rotational motion - if

RO TAT I O N A L M O T I O N 161

J. Newman, Physics of the Life Sciences, DOI: 10.1007/978-0-387-77259-2_7, © Springer Science+Business Media, LLC 2008

Once the translational motion of an object is accounted for, all the other motions ofthe object can best be described in the stationary reference frame of the center of mass. Areasonable image to keep in mind is to imagine following a seagull in a helicopter thattracks its translational motion. If you took a video of the seagull you would see quite dif-ferent motion than you would from the ground. The seagull would appear always aheadof you but would rotate and change its “shape” as it flapped its wings (e.g., see the filmWinged Migration). You’ve probably seen such wildlife videos that can track animals and“subtract” their translational motion leaving only the other collective motions about their“centers”: lions seemingly running “in place” as the scenery flies by. In physics, we’vealready shown how to account for the translational motion of the center of mass. Asidefrom a possible constant velocity drift in the absence of any forces, motion of the centerof mass is caused by external forces acting on the object. We now turn to the other motionsabout the center of mass as viewed from a reference frame fixed to the center of mass.

These collective motions are of two types: coherent and incoherent. Coherentmotions are those overall rotations or vibrations that occur within a solid in which theconstituent particles making up the object interact with each other in a coordinatedfashion. If the solid is rigid (with all the internal distances between constituent partsfixed) the only collective motion will be an overall rotation about the center of mass.For such a rigid body, a complete description of its motion includes the translationalmotion of the center of mass and the rotational motion about the center of mass.Because this nice separation of the problem can be made, we first present the descrip-tion, or kinematics, of pure rotational motion of a rigid body about a fixed axis, theaxis of rotation. In this case all points of an object rotate in circles about some fixedpoint on the axis of rotation. This type of motion occurs, for example, when a door isopened, or for the wheels of a stationary bicycle, or when you lift an object by rotat-ing your forearm about a stationary elbow. Even if the solid is not rigid, its collectivecoherent motions can be described as a rigid body rotation (of the average-shapedbody) as well as other coherent internal motions that can change the object’s shape.

We next introduce the energy associated with rotational motion and the rotationalanalog of mass, known as the moment of inertia. We show that well-placed and directedforces can produce rotational motion and we introduce the notion of torque, therotational analog of a force. For pure rotational motion there is an equation that is therotational analog of Newton’s second law that can describe the dynamics of motion.Continuing with rotational analog quantities we introduce angular momentum, the rota-tional analog of (linear or translational) momentum and learn a new fundamentalconservation law of angular momentum. Key in following the presentation of our under-standing of rotational motion is to keep in mind the strong analogy with what we havealready learned. A preview glance at Table 7.2 below shows that the important newconcepts in this chapter all have direct analogs with equations we have already studied.

One of the new and revolutionary types of microscopy, atomic force microscopy,is discussed as an application of the material in this chapter. The technique allows

7Rotational Motion

extremely high resolution maps of the microscopic surface topography, or structure,of materials and has been used extensively to study biological molecules and cells.

After briefly considering the effects of diffusion on the rotational motion ofmacromolecules, the chapter concludes with a study of the special case of objects instatic equilibrium. This is an important simplification of Newton’s laws and providesa powerful method of analyzing equilibrium situations.

The other category of collective motions is known as incoherent. These are ran-dom motions of the atoms of the material, about the equilibrium positions in a solidor with no fixed average position in a fluid. Constituent fluid particles move aboutmuch more independently, in the ideal case not interacting with their neighbors at all.In Chapters 8 and 9 we discuss the flow of ideal fluids as well as some of the com-plications that occur in complex fluids in which there are strong interactions betweenconstituents. Later in Chapters 12 and 13, we study the subject of thermodynamicsconcerned with describing the fundamental thermal properties of macroscopic sys-tems. We show that collective incoherent internal motions of an object give rise to aninternal energy that is responsible for its temperature.

1. ROTATIONAL KINEMATICS

A rigid body—one with a fixed shape—has motions that are limited to pure translationand pure rotation about its center of mass. Combinations of these can give rise tomotions that appear more complex, such as rolling, but which can be simplified to purerotations in a reference frame fixed to the center of mass. Since we’ve already learnedhow to handle translational motion in the previous chapter, here we first take up theproblem of pure rotational motion about a fixed axis of rotation, leaving their synthe-sis leading to general rigid body motions for a discussion later in the chapter.

Consider the motion of a point particle on the circumference of a circle, as shownin Figure 7.1. In order to describe its position and motion we could use its x- and y-coordinates or, better, its r and � polar coordinates. These latter coordinates are pre-ferred because r is constant if the particle remains on the circle and so in polarcoordinates there is really only one variable �, whereas both x and y change as the par-ticle moves on the circle. To describe the motion of the particle on the circle we coulduse its x- and y-components of velocity, both of which would continuously change, or,even better, we could use the �-component of velocity known as the angular velocity �,whose average value is defined as

(7.1)

In this expression, the particle has moved between two angular positions in a time �t,where the angular displacement �� must be measured in radian units and not in degrees.

The fundamental unit of angular measure is the radian, because it is defined as theratio of the arc length s to the radius r as � � s/r, and is a pure number with no units.

This definition of an angle in radians leads to the fact that there are 2� radiansin one revolution around a circle, given the circumference of s � 2�r. In onecomplete revolution there are also 360° and thus the radian is equal to 360°/2�or about 57.3°. The unit for � is, according to Equation (7.1), the radian persecond (rad/s); despite the fact that the radian unit is a pure number and wecould write the units for angular velocity as 1/s, it is useful to retain the term“rad” in the numerator. (Note: Most pocket calculators can do calculationsusing either radians or degrees and because rad must be used here, some caremust be taken when first using a new calculator.)

If our particle travels in a circle at a constant speed, executing uniform cir-cular motion, then the instantaneous value of � is constant and equal to the aver-age value. Notice that because � is an angular variable there are really only twopossible directions of travel: clockwise or counterclockwise around the circle.Just as the sign (� or �) for a linear quantity depends on the coordinate system,

v�¢u

¢t.

162 RO TAT I O N A L M O T I O N

x

y r

θ

FIGURE 7.1 A particle executingcircular motion.

we are free here to label the sign of the angular velocity in an arbitrary way, as long aswe are self-consistent within the context of any particular discussion.

In the more general case, when the particle does not travel at a constant speed, theangular velocity will vary and we need to introduce the concept of the instantaneousangular velocity, defined in a similar way to v, as

(7.2)

The distance traveled by the particle along the circumference �s is proportionalto the angular displacement �� from the relation s � r�, therefore we have that�s/�t � r ��/�t so that

(7.3)

where r is the radius of the circle. We also need to introduce the concept of an angularacceleration in order to account for a changing � in analogy with our introductionof a linear acceleration a to describe changes with time in the linear velocity v. Wedefine the average and instantaneous angular acceleration in direct analogy with theirlinear counterparts as

(7.4)

Again, because Equation (7.3) implies the change in the magnitude of the linearvelocity is proportional to the change in the angular velocity, we have a relationshipbetween the linear and angular accelerations,

(7.5)

The linear acceleration in Equation (7.5) is called the tangential acceleration andis directed parallel (or antiparallel) to the tangential velocity. It is the tangential accel-eration that is responsible for changing the speed of the particle executing circularmotion. Don’t confuse the tangential acceleration with the centripetal acceleration,

discussed earlier in connection with circular motion in Chapter 5. Even in the case of uni-form circular motion, where the speed and � are constant and so atang � 0, there is anonzero radially directed centripetal acceleration required to steer the object around thecircle. If atang is not equal to zero the particle’s speed will change as it travels in circularmotion and it will have both tangential and radial components of acceleration.

It is useful to rewrite the expression for the centripetal acceleration in anotherequivalent form. For extended objects such as a wheel, the velocity of differentparts of the wheel will be different, depending on their distance from the axis ofrotation (Figure 7.2). For this reason it is more useful to rewrite the expression forthe centripetal acceleration in terms of � using Equation (7.3)

(7.6)

Having introduced the angular variables, �, �, and , needed to describerotational motion, we are now in a position to derive a set of equations amongthese variables in the case of constant angular acceleration as we did inChapter 3 when the linear acceleration was constant (see Table 3.1). Becausewe have the proportionality of s and �, v and �, as well as a and , we canproceed by simply dividing each of the linear variables in the kinematic rela-tions of Table 3.1 by the radius of the circle r to arrive at a set of kinematicequations for the angular variables:

(7.7)v(t) � vo�at;

acent � v2

r� v2 r.

acent � v2

r,

a tang � ra.

a�¢v

¢t; a� lim

¢t:0

¢v

¢t.

v � rv,

v� lim¢t:0

¢u

¢t.

RO TAT I O N A L K I N E M AT I C S 163

FIGURE 7.2 A rotating wheel with itsincreasing velocity with increasingdistance from the axis of rotation.

v

ω

(7.8)

(7.9)

These three equations serve as a basis for describing pure rotational motion withconstant angular acceleration just as their linear counterparts were used in Chapter 3.

v 2 � vo2 �2a (u�uo).

u(t) � uo � vo t �1

2at2;


Example 7.1 A stationary exercise bicycle wheel starts from rest and acceleratesat a rate of 2 rad/s2 for 5 s, after which the speed is maintained for 60 s. Find theangular speed during the 60 s interval and the total number of revolutions thewheel turns in the first 65 s.

Solution: Using Equation (7.7) we find that the angular speed is given by

In the first 5 s, the wheel rotates through

and in the next 60 s the wheel rotates through an additional

because there is no acceleration during this interval of time. The 625 rad totalcorresponds to

The key to solving this problem was to divide the total time interval into twoportions, only one of which had an acceleration.

u�625 rad

2p rad / rev� 99.5 rev.

u� vo t � 10 # 60 � 600 rad,

u�1

2 at2 �

1

2# 2 # 52 � 25 rad,

v� 0 � at � 2 # 5 � 10 rad / s.

Example 7.2 Helical bacterial flagella drive E. coli at constant speed when theyrotate around counterclockwise (CCW, as viewed from behind the bacterium) at auniform angular velocity, appearing much like a corkscrew (Figure 7.3). Fromtime to time the flagella motor reverses to clockwise (CW) rotations, causing theflagella to disorganize themselves and the bacterium to tumble, before switching

FIGURE 7.3 (left) Coordinated flagella lead to swimming;(right) disordered flagella lead to “twiddling”.

2. ROTATIONAL ENERGY

Now that we have a set of rotational variables to describe the kinematics of rota-tional motion, we take up the description of rotational dynamics of a rigid body. Webegin our study of rotational dynamics in this section with a discussion of therotational kinetic energy for an object with a fixed axis of rotation. Recallthat in this case all parts of the object rotate about this axis in circular motion.To begin, consider a single particle when it is moving in a circle (Figure 7.4).Using our expression for kinetic energy, , and the fact that thevelocity can be written in terms of the angular velocity and the radius of thecircle, v � �r, we can write that . Defining

(7.10)

where I is called the moment of inertia (for reasons that will become clear),we can write the kinetic energy of our particle as

(7.11)KE �1

2Iv2.

I � mr2, (for a single particle),

KE � 1/2m (vr)2

KE � 1/2mv2

RO TAT I O N A L E N E R G Y 165

again to CCW rotation so the bacterium swims off in a different direction. Supposethat the flagella “motor” rotates with a frequency of 4 Hz (4 rotations per second)when in either the CCW or CW state. If the flagellum spends 98% of its time inthe CCW state and takes 5 ms to reverse its rotation (with the CCW to CW transi-tion occurring on average every 5 s), find the average angular acceleration duringa CCW to CW transition and the net angular rotation in a 10 s interval.

Solution: We are given that in a 5 ms interval, the flagellum reverses its rota-tion from an angular velocity of �o � �2�(4) rad/s to � � 2� (4) rad/s.Therefore using the equation �(t) � �o � t, we find that the average angularacceleration is

In an average 10 s interval, because the flagellum spends 98% of its time inthe CCW state, it will spend only 0.2 s in the CW state, or 0.1 s in each of twoCW intervals since the transitions occur every 5 s on average. For the 98% ofthe time in a CCW state, the flagella rotate at 8� rad/s producing a net rotationof � � �t � 8�(9.8) � 246 rad. During each of the other 0.1 s, there will be anacceleration to the CW state during 5 ms, a stay of 90 ms in that state, and anacceleration back to the CCW state for 5 ms (for a total of 100 ms � 0.1 s). Weneed to compute the net rotation during this time and multiply by 2 for the two such0.1 s intervals. But by symmetry, the two 5 ms intervals will produce exactly oppo-site net rotations, canceling their contributions, and leaving only the � � �t ��8�(.09) � �2.3 rad contribution for each of the two intervals. Adding up theangular contributions, we have

Such studies on bacteria flagella have led to an increased understanding ofthe detailed energy sources and molecular interactions necessary for motility.

unet � 246 �2(2.3) � 241 rad �241

2p rev. � 38.4 rev.

a�v � vo

t�

2p(4 � ( - 4))

5 10�3� 1.0 104 rad / s2.

v = rω

ω

r

FIGURE 7.4 A particle traveling in a circle.

Note that this equation has a form similar to that of translational kinetic energy ifwe make a correspondence between the angular velocity � and the linear velocity v andbetween the moment of inertia I and the mass m. We call the type of KE in Equation(7.11) rotational kinetic energy and discuss it further below after generalizing to therotational motion of an extended rigid object. Note that if the particle travels in a uni-form circular motion, its rotational kinetic energy is a constant, but if there is a tan-gential force acting on it as well as a centripetal force, then there will be a tangentialacceleration and the angular velocity will change as will the rotational kinetic energy.


