the symbol v indicates 'average' and the subscript x indicates the component ofvelocity in the direction corresponding to increasing values of coordinate x.)

The SI unit of velocity is the metre per second, symbol m.s-1.Q1.2 Use the data of Q1.1 to calculate the component of average velocity over the time interval between 2.0 s

and 5.0 s.

The instantaneous velocity (or just velocity if the context is clear) at any instant is the rateat which an object's position is changing at that instant. For a one-dimensional motion, the velocity

FE1: Motion 10

component is given by the slope of the position-time graph. The slope of a graph at a point is equalto the slope of the tangent to the graph at that point, which can be found as shown in figure 1.5.

Positioncomponent

Tangent tograph attime

t

t

Time

t Δ

x Δ

Figure 1.5 Finding instantaneous velocity from positionConstruct the tangent to the position-time curve at the time (t) of interest. Draw a triangle with the

tangent as hypotenuse. Read or measure off the other two sides and calculate the ratio ∆x∆t .

Q1.3 Use this definition and the data of Q1.1 to estimate the velocity component of the car at t = 2 .0 s and att = 5.0 s. Note that these values are different from the average velocity calculated in Q1.2.

In our one-dimensional example the component of instantaneous velocity is given by thetime-derivative of the position component:

vx =dxdt ... (1.2) .

In general, velocity is a vector quantity because it has direction. Note that the threecomponents of a velocity are scalar quantities. Components themselves do not have direction, butthey can take positive or negative values.

In physics the magnitude of a velocity is called speed and, since it is a magnitude, a speedcannot be negative.

1-3 ACCELERATION IN ONE DIMENSIONBy analogy with the definition of average velocity the average acceleration of an object duringsome time interval is defined as:

average acceleration = change in velocity

time taken .

This definition can be expressed in terms of components as

ax— =

∆vx∆t ... (1.3)

The instantaneous acceleration at any instant is defined as the rate at which the velocity ischanging at that instant. Acceleration is also a vector quantity so, in general, we need threecomponents to describe it. The component of instantaneous acceleration is given by the slopeof the tangent to the velocity-time graph (figure 1.6).

FE1: Motion 11

t Time

Velocitycomponent,

∆vx

∆t

vx ∆vx∆t

ax =

Figure 1.6 Finding acceleration from a velocity-time graphMeasure the slope of the tangent to the curve at the time of interest.

In calculus notation,

ax = dvxdt , ... (1.4)

i.e. acceleration is the derivative of velocity with respect to time.The SI unit of acceleration is the metre per second per second, symbol m.s-2.To find the acceleration in example 1.1, we would have to find the velocity components at a

series of instants, plot these on a velocity-time graph and find the slopes at various times.

Q1.4 Plot the velocity component of the car (example 1.1) at t = 2.0 s and t = 5.0 s on the graph paper below(figure 1.7).

0 1 2 3 4 5 6 t/s

Velocitycomponent

v /m.s -1

Elapsed time

Figure 1.7 A velocity-time graphYou should use data from figure 1.4 to construct this graph.

In this example the acceleration was constant so a straight line through the two points you have justplotted represents the velocity component at all times. Use this graph to find the constant value of theacceleration component.

FE1: Motion 12

LECTURE

1-4 MOTION IN ONE DIMENSIONIn the following one-dimensional example the velocity component remains positive, which meansthat the direction of travel remains constant. In this special case the distance travelled happens to beequal to the magnitude of the displacement, or change in position. Remember, however, that this isnot true in general.Example 1.2: Starting a car

a) A car is started by rolling it downhill out of gear.b) When it is going sufficiently fast the gears are engaged. This starts the engine successfully and thecar is driven to the bottom of the hillc) The car is then stopped.

The following position-time graph (figure 1.8) is obtained.

Figure 1.8 Position-time graph for starting a car (example 1.2)

By estimating slopes in figure 1.8 we obtain the velocity-time graph in figure 1.9.

Time/s

3

2

1

0

Velocitycomponent / m.s -1

0 2 4 6 8 10 12 14 16

Figure 1.9 Velocity-time graph for example 1.2

Then by estimating slopes on the velocity-time graph (figure 1.9) we obtain the acceleration-time graph(figure 1.10).

FE1: Motion 13

Time/s

Accelerationcomponent/

10 12 14 162 4 6 8

2

1

0

-1

-2

m.s -2

Figure 1.10 Acceleration-time graph for example 1.2

Example 1.3: A velocity-time graphSuppose that we had been given velocity-time information instead of position-time information. We canperform a reverse process and find displacement as a function of time.

time /s velocity component/m.s -10 101.0 72.0 -23.0 -174.0 -385.0 -65

We want to find the displacement component as a function of time, using the starting point as thereference point. First, we draw the velocity-time graph, figure 1.11.

Figure 1.11 Velocity-time graph for example 1.3

The displacement component after a certain time is given by the area under the velocity-time graph upto that time. By counting squares, the area under the curve up to t = 1.00 s, say, is 90 small squareswhich represents a displacement of

90 × (1.0 m.s-l) × (0.10 s) = 9.0 m.i.e. we find that the displacement after the first second is 9.0 m. We can continue this process and plot adisplacement-time graph (figure 1.12). Note that areas under the time axis in figure 1.11 are to besubtracted since the body is moving in the opposite direction; values of the velocity component arenegative, so each contribution to the displacement is negative.

FE1: Motion 14

Figure 1.12 Displacement calculated from figure 1.11

In calculus terminology this operation is an integration:

x = vx0

t∫ dt ,

i.e. displacement from the starting point is the integral of the velocity with respect to time.It is also possible to find velocities from an acceleration-time graph in a similar manner if the

starting velocity is known.

1-5 MOTION IN MORE THAN ONE DIMENSIONSo far our discussion has been restricted to one-dimensional motion. In three dimensions directionsare obviously important. Displacement, velocity and acceleration are all quantities which need boththeir magnitudes and directions to be specified in order to describe them completely. They arevectors. At any instant, the velocity is in the direction that the object is moving.Demonstrations: Ball thrown into the air

(Draw your own diagrams. They should show the path of the object together with arrows torepresent velocity and acceleration at some instant.)

When the ball is thrown vertically upwards, the acceleration is 9.8 m.s-2 downwardsalong the same vertical line, and the problem is one-dimensional.

When the ball is thrown upwards at some angle, the problem is two-dimensional sincethere is horizontal motion as well as vertical motion. The acceleration is still 9.8 m.s-2vertically downwards, but now the acceleration is not in the same direction as the velocity. Thevelocity of the ball changes in both magnitude and direction; its path curves.

The acceleration is in the vertically-down direction which means that only the verticalcomponent of velocity changes. The horizontal component of velocity remains constant.

Uniform circular motionAn object moves with constant speed, v, in a circle of radius R. The instantaneous velocity is along atangent to the circle and the direction of the acceleration is along the radius towards the centre of thecircle. This is called a centripetal acceleration. Its magnitude is

a =v2

R . ... (1.5)

In this special case the magnitude of the velocity (i.e. the speed) remains constant and only itsdirection changes. The magnitude of the acceleration is also constant but its direction keepschanging, so that it always points towards the centre of the circle. (If the speed of a circular motionchanges there is a component of acceleration tangential to the path but the radial component of theacceleration is still described by equation 1.5.)

FE1: Motion 15

1-6 FORCESHaving described motion (kinematics), we can move on to examine the causes of changes in motionor causes of acceleration (dynamics). These are forces.

Firstly, we have the law of inertia: if there are no forces acting, the motion remainsunchanged. If more than one force is acting and these forces cancel or balance each other, this hasthe same effect as no force. (This law is also known as Newton's first law of motion.)Demonstrations• Air track with glider - glider at rest, glider moving with constant velocity.• Dry ice puck.• Motion in a spacecraft cabin.

Secondly, if there is an unbalanced force acting, there is an acceleration in the direction of thenet force.

Since both magnitude and direction are needed for the complete specification of a force, forceis a vector quantity.DemonstrationsWhat is the direction of the force in each of the following cases?• A ball thrown into the air.• Circular motion examples.

POST-LECTURE

1-7 QUESTIONSMotion in one dimensionQ1.5 a) Consider the velocity and acceleration-time graph in example 1.2 above. Give an account of how the

velocity and acceleration change during each segment.b) Consider the velocity-time graph in example 1.3 above. Note that areas between the graph and the time

axis represent positive contributions to the displacement if the curve is above the axis and negativecontributions if the curve is below.

What happens to the displacement when the velocity goes through zero to become negative ?What can you say about the areas above and below the time axis when the displacement is zero (i.e.

when the object has returned to its starting point)?What does a negative displacement mean?

Q1.6 Here is a velocity-time graph (figure 1.13) for a car journey along a straight road.

0 1 2 3 4 5 6

80

40

0

-40

-80

Component of velocity / km.h -1

Time/min

Figure 1.13 Example: one dimensional motion of a car

FE1: Motion 16

Give a qualitative description of the motion of the car during the journey.How far is the car from its starting point at the end of its journey? [Hint: Accelerations for each

segment are given by slopes, distances are given by areas.]

Q1.7 The driver of a car wants to average 60 km.h-1 for a 20 km trip. The car travels the first 10 km at aconstant speed of 40 km.h-1. How fast must it go over the last 10 km?

Q1.8 This exercise is inspired by the N.S.W. Motor Traffic Handbook. It concerns stopping distances.Assume that the reaction time to apply the brakes of a car is l.0 s, and assume that the brakes give a

constant acceleration of magnitude 5 m.s-2.Compare the distances required to stop the car travelling at

(a) 9 m.s-1 and (b) 18 m.s-1. (9 m.s-1 is approximately 32 km.h-1.)(Hint: use velocity-time graphs.)

Motion in more than one dimensionQ1.9 An object is accelerating if

(a) the magnitude and direction of its velocity both change, or(b) only the magnitude of its velocity changes, or(c) only the direction of its velocity changes.

Give examples of each.ForcesQ1.10 Suppose you are standing in front of a frictionless table and an object on it is moving from left to right at

constant velocity. What happens to the motion if:(a) you apply a force directed to the right;(b) you apply a force directed to the left;(c) you apply a force directed towards yourself?

Q1.11 Consider the motion of an object on which a single force acts. What can you say about the magnitude anddirection of the forces that produce the following effects:

(a) object moving from left to right with constant velocity;(b) object moving from left to right with velocity increasing;(c) object moving from left to right irregularly (i.e. stopping and starting); (d) object moving in circle at constant speed;(e) object moving in circle at variable speed?

APPENDIX

1-8 EXAMPLES OF MATHEMATICAL DESCRIPTIONS OF MOTIONIn the lecture you have seen how motion can be described by a table of numbers or by a graph. Insome special situations a mathematical equation can be used.

For example, if an expression for the displacement is given, it may be differentiated to givevelocities and accelerations. Or, if the velocity is given, it may be differentiated to give accelerations.

Conversely, if the acceleration is given, together with some statement about what the movingobject is doing at time t = 0, it may be integrated to give velocities and displacements. Or if thevelocity is given together with some similar statement, it may be integrated to give the displacement.Q1.12 The displacement of an object is given by

a) x = A + Bt + Ct2 + Dt3 orb) x = A cos(ωt ),where A, B, C, D, ω are constants.

In each case,i) find the displacement of the object at time t = 0, andii) differentiate to find the velocity and the acceleration as functions of time.

Q1.13 At t = 0, an object is at position x = x0 and has velocity component v = v0. Its acceleration is

a) a = cb) a = - ktwhere c and k are constants.

In each case, integrate to find velocities and displacements at any subsequent time t.

17

FE2 FORCE

OBJECTIVES

AimsThis chapter will almost certainly require a re-organisation or replacement of your intuitive viewsabout the physical world! You should aim to understand the physical concept of force which isprobably quite different from the normal intuitive concept that you have already acquired. You willalso learn the principles, known as Newton's laws of motion, which underpin the scientific idea offorce. You will learn to apply the concept of force and Newton's laws to simple examples.Minimum learning goalsWhen you have finished studying this chapter you should be able to do all of the following.1. Explain, interpret and use the terms

mass, vector, component, force, newton, fundamental force, total force [resultant force, netforce], weight [gravitational force], gravitational acceleration [gravitational field], equationof motion [Newton's second law], Newton's third law, centripetal force, contact force,tension.

2. Describe the general system of classification of fundamental forces.3. State, explain and apply Newton's third law.4. Relate tensions in strings or ropes to the forces exerted by those strings or ropes.5. Draw diagrams showing the forces acting on objects.6. State the equation of motion and apply it to simple problems by considering force components

in mutually perpendicular directions.7. Use force-time graphs to find accelerations, velocities and speeds of objects.

PRE-LECTURE

IntroductionAccelerations are caused by unbalanced forces. This idea was discussed in the last chapter. Amathematical expression of this result, the equation of motion, is the main topic in this chapter.

Revise questions 1.9, 1.10 and 1.11 in FE1.

2-1 COMPONENTSThe idea of components, which was applied to the vector quantities position, velocity and accelerationin chapter FE1, can be extended to other kinds of vector quantities such as force.

The value of a vector quantity can be described by specifying its magnitude (number × unit)and its direction. An alternative, equivalent, description is in terms of three components referred tothree mutually perpendicular directions or coordinate axes. The relation between a vector and itscomponents is most easily illustrated on two-dimensional paper using two dimensions, with tworeference directions.

FE2: Force 18

Direction V

Direction H

a

θ

a cos θ

a sin θ

Figure 2.1 Components of a vector

The component of vector a in a direction at an angle θ to the vector's direction is defined to beequal to a cosθ. In the case illustrated above the component in the direction labelled H is aH = acosθ and the component in the direction labelled V is aV = a cos(90° - θ ) or a sinθ . Note that eachcomponent is a scalar quantity (i.e. one which has no direction). The value of a component mightbe positive, zero or negative, depending on the value of the angle θ .

2-2 MASSMass is a property of an object which is one way of specifying the quantity of matter in it. Theconcept of mass can be explained formally in terms of the effect of a total force on a body, using theequation of motion discussed later in this chapter. Alternatively mass can be determined in terms ofthe gravitational force on the object: at a fixed location the gravitational force on different objects(their weights) are proportional to their masses. The SI unit of mass is the kilogram (symbol kg)and the unit of weight is the newton (N).

LECTURE

2-3 THE NATURE OF FORCEPulls, pushes, compressions, friction, gravity - these are forces that we all are familiar with. All theforces occurring in nature can be classified into four fundamental types. In decreasing order ofstrength these types are:

strong nuclear,electromagnetic,weak nuclear,gravitational.

Nuclear forcesThe strong nuclear force holds neutrons and protons together in the nucleus of an atom. It is theforce associated with radioactive alpha decay and nuclear energy, including the atomic bomb and thehydrogen bomb.

The weak nuclear force holds the neutron together. It is associated with radioactive betadecay. (Beta decay of carbon-14 can be used for radioactive tracing in biological studies, and forcarbon dating of archaeological items.)

Both types of nuclear force act only over a very short range of distances comparable with thesize of an atomic nucleus, about 10-14 m (10 fm).GravityThe gravitational force is the force which causes objects to fall. It is an attractive force betweenthe Earth and the falling object. This force on a falling object is a special instance of a universalforce which acts between every pair of objects. The fundamental law of gravitational attraction says

FE2: Force 19

that the magnitude of the force between between two particles, with masses m1 and m2, at a distanced apart is

F ∝m1m2

d2 .

Note that this force extends to infinite distances and is always attractive. The attractive forcebetween Earth and Moon, Sun and planets, etc is gravitational.

The weight of an object, on the Earth or on the Moon etc, is the gravitational force acting on it.At any particular place near the Earth's surface, the weight W of a body is proportional to its mass m:

W = mg ... (2.1). The constant of proportionality g is the magnitude of the Earth's gravitational field but it is oftenloosely called the acceleration due to gravity.Electromagnetic forceThe interactions between atoms and the charged particles within atoms are all electromagnetic.These forces determine whether a substance will be a solid, liquid, gas or a plasma. They are theforces involved in chemical reactions and biological processes.

The macroscopic forces described as pulls, pushes, friction etc can be explained, on amolecular scale, as electromagnetic forces. In general, the force between molecules is stronglyrepulsive at short distances but weakly attractive at greater distances.

Separation

Force component

0

REPULSIVE

ATTRACTIVE

Equilibrium

Figure 2.2 Intermolecular forceThe equilibrium separation, where the force is zero, is typically about 10-10 m (0.1 nm).

On a macroscopic scale the forces on electric charges are electromagnetic, as are thoseinvolved in power generation and electronic communications, computers, and measuring instruments.Electromagnetic forces extend to infinite distances and can be attractive or repulsive.

2-4 PAIRS OF FORCESNote that all forces, whether they be fundamental or macroscopic, always occur in pairs. Forexample, the gravitational force exerted on the Sun by the Earth is equal in magnitude but opposite indirection to the gravitational force exerted on the Earth by the Sun.

The electrostatic force exerted on a charged particle, A, by another charged particle charge, B,is equal in magnitude but opposite in direction to the electrostatic force exerted on B by A. Thesame is true for the molecular forces between two molecules

This has macroscopic consequences. Consider a block sitting on a table. The block exerts adownward force on the table and the table exerts an upward force of equal magnitude on the block.

FE2: Force 20

These are all examples of Newton's third law of motion.

2-5 THE EQUATION OF MOTIONForces are responsible, not for motion itself, but for changes in the motion of a body. Change inmotion implies acceleration. The effect of the forces is described by the equation of motion, whichis also known as Newton's second law of motion. The equation relates total force F to accelerationa. For a body whose mass m does not change:

total force = mass × acceleration.

In symbols this equation of motion isF = m a ... (2.2).

