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9LEARNING GOALSBy studying this chapter, you will learn:
• How to describe the rotation of arigid body in terms of angular coor-dinate, angular velocity, and angularacceleration.
• How to analyze rigid-body rotationwhen the angular acceleration isconstant.
• How to relate the rotation of a rigidbody to the linear velocity and lin-ear acceleration of a point on thebody.
• The meaning of a body’s momentof inertia about a rotation axis, andhow it relates to rotational kineticenergy.
• How to calculate the moment ofinertia of various bodies.
285
ROTATION OF RIGID BODIES
?All segments of arotating helicopterblade have the sameangular velocity andangular acceleration.Compared to a givenblade segment, howmany times greater isthe linear speed of asecond segment twiceas far from the axis ofrotation? How manytimes greater is the linear acceleration?
What do the motions of a compact disc, a Ferris wheel, a circular sawblade, and a ceiling fan have in common? None of these can be repre-sented adequately as a moving point; each involves a body that
rotates about an axis that is stationary in some inertial frame of reference.Rotation occurs at all scales, from the motion of electrons in atoms to the
motions of entire galaxies. We need to develop some general methods for analyz-ing the motion of a rotating body. In this chapter and the next we consider bodiesthat have definite size and definite shape, and that in general can have rotationalas well as translational motion.
Real-world bodies can be very complicated; the forces that act on them candeform them—stretching, twisting, and squeezing them. We’ll neglect thesedeformations for now and assume that the body has a perfectly definite andunchanging shape and size. We call this idealized model a rigid body. This chap-ter and the next are mostly about rotational motion of a rigid body.
We begin with kinematic language for describing rotational motion. Next welook at the kinetic energy of rotation, the key to using energy methods for rota-tional motion. Then in Chapter 10 we’ll develop dynamic principles that relatethe forces on a body to its rotational motion.
9.1 Angular Velocity and AccelerationIn analyzing rotational motion, let’s think first about a rigid body that rotatesabout a fixed axis—an axis that is at rest in some inertial frame of reference anddoes not change direction relative to that frame. The rotating rigid body might bea motor shaft, a chunk of beef on a barbecue skewer, or a merry-go-round.
Figure 9.1 shows a rigid body (in this case, the indicator needle of aspeedometer) rotating about a fixed axis. The axis passes through point O and is
The angle u from the1x-axis specifies theneedle’s rotationalposition.
Axis of rotation passes throughorigin and points out of page.
u
x
PDirectionof needle’srotation
y
O
9.1 A speedometer needle (an example ofa rigid body) rotating counterclockwiseabout a fixed axis.
9 .1 Angular Velocity and Acceleration 287286 C HAPTE R 9 Rotation of Rigid Bodies
One radian is the angleat which the arc s hasthe same length as theradius r.
An angle u in radiansis the ratio of the arclength s to the radius r.
u 5 sr
s 5 ru
r
1 rad
s 5 r
r
(a)
(b)
9.2 Measuring angles in radians.
perpendicular to the plane of the diagram, which we choose to call the xy-plane.One way to describe the rotation of this body would be to choose a particularpoint P on the body and to keep track of the x- and y-coordinates of this point.This isn’t a terribly convenient method, since it takes two numbers (the two coor-dinates x and y) to specify the rotational position of the body. Instead, we noticethat the line OP is fixed in the body and rotates with it. The angle that this linemakes with the describes the rotational position of the body; we will usethis single quantity as a coordinate for rotation.
The angular coordinate of a rigid body rotating around a fixed axis can bepositive or negative. If we choose positive angles to be measured counterclock-wise from the positive x-axis, then the angle in Fig. 9.1 is positive. If we insteadchoose the positive rotation direction to be clockwise, then in Fig. 9.1 is nega-tive. When we considered the motion of a particle along a straight line, it wasessential to specify the direction of positive displacement along that line; whenwe discuss rotation around a fixed axis, it’s just as essential to specify the direc-tion of positive rotation.
To describe rotational motion, the most natural way to measure the angle isnot in degrees, but in radians. As shown in Fig. 9.2a, one radian (1 rad) is theangle subtended at the center of a circle by an arc with a length equal to theradius of the circle. In Fig. 9.2b an angle is subtended by an arc of length s on acircle of radius r. The value of (in radians) is equal to s divided by r:
(9.1)
An angle in radians is the ratio of two lengths, so it is a pure number, withoutdimensions. If and then but we will often writethis as 1.5 rad to distinguish it from an angle measured in degrees or revolutions.