Example 7.3 A 25 kg girl riding on the outer edge of a large merry-go-roundwith a 10 m diameter has a (rotational) kinetic energy of 20 J. Find the girl’smoment of inertia relative to the axis of rotation and find the number of revolu-tions the merry-go-round makes per minute.

Solution: The girl’s moment of inertia, calculated as if she were a point mass, isgiven as

To proceed we first calculate the angular velocity of the girl (and merry-go-round) using the expression for the rotational kinetic energy, Equation(7.11), so that

We find that � � 0.25 rad/s so that in 60 s the girl has gone around an angle� � �t of 15 rad, corresponding to 15 rad/2� � 2.7 rev.

v� A2(KE)

I.

I � mr2 � 25 (5)2 � 625 kg-m2.

FIGURE 7.5 Physics on a merry-go-round.

Although this example does not deal with a point mass, we simplified our analy-sis to that case. Next, we want to generalize our discussion to extended rigid bod-ies. As a simple model, consider a rigid collection of point masses mi attachedtogether by “massless” rods with a fixed axis of rotation. If the assembly rotatesabout this axis, each will circle about a common central axis at some radius ri. Usingthe summation convention, we can write the total kinetic energy of the collection asthe sum of the kinetic energies of all the particles

(7.12)

and, because vi � ri�, we have that

(7.13)

where the summation is only over the different masses at their corre-sponding perpendicular distances from the axis of rotation, because theangular velocity of all the particles is the same. We define the summationto be the moment of inertia of the assembly of masses about the axisof rotation

(7.14)

so that the total kinetic energy may still be written in the simple form ofEquation (7.11).

To analyze the more realistic model of a rotating extended rigidbody, rather than a collection of particles, we can follow a procedurewhere we divide the object up into small elements of mass mi, each atcorresponding distances ri from the axis of rotation (Figure 7.6). Wecan first ask how it is that a single force, applied to the rigid body ata localized point, can make the entire body rotate. When the externalforce is applied, internal forces that keep the object rigid do theappropriate amount of work on each localized mass element to main-tain the shape of the body as it rotates. These internal forces are actu-ally transmitted by electromagnetic interactions but for our purposescan be imagined to be transmitted via very stiff springs between mol-ecules, the ultimate mass elements. At this point one can imagine howthe spring properties can affect the overall rigidity of the solid andgive rise to changes in the shape of an object when external forces acton it. When the applied forces become larger, our rigid body willeventually become deformed. We have already briefly studied thedeformation of solids in Chapter 3, where we introduced the Young’smodulus as well as the shear and bulk modulus to describe differentdeformations.

The division of a solid body, rotating about a fixed axis, into smallmass elements allows a similar argument as in the case of discrete pointmasses and leads to an identical expression for the kinetic energy inEquation (7.11). The moments of inertia of various rigid objects withsome symmetry are shown in Table 7.1. (See the boxed discussion for acalculation of one case.) Note that in every case the moment of inertia isequal to the product of the total mass and the square of the pertinentdimension apart from a numerical factor that depends on the geometryas well as the axis of rotation.

I � a (mi ri2),

KE �1

2a(mi ri

2)v2,

KE �aKEi �1

2ami vi

2,


mi

ri

Vi

ω

FIGURE 7.6 A rigid body withdiscrete mass elements undergoingan overall rotation.

For a continuous rigid body the definitionof the moment of inertia given in Equation(7.14) needs to be rewritten as

,

where r is the perpendicular distance of thedifferential element of mass from the axisof rotation. If the object has a constant massdensity � then this can be rewritten as

,

where dV is the volume element containingmass dm. As an example we calculate themoment of inertia of a right circular cylinderof radius R and length L about its axis (seeFigure 7.7). We divide the cylinder into vol-ume elements that are cylindrical shells ofradius r, length L, and thickness dr. All of themass in this shell has the same r and thereforethe same I. The volume of the shell is dV �2�rLdr, so that the integral becomes

.

This expression integrates to yield I �2��R4L/4, and can be rewritten in terms ofthe total mass of the cylinder, M � ��R2L, asI � 1⁄2 MR2, giving the expression in the table.

I � rL

R

0

2prL # r2dr � 2prLL

R

0

r3 dr

I � rL

r2dV

I �L

r2dm

L

R

r

FIGURE 7.7 Construction used to calculatethe moment of inertia of a rod.


Example 7.4 Calculate the moment of inertia of the gadget shown in Figure 7.8.The small masses are attached by a light rigid rod and pivot about the left end ofthe rod. Use a value of m � 1.5 kg and d � 0.2 m. If the assembly were to pivotabout its midpoint, find the moment of inertia about this axis as well.

Solution: Using Equation (7.14), we simply add up the individual contributionsto the moment of inertia. With the pivot point at the left end, we find

and with the pivot point at the middle of the assembly, we find

I � mad

2b2

� 2mad

2b2

� 3m(2.5d)2 � 19.5md2 � 1.2 kg # m2.

I � m(2d)2 � 2m(3d)2 � 3m(5d)2 � 97md2 � 5.8 kg # m2,

SOLID CYLINDER about symmetry axis

I = MR212

I = MR225

I = MR223

I = ML213

R

I = MR212

L

SOLID CYLINDER about central diameter

HOOP about symmetry axis

I = MR2

R

HOOP about any diameter

R

SPHERE about any diameter

R

SPHERICAL SHELL about any diameter

R

LONG ROD about perpendicular axis atcenter

L

LONG ROD about perpendicular axis at end

L

R

L

I = MR2 + 14

112 ML2

I = 112 ML2

Table 7.1 Moments of Inertia of Various Symmetrical Objects

m 2m 3m

d2d 2d

FIGURE 7.8 Gadget of Example 7.4 withthree point masses attached by “massless”rods.

Now that we have introduced moment of inertia and rotational kineticenergy for an extended object that is rotating about a fixed axis of rota-tion, we are in a position to generalize these ideas to the case of rollingmotion. A wheel or other symmetric object that rolls can be shown tohave a total kinetic energy that consists of two parts: the translationalkinetic energy of the center of mass plus the pure rotational kineticenergy of the object about a fixed horizontal axis through its center ofmass (see Figure 7.10). When all the forces acting on a system of rigidbodies are conservative so that the work done by those forces can beexpressed as a potential energy difference, we can write the conservation


It should make intuitive sense, after a moment’s thought that I should besmaller in the second case, because the masses are traveling in smaller radiicircles. From Equation (7.11), for the same angular velocity in both situations,we expect there to be less kinetic energy in the second case, in agreement withthe smaller I.

Example 7.5 Find the moment of inertia of the object shown in Figure 7.9 whenpivoted about its symmetry axis. The cylinder has a mass M, radius r, and lengthL, whereas the hoop has a mass M/10 and radius 3r. Use M � 0.1 kg, r � 5 cm,and L � 25 cm.

Solution: The moment of inertia of the hoop is simply the product of its massand the square of its radius since all its mass lies at the same radius. The momentof inertia of the cylinder cannot be found so simply because its mass is distrib-uted over varying distances from the axis of rotation. Using Table 7.1 we lookup its moment of inertia and then write the total moment of inertia as the sum ofthe hoop’s and the cylinder’s as

Note that the length L of the cylinder is not in the answer, only its total massand radius.

I �1

2 Mr2 �

M

10 (3r)2 � 1.4Mr2 � 3.5 10�4 kg # m2.

FIGURE 7.9 A solid cylinder and a hoopconnected by light rods both rotatingabout their common symmetry axis.

v

ω

FIGURE 7.10 A cylinder rollingdown an inclined plane has a totalkinetic energy equal to the sum ofits center of mass translational KEand its rotational KE about the center of mass axis.

of energy equation for the system, composed of translating, rotating, or rolling sym-metric rigid bodies, as


Example 7.6 An empty bucket of 1 kg mass, attached by a light cord over thepulley for a water well, is released from rest at the top of the well. If the pulleyassembly is a 15 cm uniform cylinder of 10 kg mass free to rotate without anyfriction, find the speed of the bucket as it hits the water 12 m below.

Solution: The initial energy of the bucket–pulley system can be taken as puregravitational potential energy, measured with respect to a zero level at the watersurface. When the bucket just reaches the water the final energy is the sum of kinetic energy of the bucket (translational KE) and pulley (rotational KE). Withno frictional forces present, the initial and final energies are equal and we can write

where m is the mass of the bucket, v its velocity as it hits the water, � the angu-lar velocity of the pulley as the bucket hits the water, and I the moment of inertiaof the pulley, given by , where M is the pulley mass and r is its radius.

The bucket’s velocity and the pulley’s angular velocity are related by v � � rbecause the cord is wrapped around the pulley at radius r and does not slip, sothat we can rewrite our energy equation as

Solving for the bucket’s velocity

Substitution of numbers results in

Note that this result is independent of the radius of the pulley. If the pulleywere massless, then we would find that

v � A2mgh

m� 12gh � 15.3 m /s,

v � A2 # 9.8 # 12

1 � 0.5 # 10� 6.3 m /s.

v � A2mgh

1m � 12 M2.

mgh �1

2 mv2 �

1

2 (

1

2 Mr2)v2 �

1

2 (m �

1

2 M) v2.

I � 12 Mr2

mgh �1

2 mv2 �

1

2 Iv2,

(7.15)

(Conservation of Mechanical Energy)

The sum is the constant total mechanical energy of the system with thefirst two terms representing the total translational kinetic energy of the centerof mass of all objects in the system and the pure rotational motion about the

center of mass of each rotating object.

1

2 mv2 �

1

2 Iv2 � PE � E � constant.

If there are also nonconservative forces present, such as friction, then the right-handside of Equation (7.15) will no longer be a constant but will decrease with timebecause of the work of friction, just as in the translational motion situations we stud-ied in Chapter 5. A few examples help us to see how to apply conservation of energyprinciples in order to study problems with rotational motion.


considerably faster. Because energy is conserved in both cases, why does thebucket have a much larger KE with a massless pulley and where does the miss-ing energy go in the original problem? The smaller translational KE of thebucket is due to the relatively large rotational KE of the pulley just before thebucket hits the water.

Example 7.7 Suppose that a hoop and a cylinder, with the same radius and mass,both roll down an inclined plane, with an inclination angle �, from rest at aheight H without slipping. With what velocity does each arrive at the bottom andwhich will arrive first?

Solution: We can use the conservation of energy principle to solve this problem.The initial energy of each object is the same gravitational potential energy Ei �mgH. After rolling down the incline, the final energy of each is purely kinetic,equal to the sum of the center of mass translational KE and the rotational

, where I is the moment of inertia of each object aboutan axis through its center. Because there is no slipping, there is no loss of energydue to friction and the total energy of each object is conserved. We then canwrite that Ei � Ef or

Since there is no slipping, we can also relate the center of mass velocity v tothe angular velocity of each object through the same relation v � r�, where r is theradius of the hoop or cylinder. Then, on substituting for � (� v/r), we have that

v2 �2mgH

(m � I / r2).

mgH �1

2 mv2 �

1

2 Iv2.

KE, Ef � 1/2 mv2 � 1/2 Iv2

FIGURE 7.11 A bucket suspended from a pulley andfalling into a well.

(Continued)

3. TORQUE AND ROTATIONAL DYNAMICS OF A RIGID BODY

We turn now to the mechanism by which rotational motion is produced. In order tohave an object translate from rest, we require a net force to act. But a force, no mat-ter how large, is not necessarily able to make an object at rest rotate. Consider theexample shown in Figure 7.12a in which a door is to be opened. Pushing on thehinged side of the door with F1, no matter how hard, will not open the door; simi-larly, pushing on the edge of the opened door toward the hinges with F2 will also notresult in any rotation of the door. Thus, it is clear that a force will not produce rota-tional motion unless it is well-placed and well-directed.

To clarify what is meant by well-placed and well-directed, consider the same doorwith force F3 applied, also shown in a top view in Figure 7.12b. The force acts in thehorizontal plane at an angle � with respect to the horizontal position vector from the


Looking up in Table 7.1 that Ihoop � mr2 (it’s easy to see why this is so,because all the mass of the hoop lies the same distance r from its center) and

we have that

,

Note carefully that the final velocities do not depend on the radius of theobject, the mass, or on the inclination angle. From our expressions it is clear thatthe cylinder arrives at the bottom of the incline with a faster speed. Also, rota-tional kinematics tells us that �t � ��/�average and because they both start fromrest and accelerate uniformly, �average � �final/2, so that the object with thegreatest final angular velocity will reach the bottom fastest. With equal radii, thecylinder clearly wins the race. Without rotational motion, both of these objectswould take the same time to slide down the incline, arriving with the same speed.The rotational motion takes up some of the translational kinetic energy into rota-tional kinetic energy about the center of mass. The object with the greatermoment of inertia gains the greater rotational kinetic energy and therefore losesthe most translational energy and loses the race as well!

vcylinder � A2mgH

(m � m�2)� A

4gh

3

vhoop � A2mgH

(m � m)� 1gH

I cylinder � 1/2 mr 2,

F1

F3

F2

F1

F3

F2

A B

θ

FIGURE 7.12 (A) Door, hinged atthe left, pushed more or less effec-tively in different directions and atvarious locations. (B) A top view ofthe door.

axis of rotation to the point of application. If we imagine taking the components ofthis force along and perpendicular to the position vector, it is clear that only the per-pendicular component will result in rotation of the door. The component parallel tothe position vector, the so-called radial component, itself will not result in rotation ofthe door no matter how large it is. Furthermore, if the same force is exerted on thedoor at a closer distance to the hinge, it is less effective in rotating the door. In thelimit of applying the force directly on the hinge, no rotation at all will occur no mat-ter in what direction the force is aimed.

