Kesten Ch 8 10.21

8/16/2019 Kesten Ch 8 10.21

1/58

8

RotationalMotion

8-1 Rotational Kinetic Energy

8-2 Moment of Inertia

8-3 The Parallel-Axis Theorem

8-4 Conservation of Energy

Revisited

8-5 Rotational Kinematics

8-6 Torque

8-7 Angular Momentum

8-8 The Vector Nature of

Rotational Quantities

275

( A G E N C E N A T U R E / N a t u r a l H i s t o r y P i c t u r e A r c h i v e . )

Falling cats usually land on their feet. The physics behind this graceful motion

is a complex dance of rotations. By bending in the middle, the cat can rotate

her front and hindquarters separately. Extending her front legs and tuckingin her rear legs allows her back legs to turn more rapidly. Later in the fall, she

extends her back legs, which slows down their rotation and leaves them inthe landing position as her front legs come vertical. Each of these motions

can be described by the rotational quantities uncovered in this chapter.

Physical quantities such as kinetic energy and linear momentumdepend on speed, which we defined as the change in position divided

by the change in time. Does the windmill shown in Figure 8-1 havekinetic energy and momentum? Although its position hasn’t changedsince it was built, the rotating blades are constantly moving; there mustbe kinetic energy and momentum associated with this motion. In thischapter, we examine the physics associated with rotations.

8-1 Rotational Kinetic EnergyTo determine the kineticenergy of a windmill, con-sider the rotation of one ofits blades around the fixedcentral axis, as shown at twosuccessive times in Figure8-2. Imagine that the rigidblade is divided up into manysmall pieces, the first being atthe rotation point (fixedaxis) and each of the otherpieces farther and farther outalong the blade. We identify

one of these pieces usingthe subscript i, and specifyboth the mass mi and the dis-tance ri from the rotationpoint. In the figure, the bladeis shown rotating through anangle q in a time Dt .

Figure 8-1 Although each windmillremains in one place and thereforehas no translational kinetic energy,its blades rotate in the wind. In thischapter we add rotation to ourdiscussion of motion. (ImageState)

KESTEN-08_275-332hr.indd 275 10/21/11

8/16/2019 Kesten Ch 8 10.21

2/58

276 Chapter 8 Rotational Motion

The position of each small element of the blade is changing versus time, so each

has a defined velocity and therefore kinetic energy. The ith element, shown in thefigure, moves a distance riq in time Dt , so its speed is

vi =riq

t (8-1)

(The relationship between the distance along the arc of a circle its radius, and theangular extent of the arc is presented in Figure 3-32.) The kinetic energy of thiselement is

Ki =1

2 mi v i

2=

1

2 mia riq

t b2

=

1

2 mi r i

2aqt b2 (8-2)

Notice that the term in parentheses in the last step doesn’t depend on which ele-ment of the blade we selected, because both the angle through which the bladerotates and the time it takes are the same for all parts of the blade. For a rigidobject, this angular velocity serves as a convenient way to characterize the rotationin a way that is independent of size or shape. The Greek letter omega (v) is theconventional symbol for angular velocity:

v =q

t (8-3)

The units of angular velocity are radians per second (rad>s). As a reminder, radiansare a unit of angle; there are 2p radians in a circle. By convention, v is positivewhen q is counterclockwise and negative when q is clockwise.

At any instant, the rate at which the rotation angle changes is the same acrossall pieces of a rigid object, so angular velocity v is also the same. This does notmean, however, that v is constant over time, but rather, unlike (linear) velocity,

angular velocity does not vary across a rotating, rigid object. If the rotation ratevaries with time, then Equation 8-3 defines the average angular velocity. We definethe angular velocity at any instant, instantaneous angular velocity, by letting thetime interval Dt be infinitesimally small:

v = limt S0

q

t =

d q

dt (8-4)

√× See the Math Tutorial for moreinformation on Differential

Calculus

Figure 8-2 Each small element ofa blade of a windmill is movingaround the rotation axis andtherefore contributes to thekinetic energy associated withthe rotation of the blade.

The blade rotates through thisangle ∆q in a time ∆t.

A small element of the blade adistance ri from the rotationpoint and having mass mi travels a distance ri∆q in atime ∆t.

The velocity of this element of the

blade is v1 = =change in position––––––––––––––––

change in timeri∆q ––––∆t

The element of the blade has kinetic

energy Ki = mivi2 = mi( )

21–2

1–2

ri∆q –––––∆t

∆q

ri∆q

ri

mi

KESTEN-08_275-332hr.indd 276 10/21/11

8/16/2019 Kesten Ch 8 10.21

3/58

8-1 Rotational Kinetic Energy 277

Instantaneous angular velocity is defined as the derivative of angle with respect totime in the same way that instantaneous velocity is defined as the derivative of posi-tion with respect to time (Equation 2-16).

The relationship between speed and the radius (Equation 8-1) of an object’srotation can be rewritten as v = rv or

v=

v

r (8-5)

for an object or a piece of an object a distance r from the rotation point movingat v. Because angular velocity v is the same for any element of a rotating object,Equation 8-5 tells us that the speed of any element of a rotating object increases atthe same rate that the distance r from the rotation point increases.

The expression for the kinetic energy of the ith element of the blade(Equation 8-2) can be written in terms of v:

Ki =1

2 mi r i

2v

2

Kinetic energy is a scalar (not a vector), so the kinetic energy of the entire rotatingwindmill blade can be found by adding up the kinetic energy of each of the ele-ments of the blade:

K =

aKi =

a1

2 mi r i2v2

We need to take a bit of care in doing the sum. We must break the blade intopieces so small that there is only one value of r for each piece. If these pieceswere large, the distance from the fixed rotation axis would be different for dif-ferent parts of the piece, and we wouldn’t know what value of r to enter intothe sum.

Values that don’t change with the subscript i, such as 12 and2, can be taken out

of the sum:

K =1

2 1ami r i22v2 (8-6)

See Mathbox 8-1.

Math Box 8-1 Removing a Constant Term from a Sum

Constant terms can always be removed from a sum. Prove it to yourself by tryingit with a few numbers. For example, in the sum 2(3) + 2(4) + 2(5), the constantvalue 2 can be pulled out from each term:

2(3) + 2(4) + 2(5) = 2(3 + 4 + 5)

We could have written this expression as

a5

n=3

2n = 2a5

n=3

n = 213 + 4 + 52So for K = a 12 mi r i2v2, because 12 and v are constant, both terms can be taken outof the sum:

K =1

2 v

2ami r i2or

K =1

2 aami r i2bv2

KESTEN-08_275-332hr.indd 277 10/21/11

8/16/2019 Kesten Ch 8 10.21

4/58


We choose to write the v2 term after the sum only to make this expression have thesame form as the expression for linear kinetic energy

K =1

2 × something × 1rate22.

The term in parentheses in Equation 8-6, the rotational inertia, or momentof inertia, characterizes the motion of a rigid object that is allowed to rotate. Aswe’ll see more clearly later in this chapter, the moment of inertia plays a role inrotational motion similar to the role that mass plays in linear motion. Moment ofinertia is commonly represented by the variable I :

I = ami r i2 (8-7)Equation 8-7 requires that the separate elements of mass mi and distance ri fromthe rotation axis be small. Moment of inertia is the topic of the next section. The SIunits of I are kilogram-square meters (kg · m2).

We will call the kinetic energy of a rigid object rotating around a fixed axisrotational kinetic energy to differentiate it from the translational (or linear) kineticenergy Ktranslational =

1

2

mv2. By substituting our definition of moment of inertia(Equation 8-7) into Equation 8-6, we define rotational kinetic energy as

Krotational =1

2 I v2 (8-8)

By comparing the definitions of translational and rotational kinetic energies, noticethat mass (in the translational kinetic energy equation) corresponds to the momentof inertia (in the rotational kinetic energy equation). Whereas we interpreted massas a property of matter that represents the resistance of an object to a change intranslational velocity, the moment of inertia represents the resistance of an objectto a change in rotational or angular velocity. In the same way that we defined iner-tia as the tendency of an object to resist a change in translational motion, we candefine rotational inertia as the tendency of an object to resist a change in rotationalmotion.

Both translational and rotational kinetic energies depend on the mass of the mov-ing object. However, the moment of inertia (Equation 8-7) and therefore the rotationalkinetic energy also depends on how the mass is distributed with respect to the axis ofrotation. A bit of mass far from the rotation axis has a larger effect on the value of themoment of inertia than the same amount of mass close to the axis.

Two objects that have the same mass and move at the same translational speedshave the same kinetic energies even if their shapes are different. However, even atthe same angular velocity, these two objects will likely have different rotationalkinetic energies when rotated because the different shapes will result in differentvalues for the moment of inertia. For example, when both are rotated at the sameangular velocity, a bicycle wheel, essentially a ring with negligibly small mass nearthe center, has more rotational kinetic energy than a flat, uniform disk of the samesize and mass because more of the mass of the wheel is farther from the rotationaxis (Figure 8-3). Indeed, two identical objects rotated around different axes will

likely have different rotational kinetic energies, because again, the distribution ofmass relative to the rotation axis will be different for the two. The rotational kineticenergy of a rod rotating around an axis perpendicular to the rod and through itsend is four times larger than the rotational kinetic energy of the rod when it rotatesat the same angular velocity around an axis perpendicular to the rod and throughits center (Figure 8-4).