The circumference of a circle (that is, the arc length all the way around the cir-cle) is times the radius, so there are (about 6.283) radians in one completerevolution Therefore
Similarly, and so on. If we had insisted on meas-uring the angle in degrees, we would have needed to include an extra factor of
on the right-hand side of in Eq. (9.1). By measuring angles inradians, we keep the relationship between angle and distance along an arc as sim-ple as possible.
Angular VelocityThe coordinate shown in Fig. 9.1 specifies the rotational position of a rigidbody at a given instant. We can describe the rotational motion of such a rigidbody in terms of the rate of change of We’ll do this in an analogous way to ourdescription of straight-line motion in Chapter 2. In Fig. 9.3a, a reference line OPin a rotating body makes an angle with the at time At a later time the angle has changed to We define the average angular velocity (theGreek letter omega) of the body in the time interval as the ratio ofthe angular displacement to
(9.2)vav-z 5u2 2 u1
t2 2 t15
Du
Dt
Dt:Du 5 u2 2 u1
Dt 5 t2 2 t1
vav-zu2 .t2t1 .1x-axisu1
u.
u
s 5 ru12p/360 2 u90° 5 p/2 rad,180° 5 p rad,
1 rad 5360°
2p5 57.3°
1360° 2 . 2p2p
u 5 1.5,r 5 2.0 m,s 5 3.0 m
u 5sr or s 5 ru
uu
u
uu
uu
1x-axisu
The subscript z indicates that the body in Fig. 9.3a is rotating about the z-axis,which is perpendicular to the plane of the diagram. The instantaneous angularvelocity is the limit of as approaches zero—that is, the derivative of with respect to t:
(definition of angular velocity) (9.3)
When we refer simply to “angular velocity,” we mean the instantaneous angularvelocity, not the average angular velocity.
The angular velocity can be positive or negative, depending on the direc-tion in which the rigid body is rotating (Fig. 9.4). The angular speed which wewill use extensively in Sections 9.3 and 9.4, is the magnitude of angular velocity.Like ordinary (linear) speed the angular speed is never negative.
CAUTION Angular velocity vs. linear velocity Keep in mind the distinction betweenangular velocity and ordinary velocity, or linear velocity, (see Section 2.2). If anobject has a velocity the object as a whole is moving along the x-axis. By contrast, if anobject has an angular velocity then it is rotating around the z-axis. We do not mean thatthe object is moving along the z-axis. ❚
Different points on a rotating rigid body move different distances in a giventime interval, depending on how far the point lies from the rotation axis. Butbecause the body is rigid, all points rotate through the same angle in the same time(Fig. 9.3b). Hence at any instant, every part of a rotating rigid body has the sameangular velocity. The angular velocity is positive if the body is rotating in the direc-tion of increasing and negative if it is rotating in the direction of decreasing
If the angle is in radians, the unit of angular velocity is the radian per secondOther units, such as the revolution per minute or rpm), are often
used. Since two useful conversions are
That is, is about 10 rpm.1 rad/s
1 rev/s 5 2p rad/s and 1 rev/min 5 1 rpm 52p
60 rad/s
1 rev 5 2p rad,(rev/min(rad/s).
uu.u
vz ,vx ,
vxvz
v,
v,vz
vz 5 limDtS0
Du
Dt5
du
dt
uDtvav-zvz
Angular displacementDu of the rotating needleover a time interval Dt:
Du 5 u2 2 u1
(a)
xO
u1u2
P at t1Directionof rotation
Du
P at t2
y
(a) 9.3 (a) Angular displacement of arotating body. (b) Every part of a rotatingrigid body has the same angular velocityDu/Dt.
Du
Counterclockwiserotation positive:Du . 0, sovav-z 5 Du/Dt . 0
Clockwiserotation negative:Du , 0, sovav-z 5 Du/Dt , 0
Axis of rotation (z-axis) passes throughorigin and points out of page.
Du Du
Ox
y
Ox
y
9.4 A rigid body’s average angular veloc-ity (shown here) and instantaneous angularvelocity can be positive or negative.
(b)
9 .1 Angular Velocity and Acceleration 289
shows, the direction of is given by the right-hand rule that we used to define thevector product in Section 1.10. If the rotation is about the z-axis, then has only az-component; this component is positive if is along the positive z-axis and nega-tive if is along the negative z-axis (Fig. 9.5b).
The vector formulation is especially useful in situations in which the direc-tion of the rotation axis changes. We’ll examine such situations briefly at the endof Chapter 10. In this chapter, however, we’ll consider only situations in whichthe rotation axis is fixed. Hence throughout this chapter we’ll use “angularvelocity” to refer to the component of the angular velocity vector alongthe axis.