Having shown the need for care in defining the quantity that “drives” objects torotate, let’s first look at the work–energy theorem in the case of rotational motion.From Chapter 4 we know that the net work done by external forces on an object isequal to the change in its kinetic energy. If the object is confined to rotate about afixed axis of rotation, any change in its kinetic energy must be in its rotational kineticenergy and, in that case, we can write

(7.16)

where I is the total moment of inertia of the object with respect to the axis of rota-tion. Imagine a short interval of time �t, during which a net force does an amount ofwork �W to produce a change in angular velocity from � to � � ��. In this case wecan write the work–energy theorem as

(7.17)

Expanding the term in brackets, we can rewrite this as

Because we are interested in taking the limit as the time interval approaches zero,in which case so does ��, we neglect the second term on the right (which will bemuch smaller than the first) and rewrite our expression as

Writing ��t � �� and ��/�t � , which are correct in the limit as �t approacheszero, we have

(7.18)

Now for our case of pure rotational motion we know that all points of the objectrotate in circles. From the general definition of work, �W � (Fnet,ext)x�x, a net exter-nal force acting on a particle that is traveling in a circle will do an amount of workgiven by

where F›

is the component of the net applied force that acts along the tangential displacement direction and �x � s � r�� is the distance over which the force acts(see Figure 7.13). If we define the rotational analog of the force, known as the torque,or as the moment of the force, to be

(7.19)

we obtain the analog expression for the work done for pure rotational motion

(7.20)¢W � tnet, ext ¢u.

t� F�r,

¢W � F�r¢u,

¢Wnet, ext � Ia ¢u.

¢Wnet, ext � Iv ¢v

¢t ¢t.

¢Wnet, ext � Iv ¢v�1

2I1¢v22.

¢Wnet, ext �1

2 I1v� ¢v22 �

1

2 Iv2.

Wnet, ext � ¢KErot � ¢

1

2Iv2,

T O R Q U E A N D RO TAT I O N A L DY N A M I C S O F A R I G I D B O DY 173

r

F⊥

Fm

Δθ

FIGURE 7.13 Torque on a particlein circular motion.

The units for torque are N-m, the same units as those for work or energy. Thisfollows directly from Equation (7.19), or from Equation (7.20) since �� is dimen-sionless, but because torque and energy are different concepts, we never write atorque in units of joules, but always use N-m. Before discussing torques in moredetail, let’s first introduce the rotational analogue equation to Newton’s second law.

On comparing Equations (7.18) and (7.20), we see that

(7.21)

which is the rotational analog of Newton’s second law. Note that �, I, and are therotational analogs of F, m, and a, respectively. The moment of inertia, and not themass, enters into the rotational version of Newton’s second law, therefore not onlythe mass of the system, but also its distribution from the axis of rotation is importantin determining the response of the system to an applied torque.

With these results in hand let’s first examine Newton’s second law, in both rota-tional and translational forms, for the simplest case of the rotational motion of asingle particle of mass m. Suppose the particle is located a distance r from the axisof rotation, attached to the center of the circle by a light rod, and is set in rotationalmotion by a force F acting as shown in Figure 7.13. In that case I � mr2 and we havefrom Equation (7.21) that

From Equation (7.19), the torque on the mass is given by

so that the torque depends on three factors: the magnitude of the applied force, whereit is applied, and its orientation with respect to r, a line perpendicular from the axisof rotation to the point of application of the force. Only the perpendicular componentof F contributes to the torque’s ability to make the particle rotate around the circleand we have

The outward radial component of the force must be more than balanced by a large inward radial force supplied by the light rod that is required to keep the mass traveling in a circle. The net inward radial force is then the centripetalforce.

An alternate description of the rotational motion can be given by analyzing thetangential forces and accelerations. The tangential component of the force producesa tangential acceleration. Newton’s second law in the tangential direction lets uswrite that

in agreement with the previous equation. Although tangential forces and accelera-tions can be used in the simplest rotational problems, the first approach uses thenatural variables to describe rotational motion, angular acceleration and torque. Withmore than a single particle in the system, if the distances from the axis of rotation aredifferent for the particles, in general it will be much more difficult to analyze theproblem in terms of linear variables and much easier in terms of rotational variables.This is also true for extended real objects that are not treated as particles. Two exam-ple problems illustrating the application of the rotational form of Newton’s secondlaw help to make this discussion more concrete.

F� � ma tang � mar,

rF� � mr2a.

t� rF� ,

t� mr2a.

tnet,ext � Ia,



Example 7.8 Let’s reconsider the problem of opening a door as discussed at thebeginning of this section. Suppose the door is uniform and has a mass m � 10 kg,a height h � 2.5 m, and a width w � 1 m. The moment of inertia of a uniformrectangular slab (with dimensions w h) about a vertical axis of rotation alongone of the edges is given by , independent of h (those of you whoknow some calculus might try to derive this following the method used in the boxed calculation in Section 1). Suppose that the door is pushed with asteady horizontal force F � 5 N acting at the edge of the door and directed at aconstant 30° angle from the normal to the door as it is opened (The forcechanges direction as the door is opened to keep the angle with respect to the doorconstant at 30° as shown in Figure 7.14). Find the angular acceleration of thedoor and the time for it to swing fully open, rotating a total of 90°.

Solution: A steady torque acts to push the door, thereby producing a constantangular acceleration. The torque is given by

where the perpendicular component of F is obtained from the figure and the distancer equals w in this case. The constant angular acceleration of the door is given by

Using this acceleration, we can find the time for the door to swing by 90°(� �/2 rad) angle to be

u� 12a t2 or t � A

2ua

� A2 # 13.14/22

1.3� 1.6 s.

a�t

I�

4.3131102(1)2

� 1.3 rad /s2.

t� F�r � Fcos 30 w � 5 cos 30 � 4.3 N # m,

I � 13 mw2

w

h

F30

FIGURE 7.14 How long does it take toopen a door?

Example 7.9 An ultracentrifuge is spinning at a speed of 80,000 rpm. The rotorthat spins with the sample can be roughly approximated as a uniform cylinder of10 cm radius and 8 kg mass, spinning about its symmetry axis (so that, fromTable 7.1, ). In order to stop the rotor in under 30 s from when the I � 1

2 mr2

(Continued)

Figure 7.15 shows that the expression for the torque can be written in twoequivalent ways:

(7.22)

or, by regrouping terms,

(7.23)

We see that the torque can be calculated by either taking the product of r, thedistance from the axis of rotation to the point of application of the force, and thecomponent of the force perpendicular to r, or by taking the product of F and the com-ponent of r perpendicular to F, known as the moment (or lever) arm. The momentarm is the perpendicular distance from the axis of rotation to the line of action of theforce (the line along which the force is applied). Two additional examples clarify thecalculation of torques and their use in rotational motion problems.

t� (r sin u)F � r� F.

t� rF� � r(F sin u),


motor is turned off, find the minimum braking torque that must be applied. If nobraking torque is applied, the rotor will stop in 30 min. Find the frictional torquethat is present under normal spinning conditions.

Solution: We first need to find the minimum angular acceleration needed to stopthe rotor in under 30 s. Using

we find that to stop the rotor requires an angular acceleration of

Because �o is given as 80,000 rpm, we first must convert it to rad/s, �o �(80,000)2�/60 � 8.38 103 rad/s, where the factor 2� converts revolutions toradians and the factor 60 converts from minutes to seconds. Then we find thatwith t � 30 s,

The braking torque must have a magnitude of at least

In the absence of a braking torque, we recalculate the angular accelerationusing t � 30 min to find an acceleration 60 times smaller, � �4.7 rad/s2, sothat the normal frictional torque has a magnitude 60 times smaller as well, or � � 0.19 N-m.

t� Ia� 12 mr2a� 1

2 810.12212792� 11.2 N # m.

a� -

vo

t�

8.38 103

30� -279 rad /s2.

a� �vo

t.

v� v0 �at,

F

line of action

m

F

r

r

θ

FIGURE 7.15 Equivalent definitionsof torque: .t � r F� � r�F

Example 7.10 Calculate the forces that the biceps muscle and the upper armbone (humerus) exert on a person’s forearm when supporting a weight as shownin Figure 7.16 without any movement. The forces acting on the forearm include


its weight mg, the weight of the object held in the hand Mg, the pull of the bicepsmuscle Fbiceps, and the humerus connection at the elbow socket, Fhum. Take theweight to be 20 N, the length L of the (uniform) forearm as 40 cm, its mass as2 kg, with the biceps connecting d � 4 cm from the elbow pivot point, andassume that the arm is held at 40° with respect to the vertical.

Solution: To calculate the two unknown forces, we must realize that the netforce and net torque on the forearm must both be zero because the weight is heldat rest. This example anticipates the subject of statics, which we take up inSection 7 below. If we add up the net force and set it equal to zero and set thenet torque equal to zero as well, we will obtain two independent equations thatwill allow us to solve for the two unknown forces. Only three of the four forcesproduce a torque about the elbow because the force from the humerus acts at theelbow joint and has zero lever arm. The torque equation is

The lever arm distances were obtained from the distances along the forearmfrom the elbow pivot point by taking the horizontal components, those perpen-dicular to the vertical forces. We can then solve for the biceps force directly;canceling the common term sin 40,

The force from the humerus can be obtained by summing the forces on theforearm to zero

to find that Fhum � 260 N. Note that to lift a relatively small 20 N weightrequires very large forces on the bones and muscles of the body.

Mg � mg � Fhum � Fbiceps � 0,

Fbiceps � aMg �mg

2bL/d �

(20 � 2 # 9.8/2)0.4

0.04� 300 N.

tnet � Mg1L sin 402� mgaL

2 sin 40b � Fbiceps1d sin 402� 0.

Mg mg

Fbiceps

Fhum40°

FIGURE 7.16 A person’s arm supporting a weightand the force diagram for Example 7.10.

Example 7.11 Find the net torque about both the left end (A) and the center(B) of the uniform rod shown in Figure 7.17 with the set of external forcesshown. Use the following values F1 � 30 N, F2 � 20 N, Mg � 20 N, F3 �10 N, F4 � 15 N, a rod length L � 40 cm, with F2 acting at L/3 from theright end.

(Continued)


Solution: We first note that both F3 and F4 do not produce a torque about the leftend (A) of the rod. Adding the torques from the other three forces about the leftend, we find a net torque

In this expression we took torques tending to rotate the rod clockwise aboutthe left end as positive and used the perpendicular components of the forces inthe expressions for the torques. Substituting in the numbers, we find

where the negative sign indicates that the net torque would produce a counter-clockwise rotation about the left end.

Repeating this procedure taking torques about the center (B), note that F4and Mg do not produce any torque and we have

Substituting in numbers, we find

Why do we get these two different results when taking torques about two dif-ferent points with the same set of forces acting? First, note that calculating torquesexplicitly depends on the reference point. In fact, because the net torque is notequal to zero, the rod is not in equilibrium. So although the torque calculations areboth correct, let’s discuss which one we would use to describe the motion of therod. The actual motion of the rod can be separated into a translation of the centerof mass due to the nonzero net force (you should check that there is both anupward and leftward net force) and a rotation about the center of mass. If we wereto move with the center of mass then we would see a pure rotation of the rod aboutits center and then the net torque about the center would be equal to the momentof inertia of the rod about its center times its angular acceleration. We could thenfind this angular acceleration and combine it with the translational acceleration ofthe center of mass to describe the overall motion of the rod.

tnet, B � 10(0.2) � 20(cos 45)0.4

6� 30(sin 30)

0.4

2� � 1.9 N # m.

tnet, B � F3 L

2� F2cos 45

L

6� F1sin 30

L

2.

tnet, A � 20(0.4/2) � 20(cos 45)(2/3)(0.4) � 30(sin 30)(0.4) � � 5.8 N # m,

tnet, A � Mg L

2� F2 cos 45

2L

3� F1 sin 30 L.

30

45F1

F2

Mg

F3

F4 A B

FIGURE 7.17 A set of forces acting on a uniform rod.

As an example of rotational motion in an important biological macromolecule, let’sdiscuss some of what is known about the world’s smallest rotary motor, an enzymatic pro-tein, F1-ATPase, which helps in the efficient production of ATP in cells. Discovered in1956, this protein is found in virtually identical form in species ranging from bacteria tomammalian cells. Figure 7.18 shows both a schematic drawing and a ribbon model of theprotein structure. The central subunit acts as a shaft able to rotate within the array ofalternating and � subunits arranged in a circle. This protein is a reversible rotary motor.Normally, when driven to rotate at very high rotational speed of several thousand

revolutions per minute by energy from a proton (or hydrogen ion) membrane pump, it actsas an enzyme helping to generate huge amounts of ATP daily. When the protein is suppliedwith ATP, it can run in reverse, causing the subunit shaft to rotate just like a motor.

Recently biophysicists were able to attach a rodlike molecule to the subunit, andmeasure the torque generated by the rotary motor in turning this attached rod. Themeasurement was done using laser tweezers, discussed in Chapter 19. In fact, theywere able to lower the ATP concentration sufficiently so that individual step rotationsof 120° of the shaft were observed. The individual torque measured for each steprotation was 44 pN-nm, where the incredibly small units used are those appropriatefor the small force and step size involved. These researchers then calculated the workdone by this rotary motor in each step rotation. Using Equation (7.20) and a steprotation angle of �� 120° � (2�/3), they found that �W � (2�/3)(44 pN-nm) �92 pN-nm � 92 10�21 J. This value is very close to the energy liberated by oneATP molecule when it is hydrolyzed to ADP. Thus, this smallest of all rotary motorsis nearly 100% efficient in converting energy into rotational work. It remains to beseen to what future applications our knowledge of this protein will lead.