KESTEN-08_275-332hr.indd 278 10/21/11

8/16/2019 Kesten Ch 8 10.21

5/58

8-1 Rotational Kinetic Energy 279

Tie a cell phone to the end of a very light thread. Holding on to the other endof the thread, would you be able to swing the cell phone around in a horizontalcircle? (Careful, the answer might not be what you expect!)

?Got the Concept 8-1

Swinging a Phone

These two wheels have the same mass and size, but the first is a ring with no mass at the centerwhile the other is a uniform disk.

Sliding at the same linear velocity, both have the same linear kinetic energy.

When both rotate at the same angular velocity, the ring has greater rotational kinetic energy.More of its mass is farther from the rotation axis, resulting in a larger moment of inertiaaround that axis.

Figure 8-3 The two wheels have the same mass and size, but one is a uniform diskand the other is a ring with none of its mass in the center. When the wheels slidewith the same velocity, without rotating, they have the same kinetic energy. Whenthey rotate at the same angular velocity, the uniform disk has less rotationalkinetic energy and a smaller moment of inertia because more of its mass is closerto the rotation axis.

The kinetic energy of a rod rotating around an axisperpendicular to the rod and through its end is fourtimes larger...

The more mass there is farther from the rotation axis, the larger the moment of inertia,and the larger the rotational kinetic energy for any given angular velocity.

...than the kinetic energy of the rod when it rotatesat the same angular velocity around an axisperpendicular to the rod and through its center.

Figure 8-4 The magnitude of thekinetic energy of a rotating roddepends on the location of therotation axis relative to thedistribution of the mass of the

rod. The moment of inertia islarger when more of the mass isfarther from the rotation axis.

KESTEN-08_275-332hr.indd 279 10/21/11

8/16/2019 Kesten Ch 8 10.21

6/58


The physics which underlies the flight characteristics of birds and flying insectsis complicated. Their maneuverability has as much to do with the contribu-

tions their wings make to the moments of inertia around their roll, pitch, andyaw axes (Figure 8-5) as with the aerodynamic characteristics of the wings.A bird’s wings can be as much as 15% of the total body mass, while an insect’swings are typically considerably less. In general, would you expect a flyinginsect or a bird to be able to maneuver more quickly in flight?

z

x

y

Roll

YawPitch

Figure 8-5 Flying objects such as insects, birds, and planes can rotate around thex, y, and z axes. A rotation around the x axis is called a roll, a rotation aroundthe y axis is called a pitch, and a rotation around the z axis is called a yaw.(Dudley, R. (2002). Mechanisms and Implications of Animal FlightManeuverability. Integrative and Comparative Biology , 42:135–140.)

? Got the Concept 8-2Flight

Example 8-1 Whirl an Object

When the physicist in Figure 8-6 whirls a small red object in a nearly horizontalcircle at the end of a 0.30-m-long string, the object makes 5 rev/s. Treating theobject as if all its mass were concentrated at a single point and neglecting the massof the string, how much rotational kinetic energy must the physicist supply to causethis motion to occur? The object has a mass equal to 0.20 kg.

SET UPTo determine rotational kinetic energy,

Krotational =1

2 I v2 (8-8)

we need to know the moment of inertia I of the system around the rotation axis aswell as the angular velocity v. By neglecting the mass of the string and treating thered object as if all its mass were concentrated at a single point, we can find the

moment of inertia using Equation 8-7:I = ami r i2 = mobject r2object

SOLVEThe rotational kinetic energy of the small red object is

Krotational =1

2 mobject r

2object v

2

Figure 8-6 Physicists have morefun than physiologists. (Courtesy David Tauck)

KESTEN-08_275-332hr.indd 280 10/21/11

8/16/2019 Kesten Ch 8 10.21

7/58

8-2 Moment of Inertia 281

To compute a numeric result, convert v = 5 rev>s to rad>s:v =

5 rev

s ×

2p rad

1 rev =

10p rad

sSo

Krotational

=

1

2

10.20 kg

2 10.30 m

22

a10p rad

s b2

= 8.9 J

REFLECTThe physicist must supply nearly 9 J of energy to rotate the small red object—nota lot of energy, but not a little, either. From what height would he need to drop theobject to impart the same energy? Setting gravitational potential energy U = mgh equal to the result of 8.9 J gives h = 4.5 m. That’s probably almost two and a halftimes the physicist’s height, so it’s reasonable to conclude that it takes a modestamount of effort on his part to swing the object.

Practice Problem 8-1 When the physicist in Figure 8-6 whirls a small red objectof mass 0.20 kg in a nearly horizontal circle at the end of a 0.30-m-long string,he imparts 15 J of energy to the object. Treating the object as if all its mass wereconcentrated at a single point and neglecting the mass of the string, how many

revolutions per second does the object make?

What’s Important 8-1The kinetic energy of a rigid object rotating around a fixed axis is called

rotational kinetic energy. It depends not only on the mass and angular velocityof an object, but also on how the mass of the object is distributed with respectto the axis of rotation.

*

8-2 Moment of InertiaThe astronaut in Figure 8-7 would find that even in a weightless environment it’s

harder to cause a massive object to accelerate than it is a less massive one. A womanpulling on a massive vault door has to contend with not only the mass of the doorbut also the distribution of the mass with respect to the hinges. Inertia, the tendencyof an object to resist a change in translational motion, and the object’s mass arereally two aspects of the same physics. The moment of inertia plays the same rolein rotational physics that mass does for linear motion.

The moment of inertia of a very small object that has a mass m and rotatesaround an axis a distance r away is given by I = mr2. For an object to be consideredsmall, the distance from the rotation axis to all points on the object must be thesame; the object in Figure 8-8a is small enough, but the object in Figure 8-8b is not.The moment of inertia of a large object rotating around a given axis can be foundby imagining the object broken into many small pieces, finding the contributionI i = mi r i

2 that each piece makes to the moment of inertia, and then adding up theseparate contributions. This method leads to the same relationship we discovered

for the moment of inertia by considering the kinetic energy of rotation:

I = ami r i2 (8-7)The moment of inertia of an object can be determined only with respect to a

specific rotation axis. Except in the special case of objects which are symmetric insome way, an object likely has a different moment of inertia around each differentrotation axis.

Figure 8-7 Inertia is the

tendency of an object to resist achange in translational motion.Even in orbit around Earth, itis difficult for an astronautto cause a massive object toaccelerate because of its inertia.(NASA)

KESTEN-08_275-332hr.indd 281 10/21/11

8/16/2019 Kesten Ch 8 10.21

8/58


The moment of inertia is additive. If you know the moments of inertia of twoobjects around some rotation axis and you attach them to form a single object, themoment of inertia of the new object—around the same axis—is the sum of themoments of inertia of the two separate objects.

In Figure 8-9a, a small sphere that has a mass M1 is attached to a Styrofoamrod that has a negligible mass and a length L1. The moment of inertia of thesphere when it rotates around the axis passing through the end of the rod isI 1 = M1 L

21. Similarly, the moment of inertia of another small sphere that has a

mass M2 and rotates at the end of a rod that has a negligible mass and length L2 is I 2 = M2 L

22 (Figure 8-9b). When the ends of the two rods are attached and the

combined object is rotated around the same axis, as in Figure 8-9c, the momentof inertia is

I = I 1 + I 2 = M1 L21 + M2 L22

Moment of Inertia of a Thin, Uniform Rod Rotating around One EndThe moment of inertia of a large object is defined by

I = ami r i2 (8-7)where we have imagined the object broken into many small pieces identified by thesubscript i. Each piece must be so small that the distance from the rotation axis to allpoints on the piece is the same. To guarantee that this is always true, the pieces mustbe of infinitesimal size. To remind us that each piece is of infinitesimal mass, we writethe mass term as dm. This allows us to write the sum in Equation 8-7 as the integral:

I = dm r2

Or, to write the moment of inertia in a more standard form,

I = r2 dm (8-9)

√× See the Math Tutorial for more

information on Integrals

Figure 8-9 Moment of inertia is additive. The momentsof inertia of objects 1 and 2 are I 1 and I 2, respectively.If we combine the two objects into one single object,its moment of inertia is the sum of I 1 and I 2.

L2

L1

L1

I 1 = M1L12

I = M1L12

+ M2L22

I 2 = M2L22

L2

(a)

(b)

(c)

Figure 8-8 (a) The distance from the rotation axisto any part of a very small object is the same.(b) In contrast, different parts of a large objectwill be different distances from the rotation axis. (Ocean/Corbis)

(b)

(a)

r

r1

rotationaxis

r2r3

This object is so small thatthe distance to the rotation

axis is the same everwhere.

The distance to the rotationaxis is different at differentpoints on the larger object.

KESTEN-08_275-332hr.indd 282 10/21/11

8/16/2019 Kesten Ch 8 10.21

9/58


Again, note that in converting the sum to an integral, the mass mi of each piece ofan object becomes dm, the mass of an infinitesimally small mass element. Thebounds on the integral are set up so that every bit of mass of an object is includedin the integration.