A word about notation: symbols for vector quantities are printed in bold-face type (e.g. F) whileordinary italic type is used to indicate the magnitude of a vector (e.g. F means |F|.)

Note the following important points about the equation of motion.• Only those forces which act on a body affect the motion of that body.• All the forces acting must be included in the total force. (Total force is also called net force or

resultant force.)• Directions of forces must be considered in the calculation of the total force. (The procedure

for doing this is given below.)• The acceleration has the same direction as the total force.• The SI unit of force is the newton, symbol N.Example with one force acting: ball thrown in the air

A ball is thrown into the air (either vertically or at some angle). If the force exerted on the ball by the airis negligible, then the total force on the ball is equal to the ball's weight, W = mg downwards. Hence theacceleration a is equal to g, downwards.

Near the earth's surface, g = 9.8 m.s-2.

Uniform circular motionConsider an object moving in a circle with constant speed v. The total force in this case is calledcentripetal force. It follows from the equation of motion (F = ma ) and the formula forcentripetal acceleration (v2/R), which is towards the centre of the circle, that

F = mv2

R and the direction of F is towards the centre of the circle.

Note that centripetal force is not a new kind of force; it is just a fancy name for the radialcomponent of the total force. In the case of uniform circular motion the radial component is the onlycomponent of the total force.Calculation of total forceWhen several forces are acting on an object, they must be combined to give the total force. Ingeneral, the forces will be acting in different directions, so combining them presents a problem.Fortunately the components of the motion in each of any three mutually perpendicular directionscan be treated separately. This separation of the motion into components is helpful even when thereis only one force acting.

For example, if a ball is thrown into the air at some angle to the horizontal, the vertical andhorizontal components of the motion can be treated separately. Provided that air resistance isnegligible, the vertical motion has an acceleration vertically down, but the horizontal motion hasconstant horizontal velocity.Example: a sliding block

A block is being pulled along a frictionless table. The pulling force is directed upwards at an angle θ tothe horizontal.

FE2: Force 21

θ

N

W

P

Figure 2.3 Block pulled across a frictionless table

Forces acting on the block:• the weight of the block, W, vertically downwards;• a contact force, N, exerted by the table on the block, vertically upwards;• the pulling force, P .In this case all the forces act in one plane, the plane of the diagram, so the problem is two-dimensional.

We need only two component directions: vertically up and horizontally to the right are convenient.Components of the pulling force P:

• horizontal component PH

= P cosθ to the right;

• vertical component PV

= P cos (90°-θ ) upwards.

Note that the two component forces PH and PV are together equivalent to the pulling force P.

θ

Pis equivalent to

PV = Pcos(90° - θ)

PH = Pcos(θ)Figure 2.4 Representing a force by its components

The total forceThe horizontal component of the total force is FH = PH to the right. So the horizontal component of the

acceleration is equal to PHm to the right. (Note that components of accelerations are defined in the same

way as components of forces.)The vertical component of the total force FV = N + PV - W , upwards. This must equal zero since the

block does not accelerate vertically. We see therefore that the force the table exerts on the block,N = W - PV. Note that components, not magnitudes, are added; the component of the weight in thevertically up direction is -W.

General procedureSummarizing, the general method is as follows.(i) Draw in all the forces - make sure that none is omitted.(ii) Choose three mutually perpendicular directions and calculate the force components along these

directions. (Note that if the motion is confined to a plane as in the example above you needonly two mutually perpendicular directions.)

(iii) Calculate the component of the total force in each direction, by adding the individualcomponents, and taking account of their signs.

(iv) Calculate the acceleration component in each direction, using the equation of motion.Note that sometimes you are given the acceleration, in which case you can use the equation of

motion to give the components of the total force.

FE2: Force 22

POST-LECTURE

2-6 QUESTIONSFundamental forcesQ2.1 a) We have said that the gravitational force is weaker than the electromagnetic force. Is this your experience?

Why might the electromagnetic force appear to be weaker than the gravitational force?b) Nuclear forces were the last to be discovered. Why was this so if they are the strongest?c) How would you rate the forces in terms of their significance to biological systems?

Molecular forcesQ2.2 a) Why does a block of metal resist changes in its shape when you attempt to compress it, or when you

attempt to stretch it?b) Suppose an object is hanging from the end of a fine wire. What holds the wire together? If the load is

increased the wire stretches. How do the forces in the wire change?c) Compare the force required to pull a brick apart with the force required to lift it up. Which is greater?

What does this suggest about the relative strengths of molecular (electromagnetic) force and gravitationalforce?

WeightQ2 .3 A spring balance measures the weight of an object, i.e. the gravitational force on it. The gravitational

force on an object near the surface of the Moon is about one-sixth of that near the surface of the Earth.a) Suppose a particular object has a mass of 2 kg. According to the spring balance, what would its weight be

on Earth? On the Moon? (An approximate answer will do.)b) How could you use the spring balance to measure the mass of an object?

Equation of motionQ2.4 a) An object on the end of a string is being swung in a horizontal circle at constant speed. What provides the

centripetal force?If you want the object to swing around at twice the speed, how much greater would the centripetal force

have to be?Sketch the path the object takes if the string breaks.

b) What provides the centripetal force on the various parts of a revolving wheel?c) What provides the centripetal force on a satellite orbiting about the earth?The following dynamics problems, where several forces are involved, can all be attacked using

the method outlined in the lecture.Q2.5 A block, mass 2.00 kg, is being pulled along a frictionless table. The pulling force is 20.00 N in a

horizontal direction (figure 2.5).

F = 20.00 N

Figure 2.5

Draw in the weight of the block and the force that the table exerts on the block.Take components of forces in horizontal and vertical directions.

a) What is the horizontal acceleration?b) What is the total vertical force? (Hint: what is the total vertical acceleration?)c) Suppose now that there is a horizontal frictional force of 10.00 N opposing the motion. What is the

horizontal acceleration?

FE2: Force 23

Q2.6 a) A car is being driven along a flat road.What are the forces acting on the car?What difference does it make if the car is moving at constant velocity or accelerating?

b) The car is rounding a curve.What provides the centripetal force?

c) The car is on a hill with slope θ.What is the downhill component of its weight?

Q2.7 A block, mass 2.00 kg, is being pulled along a frictionless table. The pulling force is 20.00 N at an angleof 30° to the horizontal (figure 2.6).

30°

F = 20.00 N

Figure 2.6

Draw in all the other forces.a) What is the total horizontal force? What is the total vertical force?b) What is the horizontal acceleration? What is the vertical acceleration? (Will the block be lifted off the

table?)c) What would happen if the magnitude of the force were doubled?d) Suppose the table were not frictionless, how would this affect your answers?

Q2.8 Two forces are acting on an object (figure 2.7).

30°

90°

4.00 N

8.00 N

y

x

Figure 2.7

If these were the only forces acting, the object would accelerate.a) Find the components of the total force in the y ↑ and x → directions.b) What additional force or forces would stop the object from accelerating?

FE2: Force 24

2-7 TENSIONWhen you pull on one end of a rope which is attached to some object at the other end, the rope willexert a force on that object. Something also happens to the rope itself. To understand this, thinkabout some place inside the rope and imagine that the rope consists of two parts. Each part exerts aforce on the other part and, according to Newton's third law, those forces have the same size (figure2.8). This common magnitude of the force between the two parts of the rope is called the tension inthe rope at the particular place considered.

magnitude T magnitude T

Figure 2.8 Internal forces and tension

The force exerted on the object is equal to the tension in the rope at the point where the rope isjoined to it and the force exerted on the rope at the other end is equal to the tension there. In general,the tension in the rope will be different at different parts of the rope. In some cases where theapplied forces are much larger than the rope's weight, or if the rope is not accelerating, we can makethe useful approximation that the tension is roughly the same throughout the rope. That is the casein the next problem.Q2.9 Suppose that your car is stuck in mud at the side of the road. You have the idea that if you tie a rope, one

end to the car and the other end to a tree, and apply a force perpendicular to the rope at the middle, you willbe able to move your car (figure 2.9).

Figure 2.9 How to pull a car out of a bog

a) Why is this a good idea ?Hints

• Suppose you have started applying the force to the rope. The rope has taken on the shape shown infigure 2.10 but the car has not yet started to move. θ is a very small angle.

Tθ

θ

T

P

Figure 2.10 Forces on a point in the rope

• Consider the forces acting on the bend in the rope. The forces at this stage still balance, so takecomponents to see how the tension compares with the applied force, P. What is the magnitude of the forceon the car? On the tree?

b) Where might this good idea go wrong ?

FE2: Force 25

2-8 AN ACCELERATING SYSTEMConsider a person in a lift which has a component of acceleration, vertically upwards, of magnitude a(figure 2.11). The forces on the person are her weight W = mg downwards and a contact forceexerted by the floor of the lift, upwards.

mg

N

a

Figure 2.11 Forces on a person in an accelerating lift

The person's upward acceleration is equal to the lift's acceleration, a, soN - mg = ma .

In common usage the term weight has a different meaning from the definition given in thelecture. Commonly, the weight of an object is the force that it exerts on its support, the floor, etc.

In a lift accelerating upwards, the downward force exerted on the floor by the passenger isequal to N. Since this is greater than its usual value, mg, the passenger's apparent weight is greaterthan normal.

Q2.10 a ) Suppose the lift is accelerating downwards. How does N compare with mg now?b) When do you experience "weightlessness"?c) Why do you get a strange feeling in your stomach?

Interlude: Times 26

INTERLUDE 2 - THE RANGE OF TIMES IN THE UNIVERSEtime/seconds

age of the universe 1018 __ _ age of the Earth _1015 __ _

time since earliest humans _1012 __ _ _109 __ human life span _

1 year _106 __ 1 month

1 day _ _103 __ 1 hour _

1 minute _1 __ heart beat _ _

period of a sound wave 10-3 __ _ _10-6 __ period of a radio wave _ _10-9 __

time for light to travel 1 m _ _10-12__ period of a molecular vibration _ _

period of an atomic vibration 10-15__ _ _10-18__ time for light to cross an atom _ _

period of a nuclear vibration 10-21__ _ _ time for light to cross a nucleus10-24__

27

FE3 EQUILIBRIUM

OBJECTIVES

AimsIn this chapter you will learn the concepts and principles needed to understand mechanicalequilibrium. You should be able to demonstrate your understanding by analysing simple examplesof equilibrium. You will also learn about the interactions between fluids and solid bodies as well asthe concepts - buoyant force and pressure - used to describe the interactions.Minimum learning goalsWhen you have finished studying this chapter you should be able to do all of the following.1. Explain, interpret and use the terms

translational motion, rotational motion, rigid body, equilibrium, stable equilibrium, unstableequilibrium, neutral equilibrium, axis, torque [moment of a force], centre of gravity,buoyancy, buoyant force, Archimedes' principle, pressure, pascal, density, barometer.

2. State and apply the relation between force and torque.3. State the conditions for equilibrium and apply them to simple problems.4. Describe and explain how the centre of gravity of a body can be located.5. Describe and explain the forces acting on a body which is wholly or partly immersed in fluid.

Solve simple problems involving buoyancy.

PRE-LECTURE

IntroductionThis chapter deals mainly with the equilibrium of rigid bodies. The conclusions about rigid bodiescan also be applied to some examples of non-rigid bodies, such as bodies of fluid at rest. We startwith two simple examples of objects in equilibrium: an object at rest and one moving with constantvelocity.

All the examples and principles discussed in this chapter are restricted to systems in which allthe forces can be described in two dimensions - a plane. The extension to general three-dimensionalsystems uses the same concepts but is mathematically much more complex.

3-1 TRANSLATION AND ROTATIONThe simple descriptions of motion that we have used so far implicitly treated the motion of only onepoint in a body. That is a good kind of description provided that all points in the body followsimilar, parallel, paths. That kind of motion is called pure translational motion. As well astranslational motion, a body can also execute rotational motion (like that of a spinning wheel) andvibration (like that of a shaking jelly). A body which cannot vibrate noticeably is said to be rigid; itsshape and size do not change significantly when it is acted on by a system of forces. The mostgeneral kind of motion of a rigid body is, therefore, a combination of translation and rotation; theflight of a boomerang is a good example.

FE3: Equilibrium 28

3-2 EQUILIBRIUM OF FORCESQ3.1 A block is at rest on a table.

Figure 3.1 An object at rest. What are the forces?

a) On figure 3.1 draw in all the forces acting on the block.The block is not accelerating, so the net force on it must be zero. What does this tell you about the

vertical forces and any horizontal forces that may be present?b) Suppose that a horizontal force is applied in an attempt to push the block along the table, but the block

does not move. What is opposing the horizontal force?

Q3.2. Block moving at constant velocityA block is being pulled across the table at constant velocity (figure 3.2).

pull

Figure 3.2 An object moving with constant velocityDraw in the other forces.

A frictional force opposes the motion. Draw in this force and the other forces on the block.Again, the block is not accelerating so the net force on it must be zero. What does this tell you about

the vertical and horizontal forces?

The meaning of equilibriumThe examples in questions 3.1 and 3.2 are simple illustrations of equilibrium. In both cases thevelocity of the block is constant, i.e. its acceleration is zero, and the total force on it is zero.

However it is not enough that the forces balance in order to have equilibrium. This guaranteesonly that there is no change of translational motion, i.e. that the motion of the body as a whole doesnot change. The definition of equilibrium needs to be extended to include the requirement that therotational motion of the body also remains constant. Thus, for example, a body which is completelyat rest is in equilibrium only if it does not start to move or rotate. As another example, a wheelrotating about a fixed axle, is defined to be in equilibrium only if its rotational speed does notchange.

In order to consider this rotational aspect of equilibrium we need the concept of torque. Weshall then see that for equilibrium, torques, as well as forces, must balance.

FE3: Equilibrium 29

LECTURE

3-3 TORQUEExample 3.1. Wheel on a fixed axle

Consider a wheel which can rotate about its axle. The axle remains in a fixed position. Anobject is hung by a string from the rim of the wheel as shown in figure 3.3.

W F

F A

Figure 3.3 An unbalanced wheel

When the object is released, the wheel starts to rotate so, clearly, it is not in equilibrium.The forces acting on the wheel are shown. W is the wheel's weight, FA is the force exerted bythe axle on the wheel and F is the force exerted by the string attached to the falling object,which causes the wheel to rotate. Note that these forces are not all acting through the samepoint.

In this example, rotation occurs about an axis - a line in space - which is perpendicular to theplane of the forces involved. In such cases we define torque as follows.

The torque of a force F about a specified axis is defined asτ = Fx ... (3.1)

where x is the perpendicular distance from the axis to the force's line of action. Torque is alsoknown as the moment of a force. Note that we can, in principle, define many torques for eachforce, one for every possible choice of axis. It is not necessary for the axis to be a possible axis ofrotation.

Example 3.1 - continued

F

x

Figure 3.4 Torque on a pivoted wheel

In this example, we consider torques about the wheel's axis of rotation, its axle. Onlythe force F has a non-zero torque about the axle so there is a net torque, due to F, acting on thewheel. This net torque produces an angular acceleration, i.e. it changes the angular speed ofrotation about the axle.

FE3: Equilibrium 30

Demonstrations• Consider two wheels which have the same shape and the same total mass. One has a densemetal rim, the other has a dense metal axle. Identical loads are hung from their rims. Thisdemonstration shows that the angular acceleration depends on the distribution of mass in the object.

metalaxle

metal rim

Figure 3.5 Effect of mass distribution on angular acceleration

Identical loads were attached to the wheels and then released. The angular accelerationwas greater for the wheel whose mass is more concentrated near the axle.

• Other demonstrations show rotational motion when the axis is not fixed.

3-4 EQUILIBRIUM OF TORQUESExample 3.1 - continued

The wheel has no translational motion so the total force on it must be zero.

Torque

Fx

W F

F A

Forces Torques

Figure 3.6 Forces and torque on the pivoted wheelThe vertically upward component of the total force = FA - W - F = 0 , but the net torque, Fx, is

unbalanced. Note that this torque produces a clockwise rotation.Example 3.2: Equilibrium of the wheel

To bring our wheel back into equilibrium, a torque of the same magnitude but in the opposite(anti-clockwise) sense would have to be provided. Another object with the same mass could beattached to the opposite side of the wheel in order to achieve this. See figure 3.7. (Note that thesupporting force FA will take on a new value.)

F

F

F

W

A

Fx -Fx

Forces Torques

Figure 3.7 Forces and torques at equilibrium

FE3: Equilibrium 31

3-5 CONDITIONS FOR EQUILIBRIUMIn general, then, the conditions for equilibrium of an object which is free to rotate about a fixed axisare:(i) total force acting on the object = 0;(ii) total torque about the axis = 0 .

Note that, since force is a vector quantity, the calculation of the net force must take account ofdirections. This can be done using the method of components introduced in chapter FE2. Torque,as defined here, is a scalar quantity whose values need to be associated with either clockwise oranti-clockwise rotation. We can assign positive values to one of these two senses, and negativevalues to the other. (The usual convention makes anti-clockwise values positive.)Example 3.2 - continued

When we apply these conditions to the wheel with two objects hanging from its rim, we canchoose vertically down as a component direction for the forces (all horizontal force componentsare equal to zero).(i) F + W + F - FA = 0 .

(FA takes on a new value when the second object is attached.)The condition for balancing the torques is satisfied because the two torques have the same

magnitudes but opposite senses:(ii) F x - F x = 0 .

It is obviously inconvenient to have to draw two diagrams for each example - one showingforces and one showing torques. Henceforth both forces and torques will be shown on the samediagram. Remember, however, that forces and torques are quite different entities and must becombined separately.