4. ANGULAR MOMENTUM

In our discussion of momentum in Chapter 6 we were able to rewrite Newton’s secondlaw for a system, , in terms of its total, or center of mass, momentum sothat the net external force was equal to the rate of change of the total linear momentum

(7.24)

Recall also that in the absence of a net external force, this equation leads to thepowerful conservation of momentum principle. In this section we analyze rotationalmotion in an analogous manner and introduce the important new quantity angularmomentum and the principle of conservation of angular momentum, our third funda-mental conservation principle. (Energy, linear momentum, . . . are you counting?There are not many more in this book.)

If you had to guess how angular momentum L should be defined, based on therotational analog quantities to those defining the linear momentum, it is hoped thatyou would come up with the expression

(7.25)

Because linear momentum is defined as pS � mvS and the rotational analogs to m and v areI and �, respectively, this would be the natural candidate. Of course, such an intuitiveguess needs to be corroborated, but this is a correct expression. You may have noticed that

L � Iv.

F:

net, ext � lim ¢t:0

¢P

:

total

¢t.

F:

net, ext � ma:

cm

A N G U L A R M O M E N T U M 179

α

α

β

β

γ

FIGURE 7.18 (left) Schematic ofF1-ATPase, the world’s smallestrotary motor with three pairs ofalternating and � subunits andthe subunit shaft (right) molecularmodel with the subunit shaft inlight blue (scale bar � 2 nm).

in our analogy, Equation (7.25) has omitted vector signs on L and �, where they might beexpected. This is intentional on our part. It turns out that the vector nature of the rotationalvariables is subtle and is not needed in our basic discussions of rotational motion.

In the case of a particle of mass m constrained to rotate in a circular orbit, fromthe expression for its moment of inertia about the axis of rotation I � mr2 we canwrite an alternative expression for the angular momentum of such a particle as

(7.26)

For a system of particles or an extended body rotating about a fixed axis of rota-tion, an argument similar to the one given for the moment of inertia shows that thetotal angular momentum can also be written as

(7.27)

where the sum is over the mass elements of the system and ri and are the distances(measured from the axis of rotation) and components of momenta perpendicular to ri(or tangential to the circular trajectories for pure rotational motion).

Now that we have defined angular momentum, we turn to the rotational equationcorresponding to Equation (7.24). By analogy we should guess that this is

(7.28)

where again we omit vector signs. This can most easily be seen by noting that thetime rate of change of L in Equation (7.28) can be written as

using the definition of angular acceleration. Substituting for I from Equation (7.21)then yields Equation (7.28). We reach an important conclusion from this equation.

lim¢t:0

¢L

¢t� lim

¢t:0 I

¢v

¢t� Ia,

tnet, ext � lim¢t:0

¢Ltotal

¢t,

pi,�

Ltotal � a ri pi,�,

L � mr2v� rm(rv) � r(mv) � rp.


Example 7.12 An ice skater begins a spin by rotating at an angular velocity of 2 rad/s with both arms and one leg outstretched as in Figure 7.19. At that timeher moment of inertia is 0.5 kg-m2. She then brings her arms up over her headand her legs together, reducing her moment of inertia by 0.2 kg-m2. At whatangular velocity will she then spin?

Solution: Because there are no acting external torques (any friction is ignoredhere), angular momentum is conserved and we can write that

Iini vini � Ifinvfin.

In the absence of a net external torque on a system, its total angular momentumremains constant. This is a statement of the principle of conservation

of angular momentum.

Along with conservation of energy and of (linear) momentum, it is one of the funda-mental conservation laws in nature. For an extended body undergoing pure rotationalmotion conservation of angular momentum has the simple form

(7.29)

where the constant is the value of Ltotal at any instant of time. The following exampleillustrates the application of conservation of angular momentum.

Iv� constant, (isolated system),


In this case the skater’s moment of inertia has decreased and so her angularvelocity will increase.

We find

so that her angular velocity becomes 3.3 rad/s. The same principle controls therotational motion of a ballerina or a diver as they change their moment of iner-tia by controlling their body configuration.

0.5122� 0.3vfin,

FIGURE 7.19 An ice skater uses angularmomentum conservation.

We summarize the rotational motion equations of this chapter in Table 7.2, indi-cating the corresponding equations for translational motion. In the last examples ofthis section we integrate the concepts presented so far in solving two more complexrotational motion problems.

Table 7.2 Kinematic and Dynamic Equations for Rotational and Translational Motion

Applicability Rotational Translational Relations Between Variables

(a) � constant � � �o � t v � vo � at s � r�

(a) � constant v2 � vo2

+ 2a (u - uo) v2 � vo2 � 2a(x � xo) v � �r

(a) � constant u � uo + vo t +12 at2 x � xo � vot � 1

2 at2 atang � r

General KE � 12 Iv2 KE � 1

2mv2 I � g miri2

General�net,ext � I Fnet,ext � ma t� rF� � r�F

General tnet,ext �¢L total

¢t Fnet,ext �¢Ptotal

¢tL � Iv � rp�

General �net,ext � 0 ⇒ Ltotal � constant Fnet,ext � 0 ⇒ Ptotal � constant


Example 7.13 A hoop of mass 2 kg and radius 0.5 m has two spokes thelength of a diameter, each of mass 0.1 kg. The hoop is made to rotate fromrest by a light cord attached to a 0.02 m diameter shaft which is threaded overa frictionless pulley, and attached to a 10 kg weight (as shown in Figure 7.20).Find the angular velocity of the hoop after the 10 kg weight has fallen a dis-tance of 1 m.

Solution: The tension in the cord supplies a torque to rotate the hoop–spokeassembly at an increasing velocity. We solve this problem in two ways: usingtorques and angular accelerations and using energy concepts.

Using the first method, we first find the torque acting on the hoop and themoment of inertia of the rotating assembly so that we can substitute them intoNewton’s second law for rotations in order to find the angular acceleration. We have

where T is the tension in the rope and r is the shaft radius. The cord tension isthe only force that produces a torque on the hoop. The total moment of inertia isthat of the hoop (MR2, with M and R the mass and radius of the hoop) and thatof the two spokes (see Table 7.1 for I for a rod rotating about an axis through itsmidpoint)

where m is the mass of each spoke of length 2R. To proceed, we first need to findthe tension T which is not equal to the hanging weight. An independent equationfor T can be obtained from the equation of motion for the hanging mass m�

where m�g � T is the net force on the hanging mass and a is its linear accelera-tion. Solving for T, multiplying by r to find the torque, and inserting this intoNewton’s second law for rotations along with the expressions for I and we have

Here we have substituted � a/r because the cord unwinds with a linearacceleration proportional to the angular acceleration of the shaft. We can solvethis expression for the acceleration to find

t� m¿1g � a2r � aMR2 �2

3 mR2b aa

rb .

m¿g � T � m¿a,

I � MR2 � 2a 1

12 m (2R)2b ,

t� Tr,

FIGURE 7.20 A hoop being turned by acord tied to a hanging weight.


Now that we have a value for a we can solve for the angular velocity of thehoop. After the hanging weight has fallen 1 m, its velocity (and that of a pointon the shaft) will be given from

as

so that the angular velocity of the hoop–spoke assembly will be

where we divide by the radius of the shaft because v is the velocity of a point onthe shaft.

An alternate solution, in this case much more elegant and straightforward,uses energy conservation. We simply write expressions for the initial and finaltotal energies:

where h is the 1 m height and the initial energy is all gravitational potentialenergy of the hanging mass, and

where the final energy is all kinetic, rotational, and translational. Equating theseenergy expressions because there is no loss of energy due to friction, insertingthe above expression for I, and substituting v � r� for the velocity of the hang-ing weight, we have

Solving this for �, we find

Notice the beautiful simplicity of the conservation of energy approach!

v�

Q2m¿gh

cMR2 �2

3 mR2 � m¿r2 d

� 19 rad/s.

m¿gh �1

2 aMR2 � 2

1

12 m12R22bv2 �

1

2 m¿(rv)2.

Efin �1

2 Iv 2 �

1

2 m¿v2,

Eini � m¿gh,

v� vr

� 19 rad /s,

v � 12ax � 0.19 m/s,

v2 � 2ax,

a �m¿gr2

MR2 �2

3 mR2 � m¿r2

� 0 .019 m/s 2.

Example 7.14 A 5 m radius merry-go-round with nearly frictionless bearingsand a moment of inertia of 2,500 kg-m2 is turning at 2 rpm when the motor isturned off. If there were 10 children of 30 kg average mass initially out at theedge of the carousel and they all move into the center and huddle 1 m from the

(Continued)


axis of rotation, find the angular velocity of the carousel. If then the brakes areapplied, find the torque required to stop the carousel in 10 s.

Solution: Before the brakes are applied there are no external torques acting onthe carousel (friction is absent in the bearings) so that we know angular momen-tum is conserved. Using this guiding principle, we can first write expressions forthe initial and final angular momentum and then equate them to solve for thefinal rotational velocity. We have

where Iini � Icarousel � Ichildren � 2500 � 10(30)(5)2 � 104 kg-m2, treating thechildren as point masses located at the edge of the carousel, and �ini � 2 rpm �2(2�)/60 � 0.2 rad/s. Similarly the final angular momentum is given by an iden-tical expression with Ifin � 2500 � 10(30)(1)2 � 2800 kg-m2. Using conserva-tion of angular momentum, we then can write

so that the final angular velocity is � � 0.71 rad/s. Now, when the brakes areapplied, the frictional torque will produce an angular deceleration given by

But the angular acceleration required to stop the carousel in 10 s can becomputed from kinematics to be

Substituting this value into the previous equation and solving for the fric-tional torque gives

t� Ia� 28001� 0.0712� � 200 N # m.

a�¢v

¢t�

� 0.71

10� � 0.071 rad / s2.

a�t

I.

Lini � 10410.22� Lfin � 28001vfin2,

Lini � Iini vini,

FIGURE 7.21 Angular momentum conservation on a physicscarousel.

5. ATOMIC FORCE MICROSCOPY

As an application of the material of this chapter, we consider the functioning ofthe atomic force microscope (AFM), invented in 1986 by Gerd Binnig, who alsoinvented the scanning tunneling microscope (see Chapter 24) and shared the NobelPrize in 1986 for its discovery. The AFM provides images of the surface topogra-phy of samples with atomic resolution (see Figure 7.22). It is basically a very sim-ple instrument that uses a fine tip attached to a cantilever (a device having a“beam” extending beyond its support, like a diving board; see Figure 7.23) and israster-scanned (in a particular x � y pattern) across, while in contact with, the sur-face to be studied. As the tip encounters small surface height changes, the can-tilever is deflected proportionately due to the torque acting on it, and the heightinformation can be recorded as a function of the x � y position of the tip. Thisinformation can later be displayed in a topographical map of the surface withatomic resolution. This simple and amazing technique works because the effectivesprings acting between molecules on the surface are stiffer than the effective can-tilever spring as we discuss below.

In one method to provide extremely sensitive information about the position ofthe cantilever, a laser beam is reflected from the cantilever surface onto a position-sensitive optical detector. The detector has several segments and the relative intensi-ties recorded on the different portions of its surface allow a very sensitive measure ofthe laser beam deflection. By using a relatively long distance between the cantileverand the detector, a small angular deflection of the laser beam will result in a relativelylarge linear displacement (Figure 7.24). This scheme is called an optical leverarrangement and can be used to measure deflections corresponding to height changesof 0.01 nm (about 10% of the size of a hydrogen atom!).

How is a macroscopic tip able to measure the surface height with subatomic res-olution? The essential conditions are to have an effective spring constant for the can-tilever that is much smaller than the effective spring constant that holds the surfaceatoms together and to have the tip apply a very small (10�7 to 10�11 N) force on thesurface so that the effective contact area is extremely small. In that way the cantileverwill not distort the surface of the material, but will itself bend under the contacttorque from an area of atomic dimensions on the material surface. Effective inter-atomic spring constants are on the order of 10 N/m, whereas the effective spring con-stant of a small piece of household aluminum foil can be made to be at least ten timessmaller. Cantilevers used in AFM are usually microfabricated silicon made with inte-grated tips or with glued diamond tips with effective spring constants of 0.1–1.0 N/m.

The most common mode for imaging biological samples is the constant forcemode. In this scheme a feedback mechanism varies the sample height so that thecontact forces (or torques, because the lever arm distance is constant) can be keptsmall and constant. In this case, the small variations in sample height are tracked to

AT O M I C F O R C E M I C R O S C O P Y 185

FIGURE 7.22 Atomic resolution ofa mica surface by AFM.

FIGURE 7.23 The cantilever: the heart ofthe atomic force microscope.

FIGURE 7.24 An optical lever arrangement to measuresmall displacements of the cantilevered tip.

produce an image of the sample topography. Direct monitoring of cantilever deflec-tion without feedback varying of the sample-to-cantilever height is usually not usedsince the larger forces occurring with large cantilever deflections can damage thesurface. Biological samples are supported on a substrate, such as glass for thickersamples or cleaved mica that is flat to atomic dimensions, for thinner specimens.Figure 7.25 shows extremely high resolution images of several biological samples.


FIGURE 7.25 AFM images of (left)plasmid DNA, (center) an E. colimembrane protein crystal, (right)purple membrane (bacterial light-sensitive proteins containingan analog of rhodopsin) with high-resolution inset.