To get some hands-on experience finding the moment of inertia of an objectthat can rotate, let’s consider a thin, uniform rod of length L and mass M that

rotates around one end. Such a rod is shown in Figure 8-10. An arbitrarily selected,infinitesimal slice of the rod of mass dm is shown. Note that we exaggerated thesize of dm in the figure; in reality, even the thinnest line we could draw would betoo thick because mathematically dm must be infinitesimally small. To emphasizethis we have labeled the width of dm as dr, a differential element of distance alongthe rod.

To carry out the integral in Equation 8-9, we need to assign an upper boundand a lower bound to sweep up every possible mass element in the rod. Warning!Don’t be too quick to insert 0 to L as the bounds; the values of the bounds of anintegral must match the differential variable of integration. As it stands, the differ-ential is dm, so the differential variable is m, and 0 and L are not valid values ofmass. We can, however, directly change the variable of integration from mass tolength. Because the rod is uniform, the length of the infinitesimal slice of the rod isthe same proportion to the total length of the rod as the mass of the slice is to the

total mass of the rod:

length of slice

length of rod =

mass of slice

mass of rod

So,

dr

L =

dm

M

or

dm =M

L dr (8-10)

This relationship is valid whenever the mass of a rod—or any object we can treat

as one-dimensional—is uniformly distributed.Inserting Equation 8-10 into the expression for the moment of inertia(Equation 8-9) gives

I = r2 M

L dr =

M

L r2 dr

The distance from the rotation axis of the mass of the slice at one end of the rod isr = 0, and at the other end, r = L, so 0 and L are the lower and upper bounds ofthe variable over which the integral is evaluated:

I =M

L LL

0

r2 dr (8-11)

Solving the integral leads us to an expression for the moment of inertia:

I =M

L

r3

3 ̀ L0 = ML aL3

3 -

03

3 b = ML23The moment of inertia of a thin rod that has a mass M and a length L and rotatesaround one end is ML2>3.

r

dr

dm

L

M

=

=

length of slice–––––––––––––

total lengthmass of slice

–––––––––––––total mass

dr––L

dm–––M

=dm drM––L

Figure 8-10 The moment ofinertia of a large object is thesum of the moments of inertiaof each infinitesimally smallpiece of the object.

KESTEN-08_275-332hr.indd 283 10/21/11

8/16/2019 Kesten Ch 8 10.21

10/58


If your friend asks, “What is the moment of inertia of that object?” it could bea trick question! To determine the different values of radius in Equation 8-7requires that we first specify the axis around which the object rotates. Anyobject can be made to rotate around any number of axes, even one like aDVD that commonly rotates around a particular axis. The axis does not evenhave to pass through the object that rotates around it—imagine tying a stringto the edge of a DVD and swinging it around in a circle above your head. TheDVD would be rotating around an axis that does not pass through it. Threepossible rotation axes are shown for a DVD in Figure 8-11. Make sure youidentify the axis of rotation before determining the moment of inertia ofan object.

Figure 8-11 The moment of inertia depends on the specific rotationaxis of the object. For example, the moment of inertia of a DVDdepends on whether the rotation axis is the usual one in the center ofthe disk (top left), some other point on the disk (bottom left), or evena point outside the disk (right). You must always identify the axis ofrotation before determining the moment of inertia of an object.

! Watch Out!An object does not have “a” moment of inertia. Rather, it hasa moment of inertia defined for rotation around each specificchoice of rotation axis.

Moment of Inertia of a Thin, Uniform Rod Rotating around Its CenterA thin, uniform rod of length L and mass M rotates around its center inFigure 8-12. What is the moment of inertia of the rod around this axis, and how doesit compare to the moment of inertia of the rod rotating around one end?

The difference that the choice of rotation axis makes is evident in a comparisonof Figures 8-12 and 8-10. The values of r which identify the slices of the rod at thetwo ends are different in the two cases. When the rotation axis goes through one endof the rod, r varies from 0 to L. When the rotation axis goes through the center of therod, r must vary from -L>2 to +L>2 for the integration to cover the entire rod. Inother words, the difference between the moment of inertia of the rod rotating aroundone end and the moment of inertia of the rod rotating around the center is in the

KESTEN-08_275-332hr.indd 284 10/21/11

8/16/2019 Kesten Ch 8 10.21

11/58


bounds of the integral. We simply change the bounds in Equation 8-11 to r varying from -L>2 to +L>2 or

I =M

L L+L>2-L>2 r

2 dr

= ML

r3

3 ̀ +L

>2

-L>2 = ML a 1+

L>223

3 - 1

-L>22

3

3 b =

M

L aL3

24 - -L3

24 b = ML2

12

The moment of inertia of a thin rod of mass M and length L that rotatesaround its center is ML2>12. This value is one-fourth the value of the momentof inertia of the rod when it rotates around one end. We expect the moment ofinertia to be smaller in this case because moment of inertia varies as the square of r, andthe mass of the rod is distributed closer to the rotation axis when the rod rotatesaround its center. When the rod rotates around its center no part of it is farther thanL>2 from the axis, but when the rod rotates around one end, half of it is farther thanL>2 from the axis.

Two thin, uniform rods that each have a mass M and a length L are attachedat their centers to form an “X” shape. What is the moment of inertia when thisconfiguration is rotated around the axis which passes through their centers,perpendicular to the plane of the two rods?

? Got the Concept 8-3Moment of Inertia of Two Rods

Four thin, uniform rods that each have a mass M>2 and a length L>2 areattached at their ends to form an “X” shape. What is the moment of inertiawhen this configuration is rotated around the axis which passes through thecenter of the “X” perpendicular to the plane of the rods?

? Got the Concept 8-4Moment of Inertia of Four Rods

Moment of Inertia of a Thin, Uniform Ring Rotating around an AxisPerpendicular to the Plane of the Ring and through Its CenterA thin, uniform ring of radius R and mass M rotates around an axis perpendicularto the plane of the ring and through its center as shown in Figure 8-13. An arbitrarilyselected, infinitesimal slice of the ring of mass dm is shown. Again, for clarity weexaggerated the size of dm in the figure; dm must be infinitesimally small. To

emphasize this, we have labeled the angular extent of dm as d q, a differential elementof angle along the ring. Note that although the ring occupies a two-dimensionalspace, we can treat the mass distribution as one-dimensional because the mass isdistributed along the circumference of a circle. In effect, the ring is a thin, uniformrod which has been bent into a circle.

To solve the integral in Equation 8-9, we need to assign an upper bound and alower bound, to sum every possible mass element in the ring. By converting from

r

r =

dm

L

M

+L–––2

r = –L–––2

Figure 8-12 A thin, uniform rod oflength L and mass M rotates aroundits center. The moment of inertia isfound by adding the contribution thateach infinitesimally small mass elementdm makes to the total.

KESTEN-08_275-332hr.indd 285 10/21/11

8/16/2019 Kesten Ch 8 10.21

12/58


an integral in terms of mass to one in terms of angle, we can add upevery mass element by allowing q to vary from 0 to 2p. To make thisconversion, note that because the ring is uniform, the angular extentof an infinitesimal slice of the ring is the same proportion to the totalangle of the ring as the mass of the slice is to the total mass:

angle of sliceangle of ring

= mass of slicemass of ring

The total angular extent of the ring is 2p radians, so

d q

2p =

dm

Mor

dm =M

2p d q (8-12)

This relationship is valid because the mass of the ring is uniformlydistributed.

Inserting Equation 8-12 into the expression for the moment of inertia(Equation 8-9) gives

I = r2 M2p

d q = M2p

r2 d q

Note that the variable r is the radial distance to any particular infinitesimal pieceof the object over which we integrate. Be careful to distinguish between thevariable and the value that variable takes in any specific case. For this ring, thedistance from the rotation axis to any piece is always R, which can be taken outof the integral because it is constant. Also, to include in the integral every infini-tesimal slice of the ring, we need to integrate in a complete circle, so q rangesfrom 0 to 2p:

I =MR2

2p L2p

0

d q (8-13)

So

I = MR2

2p q ` 2p0 = MR22p 12p - 02 = MR2

The moment of inertia of a thin, uniform ring of mass M and radius R thatrotates around an axis perpendicular to the plane of the ring and through itscenter is MR2.

Moment of Inertia of a Thin, Uniform Disk Rotating around an AxisPerpendicular to the Plane of the Disk and through Its CenterA thin, uniform disk of radius R and mass M rotates around an axis perpendicu-lar to the plane of the disk and through its center. Figure 8-14 shows an arbitrarilyselected, infinitesimal piece of the disk of mass dm. Note that because the disk isthin we can treat the mass distribution as two-dimensional, which requires dm to

be two-dimensional as well. To emphasize the requirement that dm be infinitesi-mally small, we have defined the piece of the disk as being bounded by inner radiusr and outer radius r + dr, and having angular extent d q.

We need to assign an upper bound and a lower bound to the moment of iner-tia integral (Equation 8-9) so that the integration includes every possible masselement in the disk. Because the disk is uniform we can set up a proportion analo-gous to the one we used in the case of the rod and the ring. Note, however, that

√× See the Math Tutorial for moreinformation on Trigonometry

dm

R

d q

Figure 8-13 A thin, uniform ring of radius R rotates around an axis perpendicular to theplane of the ring and through its center. Themoment of inertia is found by adding thecontribution that each infinitesimally smallmass element dm makes to the total. Onearbitrarily selected mass element, which extendsover an infinitesimally small angle d q, is shown.