3-6 CENTRE OF GRAVITYExample 3.3. Centre of gravity of a flat object

A flat object is pivoted at the point P. See figure 3.8. Imagine that the object is divided up intolittle pieces. The weight of each piece provides a torque about the pivot. The object will be inequilibrium only if the torques due to all these pieces add up to zero.

P

Figure 3.8 Distribution of gravitational forces on an object

It is obviously inconvenient to have to consider the weights of all the little pieces of theobject separately. Fortunately, their total can be represented by the total weight, W , actingthrough a point called the centre of gravity.

FE3: Equilibrium 32

Centreof gravity

P

Torque

W

P

W

Equilibrium

Figure 3.9 Centre of gravity and the total weight forceThen W provides a torque about the pivot (equal to the sum of the many small torques). The object will be in equilibrium if its centre of gravity lies vertically below the fixed

pivot point. In this position the length of the perpendicular from the pivot to the line of actionof the weight is zero, and so is that from the pivot to the supporting force at the pivot.Therefore the torques of the weight and the supporting force about the pivot are both zero.

Example 3.4. Locating the centre of gravityTo locate the centre of gravity of a flat object, first mark a vertical line showing the line of

action of the weight. The centre of gravity must be somewhere on this line. Then choose adifferent pivot point and repeat the process. The centre of gravity must be also be somewhereon the new line; so it must be at the intersection of the lines.

A wheel is symmetric about its axis and the centre of gravity is at the centre of thewheel. This is easily verified.

3-7 EQUILIBRIUM OF A SYSTEM OF OBJECTS

Example 3.5: Two children balancing on a seesaw.This example is analysed in terms of forces and torques acting on a system which consists of thetwo children and the plank. The forces acting on this system are the weights of the twochildren, the weight of the seesaw's plank and a vertical supporting force at the pivot.

The weights of the children, W1 and W2 act at distances x1 and x2 from the pivot, asshown on the figure. These forces give torques W1x1 (anticlockwise) and W2x2 (clockwise)respectively about the pivot.

Suppose that the centre of gravity of the plank is directly above the pivot so that theweight WS of the seesaw plank acts downwards at the pivot. N is the upward supporting forceexerted by the pivot on the plank. Each of these two forces gives zero torque about the pivot.

N

W 2SW 1

x 1 x 2

W

W 1 x 1 W 2x 2

Figure 3.10 Balancing on a seesaw

For equilibrium the following conditions must be satisfied.(i) Total force acting on the system = 0.

Taking force components in the vertically downward direction:

FE3: Equilibrium 33

W1 + W2 + WS - N = 0.

(ii)Total torque about an axis through the pivot = 0.Taking clockwise as the positive sense:

W2 x2 - W1x1 = 0.

This analysis is essentially the same as that for a beam balance.

3-8 EQUILIBRIUM OF A FREE OBJECTA free object is one that is not pivoted. This is a more general situation than the case of a fixed axis.Demonstration

moves forward

moves forward

moves forward

spins anticlockwise

spins clockwise

no spin

push

push

push

Figure 3.11 Motion of a free objectThe centre of gravity is shown as a heavy dot.

A net force not acting through the centre of gravity of a rigid body will cause translationalacceleration of the object as well as change in its rotational motion. The resulting motion canbe described as a combination of translational motion of the centre of gravity and rotationalmotion about the centre of gravity.

3-9 GENERAL CONDITIONS FOR EQUILIBRIUMThe general conditions for equilibrium are as follows(i) The total force must be zero (as before).(ii) The total torque about any axis must be zero.

In many cases it is convenient to consider torques about axes through the centre of gravity.

3-10 BUOYANCYWhen a solid object is wholly or partly immersed in a fluid, the fluid molecules are continuallystriking the submerged surface of the object. The forces due to these impacts (which are sometimescalled pressure forces) can be combined into a single force, the buoyant force.

FE3: Equilibrium 34

Figure 3.12 Forces exerted by a fluid

Note that, for clarity, we show only the forces exerted by the surrounding fluid in this and thefollowing diagram.

We want to find the magnitude of the buoyant force and the point through which it acts.Buoyant force on a completely submerged object

Figure 3.13 Buoyant force on a submerged object

To work out how big this buoyant force is, and where it acts, we use the trick of thinking abouta 'block' of fluid, which has exactly the same shape and size as the solid object. This imaginaryportion of fluid is often called the 'displaced fluid'.

Figure 3.14 Fluid "displaced" by the submerged object

The pressure forces on this imaginary displaced fluid are exactly the same as the pressureforces on the solid object. So their total effect, the buoyant force, will be the same, irrespective ofwhat is inside the broken outline. The buoyant forces on the solid object and the 'displaced fluid' areidentical.

Now the displaced fluid must be in equilibrium. Since the only other force on it is its weight,which acts through its centre of gravity, the buoyant force must be equal to its weight, and it must actvertically upward through its centre of gravity. Hence the buoyant force on the submerged blockmust be equal to the weight of the displaced fluid and it must act vertically up through the centre ofgravity of the 'displaced' fluid body.

FE3: Equilibrium 35

Weight of displaced fluid

Buoyant force

Figure 3.15 Forces on the displaced fluid

This conclusion is known as Archimedes' principle.Buoyant force on a partly submerged object

Buoyant forceAir

Liquid

Figure 3.16 Buoyant force on a partly submerged objectNote that there is also a gravitational force (weight).

In this case the object is immersed in two fluids one of which is the air. The diagram showsonly the total force exerted by these fluids on the object.

Consider an imaginary block composed of the two fluids 'displaced' by the object.

Displaced air

Displaced liquid

Figure 3.17 Fluids displaced by the partly submerged object

This fluid block is in equilibrium and, just as before,buoyant force = total weight of the displaced fluids,

and acts vertically upward through the centre of gravity of the displaced fluids.Note that, for an object floating in a liquid, the buoyant force due to the displaced air is usually

negligibly small.

FE3: Equilibrium 36

Example 3.6: A heavy object supported in water by a string

Force exerted by string

Buoyant force

Weight

Water

String

Figure 3.18 A submerged object in equilibriumThe force exerted by the string adjusts to balance the weight and the buoyant force.

Here all the forces on the submerged object are shown.For equilibrium of this object:

force exerted by string = weight - buoyant force.We might call the weight of the block minus the buoyant force the "effective

gravitational force".This situation is demonstrated using a spring balance.If the buoyant force were greater than the weight, a force would have to be applied to hold

the object down. This happens, for example with a helium-filled balloon.Example 3.7: Floating object

Buoyant forceAir

WeightLiquid

Figure 3.19 A floating object in equilibriumThe displaced liquid adjusts so that the forces balance.

For equilibrium, taking components in the direction vertically down:weight - buoyant force = 0 .

The total torque will be zero if the buoyant force and the weight act along the same line.This means that the centres of gravity of the floating object and the displaced fluids must lie inthe same vertical line.

POST-LECTURE

3-11 MOMENT OF INERTIAThe physical quantity which describes the distribution of matter about the axis of rotation is themoment of inertia. Objects with their masses concentrated about the axis of rotation (or axle) havesmaller moments of inertia about that axis.

More precisely, if we divide the object up into small pieces each with mass Δm and at somedistance r from the axle, then the moment of inertia is

I = Σ Δm r 2

where the sum is taken over all the small pieces.

FE3: Equilibrium 37

Δm

r Axle

Figure 3.20 Defining the moment of inertia

For objects with a fixed axis of rotation, total torque about the axis equals moment of inertiaabout the axis times angular acceleration. This is the reason why the wheel with the metal axle in thelecture demonstration had the larger angular acceleration.

3-12 QUESTIONS AND PROBLEMS ON EQUILIBRIUMQ3.3 a) A tall block sitting on a table is being pulled.

F

Figure 3.21 Tipping an object

If the pull is applied near the top, the object will tip over instead of sliding along the table. Why?Hint: to get the block moving, the force F must be greater than the maximum frictional force acting at

the bottom on the block. F cannot be any smaller than this. Now consider torques. What point is theblock going to rotate about?

How could you pull the block along the table without tipping it?b) Pulling trees down with a tractor can be a dangerous occupation. Which of the methods shown

below is the less dangerous way to tie the rope to the tractor? Why?

Figure 3.22 Uprooting a tree. Which way is safer?

Equilibrium of pivoted objectsQ3 .4 Two children are balancing on a seesaw. Their weights are 200 N and 300 N . The smaller child is

1.80 m from the centre.Suppose that the weight of the seesaw is 500 N and the centre of gravity is directly above the pivot.Draw all the forces acting on the seesaw.

a) Calculate the magnitude of the force that the pivot exerts on the seesaw.b) How far from the centre is the larger child at equilibrium?

Q3.5 Explain why a beam balance gives the same value for the mass of an object on the Moon as on the Earth.

3-13 STABLE, UNSTABLE AND NEUTRAL EQUILIBRIUMWe sometimes distinguish between stable and unstable equilibrium. For example, consider a cone.

FE3: Equilibrium 38

NeutralStable Unstable

NeutralStable Unstable

Figure 3.23 Stable and unstable equilibrium

In both cases the cone is in equilibrium because the total force is zero and the total torque iszero. But the first case is stable, a slight displacement has no effect, while the second case isunstable, a slight displacement causes the cone to tip over. When the cone is lying on its side it isin neutral equilibrium.Q3.6 What forces are acting and where are they acting in each case? What torques are responsible for restoring

the cone to its original position or otherwise?Centre of GravityQ3.7 a) Does the centre of gravity always lie within an object? If not, give examples.

b) Suggest a way of locating the centre of gravity of a "lumpy" object (not a flat object).

3-14 FLUIDSPressureWhen a solid object is immersed in a fluid, the force exerted on the object by the fluid is distributedover the contact surface. For a complete description we need to look at the force acting on eachsmall part of the surface. We can define the average pressure on a flat surface to be the component,δFn, of the force perpendicular to the surface divided by the area, δA, of the surface. The limit ofthis quotient as we take smaller and smaller pieces of the contact surface (and hence also smaller andsmaller forces) is the pressure, P, at a point on the surface:

P = lim

δA→0

δFnδA

.

... (3.2)

Provided that the body and the fluid are not moving, the force on each small part of the contactsurface is perpendicular to the surface (see figure 3.12) so the interaction can be describedcompletely in terms of pressure. (On the other hand, if there is relative motion between fluid andsolid object, the force has components parallel to the surface, not described by the pressure.)

This idea of pressure can be used also to describe what goes on inside the fluid; just imaginethe fluid divided into two parts as in the argument about buoyancy. Wherever we draw the fluidboundary, we can define a pressure exerted by one part of the fluid on the other part. So we can saythat pressure exists within the fluid.

The following are important statements about pressure.• Pressure is a scalar quantity - it has no direction.• The pressure within a uniform stationary fluid is the same at all points in the same horizontalplane.• The SI unit of pressure is the pascal, symbol Pa; 1 Pa = 1 N.m-2 .

FE3: Equilibrium 39

DensityPressure variations within a fluid are affected by its density. The density, ρ, of a uniform substanceis defined as the quotient: mass ÷ volume:

ρ =mV . ... (3.3)

The barometer

h

S S'

Figure 3.24 A mercury barometerA tube, closed at one end, is filled with mercury and is then inverted over a container of

mercury as shown. If the tube is sufficiently long, the mercury falls from the top leaving anevacuated region there.

By considering the equilibrium of a part of the surface SS' of the mercury outside the tube, wecan show that the pressure in the mercury just under that surface is equal to the atmosphericpressure. So too is the pressure at the same level inside the tube.

We consider the equilibrium of a body consisting of the column of mercury in the tube whichis above the level SS'. The forces on this column of mercury are its weight, W, downward and theupward force, F, exerted by the mercury below the column. The magnitude of the upward forcemust be equal to atmospheric pressure multiplied by the cross-sectional area, A, of the tube:

F = PA.

The weight of the mercury column is equal to the product of its density, its volume and g.Since the volume is equal to the product of the column's height and cross-sectional area,

W = ρhAg .

These two forces, must have equal magnitudes, soP = ρgh . ...(3.4)

The density of mercury is 13.6 × 1 03 kg.m-3 and normal atmospheric pressure is1.01 × 1 05 Pa (101 kPa).Q3.8 a) Estimate the height of a column of mercury in a barometer.

b) The density of water is 1.00 × 103 kg.m-3. Estimate the height of the column of water in a waterbarometer.

3-15 QUESTIONS ON BUOYANT FORCESQ3.9 Density of air = 1.29 kg.m-3.

Density of helium = 0.18 kg.m-3.A balloon is filled with helium. Its volume is 1.00 m3. What downward force must be applied to stop

the balloon rising up into the sky?Q3.10 Estimate the buoyant force exerted on you by the atmosphere.Q3.11 An ice cube is floating in a glass of water. The water and the ice are at about 0°C.

Density of ice at 0°C = 917 kg.m-3.Density of water at 0°C = 1000 kg.m-3.

a) What fraction of the ice cube is submerged?

FE3: Equilibrium 40

b) Explain what happens to the level of the water as the ice melts.Q3.12 Steel is about eight times as dense as water. How can ships float?Q3.13 The density of water varies with temperature as shown in the graph below. The curve has a maximum at

about 4°C.

0 10 20 30 40995

996

997

998

999

1000

Temperature/°C

Density/kg.m -3

Figure 3.25 How the density of water changes with temperature

a) When a beaker of water is heated from below, why does the warm water at the bottom rise?b) When a lake ices over, why does freezing occur only at the top of the lake? Is there any biological

significance in that?

Interlude 3: Masses 41

INTERLUDE 3 - THE RANGE OF MASSES IN THE UNIVERSEmass/kilograms

Mass of the Universe 1050 ___

_ _ _1040 __ Mass of our Galaxy _ _ _ _

Mass of the Sun 1030 __ _ _

_ Mass of the EarthMass of the Moon _

1020 __ _ _ _ _1010 __ _ _ Ocean liner _

Person _1 __

_ _ Sugar cube _

Postage stamp _10-10__ _ _ Red blood corpuscle _ _10-20__ _

Protein molecule _ _ Oxygen molecule

Proton _10-30__ Electron

41

FE4 MOTION OF BODIES IN FLUIDS OBJECTIVES

AimsBy studying this chapter you will extend your knowledge and understanding of forces - in particularthe nature of forces exerted on solid bodies and small particles as they move through fluids. Youshould also acquire a basic understanding of the processes of sedimentation, diffusion andBrownian motion.Minimum learning goalsWhen you have finished studying this chapter you should be able to do all of the following.1. Explain, interpret and use the terms

drag force, lift force, terminal velocity, Brownian motion, diffusion, dynamic equilibrium,sedimentation.

2. Recognise and describe examples of situations involving forces exerted by fluids on movingobjects.

3. Draw force diagrams and analyse forces for systems subject to lift and drag forces.4. Describe how the forces exerted by fluids on moving bodies vary with the speed of motion.5. Describe and explain how fluids affect the motion of large objects - in particular, explain how

terminal velocity is reached in certain circumstances.6. Describe the process of sedimentation.7. Explain how fluids affect the motion of small particles and the nature of Brownian motion.8. Describe and explain the important features of the diffusion process.

PRE-LECTURE

4-1 INTRODUCTIONThe laws governing the motion of an object, as given in the previous chapters, apply only when allthe forces acting on the object are taken into account. In the real world all objects move throughfluids so we must include the forces exerted by the extremely large number of fluid molecules. Insome situations the effect of these forces can be ignored, in others they play a crucial role indetermining the motion.

We have seen in chapter FE3 that, in the special case where a large object is at rest, the forcesexerted by the fluid molecules sum to give the buoyant force, a vertical force acting through thecentre of gravity of the displaced fluid, with magnitude equal to the weight of the displaced fluid.The forces exerted by fluids on very small objects or on moving objects are, however, much morecomplicated. This lecture will give a survey of these forces and explain qualitatively how livingorganisms are affected by them. In addition to its intrinsic importance, this topic is a source ofmodels for non-mechanical phenomena.Questions to think about• When a car is driven at constant speed along a level road, a constant force is exerted on the carby the road surface. Why doesn't the car accelerate because of this force?• Oxygen gas is a denser fluid than nitrogen gas. Why doesn't the atmosphere stratify into alayer of oxygen near the earth's surface and a layer of nitrogen above it?• Revise the section on buoyant forces and try question 3.10 in FE3.

FE4: Motion in Fluids 42

LECTURE

4-2 FLUID FORCES ON MOVING OBJECTSThere are three important things to note about the force exerted by a fluid on a moving object whichis very much larger than a fluid molecule.(i) The fluid force is not just the buoyant force. For example two pieces of cardboard with thesame mass and volume (and hence with the same weight and buoyant force acting on them) do notfall at the same rate.(ii) The force is not always vertical. For example the force exerted by the air on a ‘parachute’moving horizontally at constant velocity must balance the other forces acting.

Hand force

Fluid force

Weight

Figure 4.1 Forces on a toy parachute

(iii) The force is not always along the line of motion. For example the fluid force on a marsupialglider moving at constant velocity must be vertical, balancing the animal's weight.

For convenience, the fluid force is usually divided into three parts:• the usual buoyant force,• a lift force perpendicular to the line of motion, and• a drag force parallel to the line of motion, opposing the motion.

Figure 4.2 Forces on a glider exerted by the airThe glider's weight is not shown.

The lift and drag forces come into play only when the object is moving. Usually they increasewith speed and with the size of the object.


The total fluid force does not necessarily act through the centre of gravity of the object. If itdoes not, then the object may have a rotational acceleration. (See FE3.) This complication will beignored here.

There are some objects (such as non-rotating spheres) for which the lift force is small or zero.Other objects produce large lift forces - aerofoils and birds' wings, for example.

There is always a drag force on an object moving in a fluid. For objects of diameter less thanabout a millimetre moving in water or air, or larger objects moving in more viscous (sticky) fluids:

drag force ∝ speed .