Angular AccelerationWhen the angular velocity of a rigid body changes, it has an angular acceleration.When you pedal your bicycle harder to make the wheels turn faster or apply thebrakes to bring the wheels to a stop, you’re giving the wheels an angular accelera-tion. You also impart an angular acceleration whenever you change the rotationspeed of a piece of spinning machinery such as an automobile engine’s crankshaft.
If and are the instantaneous angular velocities at times and wedefine the average angular acceleration over the interval asthe change in angular velocity divided by (Fig. 9.6):
(9.4)
The instantaneous angular acceleration is the limit of as
(definition of angular acceleration) (9.5)
The usual unit of angular acceleration is the radian per second per second, orFrom now on we will use the term “angular acceleration” to mean the
instantaneous angular acceleration rather than the average angular acceleration.Because we can also express angular acceleration as the second
derivative of the angular coordinate:
(9.6)
You have probably noticed that we are using Greek letters for angular kine-matic quantities: for angular position, for angular velocity, and for angu-lar acceleration. These are analogous to x for position, for velocity, and foracceleration, respectively, in straight-line motion. In each case, velocity is therate of change of position with respect to time and acceleration is the rate ofchange of velocity with respect to time. We will sometimes use the terms “linearvelocity” and “linear acceleration” for the familiar quantities we defined inChapters 2 and 3 to distinguish clearly between these and the angular quantitiesintroduced in this chapter.
In rotational motion, if the angular acceleration is positive, then the angu-lar velocity is increasing; if is negative, then is decreasing. The rota-tion is speeding up if and have the same sign and slowing down if and
have opposite signs. (These are exactly the same relationships as thosebetween linear acceleration and linear velocity for straight-line motion;see Section 2.3.)
vxax
vz
azvzaz
vzazvz
az
axvx
azvzu
az 5d
dt du
dt5
d 2u
dt 2
vz 5 du/dt,
rad/s2.
az 5 limDtS0
Dvz
Dt5
dvz
dt
Dt S 0:aav-zaz
aav-z 5v2z 2 v1z
t2 2 t15
Dvz
Dt
DtDt 5 t2 2 t1aav-z
t2 ,t1v2zv1z
vS
vz ,
vS
vS
vS
vS
288 C HAPTE R 9 Rotation of Rigid Bodies
Angular Velocity As a VectorAs we have seen, our notation for the angular velocity about the z-axis is reminis-cent of the notation for the ordinary velocity along the x-axis (see Section 2.2).Just as is the x-component of the velocity vector is the z-component ofan angular velocity vector directed along the axis of rotation. As Fig. 9.5av
S
vzvS,vx
vx
vz
At t1 At t2
v1z v2z
The average angular acceleration is the changein angular velocity divided by the time interval:
aav-z 5 5 v2z 2 v1z
t2 2 t1
Dvz
Dt
9.6 Calculating the average angular accel-eration of a rotating rigid body.
(a)
... your right thumbpoints in the
direction of v.S
vS
vS
If you curl thefingers of yourright hand in the direction of rotation ...
(b)
vS
v points in thepositive z-direction:
vz . 0
S v points in thenegative z-direction:
vz , 0
S
vS
z
x
y
x
y
z
9.5 (a) The right-hand rule for the direc-tion of the angular velocity vector Reversing the direction of rotation reversesthe direction of (b) The sign of forrotation along the z-axis.
vzvS .
vS .
Example 9.1 Calculating angular velocity
The flywheel of a prototype car engine is under test. The angularposition of the flywheel is given by
The diameter of the flywheel is 0.36 m. (a) Find the angle inradians and in degrees, at times and (b) Findthe distance that a particle on the rim moves during that time inter-val. (c) Find the average angular velocity, in and in (rpm), between and (d) Find the instanta-neous angular velocity at time
SOLUTION
IDENTIFY: We need to find the values and of the angularposition at times and the angular displacement between and the distance traveled and the average angular velocitybetween and and the instantaneous angular velocity at
SET UP: We’re given the angular position as a function of time,so we can easily find our first two target variables and theangular displacement is the difference between and Given we’ll find the distance and the average angular velocityusing Eqs. (9.1) and (9.2), respectively. To find the instantaneousangular velocity, we’ll take the derivative of with respect to time,as in Eq. (9.3).