KESTEN-08_275-332hr.indd 286 10/21/11

8/16/2019 Kesten Ch 8 10.21

13/58


because the infinitesimal dm is two-dimensional, the proportionality is in termsof area, or

area of piece

area of disk =

mass of piece

mass of disk

It’s helpful to know the shape of the dm piece in order to determine its area. Don’tbe fooled by the appearance of the sample piece shown in Figure 8-14—we cantreat the shape of dm as rectangular. It may appear that the outer edge of dm ismuch wider than the inner edge, but this is due to our exaggeration of the overall

size of dm. Because the angular extent d q of each piece of the disk is infinitesimallysmall, the lengths of the inner and outer sides of dm are also infinitesimally small,so we can treat them as mathematically equivalent.

Because dm is a rectangle, its area is the product of the length of the two sides.As shown in Figure 8-15, this area is (r d q)(dr), so

r d q dr

pR2 =

dm

M

or

dm =M

pR2 r dr d q (8-14)

We have been careful to distinguish between r, the radial distance to any particularinfinitesimal patch of the disk over which we will integrate, and R, the constant

radius of the entire disk. The relationship for dm in Equation 8-14 is valid becausethe mass of the disk is uniformly distributed.

Inserting Equation 8-14 into the expression for the moment of inertia(Equation 8-9) gives

I = r2 M

pR2 r dr d q =

M

pR2 r3 dr d q

dm

R

r

r+dr

d q

Figure 8-14 A thin, uniform disk of radius R rotates around an axis perpendicular to theplane of the disk and through its center. Themoment of inertia is found by adding thecontribution that each infinitesimally smallmass element dm makes to the total. Onearbitrarily selected mass element, a small sliceof an annulus of the disk, is shown in darkgray. The mass element is defined by the regionfrom a distance r from the rotation axis to r plus an infinitesimally small distance dr, and

extends an infinitesimally small distance d q along the annulus.

r+drdr

r

r dq

Figure 8-15 Because both the width of the annulus and theangular extent of the slice are infinitesimally small, we can

treat the piece shown in dark blue as if it were a rectangle oflength r d q and width dr.

KESTEN-08_275-332hr.indd 287 10/21/11

8/16/2019 Kesten Ch 8 10.21

14/58


It is necessary to write two integrals because the expression for dm involves twodifferentials. However, the r and q variables are independent of one another, so wecan treat the expression as two separate integrals or

I =M

pR2 r3 dr d q

Finally, we set the bounds to include every possible infinitesimal slice of the disk.The smallest and largest values of radius are r = 0 and r = R, respectively, and angleq runs from 0 to 2p:

I =M

pR2LR

0

r3 drL2p

0

d q (8-15)

So

I =M

pR2

r4

4 ̀ R

0

q ` 2p0

=

M

pR2 aR4

4 - 0b 12p - 02

=M

pR2

2pR4

4 =

MR2

2

The moment of inertia of a thin, uniform disk of mass M and radius R that rotatesaround an axis perpendicular to the plane of the disk and through its center is MR2

>2.

When rotating about the same axis, the moment of inertia of a thin, uniformdisk is half that of a thin, uniform ring of the same mass and radius. You shouldexpect the disk to have a smaller moment of inertia, because the moment of inertiaof an object is strongly influenced by how far the mass is from the rotation axis.All of the mass of the ring is located a distance R from the axis, while only a frac-tion of the mass of the disk is that far from the axis. Therefore, the moment ofinertia of the ring must be larger than the moment of inertia of the disk.

Math Box 8-2 Checking a Differential Area

We needed to devise an expression for the area of an infinitesimally small patch ofthe disk shown in Figure 8-14 in order to determine the moment of inertia. This

approach is the same regardless of the shape or number of dimensions of the objectin question. It is therefore useful, and often straightforward, to check whether theexpression we devise is correct. For a disk rotating around its center, for example,we integrated over infinitesimally small regions dA that, as shown in Figure 8-15,we assigned area r dr d q:

dA = r dr d q

The integral of the area elements dA over the entire disk must give the area of thedisk. To convince ourselves that the form we constructed for a tiny piece of area iscorrect, let’s integrate dA over the entire disk; we should get A = pR2. The boundson radius r and angle q are 0

8/16/2019 Kesten Ch 8 10.21

15/58


The moments of inertia of a variety of objects and rotation axes are given inTable 8-1.

What’s Important 8-2Rotational inertia is the tendency of an object to resist a change in rota-

tional motion. It depends not only on the mass of an object but also on howthe mass is distributed with respect to the axis of rotation.*

Table 8-1

Moments of Inertia of Uniform Bodies of Various Shapes*

KESTEN-08_275-332hr.indd 289 10/21/11

8/16/2019 Kesten Ch 8 10.21

16/58


8-3 The Parallel-Axis TheoremFinding the moment of inertia of an object, even one with a symmetrical shape, canbe challenging for certain choices of rotation axis. Imagine, for example, a uniformdisk rotating around an axis perpendicular to the plane of the disk but passingthrough a point near its edge, as in Figure 8-16a. In this case setting the bounds

necessary to find the moment of inertia would be cumbersome. However, a curiousrelationship exists between the moment of inertia of anobject when it rotates around its center of mass, which isoften easy to find, and the moment of inertia when theobject rotates around any other parallel axis.

Let’s say you know the moment of inertia I CM of anobject when it rotates around an axis passing through itscenter of mass. The moment of inertia for a rotation aroundany other parallel axis is

I = I CM + Mh2 (8-16)

where M is the mass of the object and h is the distancebetween the two axes. This relationship is known as theparallel-axis theorem.

To see the parallel-axis theorem in action, consider athin, uniform rod that has a mass M, a length L, and itscenter of mass at the center of the rod. In Section 8-2, wedetermined the moment of inertia of a similar rod rotatingaround an axis perpendicular to the rod and through itscenter (Figure 8-12). As summarized in Table 8-1,

I CM =ML2

12

In Section 8-2, we also found the moment of inertia for athin, uniform rod of mass M and length L that rotatesaround an axis perpendicular to the rod and through oneend (Figure 8-10) to be I = ML2

>3. Can the parallel-axis

theorem reproduce this result?The end of the rod is L>2 from the center, so the axisthrough the end is h = L>2 from the center of mass. ByEquation 8-16,

I =ML2

12 + MaL

2b2

=ML2

12 +

ML2

4

=ML2

12 +

3ML2

12

=4ML2

12

=ML2

3

R

r

r + dr

Rotation axis-in plane of page

Rotation axis-perpendicular to page

d q

q

(a)

(b)

Figure 8-16 (a) A thin, uniform disk of radius R andmass M rotates around a point near the edge of the

disk. (b) Finding the moment of inertia would becomplex without applying the parallel-axis theorem.

KESTEN-08_275-332hr.indd 290 10/21/11

8/16/2019 Kesten Ch 8 10.21

17/58

8-3 The Parallel-Axis Theorem 291

The parallel-axis theorem does indeed reproduce the correct moment of inertia forthe rod rotating about its end.

The parallel-axis theorem is particularly useful for determining the moment ofinertia when the shape of an object or the orientation of the axis makes the integralin Equation 8-9 difficult, as in the next example.

Example 8-2 The Edge of a Disk

Find the moment of inertia of a thin, uniform disk of radius R and mass M thatrotates around a pin pushed through a small hole near the edge of the disk as inFigure 8-16a.

SET UPThe center of mass of the disk is at its center. In Section 8-2, we found the momentof inertia of a thin, uniform disk rotating around an axis perpendicular to the planeof the disk and passing through its center to be I = MR2>2. We can therefore usethe parallel-axis theorem to find the moment of inertia around the axis specified inthis problem.

SOLVEThe edge of the disk is a distance h = R from the center, so by the parallel-axistheorem (Equation 8-16),

I =MR2

2 + MR2

=

3MR2

2

REFLECTFinding the moment of inertia of the disk rotating around a point on its edge wasstraightforward using the parallel-axis theorem. The integral definition of momentof inertia (Equation 8-9) would not be. Although the integral would be similar tothe one we used to find the moment of inertia for the disk rotating around its center

(Equation 8-15), determining the bounds would be challenging! As suggested bythe differential patch of area shown in Figure 8-16b, the mass dm would be thesame as that in Equation 8-15. But what, for example, are the upper bounds on r and q to cover the entire disk? As you can see from the figure, the largest value of r depends on the angle q and the largest value of q depends on the size of the disk. Asa result the integral approach is not nearly as straightforward as using the parallel-axis theorem.

Practice Problem 8-2 Find the moment of inertia of a thin, uniform disk of radiusR and mass M that rotates around a pin pushed through a small hole halfwaybetween the center and the edge of the disk.

What’s Important 8-3The parallel-axis theorem describes the relationship between the moment

of inertia of an object when it rotates around its center of mass, which is ofteneasy to find, and the moment of inertia when the object rotates around anyother parallel axis.*

KESTEN-08_275-332hr.indd 291 10/21/11

8/16/2019 Kesten Ch 8 10.21

18/58


8-4 Conservation of Energy RevisitedEvery year in the All American Soap Box Derby (Figure 8-17) thousands of boys andgirls race in homemade cars powered only by the force of gravity pulling them

down a hill. Although the organizers encourage creativity in car design, they dis-qualify competitors who do not use officially approved wheels. How would differ-ent kinds of wheels affect the outcome of the race? Conservation of energy lies atthe heart of the answer.