For objects of diameter very much greater than a millimetre moving in water or air:drag force ∝ (speed)2 .

4-3 TERMINAL VELOCITY IN A FLUIDDemonstrationA weighted balloon falls vertically through the air.ExplanationAn object falling vertically in a fluid does not accelerate at a constant rate. The downward velocityapproaches a constant value known as the terminal velocity.

Time

without drag force

with drag force

Downward velocity

Terminal velocity

00

Figure 4.3 Approach to terminal velocity

This behaviour is caused by the drag force. Consider the following force addition diagrams(figure 4.4).

B B

W WD

B

W D

Total force

Drag force

Buoyantforce

Total force

v = 0 0 < v < vT v = vT

Figure 4.4 System of forces on a falling object


When the object is at rest, the forces on it are its weight and the buoyant force. The sum ofthese forces (represented by the solid arrow in figure 4.4) produces a downward acceleration. Asthe downward velocity increases, the drag force increases from zero, reducing the total downwardforce. The downward acceleration gradually decreases to zero. The object then travels at its terminalvelocity, vT . (If there were no drag force, the total downward force and the downward accelerationwould have been constant.)

The velocity-time graph (figure 4.3) is characterised by the terminal velocity and the "risetime" - the time taken to reach terminal velocity (or more precisely some specified fraction of theterminal velocity). Both these quantities can be found from the equation of motion:

m dvdt = total force.

An object travelling at terminal velocity is in equilibrium. The terminal velocity is the velocityat which the drag force balances the other forces acting;i.e. total force = 0.

An approach to a constant velocity will always occur whenever there is a constant driving forceand an opposing force which increases with velocity until it is as large as the driving force.Examples• Birds of prey, initially travelling faster than their terminal velocity are slowed down to terminalvelocity by the drag force.• Particles of different sizes and densities have different terminal velocities in water. (See thesection on sedimentation in the POST-LECTURE.)

4-4 BROWNIAN MOTION AND DIFFUSIONWe now consider the motion of very small particles in fluids.ExamplesExamples include the circulation of nutrients and waste materials in living organisms. Particlesmove independently of the flow of blood in animals or the flow of sap in plants. Particles of mattermove through the fluids, rather than being carried along with the fluid. What are the forces exertedby fluids on the particles in these examples?ExplanationHere we are looking at very small objects in a fluid which are continually undergoing randomcollisions with fluid molecules. This situation is in contrast with that previously discussed, in whichthe objects had diameters very much larger than 10-6 m and masses more than 1012 times the massof a fluid molecule. In those cases only the effects of "averaged out" forces (buoyant, lift and drag)were detectable.

For objects smaller than 10-6 m the accelerations caused by individual collisions are larger.Such small objects also travel further between collisions (about 10-10 m to 10-7 m) since they areless likely to collide with fluid molecules because of their smaller sizes. The objects therefore makea series of random movements called Brownian motion. (See figure 4.5.)

A large collection of objects, each undergoing a random Brownian motion, can move as awhole in an organised way. This process is called diffusion. DemonstrationA one dimensional mechanical analogue of diffusion is presented. It is shown how objects canspread out by a sequence of purely random motions. The people move randomly using a codebased on the last digits from a page of telephone numbers: 0-3 means stand still; 4-6, move right: 7-9, move left.


Figure 4.5 Brownian motion

Two important features of a diffusion process(i) There is an overwhelming probability that the random movements will cause objects to diffusefrom regions of high concentration (of objects) to those of low concentration.(ii) The average time for an object to move a distance d from its starting point is proportional tod 2.

For example, glucose molecules (relative molecular mass 240) in water (relative molecularmass 18) take 10-3 s to diffuse 10-6 m but 109 s (about 20 y) for 1 m.

Diffusion is an effective mechanism of transport of materials over typical cell dimensions (lessthan about 10-5 m) in water. For example jellyfish without circulatory systems can obtain oxygenand other nutrients from the surrounding water and lose carbon dioxide. Larger organisms require acirculatory system and perhaps a respiratory system to bring the nutrients to the cells .

Diffusion occurs much faster in gases than in liquids because of the lower density of the fluid.To illustrate this, the upward diffusion of bromine gas into air, is demonstrated. A state of dynamicequilibrium is eventually reached where, at each point in space, the net rate of upward diffusion toregions of lower bromine concentration is balanced by the rate of falling.

The final demonstration shows people participating in a two-dimensional mechanical analogueof diffusion.

POST-LECTURE

4-5 SEDIMENTATIONConsider a small particle of density d and volume V falling with velocity v in a viscous fluid ofdensity ρL. The forces acting on the particle are shown in figure 4.6.

Buoyant force = ρL VgDrag force = λv

Weight = ρVg

Figure 4.6 Forces on a particle during sedimentation


The constant λ in the expression for the drag force depends on the viscosity of the fluid aswell as the size and the shape of the particle.

When the particle is travelling at its terminal velocity (or sedimentation rate) vT, these threeforces are in balance.

ρLV g + λ vT = ρ V g ;

∴ vT =Vλ (ρ - ρL) g .

Since the value of V/λ , and hence the terminal velocity or sedimentation rate, depends on thesize of the particle, sedimentation is a useful method for separating particles of different sizes. It isused, for example, in agricultural laboratories to determine the amounts of various types of soils in asample.

The sedimentation rate depends on the difference in the densities of the particles and thesurrounding fluid. It is often quite slow for biological materials in water since their density is almostequal to that of the surrounding water.

4-6 QUESTIONSQ4.1 For a spherical particle of radius r falling in a viscous fluid characterized by a coefficient of viscosity η,

the terminal velocity is

vT = 2r2

9η (ρ - ρL) g .

Find the terminal velocity or sedimentation rate of a soil particle of density 2.6 × 103 kg.m-3 andradius 0.50 × 10-6 m in water.

Viscosity of water η = 1.0 × 10-3 kg.m-l.s-l .Density of water ρL = 1.0 × 103 kg.m-3.

Q4.2 A man with a parachute is falling vertically through the atmosphere with a velocity v . The system, manplus parachute, experiences a drag force described by λv2 where λ is a constant.

i) What other vertical forces are acting on the system? At terminal velocity, the vertical forces add togive zero total force.

ii) Find an expression for the terminal velocity in terms of these forces.Assume that the man and parachute together weigh 850 N and that the buoyant force on them is

negligible. (See question 3.10 in FE3.)The constant λ depends on the shape of the falling object. When the man is falling in the sky-diving

position (on his stomach with arms extended) with his parachute unopened, λ is 0.22 kg.m-l and when hehas his parachute opened up, λ is 26 kg.m-1.

iii) Find the terminal velocity of the man (and parachute) in these two positions.If the man pulls his ripcord after sky-diving, he decelerates from the first of these terminal velocities to

the second in about 1 s .iv) What would be the average acceleration of the man during this l s interval? (Compare this with the

acceleration due to gravity.)

Q4.3 Suppose you are driving a car along a straight level road at a constant speed of about 80 km.h-l by keepingthe accelerator pedal in a fixed position. At a high speed like this, the drag force, which is proportional tothe square of the speed, is quite large. If you depress the accelerator pedal a bit more and keep it in thisnew position, the driving force on the car increases to a new constant value.i) Explain, in terms of the drag and driving forces, what happens to the speed of the car.ii) What happens if you then let the accelerator pedal come back to its original position? Sketch avelocity-time graph for the complete motion.

Q4.4 In the absence of air resistance, an object thrown vertically upwards would take the same time to rise to itsmaximum height as it does to return from it. With a drag force present this is no longer the case. Usegraphs of acceleration and velocity as functions of time to determine whether the object takes longer to riseor to fall. You may neglect the buoyant force here.

(Hint: Use slopes and the areas between these graphs and the time axis where appropriate.)


4-7 A USEFUL MATHEMATICAL MODELThe mathematical equations used to describe the motion of a small object falling in a fluid, whichexerts a drag force proportional to the velocity, can also be used to describe currents in certainelectrical circuits and nuclei undergoing radioactive decay. Try question 4.5 if you have thenecessary mathematical expertise.Q4.5 If the buoyant force is neglected, a small object falling vertically through the air has a downward

component of velocity, vx , at time t given by

vx =

mgλ

1 − e-λmt

.

i) What is the limit of the velocity (i.e. the terminal velocity) as t → ∞?

ii) What is the starting velocity at t = 0 ?iii) After what time (the rise time) is the difference between the terminal velocity and the actual velocity equal

to l/e of the terminal velocity ?iv) Sketch the velocity component as a function of time.v) Show by differentiating the expression for v that the downward component of acceleration, ax , at time

t is given by

ax = ge-λmt

.

vi) Sketch this acceleration component as a function of time.vii) Hence verify that the velocity component satisfies the equation of motion

mdvxdt

= mg − λvx

viii) Does the terminal velocity obtained directly from the equation of motion agree with that found in (i)?ix) Is the terminal velocity ever reached in this model ? How does the real world differ from the model in this

respect?x) Show by differentiating the result of (v) that the acceleration component obeys the equation

mdaxdt

= −λax

4-8 COLLOIDSInsoluble particles small enough to undergo Brownian motion will not fall in a fluid. They formwhat is known as a colloidal suspension which may be thought of as something intermediatebetween a true suspension and a solution. Common examples of colloids include smoke particles,the fat in homogenised milk and the protein molecules and protoplasm of cells. The particles arekept from clumping together, usually by some kind of electrostatic repulsion. If the surroundingfluid is altered in some way, for example by changing its acidity, temperature or chemical properties,the colloidal particles can often aggregate and settle, as for example in the setting of jellies andclotting of blood.

4-9 RANDOM NATURE OF DIFFUSION - QUESTIONSQ4.6 In the two dimensional mechanical analogue of diffusion in the video lecture, you may have noticed two

people attempting to sabotage the demonstration by making precisely the same move as one another ateach stage. At each stage, each person could move in 9 distinct ways, each with equal probability.

What is the probability that two people at the same position at the same time will make precisely thesame next move, purely by chance?

What is the probability that they will make the same next ten moves?Q4.7 Carbon dioxide gas molecules take 25 seconds to diffuse about a centimetre in air. How long would it take

them to diffuse through l metre?

Interlude 4 - Energies 48

INTERLUDE 4 - THE RANGE OF ENERGIES IN THEUNIVERSE

energy/joules1050 __

Energy equivalent* of Sun's mass _ _ _ _1040 __ _

_ _ _ Sun's annual output1030 __ _

Solar energy received by Earth _annually _

_1020 __ Annual world use of energy

Earthquake, cyclone, H bomb _ _ (Energy equivalent* of 1 g of matter _ ...(Energy from fission of 1 kg of U235 _ (Burning 7000 tonnes of coal1010 __ _

8 hours hard labour _ 1 kilowatt hour (3.6 × 106 J) _

KE of a rifle bullet _1 __

_ _ _ _10-10__ Energy equivalent of a proton

Energy of a nuclear bond _ Energy equivalent of an electron _ _

Ionization energy _Energy of a chemical bond 10-20__ 1 electron volt (1.6 × 10-19 J)

_ _ _ _10-30__

* Energy equivalent of mass is calculated using E = m c 2, where c is the speed of light in vacuum.

49

FE5 ENERGY OBJECTIVES

AimsFrom this chapter you will acquire a basic understanding of the concepts of work and energy. Youwill not understand everything about them, because energy is an idea which has no adequate simpledefinition. It is also a concept which pervades all of science, so you can expect to learn more andmore about energy throughout this course.Minimum learning goalsWhen you have finished studying this chapter you should be able to do all of the following.1. Explain, interpret and use the terms:

work [mechanical work], energy, joule, power, average power, watt, kinetic energy,conservative force, non-conservative force, potential energy, gravitational potential energy,mechanical energy, internal force, external force, system, isolated system.

2. State and apply formulas for work done by constant forces on objects moving along straightpaths.

3. Calculate work and potential energy from force-displacement graphs. Sketch and plot graphsof potential energy against position.

4. Calculate gravitational potential energy for objects near the Earth's surface.5. State the conditions for the total mechanical energy of a system to be conserved.6. Solve simple problems involving goals 2 to 5 above.

PRE-LECTURE

5-1 INTRODUCTIONNewton's three laws of motion can, in principle, be used to study the motion of any complex systemwith the following important exceptions:-(i) systems of atomic dimensions or smaller, e.g. atoms in a molecule, electrons in an atom,

protons and neutrons in an atomic nucleus (here a new form of mechanics called quantummechanics must be used);

(ii) systems containing particles moving at speeds near the speed of light (here Einstein'srelativistic mechanics, a modification of newtonian mechanics, must be used).Physicists do not fully understand how to treat systems of atomic dimensions moving at

speeds near that of light.Newton's laws are a valid approximation for systems of particles which are much larger than

atoms and move at speeds small compared to that of light. Even in these cases, however, it may notbe feasible to analyse the motion using Newton's laws because of mathematical complexityassociated with large numbers of particles. Physicists use the concept of energy to treat suchsystems. Energy also plays a central role in quantum mechanics.

Historically, the concept of energy was first applied to mechanical systems like colliding ballsand the pendulum. It was observed that the sum of quantities called ‘kinetic’ and ‘potential’energies was a constant. Later the concept was generalised to include other forms - nuclear,chemical, radiation (light), electrical, heat and sound energy. The divisions between these forms are

FE5: Energy 50

somewhat arbitrary. For example, thermal energy is now understood to be nothing more thanmechanical energy of the atoms and molecules of the system.

Energy is conserved but may be transformed from one form to another, e.g. mechanical energyto heat, etc.

This chapter is one of the key parts of this course. It describes the two forms of mechanicalenergy (kinetic and potential) and how energy is transferred to systems by means of a process calledmechanical work. Under certain circumstances, the total mechanical energy (the sum of the kineticand potential energies) of a system remains constant in time. This result can be used to study themotions of particles in very small systems such as nuclei, atoms and molecules, as well as in largescale systems.

The SI unit of energy and work is the joule (J); l J = 1 N.m = l kg.m2.s-2. To give someidea of the size of a joule: it is the energy required to raise an object weighing 1 newton (a mass ofabout 0.1 kg on Earth) e.g. an apple, by 1 metre.

Power means rate of transfer of energy. In purely mechanical systems this is equal to the rateat which work is done by some force. The SI unit of power is the watt (W); l W = l J.s-l. Forexample a 100 W light globe transfers 100 J of electrical energy to heat and radiation in 1 s. Andlifting an apple weighing 1 N a height of 1 m in 1 s corresponds to an average power of 1 W.

The buoyant force exerted on objects by the air will be neglected throughout this chapter sinceit is very small compared to the weight of the objects considered. (See, for example, question 3.10 inchapter FE3.)

LECTURE

5-2 ENERGY TRANSFERS IN THE SOLAR ENERGY CYCLEWith the exception of some nuclear fuels, geothermal and tidal power, the sun is the ultimate sourceof energy within the solar system. Hydrogen nuclei within the sun combine to release energy asheat which is transferred to radiation near the surface of the sun. Some of this radiation is convertedto chemical and thermal energy at the Earth.

Radiation is absorbed directly by molecules in the atmosphere and at the Earth's surface andappears as thermal energy. This energy is partially re-emitted as radiation; the remainder istransferred to mechanical energy, for example in atmospheric motions and in the water cycle.

The leaves of plants use radiant energy to convert carbon dioxide and water to carbohydratesby photosynthesis. This energy undergoes further chemical conversion in animal life and can alsobe recovered by burning fossil fuels such as coal and oil. (If you are interested, see the diagram inthe interlude following this chapter and Scientific American, September 1970, for further details.)

5-3 MECHANICAL WORK - A MEANS OF ENERGY TRANSFER Work is a measure of energy transfer. Work is done by a force acting on an object whenever theobject moves. If there are several forces acting on the same object then each of them does work.(This is different from the colloquial use of the word ‘work’ - in the physics definition, no work isdone if there is no movement.)Work done by a constant force for straight-line motionCase (i): Force parallel to the motionConsider first the simple case of a particle moving along a straight path, being acted on by a constantforce F, which acts in the same direction as the motion.

W = F (x l - x 0) ,

FE5: Energy 51

i.e. the work is equal to the product of the force and the displacement.Case (ii): Force at an angle to the motionIn this case we consider a constant force in a direction which makes a constant angle θ to thedirection of the motion. Work is done only by the component, Fx , of the force which is parallel tothe line of motion. The work done by the constant force as the particle goes from position x0 to apoint x l is:

W = Fx∆x = F cosθ (x1 - x0) ... ( 5.1) .

Note the following.• Case (i) is a special instance of (ii) with θ = 0.• The component of F perpendicular to the line of motion does no work.• The work done by F is the same as that done by Fx alone.• The work done is positive if the component Fx is in the direction of motion and negative if Fxis opposed to the direction of motion; i.e. if the force is ‘holding back’ the object, it does negativework.• If the constant force component Fx is plotted as a function of the position x of the object, thework done is represented by the area between the graph and the x axis from x0 to xl as shown in figure 5.1.

x x 1

W

x 0

Parallel force component

F x

Position

Figure 5.1 Work done by a constant force during astraight-line displacement

Sign convention for graphs In drawing graphs we arbitrarily choose a particular direction and call it positive. If Fx and thedisplacement (∆x = xl - x0) are both positive or both negative, then W is positive. However if Fxand the displacement have opposite signs (i.e. if the force is opposed to the direction of motion) thenW is negative.Motion in a straight line with a variable forceNow consider the case where F is an arbitrary force, so that Fx varies with position x as the bodymoves along a straight path. On a graph of Fx against x the work W is now represented by the areabetween the curve and the x axis (with the above sign convention) See figure 5.2.