Example 7.15 A microfabricated integrated tip and cantilever for an AFM hasan effective spring constant of 0.1 N/m. An optical deflection scheme is used tomeasure the deflection of the tip at the end of a 100 �m cantilever. A laser beamis reflected from the top surface of the tip and detected by a sensor 2 m awayfrom the tip. Using the relation s � r�, a small angular deflection of the tipresults in a relatively large deflection of the laser beam due to the large leverarm distance r. If the detector senses a 0.1 mm beam displacement from the“neutral,” noncontact position, calculate the applied contact force the tip exertson the sample surface.

Solution: A 0.1 mm beam displacement with a 2 m lever arm implies an angu-lar rotation of the tip corresponding to

The corresponding displacement of the tip, which has only a 100 �m lever armis s � r� � 100 10�6 5 10�5 � 5 10�9 m � 5 nm. Using the force con-stant of the assembly, assuming Hooke’s law applies, the applied force acting onthe sample is F � kx � 0.1 5 10�9 � 5 10�10 N � 500 pN. To appreciatehow small this force is, note that it is only about 100 times the force generatedby a single myosin molecule interacting with an actin filament.

#

#

u�sr

�0.0001

2� 5 10�5 rad.

A wide variety of different biological samples have been studied using AFM.Included in these are nucleic acids, under physiological conditions so that dynamicprocesses of DNA–protein interactions can be studied as they occur (in so-called“real-time,”), biological membranes, in which individual lipids can be distinguished,cell surfaces, arrays and crystals of proteins, and even isolated proteins. Great caremust be exercised to rule out artifacts in the images due to tip structure effects,scan speed artifacts, lateral forces on the tip due to frictional drag as the tip isscanned, and other problems, but the quality and the reliability of the images aresteadily improving.

6. ROTATIONAL DIFFUSION; CELL MEMBRANE DYNAMICS

In our discussion of diffusion in Chapter 2 we learned that the translational randommotion of macromolecules and microscopic objects is due to constant thermal col-lisions with the background fluid. Under the influence of numerous collisions withthe fluid, there also will be rotational motion about the center of mass occurringdue to random (in both direction and magnitude) torques acting on the molecule(see Figure 7.26). Just as in the case of translational motion, where there is a fric-tional force acting that is proportional to the velocity (see Equation (3.6)), therewill be a frictional torque acting that, to a good approximation, is proportional tothe angular velocity of the molecule

(7.30)

Even if the molecule is spherical in shape, it may be asymmetric in other wayssuch as its electrical or optical properties, and these properties may allow one to dis-tinguish different orientations. For an isolated spherical molecule of radius r, Perrinshowed that the rotational frictional coefficient, which is the proportionality constantfR between the frictional torque and the angular velocity, is

(7.31)

where � is the fluid viscosity or “stickiness” that we study in detail in Chapter 9. Ingeneral, the rotational frictional coefficients for a few other simple shapes, such asellipsoids or rods, have been calculated and the common result is a third-order depen-dence on the largest spatial dimension. This large dependence on size can be used todetermine molecular dimensions very precisely (see the example below).

Rotational diffusion of an object can be characterized by the time it takes forthe object to “randomize” its orientation or lose its “memory” of its initial orienta-tion. This time is known as the rotational relaxation time tR, and clearly is relatedto the rotational frictional coefficient, where the greater the friction, the slower thetumbling of the object and the longer its rotational relaxation time will be.Characteristic rotational relaxation times for small molecules are very fast, from psto ns (10�12 � 10�9 s), whereas larger macromolecules may have time constants of10�3 s or longer. The rotational diffusion coefficient DR has units of 1/s and isrelated to the rotational relaxation time (DR � 1/2tR). We can relate DR to the rota-tional frictional coefficient through the general relation

(7.32)

where kB is the Boltzmann constant and T is the absolute temperature. We show laterin Chapter 12 that kBT represents an average thermal energy from the collisions of allthe solvent molecules. We see that the higher the temperature is, the larger DR andthe shorter tR; the greater the rotational friction is, the smaller DR and the longer tR.These should make intuitive sense to you.

DR �kB T

fR,

fR � 8phr3,

tf �� fRv.

RO TAT I O N A L D I F F U S I O N ; C E L L M E M B R A N E DY N A M I C S 187

FIGURE 7.26 Cartoon of a macromolecule undergoing rotational diffusion due to randomcollisions with solvent molecules.

Example 7.16 A spherical virus, with electrical properties that allow one to dis-tinguish its orientation, is in a water solution at 20°C (293 K). By studying thetime-dependence of its interaction with light, the rotational diffusion time ismeasured to be 0.2 ms. Calculate the effective radius of the virus. Use a value of0.001 (SI units) for the viscosity of water.

(Continued)

One interesting area of biophysical research that involves rotational diffusion isthe study of cellular membrane dynamics. Membranes are made up of a variety oflipid molecules that have electrically charged head groups and linear hydrocarbon tailportions (Figure 7.27) and serve as a boundary for cells and other organelles. Thecharged head group is highly attracted to polar water molecules (hydrophilic) whereas

the tail groups are repelled by water molecules (hydrophobic).Biological membranes are bilayers, composed of two layers oflipid molecules arranged with the hydrophobic tails inside themembrane and with the hydrophilic head groups on the outersurface in contact with the water-based fluid inside and outsidethe cell (Figure 7.28). Synthetic bilayers can be made frompurified lipid molecules, but natural biological membranescontain large numbers of proteins in addition to other smallermolecules. Membrane proteins are classified according to theirassociation as either integral or peripheral. Integral proteins arethose that are tightly bound to the membrane, some of themeven spanning across the full width of the membrane. Theselatter proteins are important in allowing small molecules andproteins to cross the membrane barrier through channels, ormolecule-specific pores. (We study the electrical properties ofmembranes in Chapters 15 and 16.) Peripheral proteins aremore loosely bound to one of the surfaces of the membrane andcan be dissociated by changes in pH or ionic concentrations.

In the 1970s it was first discovered that the individual lipidmolecules in a membrane, as well as the embedded proteins, arequite fluid, diffusing about on the two-dimensional surface ofthe membrane at rates of several micrometers per second. Upuntil that time membranes were viewed as static structures butmeasurements in the 1970s showed that lipids actually can notonly diffuse about in their own monolayer (two-dimensionaltranslational diffusion) but even, in rare events, “translocate”from one monolayer to the other (by “flipping” in a rotationaldiffusion “event”). A model of biological membranes known asthe fluid-mosaic model was developed to describe this dynamicstructure and modified versions of it are still useful today.Proteins in the membrane are confined (to various degrees) in


Solution: From our discussion we know that the rotational time constant isrelated to the rotational diffusion coefficient by

and is further related to the sphere radius using Equations (7.31) and (7.32).Substituting for these, we find that

Solving for the sphere radius r, we have

r � a kB TtR4ph

b1/3� a1.38 10�23 # 293 # 2 10�4

4p # 0.001b1/3

� 85 nm.

tR �8phr3

2kB T.

tR �1

2DR

FIGURE 7.27 A lipid, the structuralunit of biological membranes, withpolar head and nonpolar tail.

FIGURE 7.28 Two cartoons of a cell membrane showing thephospholipid bilayer and a typical collection of associated(geometric or vegetable) proteins.

S TAT I C E Q U I L I B R I U M 189

different domains or regions of the membrane. Many proteins and other macromole-cules bind to specific cellular receptor proteins on the membrane. Often these will firstbind to the membrane surface through nonspecific binding and then diffuse on the two-dimensional membrane surface until a specific receptor is found. Two-dimensional dif-fusion greatly speeds the binding kinetics over three-dimensional diffusion and isresponsible for faster molecular recognition rates.

7. STATIC EQUILIBRIUM

An object that has both a constant linear momentum and a constant angular momen-tum is said to be in equilibrium

This definition clearly includes the special cases when and L � 0 theobject is at rest. According to Newton’s second law, at equilibrium we must thereforehave that

p:

� 0

p:

� constant as well as L � constant at equilibrium.

(7.33)

In addition to a third similar equation if the problem involves three dimensions,Fz, net � 0, the rotational form of Newton’s second law leads to another condition thatfollows from the constancy of the angular momentum (see Equation (7.28)), namely

(7.34)

where all of the torques are computed using the same arbitrary axis of rotation. If both and L are zero, then the object is in static equilibrium, whereas if and L are nonzero con-stants, it is in dynamic equilibrium. An example of dynamic equilibrium might be the(dynamic) balancing of an automobile wheel and tire so that it turns at constant angularvelocity without any wobble (due to torques) acting. Simple static balancing of the wheeland tire at rest does not always reveal whether a wheel will wobble when rotating.

In this section we focus on the conditions for static equilibrium and some exam-ple applications. Our world is full of examples of objects in static equilibrium. Allmanmade fixed structures on Earth, including buildings, bridges, tunnels, and so on,are in static equilibrium. Gravity plays a key role in most statics problems. Althoughgravity acts on all portions of an extended object, for purposes of calculating thetorque due to gravity acting on such an object we can consider the weight to act at asingle point, known as the center of gravity. For us the center of gravity is identicalto the center of mass, a distinction only occurring when the object is large enoughthat the value of g varies over the dimensions of the object. You might want to referback to the discussion on center of mass in Chapter 6 to review its calculation.

In the rest of this section we consider three static equilibrium situations and seehow to analyze the forces acting in each situation. The procedures used in these prob-lems are similar in each case.

p:

p:

tnet � 0,

Fy, net � 0.

Fx, net � 0.

Example 7.17 A two-section ladder leans against a wall at a 70° angle from theground and a man slowly climbs up the ladder as shown in Figure 7.29. Each ofthe sections of the ladder is 6 m long with the bottom section weighing 60 N andthe top section weighing 40 N. With the ladder opened so that it is 8 m in total

(Continued)


length, find all the forces acting on the ladder when the 70 kg man hasclimbed halfway up. Assume that there is no friction between the ladder andwall and that the coefficient of static friction between the ladder and groundis 0.65.

Solution: From a sketch of the situation, we construct an external force diagramshowing all the forces on the ladder, the object of interest. There are five forcesacting as shown in Figure 7.29: two normal forces, one friction force, and theweights of the man and ladder. Next we need to determine where the forces act,if not already clear. Aside from the forces acting at the top and bottom of theladder, we are told that the man stands at its midpoint. We need to find the cen-ter of mass of the ladder which is made from two uniform 6 m sections thatoverlap by 4 m. The sketch (Figure 7.30) shows the needed information to cal-culate the center of mass of the ladder. We find its position, measured from thebottom of the ladder to be

(Alternatively we could have treated the two ladder sections separately andused separate weights for each ladder section acting at their respective centers.)Now we are in a position to calculate the three unknown forces on the ladder, N1,N2, and f, when the man is at its midpoint.

Balancing vertical and horizontal forces, we can write

and N1 � f.

From the first of these we can find N2 � 790N, but another equation is needed to proceed fur-ther. An independent equation can be obtained bysumming the torques about any point and settingthem equal to zero. To simplify this equation, wechoose the bottom of the ladder as this point.Doing so eliminates f and N2 from the torque equa-tion because these forces have zerolever arm. We can write

N2 � (M � m)g,

xcm �40(5) � 60(3)

100� 3.8 m.

f

N2

N1

Mg

mgθ

FIGURE 7.29 Ladder leaning against a wall and theexternal force diagram.

40 N

60 N

3 m5 m

8 m

2 m

FIGURE 7.30 Ladderdimensions.

The steps used in the last example are appropriate for analyzing all static equi-librium problems and are summarized in Table 7.3. As you read the following twoadditional examples, note that they are approached using the outline in the Table.

S TAT I C E Q U I L I B R I U M 191

where L is the length of the ladder and the appropriate sin or cos factors areintroduced to find the lever arms of the three forces. Solving this for N1, we have

Note that the friction force is equal in magnitude to this same 140 N value,far less than its maximum value of �sN2 � 510 N.

N1 �

amgxcm � Mg L

2b cos u

L sin u�1100 # 3.8 � 70 # 9.8 # 42 cos 70

8sin 70� 140 N.

N11L sin u2 � MgaL

2 cos ub � mg1xcmcos u2� 0,

Table 7.3 Method to Solve Static Equilibrium Problems in Mechanics

Step Procedure

1. Draw an external force diagram roughly to scale and carefully label all of the forces on theobject of interest and distances.

2. Determine which are the known and unknown quantities.

3. Write the appropriate equations using , one for each relevant spatial dimension inthe problem.

F:

net � 0

4. Write the appropriate equations using �net � 0 about a convenient axis of rotation until sufficient independent equations are obtained to solve for the unknown quantities.

5. Solve the set of algebraic equations for the unknowns.

Example 7.18 Consider the situation when a person is exercising with a dumb-bell held in one arm outstretched horizontally as shown in Figure 7.31. The forcesinvolved are the weights of the arm and dumbbell, the pull of the deltoid muscleFM at an angle of 20° from the humerus bone acting at point A, and the force ofthe shoulder joint FJ acting at the axis of the shoulder joint, point O. If the arm istreated as uniform and weighs 50 N and the dumbbell weighs 75 N, find the forceexerted by both the muscle and joint tohold the dumbbell in position. Take pointA to be 1⁄4 of the distance from the shoulderjoint to the dumbbell.

Solution: On first glance it might be sur-prising that the shoulder joint exerts adownward force. To see why this must bethe case, we can imagine taking torquesabout the center of mass where the armweight acts. Then both the dumbbell andthe muscle force will act to produce aclockwise rotation about the center

20 FM

mg MgFJ

AO

θ

FIGURE 7.31 An outstretched armsupporting a dumbbell (above) withequivalent forces drawn for analysis(below).