Before the start of the race, a car at the top of a hill has gravitational potentialenergy relative to the bottom of the hill but no kinetic energy. The car gains speedas it rolls down the race course—a transformation of potential energy into kineticenergy. As we saw in Section 6-6, energy must be conserved. We write

Ki + U i = Kf + U f + W nc (6-31)

as a general statement of energy conservation, noting that a separate potentialenergy term must be included for each force an object in the system experiences andW nc is work done by nonconservative forces. Because a rotating object has rota-tional kinetic energy separate from any translational kinetic energy it possesses, we

will now augment Equation 6-31 to include both a rotational kinetic energy as wellas a translational kinetic energy term:

Ktranslational, i + Krotational, i + U i = Ktranslational, f + Krotational, f + U f + W nc (8-17)

As the car rolls down the racecourse, some of its initial gravitational potentialenergy is transformed to translational kinetic energy. The rate at which the transla-tional kinetic energy increases is directly related to the car’s linear acceleration. Butwe now see that some of the initial gravitational potential energy is transformedinto the rotational kinetic energy of the wheels. The more potential energy thatgoes into rotational kinetic energy, the less energy is available to make the car gofast (translational kinetic energy). The opposite is also true; when less energy isrequired to rotate the wheels, more of the potential energy can be transformed intotranslational kinetic energy, resulting in a higher linear velocity. Specially designedwheels could give a competitor an unfair advantage. In all cases, we assume that the

wheels roll without slipping down the hill.

Which type of wheels would allow a soapbox derby car to go faster, uniform disksor wheels that look like conventional bicycle tires where most of the mass is alongthe rim of the wheels? Assume that both types of wheels have the same mass.

? Got the Concept 8-5Faster Wheels

Figure 8-17 Competitors in thesoap box derby must useapproved wheels. A wheeldesign that reduces the amountof the car’s potential energythat goes into the rotationalkinetic energy of the wheelsresults in the car going fasterdown the hill. ( GeorgeTiedemann/NewSport/Corbis )

KESTEN-08_275-332hr.indd 292 10/21/11

8/16/2019 Kesten Ch 8 10.21

19/58

8-4 Conservation of Energy Revisited 293

Rolling, Slipping, and SlidingA moving circular or cylindrical object can slide, rollwithout slipping, or slide and roll at the same time.When sliding, as in Figure 8-18a, the same point (orpoints) on the object remains in contact with the sur-

face at all times. No rotational kinetic energy is pres-ent because the object doesn’t rotate. Rotationalkinetic energy is introduced when the object rotates,for example, when rolling. A special case of rollingmotion occurs when the edge of the disk does not sliprelative to the surface, as in Figure 8-18b. Notice thatthe colored thread that has been wrapped around thecircumference of the object unwinds as the objectrolls, marking the distance traveled. The distance thedisk moves along the surface in one rotation exactlyequals the circumference of the circle because theedge of the disk does not move relative to thesurface.

The object in Figure 8-18 could also rotate andslide at the same time, in which case a fixed point onthe edge of the disk would move relative to the sur-face as the object moved. We could in principle ana-lyze motion like this, by including a rotational kineticenergy term as well as a dissipative term due to slid-ing (kinetic) friction in Equation 8-17. In practice,however, considering the phenomenon of sliding iscomplicated by the need to know how much slidingoccurs, which in turn determines how much energy istaken up by rotation and how much is dissipated byfriction. For this reason, we will only deal with casesin which objects roll without slipping.

Objects that roll without slipping neverthelessexperience a retarding, frictional force. We tend toneglect this rolling friction, because it is usuallysmall compared to other effects, but without it arolling object would never come to a stop. Likestatic and kinetic frictional forces, the force of roll-ing friction, is proportional to the normal force act-ing on an object.

Figure 8-19a shows a completely rigid, rollingdisk. Because the disk is rigid, the contact betweenthe disk and the surface is a point or a line directlybelow the rotation axis. The normal force thereforepoints radially in toward the rotation axis and, as aresult, has no component along the direction ofmotion. The normal force has no effect on the rota-tion because it is directed toward the rotation axis.Under ideal conditions, then, a completely rigid

object experiences no rolling friction. No object (nomaterial) is perfectly rigid, however, which meansthat all objects deform when in contact with a sup-porting surface. That deformation results in thecontact between object and surface being spreadout over an area, as in Figure 8-19b. Because theleading edge of the rolling object is coming down to

(a) Sliding

(b) Rolling without slipping

R

R

2πR

Figure 8-18 A disk can slide, roll without slipping, or slide and rollat the same time. (a) An object does not rotate when it slides, so itacquires no rotational kinetic energy. (b) When a disk rollswithout slipping relative to the surface, in one rotation the diskmoves a distance exactly equal to the circumference of the disk.

(b) A deformed object

R

N

R N

(a) Completely rigid object

The leading edge of the rollingobject is coming down to thesurface, while the trailing edgeis lifting off the surface. Thenet normal force can be takenas acting in front of therotation axis.

The normal force N acting forwardof the axis counters the rotation,and has the effect of opposing themotion.

Figure 8-19 (a) A completely rigid, rolling disk only contacts the

surface at a point directly below the rotation axis. The normalforce therefore points toward the rotation axis and has noeffect on the rotation. Under ideal conditions a completely rigidobject experiences no rolling friction. (b) A nonrigid, rollingdisk contacts the surface at more than one point below therotation axis. The net normal force acts in front of the rotationaxis and opposes the motion.

KESTEN-08_275-332hr.indd 293 10/21/11

8/16/2019 Kesten Ch 8 10.21

20/58


the surface while the trailing edge is lifting off the surface, the net normal force canbe taken as acting in front of the rotation axis. This force counters the rotation, andhas the effect of opposing the motion. Rolling friction arises from the deformationof a rolling object.

Let’s compare the speed of a uniform disk to that of a hoop, when both rollingwithout slipping down a ramp. Both the disk and the hoop have the same radius R

and mass M and both traverse a vertical distance H (Figure 8-20).Total energy is conserved in both cases. The frictional forces will be negligibly

small for typical materials, so we can neglect the last term in Equation 8-17 theexpressions of energy conservation for the disk and the hoop have the same form:

Ktranslational, i + Krotational, i + U i = Ktranslational, f + Krotational, f + U f

Because both the disk and the hoop are initially at rest, Ktranslational, i and Krotational, i equal zero for both the disk and the hoop. So the previous expression for either thedisk or hoop becomes

Mghi =1

2 Mv2f +

1

2 I v2f + Mghf

Here, hi and hf are the heights of the top and bottom of the ramp, vf is the transla-tional speed at the bottom of the ramp, and vf is the rotational velocity at the bot-

tom of the ramp. The height of the ramp H is hi - hf , so

1

2 Mv2f = Mghi - Mghf -

1

2 I v2f = MgH -

1

2 I v2f (8-18)

To compare the speeds of the disk and hoop at the bottom of the ramp, we need tosolve this equation for vf in terms of the variables that define the problem: R, M,and H . This requires that both I and vf be expressed in terms of those variables aswell.

The moment of inertia of a uniform disk that has radius R, mass M, and rotatesaround an axis through the center and perpendicular to the plane of the object is

I disk =MR2

2

Figure 8-20 A disk and a hoop havethe same radius R and mass M.Both are allowed to roll from restwithout slipping down a ramp ofheight H .

H

Both disk and hoop haveradius R and mass M.

KESTEN-08_275-332hr.indd 294 10/21/11

8/16/2019 Kesten Ch 8 10.21

21/58

8-4 Conservation of Energy Revisited 295

The moment of inertia of a hoop that has a radius R and a mass M androtates around an axis through the center and perpendicular to the plane of theobject is

I hoop = MR2

The angular velocity term can be understood by using Figure 8-18b, in whicha circular object of radius R rolls at a constant rate through one full rotation with-out slipping. As we noted above, the disk moves linearly a distance equal to thecircumference when it rolls through one full rotation. The change in position of thecenter of the object is then equal to 2pR. Let the time for one full rotation be Dt ,so that the linear velocity is

v =x

t =

2pR

t

Similarly, the angular velocity is

v =q

t =

2p

t

so

v = a2p

t bR = vR (8-19)Equation 8-19 is a general relationship between the linear velocity and angularvelocity of a circular object rolling without slipping.

We can now write Equation 8-18 for the disk and the hoop separately, usingthe appropriate moment of inertia (either I disk or I hoop) and the relationshipbetween the linear velocity and angular velocity (v = v>R, from Equation 8-19).For the disk,

1

2 Mv2disk, f = MgH -

1

2 I diskavdisk, f

R b2 = MgH - 1

2

MR2

2 a vdisk, f

R b2

v2disk, f = 2 gH -1

2 v2disk, f

3

2 v2disk, f

=

2 gH

vdisk, f =4

3 gH

For the hoop,

1

2 Mv2hoop, f = MgH -

1

2 I hoop a vhoop, f

R b2 = MgH - 1

2 MR2a vhoop, f

R b2

v2hoop, f = 2 gH - v2hoop, f

2v2hoop, f = 2 gH

vhoop, f = 2 gH As we should expect, the speed of the hoop as it comes off the ramp is smaller than

the speed of the disk (

2 gH compared to

2 4 gH >3), because more of the hoop’sinitial gravitational potential energy is converted into rotational kinetic energy as

the hoop rolls down the ramp. Both objects will accelerate, but at any givenmoment, the speed of the hoop will be smaller than that of the disk. In a race, thedisk would win.