FE5: Energy 52

Parallel force component

+

- x x 1

x 0

( )F xx

Position

Figure 5.2 Work done by a variable force in straight line motionOnly the component Fx of the force which is parallel to the straight path does any work.

The procedure illustrated in figure 5.2 is equivalent to the evaluating the integral

W = Fxdx

x2

x1∫ . ... (5.2)

General caseIn the most general case of all, in which the motion is along a curved path, work can be defined byconsidering a variable coordinate direction which is everywhere parallel (tangential) to the path tracedout by the moving object. The coordinate x is replaced by a coordinate measured along the path,whose magnitude is equal to distance, s, along the curved path. Consider a graph of the tangentialcomponent, Fˆ, of the force plotted against the coordinate s. Then the work done by the force F isrepresented by the area under the curve, exactly as in the one-dimensional case.

5-4 TRANSFER OF ENERGY TO SYSTEMS BY MECHANICAL WORKA force doing a positive amount of work on a system transfers energy to that system. The energycan end up in one or more of the following forms:-

• non-mechanical energy,• kinetic energy or• potential energy

Non-mechanical energyAn example is a system consisting of a rigid object and a rough surface over which the object moves(figure 5.3). The applied force does work on the object while the object moves at constant velocity,and energy is transferred to thermal energy at the surfaces in contact. Note that there must be anequal and opposite force - friction - acting on the object which balances the applied force, since weare told that the object is moving at constant velocity.

Friction

Applied force

Figure 5.3 Object dragged across a rough surface

FE5: Energy 53

Kinetic energyAn example is a system consisting of an object and smooth (frictionless) surface on which it sits(figure 5.4). When a horizontal force is applied to the object there is no opposing force - the appliedforce increases the speed of the object and hence, also, its kinetic energy.

Applied force

Smooth surface

Figure 5.4 Object dragged across a perfectly smooth surface

The kinetic energy of an object of mass m travelling at speed v is defined to be

K = 12 mv2 . ...(5.3)

Potential energyAn example is a system consisting of a rigid object, a smooth surface on which it sits and a springconnected as shown in figure 5.5. Suppose the applied force is pulling the object to the right atconstant velocity, thereby stretching the spring. (Note that the applied force must be equal andopposite to the force exerted by the spring on the object since the velocity of the object is constant.)What happens to the energy transferred to the system as work is done by the applied force? Energyis not being transferred to either non-mechanical energy or kinetic energy, during this constantvelocity motion.

Fixedpoint

Object

Spring

Smooth surfaceFigure 5.5 Spring opposing the motion

Consider a situation where the spring is stretched with the object on the right. Suppose nowthat the applied force is reduced and kept slightly less than the force exerted by the spring. Theobject will move to the left towards the position where the spring is unstretched. During this processthe force in the spring is doing work on the pusher (the object exerting the applied force), which isnot part of the system. If the spring is ideal, the magnitude of the work done by the spring equalsthe work done by the applied force in extending the spring. The energy released by the systemcomes from the potential energy (PE) stored when the spring is stretched. Potential energy shouldbe thought of as a property of the whole system and not of individual parts of the system.

The force exerted by the spring is called conservative since the energy transferred to thesystem can be recovered directly as mechanical energy. There is always a conservative forceassociated with potential energy.

Frictional forces and drag forces are called non-conservative since mechanical energy cannotbe recovered directly by removing the applied force which produced the motion.

FE5: Energy 54

5-5 CONSERVATION OF MECHANICAL ENERGYTerminologyAn internal force is one exerted by one part of a system on another part of the system. Anexternal force is one exerted by something outside the system on some part of the system. Anisolated system is one on which no external forces are doing work, i.e one to which no mechanicalenergy is being added.

The total mechanical energy of a system is the sum of its kinetic energy and its potentialenergy. PrincipleIf there are no non-conservative forces acting within an isolated system, the total mechanicalenergy of the system is conserved (i.e. it remains constant as time progresses).

It does not matter what we take as the zero of the potential energy because in any problem weare interested only in changes in energy :

change in

total mechanical energy = change in

kinetic energy + change in

potential energy .

5-6 CALCULATING POTENTIAL ENERGYHow can the potential energy be found from the associated internal conservative force? ExampleIn the spring example above - case (iii) - we usually choose PE = 0 when the spring has its naturallength, neither extended nor compressed. Then the potential energy at extension x is equal to thework done by the force exerted by the spring in returning the object from x to 0.

Force exerted byspring at position x

O

O

x x

PE(x)

(a) (b)

Figure 5.6 Force and potential energy for an ideal spring

There is no component of F perpendicular to the motion, so in this instance Fx is the full forceacting. The PE at x indicated in figure 5.6b is therefore represented by the shaded area in figure5.6a.

FE5: Energy 55

General procedureThe general procedure for finding the potential energies of the various states of a system is asfollows.(i) Assign zero PE to one state of the system.(ii) The PE of any other state of the system equals the work done by the internal conservativeforces when the system moves from that state to the zero PE state.

5-7 WHY IS POTENTIAL ENERGY A USEFUL CONCEPT?In general the state of a system at any instant is specified by giving the positions and the velocitiesof all of the parts of the system. The PE of a state does not, however, depend on the velocities of theparticles or on how the system came to be in that state. It depends only on the positions of theparticles in the system (i.e. on their configuration). Therefore, the change in PE when the systemmoves from one state to another will depend only on the configurations of the particles in the twostates and not on how the system moves between the states.

Since the gravitational force is conservative and the electrostatic and the two nuclear forces areall conservative on atomic and nuclear scales, complicated processes can be studied comparativelyeasily using PE diagrams and the principle of conservation of mechanical energy.

POST-LECTURE

5-8 ENERGY TRANSFERRED AS WORK - QUESTIONSThe following two questions are designed to help you revise the definitions given in the lecture.Q5.1 An object of mass m slides down an incline at angle θ to the horizontal as shown.

θ

Figure 5.7 Object sliding down an inclineWhat are the forces on the object?

i) On figure 5.7 draw in all the forces acting on the object.ii) Which of these forces are doing work on the object ?iii) Which of the forces doing work on the object are conservative ? Which are non-conservative ?

For the rest of this question assume that the non-conservative forces present are negligible.iv) What work is done by the gravitational force on the object when the latter slides a distance d down the

incline ?What is the increase in kinetic energy of the object during this process?

v)

θ

F

Figure 5.8 Object dragged up an inclineDraw in the forces.

FE5: Energy 56

Suppose that a force F with magnitude F greater than mg sinθ is used to pull the object a distance d upthe incline as shown in figure 5.8.

What work is done by F ?What work is done by the gravitational force?What is the increase in the kinetic energy of the object during this process?

Q5.2 An object of mass m swings at the end of a taut, straight unstretchable rope whose other end is fixed.

Figure 5.9 A swinging objectDraw in the forces. Which forces do work?

i) On figure 5.9 draw in all the forces acting on the object.ii) Which of these forces are doing work on the object?iii) Which of the forces doing work on the object are conservative? Which are non-conservative?

Assume that the non-conservative forces present are negligible. Suppose that the object is releasedfrom rest from some raised position with the rope taut. The work done on the object by the gravitationalforce when the object moves from the point of release to the lowest point of its swing equals the kineticenergy of the object at the lowest point. The object continues its swing until it momentarily comes torest at some highest point on the other side. At this highest point the kinetic energy of the object is zero.Therefore the work done by the gravitational force when the object moves from the point of release to thishighest point is zero. (Here the object moves along a curved path so it is difficult to calculate the workdone by the gravitational force. This problem will be considered from a different viewpoint in Q5.4below.)

5-9 GRAVITATIONAL PE NEAR THE EARTH'S SURFACEConsider a system consisting of an object of mass m and the Earth. For our purposes the Earth maybe considered immovable so that all forces acting on it do no work. In any small region near theEarth's surface it is possible to choose any horizontal level and say that the system has zero PE whenthe object is on that level. If the object is at height h above this horizontal level the gravitationalpotential energy of this system is equal to mgh. If the object is a vertical distance h below thishorizontal level, the PE is equal to -mgh . (A proof of this result will be outlined in Q5.10.) |U |

= |mgh| ... (5.4)

PE =

PE =

mgh

-mgh

PE = 0

h

h

Figure 5.10 Gravitational potential energy

FE5: Energy 57

Note that these simple expressions hold only if h is small compared to the distance from thecentre of the Earth. If this condition is not satisfied, the change in the gravitational force with heightmust be taken into account.

5-10 QUESTIONSConservation of mechanical energyQ5.3 Consider the object sliding down the incline described in Q5.1. Assume that the non-conservative forces

acting on the object are negligible.The system consisting of the object and the Earth is isolated according to the definition given in the

lecture. (The incline can be treated as part of the Earth.) The gravitational force on the object is nowinternal to the system.

i) Can you explain why this system is isolated?ii) Use the idea of conservation of mechanical energy and the above expressions for the gravitational potential

energy to find the increase in kinetic energy of the object when it slides a distance d down the incline. (Doyou get the same answer as that obtained using the method of Q5.1?)

Q5.4 Consider the swinging object described in Q5.2. Assume that the non-conservative forces acting on theobject are negligible.

The system consisting of the object and the Earth (the string is not part of the system) is isolated. Thegravitational force is now internal to the system.

i) Can you explain why this system is isolated ?ii) Suppose that the object is released from rest from height h above the lowest point of its swing. Find the

maximum kinetic energy of the object and the maximum height to which it rises on the other side.iii) How would your answers to part (ii) differ if non-conservative forces were not negligible? (Note that the

system would then no longer be isolated.)Energy dissipationQ5.5 Use energy considerations to explain why an object thrown vertically into the air takes longer to fall than

it does to rise if air resistance is taken into account (c.f. Q4.4 in FE4). Hint: consider the speed of theobject at a given height on the upward and downward portions of its motion.

Energy balancesQ5 .6 Flow of sap in trees

A typical tall tree manages to raise water through a height of 20 m from roots to leaves at a rate of700 kg per day. (Approximately 90% of this water is evaporated from the leaves, the remainder is used inthe photosynthesis process.) What energy must be supplied in a day to raise this water?

It is impossible to raise water to a height of more than about 10.3 metres using a vacuum pump whichutilises atmospheric pressure. How do trees manage to accomplish this? (See Scientific American, March1963, pp 132-142.)

Q5.7

v

Figure 5.11 At what rate does the possum work?The metabolism of a possum of mass m is such that the possum is capable of doing work at a

maximum rate of P. When the possum climbs a tall vertical tree at constant speed, it must do work at arate equal to the rate of increase of the potential energy of the possum-Earth system. (We can neglect anynon-conservative forces for the purposes of this discussion.)

i) If the possum is moving at speed v, at what rate is it doing work?ii) If m = 3.0 kg and P = 60 W what is the shortest possible time in which the possum can climb a tree 20

m high ? (These ideas will be used in chapter FE8 on Scale)

FE5: Energy 58

MachinesMachines usually are devices for doing work using a small force travelling a large distance. Forexample on an inclined plane like a winding road, only a force large enough to overcome thecomponent of gravity parallel to the plane (and any non-conservative forces present) is required.However, one must travel a longer distance to reach any given height.Q5.8 When using a lever like a crowbar, one applies a force at one end to raise a load at the other (figure 5.12).

Load Appliedforce

a b

Figure 5.12 Lifting with a leverCompare forces and works.

i) By considering torques about the pivot, find the ratio of the load, mg, to the minimum applied force Frequired to raise the load, in terms of the ratio of the lengths a and b of the lever on either side of the pivotpoint. You may neglect the weight of the lever.

ii) If the load is raised a small height h , the work done by the minimum applied force must equal theincrease, mgh, in the PE of the load-Earth system. How far does the applied force have to move its end ofthe lever? (Ignore the fact that the ends of the lever move in slightly curved paths.)

Q5.9 Draw a schematic diagram of the main bones in the arm and the biceps muscle used to raise the forearm.Discuss qualitatively how this ‘machine’ in the human body differs from the machine used in Q5.8. Hint:consider the work done by the biceps muscle in raising a load held in the hand.

Gravitational force and potential energyIn questions 5.3 to 5.9 you used the result that the gravitational potential energy of an object-Earthsystem depends only on the height of the object. Question 5.10 outlines a proof of this result.Q5.10

AB

C F

ED

Figure 5.13 Work done by gravityShow that the work done by gravity does not depend on the path taken.

In figure 5.13 A is on the same horizontal level as B which is vertically above C , a distance h below.i) Find the work done by the gravitational force on an object of mass m when the object moves from A to B

as shown.ii) Find the work done when the object moves from B to C and hence the total work done when the object

moves from A to C along the path A → B → C.

iii) The path A → D → E → F → C is made up of straight vertical and horizontal segments as shown.Verify that the total work done by the gravitational force on the object when it moves from A to C alongthis path is equal to the result of (ii).

FE5: Energy 59

A

CFigure 5.14 Straight-line approximation to a curved path

Any curved path from A to C can be approximated as closely as desired by a series of vertical andhorizontal segments - see figure 5.14. The work done along the curved path is then approximately equal tothe total work done along the straight line segments, i.e. the result of (ii) .

Thus we can unambiguously define the difference in PE between the states of the object-Earth systemrepresented by A and C to be the work done by the gravitational force when the object moves from A to Calong any path, i.e. PE is equal to mgh in every case.

Other conservative forces and potential energyQ5.11

Movingobject

Wall

x Figure 5.15 Object striking a spring buffer

What happens to its kinetic energy?

Figure 5.15 shows a spring-buffer arrangement to prevent a moving object from striking a wall. The forceF exerted by the spring-buffer on the object is plotted in figure 5.16 as a function of the distance x of theobject from the wall.

8

4

00 2 31 x / metre

F / kilonewton

G E D C B A

Figure 5.16 Force exerted by the buffer spring on the block

Choose the PE of the system to be zero when the moving object is 3 metres from the wall, i.e. at thepoint labelled A .

i) Find the potential energies at points B, C, D, E and G by calculating the work done by F when the objectmoves from those points to A . (Hint: Use areas under the graph.)

ii) Sketch the potential energy diagram for the system.iii) Use the diagram to find the minimum kinetic energy of the object for it to strike the wall.iv) How would the answers to (i) - (iii) be altered if the PE of the system had been chosen to be zero at

x = 0?

FE5: Energy 60

Example: Atoms in a diatomic moleculeThe force between atoms is a complicated one due to individual electrical forces exerted by thenucleus and the electrons of one atom on those of the other atom. There is an infinite repulsive forceat zero separation and the force decreases and becomes attractive at larger separations (figure 5.17).

Component offorce exertedby atom at0 on atomat positionx

x Separation

Repulsive force

Attractive force

0

Figure 5.17 Force between atomsThe graph shows the force exerted by one atom on the other. The component of the force is taken inthe direction of the line joining the atoms. Positive values of the component correspond to repulsion;

negative values represent attraction

The zero of PE for the two atoms is conventionally defined to occur when the atoms are a verylarge (infinite) distance apart. Can you see why it would not be sensible to define PE = 0 when x =0?

The PE is then the work done by F(x) when the separation is increased from x to ∞ , i.e. theshaded area in figure 5.17. Although the curve extends to very large distances (∞) this shaded areais finite.

PE(

x Separation

0A

B

C

x )

0

Figure 5.18 Potential energy for the system of two atomsBy convention the PE is taken to be zero when the separation is infinite.

Q5.12 Use the section, Sign convention for graphs, on page 51 to explain why the PE is negative at thepoint marked A in figure 5.18 .

At point B , the PE is minimum. To which point on the force-separation curve does this correspond?How does the force behave on either side of this point ? What happens to a stationary object placed at thatpoint? At point C, the PE is zero. To which point on the force-separation curve does this correspond?

FE5: Energy 61

5-11 FINDING THE CONSERVATIVE FORCE FROM THE PE CURVEQuestions 5.10, 5.11 and 5.12 were practice in finding the PE curve for a system from theappropriate conservative force. It is worth noting in passing that since we obtain the PE curve fromthe conservative force by integration (areas under curves) we may also find the force bydifferentiating (finding the slope) of the PE curve (c.f. velocity and distance).

Conservative force component = - d dx (potential energy)

= - (slope of PE vs x curve).The minus sign is needed to give the correct direction for the conservative force.

This result can be used to give a quick check on whether the PE curve obtained from aconservative force is correct : the force should be negative if the PE increases as x increases, positiveif the PE decreases and zero if the PE is constant. Try this out on the PE curves you have met sofar.

5-12 CONCEPTUAL MODELS FOR POTENTIAL ENERGYIf you are finding PE curves to be rather abstract, you may like to think of them as depicting asuccession of hills and valleys viewed from the side (see figure 5.19).

x x 1

PE( )x 1

x2

PE(x)

2PE(x )

x2

0

Figure 5.19 Potential energy hills and valleys

The difference between the potential energies at x1 and x2 is just like the difference betweenthe gravitational potential energies when the object is at different heights on the hills above x1 and x2.

Consider an object sliding without friction up and down the ‘hills’. When viewed fromdirectly above, only the horizontal motion in a straight line can be seen. The object appears to speedup and slow down so there must be a horizontal force acting on it. This force at any point isobviously related to the slope of the hill at that point (c.f. previous section). The PE surface seen inthe TV lecture was a generalisation of the PE curve to a case where the object can undergo a twodimensional motion.

5-13 POWERPower, the rate at which energy is supplied to, released by, or dissipated within a system is often aquantity of interest.

The average power during a time interval is the energy transferred during that time intervaldivided by the time interval.Q5.13 Calculate the average power required to raise the water in the tree of Q5.6.

You will meet energy again in all of the other units of this course. The unit Thermal Physics inparticular is concerned mainly with transfers of energy to and from systems.

62

INTERLUDE 5 - EARTH'S ENERGY BALANCE AND FLOW

Note:1 TW = 1 × 101 2 W.