(Continued)


of mass and the shoulder joint must supply a torque tending to produce a coun-terclockwise rotation, hence a downward force. There are three unknown quan-tities in this problem: FM, FJ, and �. We can obtain two equations by writing

and

To proceed further we can write an additional equation, taking torques aboutpoint O,

This last equation can be directly solved for FM to find

Substituting this back into the force balance equations, we can write

and

Solving first for �, we find, by dividing one equation by the other,

,

and then by substituting into either force balance equation,

Note the relatively large forces needed to support a modest weight. Theselarge forces make muscles and joints very susceptible to injury.

FJ � 1100 N.

tan u� 0.25 or u� 14°

FJ cos u� FM cos 20 � 1100 N.

FJ sin u� FM sin 20 - mg - Mg � 280 N

FM � 1mg / 2 � Mg2a 4

sin 20b � 1200 N.

a to � mg1L / 22� MgL � FM 1L / 42 sin 20 � 0.

aFvert � FM sin 20 � FJ sin u � mg � Mg � 0.

aFhoriz � FJ cos u � FM cos 20 � 0

Example 7.19 Suppose that a 50 N uniform crate at rest is pushed with a hori-zontal force of 30 N applied at the top of the crate with dimensions as shown inFigure 7.32. If the coefficient of static friction is 0.7, will the crate slide alongthe surface or pivot at point O? If it will pivot, find the minimum applied forcethat will make the crate pivot about O.

Solution: For the crate to slide, the external force F must be greater than themaximum static friction force given by �sN. The normal force is equal to theweight, although if the crate is about to pivot, the normal force will act at pointO and not through the center of mass. In either case, we find the friction force toequal 35 N, more than the external force F, and so the crate will not slide. For

C H A P T E R S U M M A RY 193

the crate to pivot about point O, the torqueproduced by F must overcome that of theweight of the crate (note again that whenthe crate is about to pivot, the normal forcemust act at point O and therefore produceno torque; similarly the friction force pro-duces no torque). We can therefore write

But this will be a positive quantity sothat the crate will, in fact pivot about pointO. We find the minimum force needed topivot about O by noting that in that case�net � 0, so that F � Mg/2 � 25 N.

tnet � FL � Mg1L/22.

L

L

O

F

N

f

Mg

•

FIGURE 7.32 A heavy crate aboutto pivot (?) or slide (?) along arough surface with external forcediagram below.

CHAPTER SUMMARYTable 7.2 provides a useful summary table of rotationalkinematical and dynamical equations and a comparisonwith corresponding linear equations.

The rotational kinematical equations analogous tothose studied for linear motion (with �, �, and equal-ing the rotational angle, velocity, and acceleration,respectively) are

(7.7)

(7.8)

(7.9)

Rotational kinetic energy is defined as

(7.11)

where the moment of inertia, I, is given by

(7.14)

When the forces acting on an object are conserva-tive, so that the work they do can be expressed in termsof a PE, conservation of energy can be expressed as

I � a (mi ri2).

KE �1

2 Iv2,

v2 � vo2

+ 2a (u - uo).

u(t) � uo � vo t �1

2 at2;

v(t) � vo �at;

(Continued)

(7.15)

The torque produced by a force F acting on anobject can be calculated in either of two equivalent ways:

(7.22)

or

(7.23)

When a net external torque produces a rotation of an objectabout a fixed axis, the amount of work done is given by

(7.20)

The angular momentum of a system can be writtenas either

(7.25/7.27)

Newton’s second law has a rotational form whichcan be written in two forms, analogous to F � ma andto F � dp/dt:

(7.21)tnet, ext � Ia,

L � Itotal v or L � a ri pi,�.

¢W � tnet, ext ¢u.

t� ( r sin u)F � r�F.

t� rF� � r(F sin u),

1

2 mv2 �

1

2 Iv2 � PE � E � constant.

QUESTIONS1. Describe the possible overall motions of a slinky

thrown into the air. How does the motion depend onthe initial conditions as it is released?

2. Compare angular velocity as measured in units ofrad/s and rev/min (rpm).

3. A piece of gum is stuck to the tire of a bicycle. As agirl starts to ride the bike from rest, does the gumhave an angular velocity? A tangential velocity? Anangular acceleration? A tangential acceleration?Answer these four questions again when the girlnow coasts along at a constant translational velocity.

4. Explain why as a potter’s wheel spins, the clay potbeing made tends to expand radially outward.

5. If an object has a constant angular velocity, does itundergo any acceleration?

6. Picture two horses, side by side, on a merry-go-round. Which has the greater angular velocity? Thegreater linear velocity?

7. What is the difference between average angular accel-eration and instantaneous angular acceleration? Canyou give an example where they are not equal?

8. Explain why rotational velocity rather than linearvelocity is the natural variable to use when describingpure rotational motion. Illustrate your argument withan example.

9. In the following examples state which of the twoobjects has the larger moment of inertia (or are theythe same), measured about their symmetry axis:(a) Two balls of equal mass and radius: one solid and

one hollow

(b) Two solid cylinders with the same mass andradius: one twice as long as the other

(c) A solid cylinder of mass M and radius 2R, or oneof mass 2M and radius R

10. Explain in words. The expression for the moment ofinertia I of a long rod (see Table 7.1) depends on theaxis about which the rotation occurs. Why is thenumerical value of I for any such rod four timesgreater for rotation about the end as compared withthe middle?

11. Discuss the definitions of “line of action” and “leverarm” in Figure 7.15. For a given force and pivot point,how should the force be oriented to maximize thetorque applied to an object?

12. Discuss the equivalence of the two expressions for the torque , carefully defining theterms.

13. In the following diagram state whether each forcewould produce a clockwise or counterclockwiserotation about each of the three labeled pivotpoints.

t� F�r � r�F


or

(7.28)

From the previous equation, we see that if there isno net external torque acting on a system then the totalangular momentum of the system is conserved(remains a constant in time).

Atomic force microscopy is an imaging techniquethat uses a cantilevered microfabricated tip that isscanned over a surface to produce an atomic image ofthe surface. The contact torque bends the cantilever anda feedback system moves the sample height in order tomaintain a constant deflection as the cantilever isscanned over the surface.

Diffusion of asymmetric molecules results in notonly translational diffusion of the center of mass, butalso rotational diffusion about the center of mass. The

tnet, ext � lim¢t:0

¢Ltotal

¢t.

rotational diffusion coefficient DR, similar to thetranslational diffusion coefficient discussed inChapter 2, describes the time it takes for a molecule torotate, or tumble, in solution. The corresponding rota-tional times can be very fast (ps to ns) for small mol-ecules and much slower (~ms) for largermacromolecules.

In the special case when there is no motion of asystem, said to be in static equilibrium, then the netforce and torque on the system must both equal zero:

(7.33) Fy, net � 0,

Fx, net � 0,

and also

(7.34)tnet � 0.

A

B

C

F1

F2

F3

14. A ball rolls from rest down a steep incline, up asmaller hill and collides with and compresses a springlocated at the hilltop as shown. Describe the differenttypes of energy associated with the ball at eachlabeled point.

Q U E S T I O N S /P R O B L E M S 195

(b) Imagine the lever represents your forearm in ahorizontal position with a handheld weight W andelbow at P. Does your biceps function as if itworks at P1 or at P2? Discuss the relative merits ofarm design and function in light of your answer.

23. A centrifuge is a laboratory appliance useful for sep-arating dissolved solid particles from liquid. When asample is placed within a chamber, a dummy sampleof approximately equal mass should always be placedinto the chamber diametrically opposite. Explain thereason for this procedure.

MULTIPLE CHOICE QUESTIONSQuestions 1–3 refer to a CD, with its information storedstarting at an inner radius of 2.2 cm and out to an outerradius of 5.7 cm, that spins at 5500 rpm.1. What is the fastest velocity at which information

can be read? (a) 1970 m/s, (b) 33 m/s, (c) 256 m/s, (d) 3300 m/s.

2. If it takes 4 s to get up to speed from rest, what is theangular acceleration of the CD? (a) 1400 rad/s2, (b) 23 rad/s2, (c) 144 rad/s2, (d) 8600 rad/s2.

3. In getting up to speed from rest (see previousquestion), the CD makes (a) 730, (b) 1750, (c) 1150,(d) 180 revolutions.

4. The moment of inertia of a 20 cm uniform rod of 2.4 kg mass rotating perpendicular to its long axis aboutthe rod center is (a) 0.008 kg-m2, (b) 0.032 kg-m2, (c) 80 kg-m2, (d) 4.0 kg-m2.

A

B

C D

15. Define angular momentum in words giving an exam-ple to illustrate your definition.

16. Discuss the differences and similarities of the conser-vation of momentum and the conservation of angularmomentum. Can you find examples where one andnot the other quantity is conserved?

17. In the game “crack the whip”, a number of partici-pants join hands and run along the ground (or skatealong the ice). Usually this human chain begins to“whip,” with the trailing end of the line beginning tofishtail. Those at the end are soon flung free. Explainthe mechanics of motion considering angular andlinear speed.

18. Consider an individual riding in an automobile whileusing a lap belt but no shoulder harness. Suppose the caris brought to stop in a frontal collision. Describe the sub-sequent motion of the body and the potential for bodilyinjury in terms of inertia and angular momentum.

19. If one attempts to carry and transport an object such asa ladder, a large storm window, or a sheet of plywood,one finds that it is relatively easy to do so if the load islifted at the proper point. Attempting to lift and carryat other points is difficult if not impossible. Discussthis matter considering torque and center of gravity.

20. Why should a house painter, when shifting a raisedladder sideways, attempt to keep the ladder as close tovertical as possible?

21. Applying what you know about the nature of inter-atomic forces, explain why any force F, howeversmall, applied to the end of a cantilevered bar asshown, must result in some amount of sag to the bar.

FO

A

B C

D

P P1 P2

W

265

22. Consider the lever with fulcrum P and weight W asshown. The lever arm is pinned at P but is free to pivot.(a) If you had to hold this lever arm horizontal by

exerting an upward pull to counteract the down-ward force of the weight W, which would be eas-ier, a pull at P1 or at P2?

5. In the above diagram, which of the equal magnitudeforces produces the largest torque about point O?(a) A(b) B(c) C(d) D

6. As a particle traveling in a circle speeds up at aconstant rate, its net acceleration (a) increases andpoints more and more toward the tangential direction,(b) increases and points more and more toward theinward radial direction, (c) increases and points more

and more toward the outward radial direction, (d) decreases and points more and more toward theinward radial direction, (e) none of the above.

7. The net torque exerted by the forces shown aboutpoint O is


(c) The hoop because it has the smaller mass and willtherefore have less rotational and more transla-tional speed compared to the cylinder

(d) The hoop because it has the larger moment ofinertia and for the same torque will have the largerangular acceleration

13. Consider two equal mass cylinders rolling with thesame translational velocity. The first cylinder (radius R)is hollow and has a moment of inertia about its rota-tional axis of MR2, and the second cylinder (radius r) is solid and has a moment of inertia about its axis of 0.5 Mr2. What is the ratio of the hollow cylinder’s angu-lar momentum to that of the solid cylinder? (a) r2/2R2,(b) 2R2/r2, (c) r/2R, (d) 2R/r.

O

F

F’

R1

R2

R3

A

60°

y

x

r

F

5 N

2.0 m

4 N 1.0 m

30°

6 N

1.0 m

0.2 m

(a) 15 N-m, (b) 18 N-m, (c) 6 N-m, (d) 14 N-m, (e) noneof the above.

8. A 5 N force acts alone to slow down a 2 m radius uni-form rotating circular platform. The force acts at theedge of the platform and is directed at 30° to the out-ward radial direction. The applied torque is equal to(a) 8.7 Nm, (b) 5 Nm, (c) 0, (d) 10 Nm.

9. If the platform in the last question has a mass of 20 kg, the angular deceleration of the platform is (a) 0.25 rad/s2, (b) 0, (c) 0.125 rad/s2, (d) 0.5 rad/s2.

10. From the diagram, the magnitude of the torque of forceF about point A, the lower left corner, is given by(a) Fr(b) Fx(c) Fy(d) Fr sin 60(e) None of these

11. A particle is speeding up while traveling clockwise ina vertical circle. When it is at the 3 o’clock position,its net acceleration might point (a) toward 9 o’clock,(b) vertically downward, (c) toward noon, (d) toward6 o’clock, (e) none of these are possible correctchoices.

12. Which of the following uses correct logic? A cylinderof mass 2M and radius R has a race down an inclinewith a hoop of mass M and radius R. The winner is(a) The cylinder, because it has the larger mass and

therefore will accelerate faster(b) The cylinder, because its mass is distributed

throughout its volume and not all concentrated atthe radius so it will travel faster even though it hastwice the mass

14. A force F acts on a circular disk as shown. F� is thetangential component of the force at the point of appli-cation. The torque that the force F produces aboutpoint O is given by (a) F� R2, (b) FR3, (c) FR1, (d) FR2.

15. An Atwood machine is a real pulley mounted on a realshaft used to help lift a heavy weight by attaching it toanother weight by a rope strung over the pulley. Oncethe weights leave the ground (a) the sum of the kineticand gravitational potential energies of the two weightsis constant, (b) the sum of the kinetic and gravitationalpotential energies of the two weights increases withtime, (c) the sum of the kinetic and gravitationalpotential energies of the two weights and the rota-tional kinetic energy of the pulley decreases with time,(d) the sum of the kinetic and gravitational potentialenergies of the two weights and the rotational kineticenergy of the pulley is constant.