KESTEN-08_275-332hr.indd 295 10/21/11

8/16/2019 Kesten Ch 8 10.21

22/58


A hoop and a block are released from rest down adjacent ramps that haveidentical angles. The hoop rolls without slipping and the block slides without

friction, both starting from the same height. Which one reaches the bottom ofthe ramp first?

? Got the Concept 8-6Hoop and Block

Estimate It! 8-1 Rolling Spider

To escape predators, the golden wheel spider of the Namib Desert extends its legslike spokes on a wheel and rolls rapidly down sand dunes. Estimate the speed sucha spider might attain after rolling 1 m down a 15° slope.

SET UPWe can treat the spider as an object that rolls without slipping down a ramp of

angle 15°. As shown in the sketch in Figure 8-21, the spider travels a distance L while dropping a vertical distance H . Energy is conserved as the spider rolls downthe sand dune, so we start from Equation 8-17. Note that the spider has nokinetic energy at the instant it starts to roll. We’ll also neglect frictional forces,which we would expect to be small. If we declare the initial height of the spiderto be H , then the final height is zero; the initial potential energy U i then equals thesum of the final translational and rotational kinetic energies Ktranslational, f andKrotational, f :

U i = Ktranslational, f + Krotational, f

or

Mspider gH =1

2 Mspider v

2 +1

2 I spider v

2

where Mspider is the mass of the spider, v is the translational speed of the spider, andis its angular velocity.

We can approximate the spider as a uniform disk rotating around its centralaxis. Even when the spider sticks out its legs as spokes, most of its mass is concen-trated near the rotation axis, and the spider is also relatively flat. The effectiveradius Reff of the disk is equal to that of the main part of the spider’s body. FromTable 8-1, the moment of inertia of the spider is then

I spider =Mspider R

2eff

2

SOLVEConservation of energy therefore leads to

Mspider gH =1

2

Mspider v2+

1

2

Mspider R2eff

2

v2

Notice that the mass of the spider appears in each term, so it can be canceled. Also,from Equation 8-19, v is equal to vR, where R is the radius of the spider including its spokelike legs. So

gH =1

2 v2 +

1

2

R2eff

2

v2

R2

L

H = L sin q

H

q

Figure 8-21

KESTEN-08_275-332hr.indd 296 10/21/11

8/16/2019 Kesten Ch 8 10.21

23/58

8-5 Rotational Kinematics 297

or

v =B 2 gH

1 + R2eff >2R2From Figure 8-21, H equals L sin 15°, where L is given as 1 m. We want to useround, but reasonable, values in doing estimations; based on our experience with

spiders let’s take Reff equal to 0.2 cm and R equal to 0.5 cm. To the level of signifi-cance of these values, we can also use 10 m>s2 as an approximate value for g . Then

v 2110 m>s22 11 m sin 15°2

1 + 10.002 m22>210.005 m22 = 2.2 m>sTo one significant figure, we estimate the speed of the spider to be 2 m>s.REFLECTOur estimate certainly seems reasonable—golden wheel spiders can attain speedsof approximately 1 m>s.

What’s Important 8-4A rotating object has rotational kinetic energy, a quantity separate from

any translational kinetic energy it possesses. An application of the statement ofconservation of energy therefore includes terms for both rotational kineticenergy as well as translational kinetic energy.

*

8-5 Rotational KinematicsVideo information recorded on a DVD zips around in a circle as the disk rotates.Although the data recorded near the outer edge of the DVD travels farther in onerevolution of the disk than data recorded near the disk’s center, both parts of thedisk make the same one revolution. In a similar way, the speed of information atthe outer edge passing the reader in the DVD player is higher than the speed atwhich information recorded near the center passes the reader. Yet both parts of thedisk make the same number of revolutions in any given time. In this section, we willaddress the correspondence between the quantities that describe linear motion,such as distance and speed, and the quantities that describe rotational motion, suchas angle and angular velocity.

You probably noticed the similarities between the equations that define trans-lational kinetic energy (Equation 6-10) and rotational kinetic energy (Equation 8-8):

Ktranslational =1

2 mv2 =

1

2 madx

dt b2

Krotational =1

2 I v2 =

1

2 I ad q

dt b2

The form of these equations reveals both a similarity and a difference between massand moment of inertia. Both are properties of an object, and moment of inertia isrelated to mass. But while mass quantifies the tendency of an object to resist a

change in linear motion, moment of inertia quantifies the tendency of the object toresist a change in rotational motion. This is inferred from the presence of m andlinear velocity dx>dt in Ktranslational, but I and angular velocity in Krotational. The rela-tionship between x and q is also apparent: x measures linear displacement and q measures angular displacement, the angle through which an object has rotated.

We can extend the relationship between x and q to the derivatives of thesequantities with respect to time. The rotational equivalent of linear velocity v is

√× See the Math Tutorial for moreinformation on Differential

Calculus

KESTEN-08_275-332hr.indd 297 10/21/11

8/16/2019 Kesten Ch 8 10.21

24/58


angular velocity v; this result can be seen by comparing Ktranslational and Krotational, orby taking the derivatives of x and q with respect to time:

v =dx

dt

v =d q

dt And finally, as linear acceleration is the derivative of linear velocity with respect

to time, we can define an analogous quantity angular acceleration a for objects thatare rotating:

a =dv

dt

a =d v

dt (8-20)

Angular displacement is measured in radians. The SI units of v are

3v4 =rad

s

Therefore, the units of a are radians per square second (rad >s2). We can alsodescribe both angular displacement and angular velocity in terms of the number of

revolutions an object makes. Another useful set of units for v is therefore

3v4 =rev

s

A full circle is 2p rad, so a rate of 1 rev>s is equal to 2p rad>s. In addition, angularvelocity can be converted into an equivalent linear speed by recognizing that apoint at radius r from the rotation axis travels a distance 2pr in one revolution. Soan angular velocity of 1 rev>s is also equal to 2pr m>s.

The correspondence between the linear variables x, v, a, and m and the rota-tional variables q, v, a, and I enable a translation of sorts between the equationsthat describe linear kinematics and equations that describe rotational kinematics.In Chapter 2, we developed two fundamental equations that describe motion inone dimension under constant acceleration:

v = v0 + at (2-23)

x - x0 = v0 t + 12

at 2 (2-26)

We can now use similar equations to describe rotational motion under constantangular acceleration:

v = v0 + at (8-21)

q - q0 = v0 t +1

2 at 2 (8-22)

As we did for linear kinematics, we set the initial value of angular velocity to be v0.We let the time variable “start” at t 0 = 0 in accordance with the standard convention.We also define an initial angular position q0, so that angular displacement is q - q0.

Information recorded on a DVD is evenly spaced along a long spiral that spansmost of the surface of the disk. DVD players read the information at a constantrate. Should the disk rotate faster, slower, or at the same rate as the player readsinformation recorded closer and closer to the center of the DVD?

?Got the Concept 8-7DVD

KESTEN-08_275-332hr.indd 298 10/21/11

8/16/2019 Kesten Ch 8 10.21

25/58

8-5 Rotational Kinematics 299

Figure 8-22

Timet

w

w 0

Time

Height = w ′

Base = ∆t

t

w

w 0

∆t

w ′

Equations 2-23 and 2-26 completely describe linear motion when accelerationis constant and enable us to solve a wide range of problems. Using just the twoequations we could, for example, find the distance a hockey puck travels given itsinitial velocity or we could find the constant acceleration required for a car toattain a certain velocity after starting from rest. In the same way, Equations 8-21and 8-22 fully describe rotational motion when angular acceleration is constant.

The variables are different, but the physics is exactly the same as that of linearmotion discussed in Chapter 2.

Example 8-3 A Stopping Top

A top spinning at 25 rad>s (4.0 rev>s) comes to a complete stop in 42 s. Assumingthe top decelerates at a constant rate, how many revolutions does it make beforecoming to a stop?

SET UPThe two fundamental linear kinematics equations (Equations 2-23 and 2-26) con-tain five variables; solving any problem requires identifying three known variablesand then eliminating a fourth to find the value of the unknown variable. Themethod of creating a Know/Don’t Know table, useful for determining a value forthe unknown variable in linear kinematics, can also be applied here. As you can seefrom the Know/Don’t Know table in Table 8-2, we need to eliminate angular accel-eration by solving one of the rotational motion equations for a and then substitut-ing the resulting expression into the other equation. The final equation will includethe three variables of known value (v0, v, and t ) and q - q0. We can solve for q - q0 in terms of v0, v, and t , from which we can determine the number of revolutionsthe top makes before stopping.