EXECUTE: (a) We substitute the values of t into the given equation:
5 1250 rad 2
360°
2p rad5 14,000°
u2 5 12.0 rad/s3 2 15.0 s 2 3 5 250 rad
5 116 rad 2
360°
2p rad5 920°
u1 5 12.0 rad/s3 2 12.0 s 2 3 5 16 rad
u
Duu2 .u1Du
u2 ;u1
u
t2 .t2 ,t1
t2 ,t1Dut2 ,t1
u2u1
t 5 t2 5 5.0 s.t2 5 5.0 s.t1 5 2.0 s
rev/minrad/s
t2 5 5.0 s.t1 5 2.0 su,
u 5 12.0 rad/s3 2 t 3
u(b) The flywheel turns through an angular displacement of
The radius r is halfthe diameter, or 0.18 m. Equation (9.1) gives
To use Eq. (9.1), the angle must be expressed in radians. We drop“radians” from the unit for s because is really a dimensionlesspure number; s is a distance and is measured in meters, the sameunit as r.
(c) In Eq. (9.2) we have
(d) We use Eq. (9.3):
At time
EVALUATE: Our result in part (d) shows that is proportional toand hence increases with time. Our numerical results are consis-
tent with this result: The instantaneous angular velocityat is greater than the average angular velocityfor the 3.0-s interval leading up to that time (from tot2 5 5.0 s).
t1 5 2.0 s78-rad/st 5 5.0 s
150-rad/st 2
vz
vz 5 16.0 rad/s3 2 15.0 s 2 2 5 150 rad/s
t 5 5.0 s,
5 16.0 rad/s3 2 t 2
vz 5du
dt5
d
dt 3 12.0 rad/s3 2 t 3 4 5 12.0 rad/s3 2 13t 2 2
5 178 rad
s 2 1 1 rev
2p rad 2 1 60 s
1 min 2 5 740 rev/min
vav-z 5u2 2 u1
t2 2 t15
250 rad 2 16 rad
5.0 s 2 2.0 s5 78 rad/s
u
s 5 ru 5 10.18 m 2 1234 rad 2 5 42 m
u2 2 u1 5 250 rad 2 16 rad 5 234 rad.Du 5
9 .2 Rotation with Constant Angular Acceleration 291290 C HAPTE R 9 Rotation of Rigid Bodies
Test Your Understanding of Section 9.1The figure shows a graph of and versus time for a particular rotating body. (a) During which time intervals is the rotation speeding up? (i)(ii) (iii) (b) During which time intervals is the rotation slowing down?(i) (ii) (iii)
❚4 s , 5 , 6 s.2 s , t , 4 s;0 , t , 2 s;
4 s , t , 6 s.2 s , t , 4 s;0 , t , 2 s;
azvz
1 2 3 4 5 6t (s)
O
az vz
Angular Acceleration As a VectorJust as we did for angular velocity, it’s useful to define an angular accelerationvector Mathematically, is the time derivative of the angular velocity vector
If the object rotates around the fixed z-axis, then has only a z-component;the quantity is just that component. In this case, is in the same direction as if the rotation is speeding up and opposite to if the rotation is slowing down(Fig. 9.7).
The angular acceleration vector will be particularly useful in Chapter 10 whenwe discuss what happens when the rotation axis can change direction. In thischapter, however, the rotation axis will always be fixed and we need use only thez-component az .
vS
vS
aS
az
aS
vS .
aS
aS.
equal to the average value for any interval. Using Eq. (9.4) with the interval from0 to t, we find
(constant angular acceleration only) (9.7)
The product is the total change in between and the later time t; theangular velocity at time t is the sum of the initial value and this totalchange.
With constant angular acceleration, the angular velocity changes at a uniformrate, so its average value between 0 and t is the average of the initial and finalvalues:
(9.8)
We also know that is the total angular displacement divided by thetime interval
(9.9)
When we equate Eqs. (9.8) and (9.9) and multiply the result by t, we get
(constant angular acceleration only) (9.10)
To obtain a relationship between and t that doesn’t contain we substituteEq. (9.7) into Eq. (9.10):
or
(constant angular acceleration only) (9.11)
That is, if at the initial time the body is at angular position and has angu-lar velocity then its angular position at any later time t is the sum of threeterms: its initial angular position plus the rotation it would have if theangular velocity were constant, plus an additional rotation caused by thechanging angular velocity.