16. Two point masses, each 5 kg, lie at either end of a lightrod of length 2 m. What is the moment of inertia of thesystem about the left end of the rod (in kg-m2)? (a) 10,(b) 5, (c) 40, (d) 20, (e) none of the above.

17. A cylinder with 2 kg mass and 0.01 kgm2 moment ofinertia (I � 1⁄2 MR2 for a cylinder) is rolling down aninclined plane with 30° inclination. At a point where itscenter of mass velocity is 1.0 m/s and its height fromthe ground is 0.1 m, what is its total mechanical energy(with respect to the ground)? (a) 1.5 J, (b) 1.96 J, (c) 2.96 J, (d) 3.46 J, (e) none of the above.

18. A 3 kg point mass is at the end of a light 2 m rod hang-ing vertically and hinged at the other end. If a 5 N force is exerted at the midpoint of the rod at a 45° angle below the horizontal, the initial angular accel-eration of the mass is (a) 0.83 rad/s2, (b) 0.42 rad/s2, (c) 0.29 rad/s2, (d) 0.59 rad/s2.

19. What physics principle does a high diver use inexecuting a dive? (a) Conservation of momentum, (b) conservation of angular momentum, (c) conservationof moment of inertia, (d) conservation of torque.

20. An isolated horizontal circular platform is spinning ona frictionless axle with a person standing at its edge. Ifthe person walks halfway in toward the center of theplatform, the principle that allows you to find the newangular velocity is (a) conservation of energy, (b) con-servation of momentum, (c) conservation of angularmomentum, (d) conservation of torque.

21. An ice skater is spinning with her hands overhead andlegs straight, with a moment of inertia of 0.3 kg-m2,so that she has a 2 s rotational period. She extends her arms sideways, increasing her moment of inertiato 0.4 kg-m2. Her final rotational period is (a) 2.7 s,(b) 1.5 s, (c) 2.3 s, (d) 1.7 s.

22. A spinning ice skater pulls in her outstretched arms.What happens to her angular momentum about theaxis of rotation? It (a) does not change, (b) increases,(c) decreases, (d) changes but it is impossible to tellwhich way.

23. Answer each of these by choosing Yes or No.(a) If the net force on a rigid body is zero, can it have

an angular acceleration? Yes or No(b) If the net torque on a rigid body is zero, can it

have a linear acceleration? Yes or No(c) Is angular momentum necessarily conserved in

part (a)? Yes or No(d) Is angular momentum necessarily conserved in

part b)? Yes or No24. Effective interatomic spring constants are on the order

of (a) 0.1 N/m, (b) 10 N/m, (c) 103 N/m, (d) 105 N/m.25. A uniform ladder is leaning against a rough vertical

wall. The ladder makes an angle � with the horizontalground. Which of the following statements is false?(a) The weight of the ladder equals the normal force atthe ground; (b) the normal force at the wall equals thefrictional force at the ground; (c) the torque producedabout the contact point with the wall by the weight, bythe frictional force and by the normal force at theground must all add to zero; (d) the weight of the lad-der can be considered to act at the center of the ladder.

26. A ladder is leaning against a wall with a man stand-ing at its midpoint. Which of the following is a falsestatement? (a) The net vertical force on the ladder iszero; (b) the net horizontal force on the ladder is zero;(c) the net torque about the bottom of the ladder iszero; (d) the net torque about the top of the ladder iszero; (e) none of the above is false.

27. In an Atwood machine (two unequal masses hungover a real pulley), the tension in the string attachedto both masses is not the same because (a) the twomasses are not the same, (b) the pulley has a nonzeromoment of inertia, (c) there is friction in the pulley’sbearings, (d) the acceleration of the two masses is dif-ferent, (e) none of the above.


28. A uniform plank is used as a seesaw, but the fulcrumis not placed at the center, but at a point 1/3 of thelength from one end. If the plank has a mass M, whatmass must be placed at the end of the shorter side inorder to keep it balanced? (a) M, (b) 2M, (c) M/2, (d) M/3.

29. A 10 kg child and a 20 kg child sit balanced on theends of a teeter-totter. The teeter-totter is a uniformplank of mass 5 kg which is placed on a fulcrum.Suppose now that each child moves halfway in towardthe fulcrum. Will the teeter-totter remain in balance?(a) No, the end with the heavier child will go down.(b) No, the end with the lighter child will go down. (c) Yes, the teeter-totter will remain in balance. (d) Itis not possible to tell from the information given.

30. Suppose you wish to tip a large packing crate so thatyou can put a hand truck under it. Assume the cratedoes not slide along the floor and that it tips about pointP. Should you push or pull on the crate (or does it mat-ter?) And where should you apply your force in order touse the smallest force to tip the crate? (a) Push on pointX in the direction A. (b) Push on point X in the direc-tion B. (c) Pull up on point X in the direction C. (d) Pullon Point Y in the direction E. (e) More than one of thesechoices will work equally well.

A

B

C

D

E

X Y

P

.

..

PROBLEMS1. In 1986 the Voyager plane was the first to circumnav-

igate the Earth without refueling, taking just over 9 days to travel 24,987 miles around the Earth overboth poles. Find its average speed and averageangular velocity.

2. (a) How long does a centrifuge take to get up to arotational speed of 80,000 rpm from rest with anacceleration of 40 rad/s2?

(b) When shut down after spinning at that speed for 2 h, the rotor slows at a rate of 4 rad/s2 without thebrakes being applied. How long does it take tocome to a stop after the two hour spin?

(c) Find the total number of revolutions the rotor hasspun during the entire centrifugation run.

3. If the rotor of the previous problem is modeled as auniform cylinder of 20 kg mass and 25 cm diameter,find its kinetic energy when spinning at its top speed.

4. A motor rotor turns at 1800 rpm. What are the angu-lar and linear velocities of a point on the motor wind-ing 20 cm away from the rotor axis?

5. A certain car tire is guaranteed to give 40,000 miles(64,400 km) of use.(a) If the tire radius is 25 cm, how many revolutions

does this amount of use correspond to?(b) What is the angular speed for the tires (radius 25

cm) of a car traveling at 100 km/hr?6. The world’s largest clock face has a radius of 15.4 m.

If that is the length of the minute hand find the linearspeed of the tip of the minute hand.

7. What is the linear speed of the minute hand of a wrist-watch if a 1 cm length is assumed?

8. A maple seed wing pair falling to the ground whirlsaround at 3 revolutions per second.(a) If the overall length of the wing pair is 6 cm, what

is the horizontal linear speed of the wing tip?(b) Assume that because of air resistance, the wing

pair falls at a constant speed of 80 cm/s. What isthe total linear speed of a wing tip during fall?

9. A large tree is blown to and fro in a strong wind. Thetree swings through an arc of 12° taking one secondto swing from one way to the other. If a bird’s nest is24 m above the ground in the tree’s branches, what isthe average speed of the nest with respect to theground as it moves back and forth with the tree?

10. Because of the rotation of the Earth, a person standingat the equator is moving through space at considerablespeed with respect to another who stands at either pole.Compute this speed, considering the Earth as a sphere.

11. A compressor motor for a cooling system, respondingto a thermostatic control, turns on and is brought upto its operating speed of 1200 rpm is 1.4 s.(a) What is the angular acceleration of the motor shaft?(b) If the motor assembly of the previous part has a

mass of 9 kg and is modeled as a solid cylinder ofradius 20 cm, what is the angular momentum ofthe motor at operating speed?

12. A dormant bacterium responds to stimulus and beginsto move, via rotary motion of its flagellum.(a) If it takes 2.5 ms to attain the normal rotational

speed of 4 Hz, what is the angular acceleration ofits flagellum?

(b) Suppose in response to an environmental stimulus,the rate of rotation of a bacterium flagellumdecreases from 4 Hz to 3 Hz. The change isobserved to occur slowly, over a 15 s interval. Whatis the angular acceleration of the flagellum?

13. A recording tape is wound up onto a take-up spool. Inorder to achieve sound fidelity during playback, thetape movement must be such that its linear speed is con-stant throughout time. From start to finish, the spoolradius ranges from 1 cm to 0.5 cm. If the linear speed ofthe tape is 5 cm/s, what are the rotational speeds of thetake-up spool at the beginning and at the end of play?

14. A clock escapement wheel oscillates back and forthwith each swing in a single direction amounting to1/8 turn and taking 1/2 s. What is the average angularspeed of the wheel during each swing?


15. A 50 cm outer diameter tire on a bicycle has a 0.05 kgpiece of chewing gum stuck to its edge.(a) If the bike starts from rest and attains a linear

speed of 6 m/s in 30 s by a uniform acceleration,what is the angular acceleration of the gum?

(b) How many revolutions did the wheel make in thattime?

(c) What were the tangential and radial componentsof the gum’s acceleration at the end of the 30 s?

(d) How large must the force from the tire on the gumhave been for it to remain stuck on the tire duringthe entire acceleration?

16. A honeybee flaps its wings about 200 times per sec-ond. Assume a wing is 0.7 cm in length and swingsthrough an arc of 100°. What is the average speed ofa wing tip during flight of the bee?

17. Find the net torque on the object shown about thepivot point O. (Hint: Look at the two components ofthe 6N force separately.)

0.1 m8 N

2 N

4 N

6 N

0.1 m

0.2 m

0.05 m 30°O.

18. A space station consisting of a ring with radius R � 20 mand mass M � 100,000 kg is spun around its center at arate of � � 0.7 s�1 in order to produce artificial gravity.The moment of inertia is I � MR2 for a ring.(a) What is the centripetal acceleration of a point on

the outside of the ring?(b) What is the linear velocity of a point on the out-

side of the ring?(c) What is the kinetic energy associated with the

rotation?A spherical asteroid of mass 50,000 kg and radius 15 m collides with the station at a speed of v �10 m/s and lodges in the center of the ring.

(d) What is the linear velocity of the combined stationand asteroid after the collision?

(e) What is the rotational velocity of the combinedstation and asteroid after the collision, assumingthat the asteroid was not initially rotating?

19. As a publicity stunt, a toy company constructs theworld’s largest yo-yo, consisting of a sphere 4 m indiameter with a mass of 1000 kg, with a steel cablewrapped around the middle of the sphere. Theydemonstrate it by dropping it off the GeorgeWashington Bridge, in New York City, using a craneto hold the free end of the cable. The yo-yo rollsdown the cable without slipping.(a) When the yo-yo has fallen a distance of 10 m, how

many radians has the sphere turned through?

(b) What is the angular velocity of the spinningsphere when it has fallen a distance of 10 m?

(c) What is the linear velocity of the center of the sphere?20. A cylinder of moment of inertia I1 rotates about a

vertical frictionless axle with angular velocity �i. Asecond cylinder that has moment of inertia I2 and ini-tially not rotating is dropped onto the first cylinder.Because friction exists between the two surfaces ofthe cylinders, they eventually reach the same finalangular speed, �f.(a) What is the expression for the magnitude of �f?(b) Show that the kinetic energy of the system decreases

in this interaction and calculate the ratio of the finalrotational energy to the initial rotational energy.

(c) Why does the kinetic energy of the system decrease?21. The Pumpkin on the Nott revisited! Suppose that the

Nott Memorial is topped with an approximatelyhemispherical dome of radius R � 89 feet. Somehowan individual has balanced a spherical pumpkin at thetop of the dome at an angle of �i � 0° with the verti-cal. Suppose that a gust of wind starts the pumpkinrolling from rest. It loses contact with the dome whenthe line from the center of the hemispherical dome tothe pumpkin makes a certain angle with respect to thevertical. At what angle does this happen? Compareyour results with those of Chapter 5, problem 28.

22. During most of its lifetime a star maintains an equi-librium size in which the inward force of gravity oneach atom is balanced by an outward pressure forcedue to the heat of nuclear reactions in its core. Afterall of the hydrogen “fuel” is consumed by nuclearfusion, the pressure force drops and the star under-goes a gravitational collapse until it becomes a neu-tron star. In a neutron star, the electrons and protonsare squeezed together by gravity until they fuse intoneutrons. Neutron stars spin very rapidly and emitintense radio pulses, one pulse per rotation.(a) Our sun has a mass M � 2 1030 kg and radius

R � 3.5 108 m and rotates once every 27 days.What is the initial magnitude of the angular veloc-ity of the Sun?

(b) Suppose that the sun after undergoing gravitationalcollapse, forms a pulsar that is observed to emitradio pulses every 0.1 s. What is the magnitude ofthe angular velocity of the pulsar? (The sun wouldnot actually form a neutron star as it is well belowthe minimum mass limit of 4 solar masses.)

(c) If the sun does not lose any mass in the collapse,what is the radius of the neutron star after the col-lapse? (Hint: Consider the sun before and after thecollapse to be a solid sphere with moment of iner-tia .)

(d) Is there a change in kinetic energy of the collapsingsun? If your answer is yes, how much work did grav-ity do in collapsing the sun and why is work donecollapsing the sun? If your answer is no, then explainwhy gravity does no work in collapsing the sun.

Istar � 25 MR2


23. A rotating space ship has a mass of 1,000,000 kg, mostof it due to a large cylindrical tank of water (radius 10 m)on the central axis of the ship (with outer hull radius 20 m). While making its way to Alpha Centauri, the shipspins about this axis to generate the illusion of gravity.(a) Initially the rotation rate is set so that the cen-

tripetal acceleration of a person just inside theouter hull is equal to the normal acceleration ofgravity on the surface of the Earth. What is the lin-ear speed of a point just inside the outer hull?

(b) What is the angular velocity of a point just insidethe outer hull?

(c) What is the angular momentum of the ship (Forthis part, ignore the mass of the ship outside thewater tanks.)