SOLVEWe can choose to solve either of the rotational motion equations for a to eliminateangular acceleration, but it’s more straightforward to solve Equation 8-21. We canalso set the final value of angular velocity (v) to zero, as indicated in the Know/ Don’t Know table, leading to

a= -

v0

t Substituting this expression into Equation 8-22 gives

q - q0 = v0 t +1

2 a-v0

t bt 2 = v0 t - 1

2 v0 t

=1

2 v0 t

All of the variables on the right side of the resulting equationare known, so we can compute a numeric answer:

q - q0 =1

2 a25 rad

s b 142 s2 = 5.3 × 102 rad

The question asks us to express our answer in terms of revo-lutions, so we also write

q - q0 =1

2 a4 revs b 142 s2 = 84 rev

REFLECTThe result, q - q0 =

12v0 t , can be interpreted graphically.

Because the angular deceleration is constant, the angularvelocity decreased linearly, as shown in Figure 8-22a. In anyshort time interval Dt such as the one shown in the figure,

Table 8-2

Variable Know/Don’t Know

q – q0 ?

ω 0 25 rad /s

ω 0 rad /s

a X

t 42 s

(a)

(b)

KESTEN-08_275-332hr.indd 299 10/21/11

8/16/2019 Kesten Ch 8 10.21

26/58


the number of revolutions the top makes is the product of Dt and the angularvelocity v during that time (or the average angular velocity, if the time interval isnot infinitesimally short). As you can see, the product is the area of a narrow rect-angle, and adding the areas of the rectangles over the total stopping time of the top(Figure 8-22b) gives the area under the v versus t line. The shape of that area is atriangle, the area of which is 12 × base × height, or

12 v0t .

Practice Problem 8-3 A top spinning at 25 rad>s (4.0 rev>s) slows down and comesto a stop after making 36 revolutions. Assuming the top decelerates at a constantrate, how long does it take to come to a stop?

What’s Important 8-5Every quantity we use to describe linear (translational) motion has a

rotational analog. For example, the angular acceleration of a rotating object isthe derivative of angular velocity with respect to time, so it is analogous to thelinear acceleration of an object moving along a straight line.

*

8-6 TorqueWhere do you push on a door to open it easily? Perhaps without knowing why,you’ve learned that it’s easiest to open a door by pushing on a spot far from thehinges. You would never push near the hinges, such as the person in Figure 8-23a,because you learned a long time ago that it’s nearly impossible to open a door thatway. In which direction do you push on a door? The best choice is perpendicular tothe plane of the door and far from the hinges (Figure 8-23b). In this section, we willsee why opening the door depends not just on the magnitude of the applied force,but also on both the direction of the force vector and the distance between the rota-tion axis and the position at which the force is applied.

Torque t is the rotational analog of force and takes into account the distance r between a force F and the rotation axis as well as the angle j between the r s vectorand the F s vector. (The r s vector points from the rotation axis to the point at whichthe force is applied.) The magnitude of torque is

t = rF sin w (8-23)

The quantities on which torque depends are shown in Figure 8-24.Notice that F sin w on the right side of the relationship is the component of

the applied force in the direction perpendicular to the r s vector. The r and sin w

Figure 8-23 (a) It’s hard to opena door by pushing it close tothe hinges. (b) The door openseffortlessly when force isexerted far from the hinge, andperpendicular to the plane ofthe door. (a) (b)

KESTEN-08_275-332hr.indd 300 10/21/11

8/16/2019 Kesten Ch 8 10.21

27/58

8-6 Torque 301

terms taken together describe the perpendicular distance from the axis of rota-tion to the line along which the force acts. Applying a given force for a large valueof this distance, called the lever arm or moment arm, provides a mechanicaladvantage that, for example, allows you to open a massive door with relativelylittle effort.

The SI units of torque are evident from Equation 8-23:

t = r F sin w = m # N

By convention, the units are written in the opposite order, as N · m, and referred toas newton-meters.

A socket wrench can be used to loosen a bolt. A common trick to turn a boltthat has become frozen in place is to slide a section of pipe over the handle ofthe wrench and turn the bolt while gripping the end of the pipe. Why does thiswork? Why do many experienced mechanics tend to avoid using this trick?

? Got the Concept 8-8Socket Wrench

Using the definition of torque (Equation 8-23), we see that the lever arm caneither amplify or reduce the effect that an applied force has on rotating an object.When the lever arm is large (for example, when the application of the force is rela-tively far from the rotation axis), a small force generates a large torque. Humansand other animals take advantage of the power of the lever arm in the arrange-ment of muscles and bones as shown in Figure 8-25. The point at which the mus-cle is attached to the lower jawbone, for example, is far from the joint aroundwhich the jaw rotates, resulting in a torque large enough to crack a nut betweenyour back teeth. In contrast, one end of the biceps muscle attaches to the bone ofthe upper arm and the other to the lower arm just below the elbow (Figure 8-26).This muscle–joint arrangement doesn’t generate a large torque relative to the sizeof the muscle because the length of the lever arm is relatively small. The anatomy

Figure 8-25 The arrangement ofmuscles and bones takesadvantage of the power of thelever arm. The masseter muscleconnects the upper jawbone to

a point on the lower jawbonethat is rather far from the jointaround which the jaw rotates.This orientation allows you togenerate a torque large enoughto crack a macadamia nutbetween your back teeth.

Upper jaw

Lower jaw

Rotationaxis of jaw

Massetermuscle

Lever arm

φ

Rotation axis

F

r

Figure 8-24 Torque t is therotational analog of force andtakes into account the distancer between where a force F isapplied and the rotation axis.Torque also takes into accountthe angle j between the forcevector F s and the r s vector that

points from the rotation axis tothe point at which the force isapplied.

KESTEN-08_275-332hr.indd 301 10/21/11

8/16/2019 Kesten Ch 8 10.21

28/58


does provide an advantage, however; a small change in the length of the biceps

produces a large, fast movement of the hand at the end of the arm.In Section 8-5, we described a translation between quantities necessary todescribe linear motion and those used to describe rotational motion. The quanti-ties mass m and moment of inertia I play analogous roles, as do linear and angulardisplacements (x and q, respectively), linear and angular velocities (v and v),and linear and angular accelerations (a and a). We now see that force and torqueare also analogs. This conclusion allows us to write a rotational equi valent ofNewton’s second law by substituting rotational analogs for the linear quantities:

a F sdir = masdir (4-1)becomes

a t s = I αs (8-24)Don’t treat this expression as something new; the physics supporting it is stillNewton’s second law. We have simply applied Newton’s second law to rotationalmotion.

You may be surprised to find that both torque and angular acceleration arevectors. We will look more carefully at the vector nature of rotational quantities inSection 8-8; until then, we will consider only the magnitudes of the quantities.

Figure 8-26 One end of thebiceps muscle attaches to thebone of the upper arm and theother end attaches to the lowerarm just below the elbow.This muscle–joint arrangement

doesn’t generate a large torquerelative to the size of the musclebecause the length of the leverarm is relatively small. A smallchange in the length of thebiceps, however, results in alarge, fast movement of the hand.

Biceps

Lever arm

Rotation axisof lower arm

KESTEN-08_275-332hr.indd 302 10/21/11

8/16/2019 Kesten Ch 8 10.21

29/58

8-6 Torque 303

Example 8-4 A Simple Pulley

A block has a mass Mblock and is attached to one end of an inelasticthread that has negligible mass. The other end of the thread iswrapped around the circumference of a thin, uniform cylinder that

has a radius R and a mass Mcylinder and is allowed to rotate aroundits central axis as shown in Figure 8-27. At what rate, relative to g ,does the block accelerate as the thread unwinds without slipping?

SET UPThe block exerts a force on the thread which in turn exerts a forcethe edge of the cylinder. This force results in a torque on the cylin-der which causes it to rotate.

Because this example is fundamentally a force problem, let’suse the strategy developed in Chapter 4 for problems involv-ing forces. The first step is to make a free-body diagram, shownin Figure 8-28. The second step is to write an expression for thesituation using Newton’s second law and the free-body diagram. Interms of the magnitudes, the sum of the forces on the box in they direction is

Mbox g - T = Mbox a (8-25)

If we can find the tension T , this expression will provide a solutionfor the acceleration of the box.

The tension T exerts a torque on the cylinder, and we know twoways to express this torque. Using Equation 8-23,

t = RT sin w (8-26)

and using Equation 8-24,

t = I a (8-27)

We now have three equations, all of which depend on either tension, acceleration,or both. Yes, the last equation depends on angular acceleration rather than linearacceleration, so we will also need to find a way to relate these two quantities.

SOLVEBecause the thread unwinds without slipping, the linear acceleration of the edge ofthe disk is equal to the downward acceleration of the box. We can find a relation-ship between the linear acceleration and angular acceleration of the cylinder by usingEquation 8-5, which directly relates a linear quantity to a rotational quantity:

v =v

r

Using the definition of angular acceleration (Equation 8-20),

a =d v

dt =

d

dt a v

Rb = 1

R

dv

dt =

a

R

The second of the two torque equations (Equation 8-27) can now be writtenin terms of linear acceleration using a = a

>R. Together with Equation 8-26 and

Equation 8-25, we now have three equations and two unknowns, a solvable prob-lem. For convenience, here are the three equations:

t = I a

R (8-28)

t = RT sin w

Mbox g - T = Mbox a

Cylinder

Thread

Block

Figure 8-27 A block is attached to one end of aninelastic thread that has negligible mass. Theother end of the thread is wrapped around thecircumference of a thin, uniform cylinder ofnonnegligible mass that can rotate around itscentral axis.

y

R

T

T

boxM g

Figure 8-28 A free-body diagramshows the forces acting on ablock hanging from a threadwrapped around a pulley of

nonnegligible mass.