Short wavelengthradiation

Long wavelengthradiation

Tidalenergy

Solar radiation178 TW

Direct reflection62 000 TW (35%) Tides, tidal

currents etc.3 TW

Direct conversionto heat

76 000 TW (43%)

Storage:water & ice

Evaporation, precipitation, runoff etc.

40 000 TW (22%)

Winds, waves, convection and curre nts

Storage:plants

Photosynthesis40 TW

Decay

Animals

Fossil fuelsNuclear, thermal &gravitational enrgy

Terrestrial enegy

Conduction32 TW

Convection,volcanoes,hot springs

0.3 TW

63

FE6 ROTATION

OBJECTIVES

AimsA careful study of this chapter should equip you with a basic understanding of the concepts andprinciples used to describe rotational motion. You will learn how to apply these ideas to fairlysimple cases where things rotate about an axis whose direction in space remains fixed. You shouldunderstand the idea of a frame of reference and the kinds of modifications to physical theory whichare adopted in order to cope with situations observed from an accelerating frame of reference.Minimum learning goalsWhen you have finished studying this chapter you should be able to do all of the following.1. Explain, interpret and use the terms

angular velocity, angular acceleration, centripetal force, translational kinetic energy,rotational kinetic energy, frame of reference, pseudoforce, centrifugal force, centrifuge,ultracentrifuge.

2. Solve simple problems on uniform circular motion involving centripetal force, centripetalacceleration, mass, radius, speed and angular velocity.

3. Explain how the total kinetic energy of a rigid body can be expressed as the sum oftranslational and rotational kinetic energies and apply this result to simple problems.

4. Explain why and how the equation of motion for an accelerating frame of reference ismodified by including pseudoforces

5. State and apply the formula for centrifugal force in terms of mass, angular velocity and radius.6. Explain the operating principles of the centrifuge and ultracentrifuge.

PRE-LECTURE

The theme of this chapter is rotation. We shall examine two quite different aspects. Firstly,extending the discussion on energy in the previous chapter, we shall look at the kinetic energy ofrotating objects. Secondly, we shall look at the centrifuge. In order to talk about the centrifuge, wedigress and see how the equation of motion may be written in different frames of reference; inparticular, how it may be written in a rotating frame of reference.

FE6: Rotation 64

6-1 CIRCULAR MOTIONConsider a small object moving with constant speed v in a circle of radius R.

R

v

v dθ

Figure 6.1 Circular motion

In a short time interval dt it sweeps out an angle dθ . The angular velocity of the object isdefined to be

ω =dθd t . ... (6.1)

The angular velocity and the speed of the object are connected by the relation

ω = vR . ... (6.2)

If the angular velocity changes with time the object also has an angular acceleration:

α =dωd t . ... (6.3)

The SI unit of angular velocity is the radian per second, symbol rad.s-l. The unit of angularacceleration is radian per second per second, rad.s-2.

6-2 ROTATION OF A RIGID BODY ABOUT A FIXED AXISThe reason for introducing these angular variables is to simplify the description of rotating bodies.If a rigid body rotates about an axis which is fixed in space, then different parts of the body havedifferent linear velocities and accelerations. On the other hand, all the particles in the body have thesame angular velocity and the same angular acceleration. This description in terms of angularvariables is useful even if the axis of rotation moves.

Q6.1 In chapter FE1 we quoted the result that the centripetal acceleration of an object moving with speed v in a

circular path was equal to v2

R . Re-express this in terms of angular velocity. (We need this result later.)

PreparationRead the sections in chapter FE3 about rotational motion and moments of inertia. Revise questions2.4 and 2.6(b) in FE2.

FE6: Rotation 65

LECTURE

6-3 ROTATIONAL KINETIC ENERGYWhen a rigid body is rotating about a fixed axis, different parts of the object move with differentspeeds. Parts near the edge have greater speeds than those near the axis of rotation. Consider onesmall part of the object, with mass Δm, moving at speed v. Its kinetic energy is 1

2 (Δm) v2, so thetotal kinetic energy of the whole body is the sum of the kinetic energies of all its parts, which we canwrite as Σ 1

2 (Δm) v2. Note that there will be many different values of v in this sum. However, byusing the common angular velocity, ω, instead of speed we can write a simple formula for therotational kinetic energy:

KE = 12 Iω 2 ... (6.4)

where I is the moment of inertia of the object, mentioned in chapter FE3. The moment of inertiadepends on the distribution of mass in the body. If you consider two objects with the same totalmass, the one with more of its mass further from the axis of rotation has the higher moment ofinertia. Note also that a body does not have a unique moment of inertia; the value of I depends onthe location of the axis of rotation.

In general a rigid body (e.g. a boomerang) has both rotational and translational motion. Eachpart still has kinetic energy 12 (Δm) v2, but some of that energy is now associated with the motion ofthe body as a whole and the rest of it is associated with the rotational motion. Consequently the totalkinetic energy of the body can be expressed as a sum of translational kinetic energy, associatedwith the motion of the centre of gravity of the body, and kinetic energy of rotation. Provided that thebody is rigid and that its axis of rotation through the centre of gravity stays in the same direction inspace, the total KE can be written in the simple form:

total KE = 12 m vC2 +

12 Iω 2 ,

where vC is the speed of the centre of gravity and m is the total mass of the body.

Demonstrations• A block of dry ice slides down a slope more quickly than a round object rolls down the slope.

Gravitational force does work on the objects. All the PE of the sliding object is converted intoKE of translation. But the PE of the rolling object must be shared between KE of translationand KE of rotation which means that the translational motion takes only a fraction of the totalKE. (The value of that fraction depends on the shape of the body, but not its size.) At a givendistance down the slope, the speed of the centre of gravity must be less for the rolling object.

• A sphere rolls more quickly than a solid cylinder which in turn rolls more quickly than ahollow cylinder, even though all three objects have equal radii. This happens because the massis distributed differently in each object and the sharing between the KE of translation and theKE of rotation is different.

• In a model car energy is stored as rotational KE of a flywheel. It is converted into translationalKE of the car and eventually into non-mechanical energy which is lost from the system.

FE6: Rotation 66

6-4 ACCELERATED FRAMES OF REFERENCE AND PSEUDOFORCESFrames of referenceIf you want to describe the motion of an object you need a coordinate system, or somethingequivalent, to which you can relate your measurements of position; it doesn't make sense to giveposition coordinates (x, y, z) unless you know where the origin is and the directions of the axes.Specifying the origin and the axes is a formal, but somewhat abstract, way of describing a frame ofreference. In practice a frame of reference is anchored to some physical objects, and when we talkabout an observer we imagine a person at rest in his or her own reference frame. We now exploresome aspects of how the choice of a reference frame can affect the observed motions of objects andeven the laws of motion themselves.Demonstration: Acceleration in a straight line• A truck is moving at constant velocity (figure 6.2). An object is dropped and is seen to fall toa point on the floor of the truck, directly below the point of release. To an observer in the truck'sframe of reference, the object falls vertically along a straight line path. The truck's frame of referenceis not accelerating, and nothing unusual occurs.

... strikes floor here

Object released here ...

Observer moving with truck

Truck moves at constant velocity

No Acceleration

Figure 6.2 Dropping an object in a moving frame of referenceThe frame of reference attached to the truck is moving with constant velocity.

• In the second part of the demonstration the object is dropped while the truck is slowing downto stop (figure 6.3).

Figure 6.3 Dropping an object in an accelerated frame

In this case the truck and its frame of reference are accelerating. To an observer in the truck'sframe of reference it looks as if there is a force pushing the object forward as it falls. The observercannot explain this using the equation of motion. There is no known kind of forward force to

FE6: Rotation 67

account for the path of the falling object. The problem can be resolved by introducing a fictitiousforce, or pseudoforce, to make the equation of motion work in the accelerated frame of reference.Demonstration: Measurement of weight in a liftA body is weighed in a lift. (This problem was described in Q2.10, FE2, where a fixed frame ofreference, outside the lift, was used.) If the frame of reference is attached to the accelerating lift, thenthe forces on the body being weighed are the gravitational force, the supporting force exerted by theplatform of the scales, plus a pseudoforce, associated with the accelerated frame (figure 2.11). Thesupporting force, whose magnitude is equal to the apparent weight recorded on the scales dependson both the gravitational force and the pseudoforce.Motion in a circleImagine an object which has a string tied to it and is swung around in an approximately horizontalcircular path (figure 6.4). Think of an imaginary observer riding on the object. The frame ofreference for this observer is accelerating. Now the observer knows that the string exerts a force tothe left, but cannot explain why the object does not go off in that direction, although the equation ofmotion predicts that it should. To make the equation of motion work the observer introduces apseudoforce - in this case a force to the right in order to counteract the force exerted by the string.

Forceexerted bystring

Observer moving withthe object

?

Figure 6.4 Observer in a rotating frame of reference

We know from our bird's eye view of the object moving in the circle that the force exerted bythe string is equal to mω 2R. Therefore the moving observer's pseudoforce must also be equal tomω 2R, but it is in the opposite direction - outwards. This kind of pseudoforce is often calledcentrifugal force.

Forceexerted bystring

Pseudoforce2ω m R

Figure 6.5 Forces in a rotating frame

In summary, the equation of motion works properly only in a frame of reference that is notaccelerating. However it is often convenient, from the point of view of mathematical simplicity, touse an accelerated reference frame and non-physical pseudoforces which are invented only in orderto preserve the equation of motion. We say that the pseudoforces are non-physical firstly becausethey violate the law that forces always occur in pairs and secondly because it is not possible toidentify a physical object which is the source of the force.

FE6: Rotation 68

6-5 THE CENTRIFUGEThe centrifuge is used extensively to separate materials of differing densities. A tube containingthe materials swings around the axis of the centrifuge at a high angular velocity (figure 6.6).

Axis

Figure 6.6 Principle of the centrifuge

If a suspended particle is going to remain at a given radius as the centrifuge rotates, the forceson it due to the surrounding material must combine to provide a centripetal force mω 2R. If thesurrounding material cannot provide sufficient force the particle moves to a greater radius where thesurrounding material or the bottom of test-tube can provide the necessary centripetal force. (Seefigure 6.7.)

Figure 6.7 Path of a particle in a centrifugeThis is a bird's eye view of the path. The particle moves to greater radius

Our discussion of the centrifuge can be greatly simplified by looking at things from the pointof view of an imaginary observer who rides around with the test tube rather than from a bird's eyeview. An observer riding around with the tube in the centrifuge is accelerating so we can use theequation of motion (F = ma) only by introducing a fictitious centrifugal force.

Consider the motion of a suspended particle as described in the accelerated frame of reference.In figure 6.8 the force to the left is provided by the surrounding material.

R

Contact force Centrifugal force

Axis of rotation

ROTATING FRAME

Figure 6.8 Real force and pseudoforce on a particle in a centrifugeIn this case the particle is suspended at a fixed radius, so it is at rest in the rotating frame. The

centrifugal force is a pseudoforce. The buoyant-like contact force is a real pressure force exerted by thesurrounding fluid.

FE6: Rotation 69

The pseudoforce is still equal to mω 2R. If the real contact force provided by the surroundingmaterial is less than or equal to mω 2R then the particle moves further down the test-tube. Thecentrifugal pseudoforce is rather like a gravitational force which causes suspended particles to settleout. By making the angular velocity (ω) very high the pseudoforce (mω 2R) can be 104 to 105 timesgreater than the weight of the particle. So we can ignore the weight, giving a total force along theradius. Seen from the rotating frame, the motion of a particle that starts from rest will therefore be astraight line, not the spiral seen by the outside observer.

6-6 THE ULTRACENTRIFUGEThe ultracentrifuge in the Department of Biochemistry is often used to measure the molecularweights of protein molecules. The rotor shown in the lecture runs at speeds up to 60 000revolutions per minute giving a maximum centrifugal force of 260 000 times that of gravity. Rotors for ultracentrifuges must be carefully designed so that they do not fly apart under the largecentrifugal forces. These considerations set a limit on the radius of a rotor and its speed ofoperation. The chamber containing the rotor is evacuated to eliminate heating due to air friction.

Cell Cell

Figure 6.9 An ultracentrifuge rotorA cut-away view showing the location of cells containing protein samples.

In this case, the particles of interest are very small - small enough for diffusion to occur; sothey are also small enough to be influenced by the forces exerted by individual liquid moleculesrather than by the ‘averaged out’ forces (the drag and the buoyant forces). Sedimentation of theprotein molecules cannot take place under gravity because their weight is not sufficient to overcometheir diffusion upwards. If we centrifuge the sample, we can increase the speed of rotation until thecentrifugal (pseudo) force overcomes the diffusion and drives the protein molecule to a greaterradius.

The molecular weight of the protein molecules is determined from the sedimentation rate, thedensity of the molecules and the density of the surrounding fluid. Diffusion effects must besubtracted out. The size of the diffusion effects is obtained from measurements of thesedimentation rate taken at different speeds of rotation.

The measurements are all made while the sample is spinning around inside the machine. Thisis the difference between an analytical ultracentrifuge and the ordinary type of centrifuge which isused for separation. The radial movement of the protein molecules is monitored by opticaltechniques. Use is made of the fact that the refractive index of the protein molecules is differentfrom that of the surrounding liquid.

6-7 CORIOLIS FORCEAnother kind of pseudoforce, the Coriolis force, appears in a rotating frame of reference. Ademonstration shows a two-dimensional version of the Coriolis effect. A container moving along astraight path drops sand onto a rotating platform. The track of the falling sand in the frame of the

FE6: Rotation 70

platform is a curve rather than a straight line. To explain this curved path another fictitious force, theCoriolis force, can be introduced.

A similar effect occurs with global wind patterns. Air masses moving towards the equatorare deflected from a true south-north path because the Earth rotates underneath. In a rotating frameattached to the Earth's surface, the winds appear to be deflected. In order to preserve the equation ofmotion in the rotating frame we invent two fictitious forces - centrifugal and Coriolis.

POST-LECTURE

6-8 QUESTIONSRotationQ6.2 A gramophone record is rotating at a rate of 33 l/3 revolutions per minute. What is its angular velocity in

rad.s-l?Q6.3 The dental drill probably reaches higher speed than any other production micro-mechanism. The angular

speed is about 60 000 rad.s-l. If the dental burr has a diameter of 1.5 mm, how fast is the outside of theburr moving?

Rotational kinetic energyQ6.4 A flywheel is rotating with an angular velocity of 5 revolutions per second. What does its angular

velocity become when its rotational kinetic energy increases 16 times?Q6.5 Could a sphere roll down a slope if friction were absent?Q6.6 How can you tell the difference between a raw egg and a hard-boiled egg by spinning them?Accelerated frames of reference and pseudoforcesQ6 .7 Refer to the demonstrations of motion observed in a frame of reference attached to a moving truck.

a) If the truck were speeding up (figure 6.10) where would the object strike the floor?

Object released here

Acceleration

Truck is speeding up

Figure 6.10 An accelerating frameWhat happens to the falling object?

b) Can an observer standing at the side of the road explain the motion using the equation of motion? (Ordoes the observer need to invoke pseudoforces?)

Q6.8 Discuss the way you are thrown around in a car when it stops suddenly and when it turns a corner sharplyi) from the point of view of a bird watching from its perch on a power line, andii) from the point of view where all measurements are made with respect to the car.

6-9 MORE ABOUT THE CENTRIFUGEThe centrifuge is used to increase the settling speed of particles which fall very slowly through thesurrounding liquid. (See §4-5 on Sedimentation). In a frame of reference rotating with the test-tubethe horizontal forces on a typical particle are as shown in the horizontal plan view in figure 6-11. Incomparison with these forces, the vertical forces on the particle are very small and may be neglected.(The Coriolis force can also be neglected in this argument.) The two forces acting to the left areexerted by the surrounding liquid. The drag force is the same as that introduced in FE4. The

FE6: Rotation 71

‘buoyant-like’ force is the analogue of the familiar buoyant force - its magnitude is independent ofthe motion of the particle relative to the liquid. Both are real forces.

Figure 6.11 Real forces and pseudoforce on a particle in a centrifugeHere the particle is moving ‘down’ the tube to the right. The ‘buoyant like’ contact force and the drag

force, both exerted by the fluid, are real forces. The centrifugal force is a pseudoforce.

The magnitude of the buoyant-like force can be found by the method used for the buoyantforce in FE3. Consider the forces which would act on a portion of fluid, mass mL, corresponding tothe liquid displaced by the particle. See the upper part of figure 6.11. Since this portion of liquiddoes not move along the radius (why?), the ‘buoyant like’ force balances the centrifugal force mLω 2R. Using ρL to denote the density of the liquid and V for the common volume of the particle andthe displaced liquid we get:

buoyant-like force = mL ω 2R = ρLVω 2R.

Returning to the forces on the suspended solid particle (lower diagram in figure 6.11), thebuoyant like force on the particle is equal to that on the displaced liquid. (Why?) Also

centrifugal force on the particle = ρVω 2R ,where ρ is the density of the particle.

The terminal velocity is the velocity at which the drag force λv balances the other forces acting.So the equation of motion for the suspended particle becomes:

ρLV ω 2R + λvT = ρVω 2R. We can rearrange this to get an expression for the settling speed:

v =Vλ (ρ - ρL) ω 2R.

Compare this with the equation for gravitational settling worked out in §4-5 :

vT

=Vλ (ρ - ρL) g .

The angular speed ω of the centrifuge can be increased to make the centripetal acceleration, ω2R , very much greater than g and to provide very much greater settling speeds.Q6.9 A general laboratory centrifuge rotates at its maximum angular speed of 8 000 revolutions per minute.