(d) A year or so into the trip they realize that the rota-tion is making the pilots seasick. Not wanting towaste fuel using rockets to slow the rotation, theydecide to use angular momentum to their advan-tage, and instead pump the water out of the centraltanks into a thin shell around the outer hull. Whatis the new angular velocity after this operation?(Again, consider only the mass of the water.)

(e) What is the acceleration of a person just inside theouter hull after the operation in part (d)?

24. A 200 kg playground merry-go-round with a 2 m radiusis subject to a frictional torque of 40 Nm.(a) If the merry-go-round goes round with a linear

velocity of 6 m/s on the outside edge, what is itsangular velocity?

(b) What force must be applied by one of the child’sparents pushing on the outside edge to keep themerry-go-round moving at a constant angularvelocity? (Note for this system.)

(c) When the parent gets tired and lets go, how longdoes the merry-go-round take to stop?

(d) At a point when the merry-go-round has lost halfof its initial angular velocity, how much energyhas been lost to frictional heating of the system?

25. A student sits on a freely rotating stool holding twoweights, each of which has a mass of 3.00 kg. Whenhis arms are extended horizontally, the weights are1.00 m from the axis of rotation and he rotates with anangular speed of 0.750 rad/s. The moment of inertia ofthe student plus stool is 3.00 kgm2 and is assumed tobe constant. The student pulls the weights inward hor-izontally to a position 0.300 m from the rotation axis.(a) Find the new angular speed of the student.(b) Find the kinetic energy of the rotating system

before and after he pulls the weights inward.26. A uniform cylinder of 0.5 kg mass and 5 cm radius lies

on an inclined plane with a 30° angle of inclination.

I �1/2 MR2

F

(a) Draw a carefully labeled free-body diagram for thecylinder when at rest at its initial height of 1.5 m andcalculate the external force F that must be applied tothe cylinder as shown to keep it from rolling.

(b) If this external force is now removed, use conser-vation of energy principles to find the speed of thecylinder’s center at the bottom of the incline.

(c) What is the cylinder’s angular momentum at theinstant that it reaches the bottom of the incline?

27. A physics professor lecturing about rotational motionuses as a prop a weighted bicycle wheel with a radius of0.2 m and a mass of 5 kg, concentrated at the rim (i.e.,ignore the hub and spokes when considering its motion).(a) If the wheel is set to spinning at 150 revolutions

per minute, what is the angular velocity of the wheel?(b) If it takes 5.0 s to get the wheel up to speed, what

torque was applied? What force does this requirethe professor to exert on the rim?

(c) Having stayed out late the night before, the pro-fessor drops the wheel on the floor. Assuming thewheel continues spinning at the same 150 rpm, howlong does it take to roll into the wall, 10 m away?

(d) What is the kinetic energy of the rolling wheel?28. A thin hoop with 2 kg mass and 1.5 m radius rolls

down a 5 m long 30° inclined plane from rest.(a) Find the center of mass velocity of the hoop at the

bottom of the incline.(b) Find the acceleration of the center of mass down

the incline.(c) How long does it take to get to the bottom?

29. A centrifuge rotor, initially at rest, has a constantapplied torque of 500 Nm causing it to speed up.Approximate the rotor as a uniform cylinder of 20 cmradius and 15 kg mass.(a) If the friction force is negligible, find the angular

acceleration of the rotor.(b) How long does it take the rotor to reach 80,000 rpm?(c) Suppose the applied torque is removed immedi-

ately upon getting up to speed and a small 30 Nmbraking torque slows the rotor. How long does ittake to stop?

(d) Find the total number of revolutions recorded onthe centrifuge meter for this centrifuge run.

30. A custodian raising a bucket of coins from the bottomof a wishing well turns the handle attached to a spoolof rope at a constant rate of 20 revolutions per minute.(a) What is the angular velocity of the spool in rad/s?(b) If the spool has a radius of 0.10 m, how much time

is required to raise the bucket by 10 m?(c) If the weight of the bucket exerts a torque of �2 Nm

on the spool, what force must the custodian applyto the end of the 30 cm handle to keep the spoolturning at a constant angular velocity?

(d) If the bucket is raised to the top, emptied, andallowed to drop back into the well, what is theangular velocity of the spool after the bucket hasfallen 10 m, if the mass of the bucket is 10 kg and


the cylindrical spool has a mass of 20 kg?(Assume frictionless bearings in the spool.)

31. What must the mass of m3 be in order to balance outthe individuals of masses m1 and m2 situated on theseesaw as shown (Take mseesaw � 76 kg, m1 � 18 kg,m2 � 16 kg, x1 � 1.2 m, x2 � 1.4 m, x3 � 1.4 m)?

m2 m1 m3

x2

x1 x3

2 m

0.6 m

m2 m1

x2

x1 x3

30°

32. Suppose in the previous problem m3 balances the see-saw by pulling now on the end 1.5 m from the ful-crum at an angle of 30° from the horizontal as shown.What force is necessary for balance?

33. A housepainter who weighs 750 N stands 0.6 m fromone end of a 2.0 m long plank that is supported at each end by ladder anchors. If the plank weighs100 N, what force is exerted upon each anchor?

34. A car (14,500 N) travels across a simple truss bridge(230,000 N; 16 m long), that is supported by pylonsat each end.(a) What are the maximum and minimum forces

exerted on each pylon due to the crossing?(b) Suppose, all the while, a road crew (total weight

22,000 N) is situated 4 m from one end. Whatnow are the maximum and minimum forces exertedon each pylon due to the crossing?

4 m

16 m

35. A carpenter places a 2 kg, 4 m long plank atop a work-bench surface, with one end of the board overhangingthe benchtop by 1⁄4 of its length. A curious 8 kg cat thenleaps up to the bench and begins to creep out along theoverhang. How far can it go before the board tips,sending both cat and board to the floor?


38. A specification chart notes that a certain machine boltshould be tightened to 85 N-m of torque. Because ofinterference with other adjacent machine parts, themechanic can only grasp a wrench at the handle mid-dle, and can pull along the direction shown, at anangle of 60° from the handle’s long axis. If thewrench is 30 cm long, how much force must beapplied in order to attain the specified torque?

36. A 50 kg sign boom PQ is held horizontal by a wire cablethat runs from Q to an anchor point on a building wall,R. The boom is free to pivot in the vertical plane aboutthe hinge at P. If a sign suspended from the boom weighs340 N, what is the tension in the cable necessary to holdthe boom horizontal?

R

P Q60°

JOE'S EATS

37. Lifting an object from a forward-leaning position.Consider the biomechanical stress that is sustainedwithin the backbone and muscles of an individual whoattempts to lift a weight from a stance in which thebody leans forward. A represents the hinge pointbetween back and hips. B represents a point in the lum-bar region of the back. C is the hinge point betweenupper back and arms (with the shoulder bones inter-vening). Assume the individual pulls directly upward,from D to C. �1 indicates the amount of forward tilt ofthe spinal axis. Except for this tilt, an otherwisestraight spine is assumed. �2 is the angle between thespinal axis and the line along which a set of lower backmuscles exert tension. If the mass to be lifted is 2.5 kg,determine the magnitude of F, the amount of tensionsupplied by the lumbar muscles. Determine the com-pressive force on the lower spinal vertebrae, between Aand B. Take , and �2 � 8°.AB � 1

4 AC, u1 � 45°

F

A

BD

C

m, mass

θ2

θ1

39. A small 5 kg lead ball is on the end of a 0.5 m lightrod that is hinged at one end so it is free to pivot andis held on the other end so that it is horizontal. Whenlet go from rest find the(a) Initial torque on the rod about the hinge(b) Initial angular acceleration of the rod(c) Angular velocity of the ball at its lowest point(d) Angular momentum of the ball at its lowest point

40. An old-fashioned child’s toy top is set spinning byfirst winding string around it, and then tossing the topforward toward a smooth surface and then immedi-ately pulling back sharply on the string. Suppose thestring pull exists for 0.8 s during which time the topis set spinning at 15 turns per second. Treat the top asa solid sphere of radius 2 cm and of mass 100 gm.What is the pulling force that must be applied to thestring to attain the motion described?

41. A uniform board, hinged at one end, is just barely sup-ported in a horizontal position by an 8 N force appliedat the other end and acting at a 30° angle with thehorizontal.(a) What is the weight of the board?(b) What is the minimum force acting at the far end of

the board that can keep it in a horizontal position?

42. A water wheel with an 8 foot radius collects waterspilling from a millrace. The resulting weight imbal-ance produces a torque which turns the wheel.Suppose that only the five labeled buckets hold appre-ciable amounts of water, as follows: A1 � A2 � 0.5cubic ft; B1 � B2 � 1 cubic ft; C � 1.5 cubic ft.(a) What is the resulting torque on the wheel due to

the water?(b) If a capstan reel of radius 6� is also mounted along

the wheel axle, what is the maximum weight thatcan be raised by a rope wound round the reel?


A1

B1

C

B2

A2

43. A marching baton, twirled and tossed aloft, rotatesend over end about its center of mass. Suppose itsshaft is 45 cm long and has mass of 400 gm, and itsrubber end caps are 40 gm each. Consider the shaft asa uniform rod and the endcaps as point masses situ-ated at either end. If such baton rotates 3 times persecond, what is its angular momentum?

44. A competitive diver executes a forward flip from a 3 m board. Suppose upon leaving the board thediver’s body is assumed to be fully extended and istilting forward at 0.5 turn per second. At the peak ofthe jump the diver tucks, bringing knees to chest andfolding arms around knees. The maximum heightreached by the diver is 1 m above the board.Representing the diver’s body as a cylinder andassuming that tuck position reduces to one-half theoverall length of the cylinder/body, show that it ispossible for the diver to complete more than onecomplete somersault before falling to the watersurface.

45. A 20 m long uniform beam weighing 600 N issupported on two 3 m long concrete columns A and Beach having a cross-sectional diameter of 10 cm asshown.

(a) Find the maximum weight a person can have and stillwalk to the extreme end D without tipping the beam.

(b) Find the forces that the columns A and B exert onthe beam when the same person is standing at apoint 2 m to the right of B.

46. A crate with rectangular faces (height h, depth d,width w) having roughly the outline of an uprightrefrigerator is slid sideways across the floor as indi-cated. Suppose that the leading edge of the cratestrikes an irregularity in the floor and begins to tip.Determine an algebraic expression for the maximumangle through which the crate can tip about its bottomforward corner edge and still fall back to the uprightposition. Assume the crate and its contents uniformlyoccupy its volume.

A BD

12 m3 m

h

w

d

floor bump

47. A light 4 m long rod is hinged (frictionless) at oneend, has two weights attached (one of 2 kg fastened atits center and one of 4 kg fastened at its other end),and is held in place at a 30° angle by a horizontalcable fixed at 3⁄4 of the way along the rod from thehinge as shown.

(a) Find the tension in the cable(b) If the cable is cut, find the initial angular acceler-

ation of the rod.48. Four children situate themselves on a playground

whirly-go-round, a large metal platter that can rotateabout a central vertical axis. The whirly-go-roundplatter has a mass of 90 kg and a radius of 1 m. Eachchild has a mass of 20 kg and sits 75 cm away fromthe center (25 cm in from the outer edge). After 15 sof sequential tugs at the attached metal bars, an adultsets the whirly-go-round spinning at 1 turn per second.(a) What is the average force supplied by the adult?(b) With the whirly-go-round now spinning at 1 turn

per second, if the children allow themselves eachto now move out to the edge of the whirly-go-round, what will be its new rotational speed?

49. A certain church bell weighs 800 lb (3600 N). Inorder for the clapper to contact the bell, the bell mustbe tilted 24°. Suppose the bell is mounted at M uponan axis 15 cm above the bell center of mass C. Thewinding wheel (diameter 1 m) for the bellpull has itscenter at the mounting axis of the bell. What forcemust be applied to the bellpull in order to impart tiltsufficient to ring the bell?


Pm1

m2

L

r45°

60°

R

Q

C

(a) What is the angular momentum of the mass about C?(b) Suppose an identical mass m2 travels, also at a

constant 2 m/s, along the straight line l which is1.5 m from C. Calculate the angular momentumof the mass when it is at point P, the point of inter-section of the circular path and the line.

(c) Calculate the angular momentum of the mass m2about C when m2 is at point Q.

(d) Calculate the angular momentum of the mass m2about C when m2 is at point R.

Your findings should help you to see that the magni-tude of angular momentum calculated for any massdepends, as does torque, on the point about which thevalue is calculated. Furthermore, different masses,with apparently different motions, can have the sameangular momentum about a given point.

51. A 5.0 m long ladder with mass 100 kg is laid againsta frictionless wall at an angle � with respect to thefloor as shown below. Suppose that the coefficient ofstatic friction between the floor and ladder is 0.09 andthat a painter of mass 60 kg has climbed up the ladderand has made it to a point 70% of the length of theladder when the ladder begins to slip.(a) In your own words, write a brief description of the

problem stating the main physical principle(s)behind the problem.

(b) Draw a carefully labeled free-body diagram show-ing all of the forces that act on the ladder.

(c) From your free-body diagram, determine expressionsfor the normal forces due to the wall and the floor.

(d) Write an expression for the sum of the torquesabout the origin O (shown above) in terms of theangle � and then evaluate your expression using theinformation given. (Hints: You will need the factthat sin(90 � �) � cos � and for counter-clockwiserotations choose � for the direction of the torque.)

24°

M

C

15 cm

bellrope

windingwheel

50. A mass m1 of 400 gm travels a circular path around acenter C at a radius r � 1.5 m at a constant speed of2 m per second.

floor

person

O wall

θ

rotational motion - if

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