KESTEN-08_275-332hr.indd 303 10/21/11

8/16/2019 Kesten Ch 8 10.21

30/58


The first equation gives t in terms of a. The second equation gives T in terms of t(step 2 below), and the last equation gives a in terms of T (step 1 below). We’redone, except for the algebra!

1: a =Mbox g -

Mbox= g -

T

Mbox

2:

T = t R sin w

Using the expression for t from Equation 8-28, the relationship in step 2 becomes

T =I a

R2 sin w

Substituting this expression into the step 1 equation gives

a = g - I a

R2 Mbox sin w

All of the variables other than a are known. From Table 8-1, the moment of inertiaof the rotating cylinder is McylinderR

2

>2. The angle w between the direction in which

the thread is pulled and the vector that extends from the center of the cylinder tothe point at the thread comes off the cylinder is 90°. As shown in Figure 8-29,whenever an object that has been wrapped around a circular surface is pulled, thetension force is always tangential to the edge, so sin w equals 1 regardless of thedirection in which the thread is pulled. By using all this information,

a = g - aMcylinder R22 b a a

R2 Mboxb

or

a = g -Mcylinder a

2Mbox

Thread is pulled in this direction.

Thread is pulled in this direction.

The thread is tangential tothe edge of the cylinder.

If you try to change the angle that the

thread makes with the edge of the cylinder,the contact point between the thread and the edgechanges so that the thread remains tangential.

Figure 8-29 A thread is wrappedaround the circumference of acylinder. Regardless of thedirection in which the threadis pulled, the tension force isalways tangential to the edgeof the cylinder.

KESTEN-08_275-332hr.indd 304 10/21/11

8/16/2019 Kesten Ch 8 10.21

31/58

8-6 Torque 305

Gather the terms that contain a on the left side of the equation

aa1 + Mcylinder2Mbox

b = g and then solve for a

a = 2Mbox2Mbox + Mcylinder

g

REFLECTThe acceleration of the box is proportional to g . Let’s test the result for limitingcases of the mass of the cylinder. When the cylinder has negligible mass, the multi-plicative factor is 1 and a g. That is surely reasonable; when the cylinder istreated as having no mass, the box undergoes free-fall motion. A consideration ofenergy supports this result; no kinetic energy is required to rotate a cylinder thathas no mass, so all of the box’s initial potential energy transforms into translationalkinetic energy as the box falls. When Mcylinder is very large compared to Mbox, thedenominator of the multiplicative factor overwhelms the numerator and results ina fraction tending to 0, so a 0. This result also makes sense. A large mass resultsin the cylinder having a large moment of inertia around the rotation axis, so the

torque (t = I a) produces only a small angular acceleration.

The human jaw can rotate around three axes as shown in Figure 8-30. Chew-ing is accomplished primarily by rotations around the axis labeled y. The mo-ments of inertia around each of the axes have been measured. For a lower jawthat has a mass approximately equal to 0.4 kg, typical values for the momentsof inertia are I y = 3× 10

-4 kg · m2 and I z = 9× 10-4 kg · m2. Does the anato-

my of the jaw favor up–down motions or side-to-side motions? Is more or lesstorque required to open your jaw or to rotate it from side to side?

x

y

z

Figure 8-30 The human jaw rotates to some degree about the x, y, and z axes.

? Got the Concept 8-9Human Jaw

KESTEN-08_275-332hr.indd 305 10/21/11

8/16/2019 Kesten Ch 8 10.21

32/58


What’s Important 8-6Torque is the rotational analog of force. Torque depends on the magni-

tude of the applied force, and also on the lever (or moment) arm, the per-pendicular distance between the axis of rotation and the line along which theforce is applied. Torque is maximum when the force is applied in the direction

perpendicular to the direction of the lever arm.

*

8-7 Angular MomentumWhen a child runs across the playground and jumps onto a merry-go-round, theyboth rotate together (Figure 8-31). This situation seems very much like the inelasticcollisions we discussed in Section 7-3, in which a moving object collided with aninitially stationary one and the two stuck together afterward. We approached thoseinelastic collisions by demanding that linear momentum is conserved. In the colli-sion that takes place on the playground, however, the merry-go-round has zerolinear momentum both before and after the girl jumps on. It is certainly movingafterward, however. How can momentum be conserved? To resolve this issue, wewill need to define a new quantity, one similar to linear momentum that applies to

rotating objects.The rotational analog of linear momentum is angular momentum, traditionally

represented by either a lowercase or uppercase L. We will use L to represent angu-lar momentum. Using the translation between linear quantities and rotationalquantities described in Section 8-5, we can write a rotational equivalent of linearmomentum by substituting rotational analogs for the linear quantities. Just as lin-ear momentum is defined by

ps = mv s

we define angular momentum as the product of angular velocity and moment ofinertia:

Ls = I Vs (8-29)

As with torque and angular acceleration, angular momentum and angular velocity

are vectors. We will look more carefully at the vector nature of rotational quanti-ties in Section 8-8; until then, we will consider only the magnitudes of thesequantities.

To understand angular momentum L, let’s extend the analogy between linearmomentum and angular momentum to other linear quantities. We determined lin-ear momentum through its relationship to force, for example,

F s =d ps

dt (7-23)

We have seen that torque τ s is the rotational analog of force F s, so torque and angu-lar momentum are related by

τ s =d Ls

dt (8-30)

In cases in which the net torque is zero, d Ls

>dt is zero and angular momentum isconstant. In the same way that linear momentum is conserved when the net forceon a system is zero, angular momentum is conserved when the net torque on asystem is zero. In particular, angular momentum is conserved in rotating systemsthat experience no external forces or torques.

Figure 8-31 A girl has run acrossthe playground and jumped ona merry-go-round, which is freeto rotate around its center.(Lane V. Erickson/Shutterstock.

com)

KESTEN-08_275-332hr.indd 306 10/21/11

8/16/2019 Kesten Ch 8 10.21

33/58

8-7 Angular Momentum 307

In the children’s game of tetherball, a rope attached to the top of a tall pole istied to a ball. Players hit the ball in opposite directions in an attempt to wind

the rope around the pole. As the ball circles the pole, does the speed of the balldecrease, stay the same, or increase? Explain in terms of rotational kinematics.Treat the rope as having negligible mass and neglect any resistive forces.

? Got the Concept 8-10Tetherball

Example 8-5 Figure Skater

A figure skater executing a “scratch spin” gradually pulls her arms in toward herbody while spinning on one skate. During the spin her angular velocity increasesfrom 1.5 rad>s (approximately 1 revolution every 2 s) to 19 rad>s (approximately3 revolutions per 1 s). By what factor does her moment of inertia around her cen-tral axis change as she pulls in her arms?

SET UPWe treat the ice as frictionless so that the ice skater’s angular momentum can beconsidered constant. Her moment of inertia is not constant, however, because thedistribution of her mass around the rotation axis changes as she brings in her arms.The magnitude of angular momentum is then L = I startvstart (from Equation 8-29)at the start of the spin and L = I endvend at the end. The two expressions are equalbecause angular momentum is conserved.

SOLVEWe begin by setting the expressions for angular momentum at the start and end ofthe spin equal:

I start vstart = I end vend

The ratio of the moments of inertia is

I end

I start=

vstart

vend

So, the factor by which her moment of inertia around her central axis changes asshe pulls in her arms is

I end

I start=

1.5 rad s

19 rad>s 113REFLECTThe skater’s moment of inertia decreases by a factor of approximately 13. Thismight seem large, because her arms (and one leg) are a relatively small fraction ofher total mass. Moment of inertia depends on the square of distance (Equations 8-7and 8-9), however, so holding her arms and leg close to her body rather thanextended has a significant effect on the skater’s moment of inertia.

Practice Problem 8-5 A figure skater executing a scratch spin starts spinning onone skate at 1.5

rad>s (approximately 1 revolution every 2 s) and gradually pullsher arms in toward her body. If this reduces her moment of inertia by a factor offour (from I 0 to I 0>4), what is her angular velocity after she has pulled her arms incompletely?

Figure 8-32 In the children’sgame of tetherball, players hita ball at the end of a ropeattached to the top of a tallpole. Players try to wind therope around the pole by hittingthe ball in opposite directions.(Michael Newman/PhotoEdit)

KESTEN-08_275-332hr.indd 307 10/21/11

8/16/2019 Kesten Ch 8 10.21

34/58


We can extend the description of angular momentum as a rotational analog oflinear momentum. From Equation 8-30, the magnitude of torque is

t =dL

dt

Torque is generated by a force applied at a distance r from the axis around which

an object can rotate, t =

rF

sin w (Equation 8-23), so,dL

dt = rF sin w

Finally, we can apply the definition of force as the derivative of momentum withrespect to time by using Equation 7-23 to relate angular momentum to linearmomentum:

dL

dt = r

dp

dt sin w

or

dL = r dp sin w

where w is the angle between the r s and ps vectors. Integrating both sides gives

L = rp sin w (8-31)

The angular momentum that an object has as it moves with respect to a rotation

axis depends on the distance from the axis and direction

Documents

Kesten Ch 8 10.21