Consider a particle at a rotation radius of 110 mm.a) Compare the centrifugal force on the particle with its weight.b) In Q4.l, FE4, the sedimentation rate for a particle settling under gravity was several millimetres per

hour. What would the sedimentation rate be if this centrifuge were used?

72

SUMMARY: GRAPHICAL PRESENTATION OF INFORMATION

The table below shows how slopes and areas of various kinds of graphs can be used to calculatevalues of physical quantities.

Ordinate (‘y’) Abscissa (‘x ’) Slope Area between curveand ‘x’ axis

distance travelled time speed --------

displacement* time velocity* --------

speed time --------- distance travelled

velocity* time acceleration* displacement*

acceleration* time --------- change in velocity* (v)

total force* time ---------- mass × (change invelocity*)

force* displacement* --------- work

conservative force* displacement* --------- - (change in potentialenergy)

potential energy displacement* -(conservative force*) ------------

73

FE7 OSCILLATIONS

OBJECTIVES

AimsBy studying this chapter you can expect to understand the nature and causes of oscillations. Youwill need to learn a fair number of new terms, but some care and effort in doing that will be wellrewarded later because the ideas and principles introduced here can be used to understand a widerange of natural phenomena. In particular, the concepts and language of oscillations form the basisfor understanding waves and optics, which come later in this course.Minimum learning goalsWhen you have finished studying this chapter you should be able to do all of the following.1. Explain, interpret and use the terms

oscillation, simple harmonic motion [SHM], amplitude, period, frequency, angularfrequency, phase, initial phase, free oscillation, damped oscillation, forced oscillation,natural frequency, restoring force, force constant [spring constant], driving force, dampingforce, transient motion, steady state oscillation, resonance, resonance peak, resonantfrequency.

2. Explain free, damped and forced oscillations in terms of forces and energy transfers.3. State and apply the equations and formulae of one-dimensional SHM for

(a) displacement, velocity and acceleration in terms of time;(b) restoring force and acceleration in terms of displacement;(c) frequency and angular frequency in terms of force constant and mass for an object on aspring.

4. Describe the important properties of SHM and explain how SHMs can be combined to givemore general oscillations.

5. Describe and explain the phenomenon of resonance and recognise examples of resonance.

PRE-LECTURE

7-1 SIMPLE HARMONIC MOTION (SHM)Oscillations (periodic to and fro motions) ofobjects occur frequently both in nature and technology. For example, sound waves, water waves,elastic waves in stretched strings etc. are made up of large numbers of objects oscillating in acoordinated manner. This chapter will describe the various types of oscillations in terms of theforces and energy transfers necessary for their existence.

Consider a particle moving in one dimension so that its displacement from a fixed point at timet is given by

x = A cos(ω t)where A and ω are constants. This equation describes a simple harmonic motion. The followingquestions are designed to lead you through some important characteristics of SHM.

FE7: Oscillations 74

Q7.1 i) Where is the particle at t = 0 ?ii) Sketch a displacement-time graph of the motion. Note that the graph would look the same if you shifted

it either to the right or to the left by a time interval T = 2πω .

iii) What is the largest value of |x | ?At what times is the distance from the centre, i.e. |x|, largest?

iv) Verify that the x-component of velocity, vx = dx d t , at time t is equal to - Aω sin(ω t) .

v) Hence verify that the speed is smallest when |x| is largest and largest when |x| is smallest. (Do theseresults agree with those obtained directly from the slopes of the displacement-time graph?)

vi) Verify that the acceleration component, ax = dvxd t , at time t is -Αω2cos(ωt).

• The acceleration is proportional to the displacement x and is always directed towards thecentre, x = 0. The proportionality constant is -ω2; that is, the acceleration component is given by:

ax = -ω2x ...(7.1)

vii) Show that the displacement, velocity and acceleration will all have the same values again after a furthertime T = 2π

ω .

• The motion is cyclic, continually repeating itself.As you will have seen in (i), at time t = 0, the displacement is A ; the particle is at one of the

extreme points of the oscillation. It is often useful to start the description of the oscillation at someother part of the cycle. Mathematically this corresponds to writing

x = A cos(ωt + φ ) ... (7.2)

= A cos[ω (t + φω )]

or shifting the displacement-time graph a distance Δt = φ /ω to the left. Since the velocitycomponent at any time t is just the slope of the displacement-time graph, and the accelerationcomponent is the slope of the velocity-time graph, the velocity-time and acceleration-time graphs arealso shifted to the left by the same amount, φ /ω. The relationships between these three quantities, x, vx, and ax are unaltered by the shift. (The same initial phase φ appears in each.)

Equation 7.2 is the most general expression for the displacement of a particle undergoingsimple harmonic motion. As the name suggests, it is the simplest form of oscillatory motion; itsimportance lies in the fact that more complex oscillations can be studied in terms of it.

Terminology

Symbol Name SI unitA amplitude mω angular frequency s-l (or rad.s-l)

ω t + φ phase no unit (or rad)φ initial phase (i.e. phase at t = 0) no unit (or rad)T period sf frequency Hz


Frequency, angular frequency and period for any kind of oscillation are related as follows:

f = 1T =

ω2π ... (7.3).

The SI unit for frequency is the hertz (Hz); 1 Hz = 1 cycle per second. Phase is a dimensionlessquantity, so it needs no unit. However, since it is often helpful to visualise phases as angles, they areoften given the unit radian. Note that there are 2π radians in each cycle, so a frequency of 1 Hzcorresponds to an angular frequency of 2π s-l or 2π rad.s-l.

Two oscillations of the same frequency are said to be in phase at any given time if they havethe same phase at that time. Otherwise they are out of phase by the difference of their phases(φ2 - φ1).

Out of phase by π/2

In phase

Figure 7.1 Oscillations in phase and out of phase

LECTURE

7-2 FREE OSCILLATIONSIf the oscillating system is isolated (i.e. if no energy is being added to or taken away from thesystem) the oscillations are called free oscillations. Oscillations (free or otherwise) can occurwhenever a restoring force capable of transforming potential energy (PE) to kinetic energy (KE) andvice versa is present. In a free oscillation, since the sum of the PE and KE cannot increase, the PEmust be largest at the extreme points of the oscillation where the KE is zero.Examples• Liquid sloshing mode - the restoring forces are due to gravity.• A vibrating metal plate - elastic restoring forces.• Stretched string - the restoring force is provided by tension in the string.

In each of these three examples all the oscillating particles together formed a standing wavepattern.


Example of a single oscillating objectA ball rolls in a curved track. When viewed from above, the ball is executing a one-dimensionaloscillation. The restoring force for the horizontal oscillation is the horizontal component of the forceexerted on the ball by the track. (The ball also oscillates vertically.)

W

TorqueC

W

CV

CH

Contact force

Torque

W

TorqueC

W

CV

CH

Contact force

Torque

Weight

Figure 7.2 A ball oscillating on a curved trackThe ball is acted on by two forces (left), a contact force C and the gravitational force W. The force C

produces a torque about the centre of the ball because its line of action does not pass through the centre.An equivalent representation (right ) shows the single contact force replaced by a vertical contact force

and a horizontal contact force.

Since the track can have any shape (provided that every point in the middle is lower than theends) there are infinitely many possible restoring force functions and therefore there are infinitelymany different one-dimensional oscillations.

SHM - the simplest free oscillationAny oscillation can be represented as a combination of SHMs for the following reasons.• Any one-dimensional oscillation (free or otherwise) can be represented mathematically as acombination of SHMs with appropriate frequencies, amplitudes and initial phases. An example isthe synthesis of the displacement-time graph of a sawtooth oscillation from SHMs.• Any two- or three-dimensional oscillation can be treated as two or three one-dimensionaloscillations at right angles to one another, since the motions in those directions are independent.(Some examples, Lissajous figures in two dimensions, are shown in §7-7, figure 7.10.)

When an object is undergoing SHM, its acceleration and the total restoring force at any timeare both proportional to the displacement from the centre and they are both directed towards thecentre.Example: Body on a springConsider a body on the end of a spring, moving on a frictionless surface.


F Fixedpoint

x

Figure 7.3 A system which does SHM

For an ideal spring, the x-component F of the restoring force is equal to -k x , where k is aconstant; i.e. F is proportional to displacement and is directed towards the equilibrium point where x= 0.

The equation of motion is max = −kx or ax = −kmx .

Comparison with equation 7.1 gives the natural frequency of oscillation:

ωN =

km or fN =

12π

km

. ... (7.4).

EnergyIn a free oscillation, since the only force doing work is conservative; the total mechanical energy (KE+ PE) of the system is a constant.

KE

PE

KE = 0

PE(x)

Totalmechanical

energy

-A A0x

Figure 7.4 Energy of a free oscillationThe curve shows the potential energy (PE). The line across the top represents the level of the constantmechanical energy which is also the maximum value of PE. As PE increases, KE decreases. When

displacement is an extreme value, x = A or x = -A, KE is zero and PE is at its maximum. For SHMthe curve is a parabola; for other oscillations the shape is different but the energy relations shown here

are the same.

At the extreme points of the oscillation, |x | = A , the KE is zero and the total mechanicalenergy is equal to the PE. In the special case of SHM the restoring force is proportional to thedisplacement so the force-displacement curve is a straight line (see §5-6, figure 5.6). From the areaunder the curve, a triangle, the PE when x = A is:


maximum PE = 12 A (kA) = 12 kA2 = 12 mωN2A2 .

The energy of the oscillation depends on the square of the amplitude A. This result is true forall SHMs.

7-3 DAMPED OSCILLATIONSIf the total mechanical energy of an oscillating system were conserved, the system would oscillateindefinitely with the same amplitude. In any real situation, however, there are always non-conservative forces such as friction or drag forces present. These dissipative forces decrease thesystem's total mechanical energy and can be either internal or external to the system. The amplitudewill gradually decrease and the oscillations will die out. An oscillation which dies away is anexample of a transient motion. Examples include pendulums with damping forces.

Displacement

Time

Figure 7.5 A damped oscillationThe amplitude decreases with time and the system loses mechanical energy.

The frequency, fD, of a damped system is always less than fN, the natural frequency that thesystem would have if the damping forces could be removed, since the damping forces always act toretard the motion.

7-4 FORCED OSCILLATIONSWhen non-conservative forces are present, an oscillation can die out unless energy is continuallysupplied to the system by a driving force. An oscillation which is kept going by a periodic drivingforce is called a forced oscillation. An example is a car engine.

If the driving force is sinusoidal (i.e. if a graph of the driving force against time looks like asine curve, perhaps shifted along the time axis), there will be a steady state oscillation set up at thefrequency, fF, of the driving force. In general, fF will be quite different from fD.

For a given amplitude of driving force, the amplitude of the steady state oscillation depends onthe driving frequency fF. An example is a vibrating metal rule driven by an electromagnet.

Each time the frequency of the driving force is changed, the new oscillation takes a while tobecome stable, while the transient oscillations die away. As the driving frequency is increased fromzero, the amplitude of the steady state oscillation gradually increases until it reaches a peak and thendies away again (figure 7.6). This phenomenon is called resonance. The frequency (fF) of thedriving force at which the amplitude of the oscillation is a maximum is called the resonancefrequency, fR, and the region of the graph near fR is called the resonance peak.


Amplitude ofthe steady stateoscillation

Resonance peak

ff

RF

Figure 7.6 Response curve for a resonance

For lightly damped systems, fN , fD and fR are all approximately equal. At resonance thedriving force adds to the restoring force in the damped system, producing large accelerations andoscillations.Examples of forced oscillations• Shivering to heat the body in response to cold.• Expansion of the chest in breathing.• Beating of the heart and dilation and contraction of the arteries.

In each case the driving force is provided by the muscles. When these systems relax fromhigh to low potential energy states, their mechanical energy is decreased by the work done by non-conservative forces. The systems are highly damped and require the driving force to cause themotion.

POST-LECTURE

7-5 QUESTIONSQ7.2 List the forces that are acting on the following oscillating objects and in each case state whether the forces

are restoring, damping or driving forces:i) a tree swaying in the breeze,ii) the earth vibrating after an earthquake,iii) a child on a swing,iv) a vibrating eardrum,v) a marker buoy bobbing in the water,vi) the atoms in a solid when a sound wave passes through.

Simple harmonic motionIn general one can obtain an order of magnitude estimate for the square of the natural angularfrequency (ω) of an oscillation by dividing an order of magnitude estimate of the restoring force per


displacement (k) by the mass (m) of the object in motion. This relation is exact for SHM (seeequation 7.4) but it also gives results which are roughly OK for other systems.Q7.3 When a spring is stretched by a pair of 2.4 N forces its length increases by 62 mm. Predict the period of

oscillation when an object of mass 0.32 kg is hung from the spring and set vibrating. What will happento the period if the object is replaced by one with 10 times the mass?

Q7.4

A wooden rectangular pontoon of densityρ = 0.50 × 1 03 kg.m-3 floats in water(density ρW = 1.00 × 1 03 kg.m-3) so that twofaces, each of area A = 20 m2, are horizontal.The vertical sides each have a length L = l.0 m.

Figure 7.7i) How far is the bottom face below the water at equilibrium? (You may neglect the weight of the displaced

air in this problem.)ii) What is the total force acting on the pontoon when it is a small distance x below the equilibrium

position with the same two faces horizontal?iii) What would this force be if the pontoon were a distance x above the equilibrium position?iv) Hence show that if the pontoon were released from a position other than its equilibrium position it would

execute SHM in the vertical direction if no non-conservative forces were present. (What non-conservativeforces would you expect to be present and what effect would they have?)

v) What is the natural frequency of the above SHM ?Q7.5 Use energy considerations to find the maximum speed of an object undergoing SHM in terms of the

amplitude and natural (angular) frequency of the motion.Q7.6

A somewhat bizarre proposal for a rapidtransit system between Boston and Washington(a distance of 630 km) made use of the Earth'sgravitational attraction. (Scientific American,August 1965, pp 30 - 40.) Stripped of itstechnical complications, the proposal had acapsule travelling through an evacuated tunneldrilled in a straight line between those cities.

For the first half of its journey, the capsulewould gradually be getting closer to the centre ofthe earth (i.e. sliding down an incline); for thesecond half it would be travelling away from theEarth's centre.

Boston Washington

Centre ofthe Earth

Figure 7.8

In the idealised case where non-conservative forces are neglected, the potential energies at the beginningand the end of the journey would be equal. (In the actual proposal, the capsule was helped along bycompressed air to overcome the non-conservative forces.) In this idealised case, the acceleration of thecapsule would be

a = -C xwhere x is the displacement from the midpoint of the journey and C is a constant equal to1.55 × 10-6 s-2. The capsule would therefore be undergoing part of a SHM.

i) How long would the one way trip from Boston to Washington take ?ii) What would be the maximum speed of the capsule on the journey ?

7-6 FURTHER DISCUSSION OF RESONANCEIn the video lecture it was stated that the Tacoma Narrows Bridge collapse was caused by resonance.The driving force in that case, however, was not sinusoidal nor did it oscillate at the frequency of thesteady state oscillations of the bridge. In fact the driving force was provided by a wind blowing inone direction for a time long compared to the period of the bridge's oscillation. How then doesresonance occur ?


Any driving force whatever may be considered mathematically to be a combination ofsinusoidal driving forces with appropriate frequencies, amplitudes and initial phases. The procedurefor splitting up a force in this way is called Fourier analysis (after the French mathematicalphysicist who developed it to study the flow of heat through metals). Each of these sinusoidalforces can be thought of as acting on the damped oscillating system independently of the othersinusoidal components setting up a sinusoidal steady state oscillation at its own frequency. Theseoscillations add up to give the actual oscillation caused by the full original driving force.Schematically, this procedure can be summarised as shown in figure 7.9.

+

+

+

+

+

+ . .

. . . .

Fulldrivingforce

Fulloscillation

≡ ≡

Sum of sinusoidaldriving forces

Sum of sinusoidalsteady state oscillations

is equivalent to

Figure 7.9 Resonant response to a complex driving force

The ratio of the amplitude of a steady state oscillation to that of the sinusoidal driving forcethat caused it is largest when the driving force has a frequency equal to the resonance frequency.Quite often therefore, the combined oscillation is dominated by the steady state oscillations withfrequency approximately equal to fR, particularly if the resonance curve is very sharply peaked.

These ideas are the basis of radio and TV receivers, for example. The driving force is thecomplicated electrical signal (the sum of all the signals sent out by the various stations plusinterference from fluorescent lights, spark plugs etc.) from the aerial or antenna. The tuning circuitof a radio or TV set is an oscillating system whose resonance frequency can be varied to select thesteady state oscillations at the frequency of the transmitting station. The resonance peak for thesetuning circuits has to be sufficiently narrow that only one station is transmitting at a frequencywithin the peak at any given setting.

In the bridge collapse, the driving force provided by the wind had a large sinusoidalcomponent at one of the natural frequencies of oscillation of the bridge. This component caused thebridge to resonate and ultimately break apart.

Q7.7 Can you think of any other situations in which resonance plays an important role ?


7-7 APPENDIX: LISSAJOUS FIGURESWhen two SHMs at right angles are combined, the path traced out will be a Lissajous figure if theratio of the two frequencies is equal to a ratio of two integers. The shape is also affected by thephase difference between the oscillations.

1

1/2

1/3

2/3

3/4

3/5

4/5

5/6

Figure 7.10 Some Lissajous figuresThe number at the left of each row is the ratio of the frequency of the vertical oscillation to that of thehorizontal oscillation for the figures in that row. The figures in each row correspond to different initial

phases of these two oscillations.

83

FE8 SCALE OBJECTIVES

AimsFrom this chapter you should learn to appreciate the great power of physical reasoning using veryrough approximations to more exact principles. You should become reasonably adept at applyingthis kind of reasoning and the physical principles that you already know to problems of scaling -comparing the behaviour of things, including animals, which have roughly the same shape andcomposition.Minimum learning goalsWhen you have finished studying this chapter you should be able to do all of the following.1. Explain, interpret and use the concept of scale factor.2. Explain how the laws of physics can be applied to a wide range of similar problems by

concentrating on the similarities and ignoring particular idiosyncrasies.3. Apply the concept of scale factor to analyse the way the sizes of different animals determine

their ability to perform simple mechanical operations such as running and jumping.4. Describe and explain the idea that bone strengths and muscular forces per cross-sectional area

are much the same for all animals.5. Recall and use the fact that the rate of energy expenditure by an animal is equal to the rate of

energy supply, which is controlled by the amount of oxygen available to the cells. 6. State and apply the scaling laws listed in the table of §8-7.7. Describe some of the pitfalls to be avoided in scaling problems.

PRE-LECTURE

8-1 INTRODUCTIONThe earlier parts of this unit have mainly been concerned with the application of physical concepts toexact calculations, using carefully measured data. This technique is used, for example, by engineersin designing bridges and power stations, by physicians in radiation dosimetry and by optometrists ingrinding lenses.

There is an equally important, although less well-known, way of using the concepts of physicsin non-exact calculations just to see whether a particular idea is valid. The arithmetic used may berough and ready - of the 3 × 3 = 10 variety. Frequent use may be made of statements involving thewords ‘about equal’. This is the style of approach used, for example, in forecasting trends in worldpopulation and consumption of raw materials.

This second type of application of physics is the theme of this chapter. The concepts of forcesin equilibrium, work done by forces and energy balances will be used to examine some of thefeatures of the animal world - the strengths of skeletons, running, jumping and diving. The interestis not in the subtler details of the mechanisms of these things, e.g. how a grasshopper bends its legsprior to jumping or how a whale stocks up on oxygen before diving, but in broader considerations,e.g. why grasshoppers, frogs and kangaroos all raise their centres of gravity by roughly the sameamount when jumping, or what advantages large mammals have over small mammals in diving.

FE8: Scale 84

In the spirit of the ‘equality’ 3 × 3 = 10, the physics used will be stripped to its bareessentials. For the purposes of this chapter a force whose average magnitude is F moving through adisplacement d does work Fd - we will neglect niceties like including the cosine of the anglebetween the force and the displacement. All such extraneous considerations will be lumped togetherin proportionality constants which will have no bearing on the conclusions we draw from thephysics.

The primary aim of this chapter is to show how these perfectly respectable procedures can beused to extract information. The results obtained simply illustrate the power of the method; you arenot expected to memorise these results for examination purposes. You will be expected to be able toreason along the same lines.Q8.1 There are two solid cylinders made of the same material. All the linear dimensions of the larger one are

exactly twice those of the smaller.Compare their heights, diameters, base areas, curved surface areas, volumes and masses.

LECTURE

8-2 SCALE FACTORThe question, ‘how much bigger is one tree than another?’ has no unique answer unless we knowon what basis the comparison is to be made - on their heights, on the cross-sectional area of theirtrunks or on the volume of wood in their trunks, for example.

If the trees have roughly the same shape, a comparison can be made in terms of a measurebased on length - the scale factor, L. Then if one tree were L times taller than the other, the radiusof its trunk would be L times bigger, the cross-sectional area of its trunk L2 times bigger and thevolume of its trunk L3 times bigger than the other. The symbol L has been chosen deliberately as areminder that our basic comparison is being made on corresponding lengths. However L is a ratioof lengths; it has no unit; it is a scale factor, not a scale length.

A scale factor can be assigned to any class of objects which have essentially the same shape.In the following we will be comparing the properties of different species of animals which haveroughly similar basic shapes, but which differ in size, e.g. dogs, sheep and horses.

8-3 BONE LOADS AND MUSCULAR FORCESWe make the assumptions that the strength of the materials that bones are made of and the way thebones are constructed are much the same for all animals. By strength we mean the maximummagnitude of the force per cross-sectional area that the bone is able to withstand under acompressional or an extensive load (the two values need not be the same). This means that themaximum force that can be applied along the length of a bone without breaking it depends on thecross-sectional area of the bone, i.e. on the square of the bone's diameter.

This assumption is supported indirectly by two sets of experimental results on the breaking oftibia bones of different sizes. (Notice that in this experiment the loads are applied at right angles tothe bones - they are not along the bones themselves.)

FE8: Scale 85

• In the experiment shown on TV, 37 force units were required to break a dog's tibia of diameter1.66 length units while 50 force units were required to break a sheep's tibia of diameter 1.84 lengthunits.• The following results were obtained for the tibia bones of different breeds of dog (figure 8.1).

Diameter/arbitraryunit

Force/arbitrary unit

2.0

1.0

1.5

20 30 40

•

••

•

•• •

•

•••

••

••• ••

•• •• ••

••••

•

•••

•

Figure 8.1 Force required to break a dog's bone

The same type of assumptions about the maximum forces supplied by muscles or tendonslead to the result that these forces also scale as the square of the diameter.

ApplicationsAnimals standing stillThe leg bones of a stationary animal must support a compressional load proportional to the animal'sweight. The weight of the animal is proportional to its volume and so, between different sizedanimals, varies as L3. We saw above that the maximum compressional force which the bones cansupport varies as the square of the diameter. The diameter of the leg bones of similar animals musttherefore increase faster than the other linear dimensions of these animals. Thus, for example, theleg bones of elephants are comparatively much thicker than those of mice.Moving animalsThe legs of an animal are also subject to bending forces, particularly when they are moving.Consider for example the forces acting on the humerus bone of a mammal.

T

T

F

F

Figure 8.2 Forces on a bone

The adjacent bones exert vertical forces F proportional to the weight of the mammal (to L3).To keep the bone from twisting (rotating) there must be a pair of equal and opposite forces T

FE8: Scale 86

exerted by the muscles and the tendons along the humerus bone so that there is no net torque: theseforces provide a bending (or shearing) load on the bone.

d

l

Figure 8.3 Dimensions of a bone

If the length of the humerus bone is l and its diameter is d, since there is no net torque,Fl - T d = 0

so T =Fld .

Since l and d both vary as L , then T , like F , varies as L3.The bending forces T act over areas A near the surface of the bone as shown in the cross-

sectional view below. These areas, and hence the capacity of the bone to withstand the bending, varyas the square of the diameter.

d

bone

marrow

A

A Figure 8.4 Location of bending forces in a leg bone

Thus the bending forces in the humerus increase more rapidly with L than the capacity of thebone to withstand them - essentially the same result as in the case of standing still. These bendingforces are the reason why giraffes keep their legs straight while running.

8-4 SUPPLY OF CHEMICAL ENERGY IN THE BODYThe maximum continuous rate at which an animal can do work while engaging in various activities islimited by the rate at which the cells in the body can supply energy to the muscles. The latter rate iscontrolled by the amount of oxygen available to the cells; it is shown in the POST-LECTURE to varyas L2 .

Thus, for animals of similar shape,the maximum rate of energy supply = k1L2 , .... (8.1)

where k1 is a constant independent of the size of the animal and hence of L.ApplicationsRunning on the flatDoes the size of an animal determine the speed at which it can run on the flat? While the animal isrunning at a fixed speed, here its maximum speed v, its kinetic energy is constant. Since it is

FE8: Scale 87

running on flat ground its gravitational potential energy is also constant. Hence the energy supplied(by the cells) and the (negative) work being done by the drag force exerted by the air must total zero.

The drag force is proportional to the cross-sectional area of the animal and to the square of thespeed. In symbols

drag force = k2L2v2,where k2 is independent of L and v .

In a time interval t the animal travels a distance vt so the work done by the drag force duringthis time is -k2L2v3t.

Equating the magnitude of the rate of doing this work with the maximum rate of energy supply(from equation 1 above),

k2L2v3 = klL2,

or v3 =k1k2 = constant.

This means that, on the basis of this scaling argument, the maximum speed of an animalrunning on the flat is expected to be independent of its size. This expectation agrees with theobservation that, whilst there are slight individual differences between the maximum speeds ofanimals, there is no systematic trend with size. For example, dogs and horses can run at about16 m.s-1 while humans, intermediate in size between these, run at about 10 m.s-1.Running up a steep inclineThe dominant force doing work on the animal is the gravitational force. This force (the weight) isproportional to L3. As in the previous application, if the animal is travelling at constant speed theenergy supplied (by the cells) and the work being done by the gravitational force must total zero.

Suppose that the animal is travelling at its maximum speed v up the incline. In a time interval tthe animal travels a distance vt so the work done by the gravitational force during this time is -k3L3vt. (k3 is also independent of L and v .)

Equating the magnitude of the rate of doing this work to the maximum rate of energy supply,k3L3v = k1L2

or v =klk3

L-1 .

In other words, the maximum speed of an animal travelling up an incline would be expected tobe inversely proportional to its size. For example a greyhound would be expected to travel fasterthan a horse up a hill.

8-5 JUMPINGAn animal standing on the ground exerts a force on the ground equal in magnitude to the weight ofthe animal. In order to jump, the animal first crouches, lowering its centre of gravity by a smalldisplacement d. It then pushes against the ground with an additional force of average magnitude F.This means that there is an unbalanced force of this average magnitude upwards on the animal aslong as it remains in contact with the ground. As the animal leaves the ground it accelerates upwardsand its kinetic energy increases as its centre of gravity returns to its original position. This increasein KE actually comes from work done by the leg muscles but its value can be calculated using theproduct of the net external force (F) and the displacement (d) of the centre of gravity. So the workdone on the animal is equal to Fd.

After the animal leaves the ground its centre of gravity rises to a height h above its positionwhen it was crouching.

FE8: Scale 88

d h

Figure 8.5 Location of the centre of gravity during a jump

The associated increase in gravitational potential energy can be equated to the kinetic energy attake-off or the work done during the launch.

The increase in gravitational potential energy is k5L3h where k5 is independent of L and h.The average force F provided by the muscles is proportional to L2 (see earlier) and the

crouching distance d is proportional to L, so the work done is equal to k4L3 where k4 is independentof L .

Equating these two quantities:k5L3h = k 4L3;

hence h = k4k5

.

This means that the maximum height through which the centre of gravity can be raised isexpected to be independent of the animal's size. Observation shows that this is so; for most animalsthe height involved is a little over 1 metre.

8-6 DIVINGWhen a mammal dives below the surface of the water, it consumes oxygen at a rate proportional toL2 (see the section in the post-lecture on the rate of supply of energy); in a time interval t themammal consumes a quantity k6L2t of oxygen (k6 is independent of L and t).

The total oxygen supply depends on the volume of the lungs and hence can be written as k7L3.(k7 is independent of L.) Hence the maximum duration of the dive is given by

t =k7k6

L ,

which, since k6 and k7 are independent of L, indicates that larger mammals can stay under water forlonger periods. Again this is confirmed by observation.

POST-LECTURE

8-7 APPLICATION OF THE CONCEPT OF SCALE FACTORApplications involving scale factors fall into two classes.• The first, rather straight-forward, class is that in which measured or given data lead directly orindirectly to a value of the scale factor L and this value can be used to estimate other data pertinent tothe system. In dealing with these problems it is useful to be able to recollect the way certainfundamental quantities are related to the scale factor. Some of these relations are summarised herein tabular form.

FE8: Scale 89

L L2 L3

length surface area volumegirth cross-sectional area mass

height rate of energy supply weightetc. etc. etc.

An example of this class of problem is as follows. Consider two oranges, both roughlyspherical, one of mass 0.20 kg, the other of mass 0.40 kg and ask the question as to how the surfaceareas of the two oranges compare. To answer this one uses the mass ratio (2:1) to find the value ofL3 and proceeds to compare surface areas through the derived value of L2.• The second class is where it is required to derive the way in which the variation of somequantity among things of similar shape is related to the scale factor. Such problems usually aremore involved and require the use of some linking factor known from everyday life or from theformal study of physics. Many of the illustrations in the lecture material were of this type: running,jumping and diving animals, to name a few.Q8.2 Suppose that in order to make Australians champions in all sporting events we breed fair dinkum sun-

bronzed Aussies who are scaled up × 2 in all linear dimensions. They are made of the same stuff asconventional × 1 people. How would a × 2 person's performance compare with that of a × 1 person inthe following events:

a) a form of high jumping where, to even things up, the judges measure the increases in heights of thejumpers' centres of gravity,

b) weight-lifting,c) sprinting (assume that the forward acceleration from rest at the beginning of the sprint is the dominant

influence on the sprinters' performances),d) putting a shot made of the usual material but whose linear dimensions are proportional to the height of the

shot putter? (For simplicity assume that the shot is launched horizontally.)

8-8 BREAKING OF DOGS' BONESThe experimental results shown in the lecture for the tibia bones of different breeds of dog arereproduced in graphical form below.

Force/arbitraryunit

Diameter/arbitrary unit50 10 15 200

10

30

40

20•••• ••• •

•• ••••••

•

•••

•

•••• •

••

• •

•

• ••

Figure 8.6 Linear plot of the force required to break a dog's bone

This graph differs from figure 8.1 in that the origin is shown on the graph. This point is adata point on logical grounds rather than experimental evidence. You can now see that therelationship between breaking force and bone diameter is not linear. It may follow some sort ofpower law. The axes of the graph have been swapped to make the following question simpler.

FE8: Scale 90

Q8.3 Suppose the relationship between diameter d and force F is of the formF = k d n

where k and n are constants.Re-plot the data on a log-log graph to see if the data support the view, held in the lecture, that n = 2.

5

4

3

2

1 9 8 7 65

5 6 7 8 9 1 2 3 4 5

Figure 8.7 Logarithmic plot of the force needed to break a dog's boneYou should replot the data from figure 8.6.

8-8 RATE OF ENERGY SUPPLY AND PULSE RATERate of energy supply in an animalThe rate of supply of energy to muscles depends on the rate of oxygen supply which, in turn,depends on the volume rate of flow of blood passing through the muscle tissues. This volume rateof flow of blood depends on the cross-sectional area of the aorta (proportional to L2) and the speedof the blood flow.

The first question then is whether the speed of the blood flow depends on size. This questioncan be answered in a round-about fashion. Since the speed of the blood flow is itself proportional tothe blood pressure the question becomes: does the blood pressure depend on size?

A

A

1

2

Figure 8.8 Cross-section through an artery

The muscular action in the ventricle of the heart keeps the blood moving with more or lessconstant volume rate of flow so the force exerted by the muscles equals that arising from thepressure. If P is the blood pressure, S the muscular force per cross-sectional area of the muscles, Alis the cross-sectional area of the ventricle and A2 is the cross-sectional area of the muscle,

P A1 = S A2,

that is P = S A2Al

.

FE8: Scale 91

The two areas depend on L2 while S is independent of L. This means that the pressure, andhence the speed of blood through the aorta, does not depend on size.

Therefore both the rate of flow of blood delivered by the heart and the rate at which energy issupplied are proportional to the cross-sectional area of the aorta and hence depend on L2 .

Pulse rateQ8 .4 We have seen that the rate of blood supply from the heart varies as L2. The volume of the heart will

naturally vary as L3. What conclusion can you reach regarding the pulse rate of animals of various sizes?

8-9 A CAUTIONARY TALE ABOUT DRUG DOSAGESScaling problems can often present traps for the unwary, even for experimenters who should knowbetter. A few years ago some scientists wished to test the reaction of elephants to LSD dosage.They calculated the dose from that already found to put a cat into a rage, (0.1 mg of LSD per kg ofbody weight) and scaled up by weight to get a dose of 297 mg. On receiving this dose the elephantimmediately started trumpeting and running around, then it stopped, swayed and, after five minutes,collapsed, went into convulsions, defecated and died. The scientists concluded that elephants arepeculiarly sensitive to LSD.

Was the scaling done correctly? There are several approaches to this problem, leading to alarge range of suggested doses.(a) Equal concentrations of LSD in body fluids for elephant and cat, i.e. scaling according to

weight. Dose: 297 mg.(b) Scaling based on metabolic rate, which will control detoxification of the drug and its excretion.

Dose: 80 mg.(c) Scaling based on an animal which is not as notoriously tolerant to LSD as a cat, for example

we could use a human. A dose of 0.2 mg LSD produces severe psychotic symptoms inhumans. Using a scaling based on weight, the elephant dose is 8 mg.

(d) Scaling on metabolic rate, using the human dose as a base. Dose: 3 mg.(e) Scaling on brain size, since LSD could be more concentrated there, again using human dose as

base (human brain 1.4 kg, elephant brain 3 kg ). Dose: 0.4 mg.

8-10 EARLY ATTEMPTS AT SCALINGThe idea of scaling in physics is not new. You may be interested in reading the discussion inGalileo's work Concerning the Two New Sciences, pp 184-188, vol 28, Britannica Great Bookswhich is available in the Fisher Library Reading Room. Jonathan Swift encountered the problem inGulliver's Travels where the Lilliputian emperor decided that Gulliver needed (12)3, that is 1728,Lilliputian portions of food. Actually the emperor overestimated the situation by scaling accordingto weight rather than to surface area.

8-11 SCALING APPLIED TO MOTOR VEHICLESA feature of scaling problems involving animals is that the power available varies as L2. Thebuilders of machines are not bound by this constraint and can adjust the scaling factors to best suittheir requirements. For example, as the size of vehicles is increased from cars to semi-trailers theprime consideration is to increase the carrying capacity, even if the motive power does not increase inthe same proportion as in the animal world.Q8.5 A family car is about 3 m long, has a mass of about 1000 kg and develops a power of about 75 kW. A

semitrailer is about 9 m long, has a laden mass of about 30000 kg and develops a power of about 225 kW.How much faster can the car climb a hill than the semitrailer ?

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