Following the same procedure as for straight-line motion in Section 2.4, wecan combine Eqs. (9.7) and (9.11) to obtain a relationship between and thatdoes not contain t. We invite you to work out the details, following the sameprocedure we used to get Eq. (2.13). (See Exercise 9.12.) In fact, because of theperfect analogy between straight-line and rotational quantities, we can simplytake Eq. (2.13) and replace each straight-line quantity by its rotational analog.We get
(constant angular acceleration only) (9.12)
CAUTION Constant angular acceleration Keep in mind that all of these results arevalid only when the angular acceleration is constant; be careful not to try to apply themto problems in which is not constant. Table 9.1 shows the analogy between Eqs. (9.7),(9.10), (9.11), and (9.12) for fixed-axis rotation with constant angular acceleration and thecorresponding equations for straight-line motion with constant linear acceleration. ❚
az
az
vz
2 5 v0z
2 1 2az 1 u 2 u0 2
vzu
12 az t 2
v0z tu0 ,uv0z ,
u0t 5 0
u 5 u0 1 v0z t 11
2 az t 2
u 2 u0 51
2 3v0z 1 1v0z 1 az t 2 4t
vz ,u
u 2 u0 51
2 1v0z 1 vz 2 t
vav-z 5u 2 u0
t 2 0
1 t 2 0 2 : 1 u 2 u0 2vav-z
vav-z 5v0z 1 vz
2
v0zvz
t 5 0vzaz t
vz 5 v0z 1 az t
az 5vz 2 v0z
t 2 0 or
vS
aS
a and v in the samedirection: Rotationspeeding up.
SS
a and v in the oppositedirections: Rotationslowing down.
SS
vS
aS
9.7 When the rotation axis is fixed, theangular acceleration and angular velocityvectors both lie along that axis.
9.2 Rotation with Constant AngularAcceleration
In Chapter 2 we found that straight-line motion is particularly simple when theacceleration is constant. This is also true of rotational motion about a fixed axis.When the angular acceleration is constant, we can derive equations for angularvelocity and angular position using exactly the same procedure that we used forstraight-line motion in Section 2.4. In fact, the equations we are about to deriveare identical to Eqs. (2.8), (2.12), (2.13), and (2.14) if we replace x with with
and with We suggest that you review Section 2.4 before continuing.Let be the angular velocity of a rigid body at time and let be its
angular velocity at any later time t. The angular acceleration is constant andaz
vzt 5 0,v0z
az .axvz ,vxu,
Example 9.2 Calculating angular acceleration
In Example 9.1 we found that the instantaneous angular velocityof the flywheel at any time t is given by
(a) Find the average angular acceleration between and(b) Find the instantaneous angular acceleration at time
SOLUTION
IDENTIFY: This example uses the definitions of average angularacceleration and instantaneous angular acceleration
SET UP: We’ll use Eqs. (9.4) and (9.5) to find the value of between and and the value of at
EXECUTE: (a) The values of at the two times are
v2z 5 16.0 rad/s3 2 1 5.0 s 2 2 5 150 rad/s v1z 5 16.0 rad/s3 2 1 2.0 s 2 2 5 24 rad/s
vz
t 5 t2 .azt2t1
aav-z
az .aav-z
t2 5 5.0 s.t2 5 5.0 s.
t1 5 2.0 s
vz 5 16.0 rad/s3 2 t 2
vz
From Eq. (9.4) the average angular acceleration is
(b) From Eq. (9.5) the instantaneous angular acceleration at anytime t is
At time
EVALUATE: Note that the angular acceleration is not constant inthis situation. The angular velocity is always increasing because
is always positive. Furthermore, the rate at which the angularvelocity increases is itself increasing, since increases with time.az
az
vz
az 5 112 rad/s3 2 15.0 s 2 5 60 rad/s2
t 5 5.0 s,
5 112 rad/s3 2 t az 5
dvz
dt5
d
dt 3 16.0 rad/s3 2 1 t 2 2 4 5 16.0 rad/s3 2 12t 2
aav-z 5150 rad/s 2 24 rad/s
5.0 s 2 2.0 s5 42 rad/s2
7.7 Rotational Kinematics
O N L I N E
9 .3 Relating Linear and Angular Kinematics 293292 C HAPTE R 9 Rotation of Rigid Bodies
9.3 Relating Linear and Angular KinematicsHow do we find the linear speed and acceleration of a particular point in a rotat-ing rigid body? We need to answer this question to proceed with our study ofrotation. For example, to find the kinetic energy of a rotating body, we have tostart from for a particle, and this requires knowing the speed for eachparticle in the body. So it’s worthwhile to develop general relationships betweenthe angular speed and acceleration of a rigid body rotating about a fixed axis andthe linear speed and acceleration of a specific point or particle in the body.
Linear Speed in Rigid-Body RotationWhen a rigid body rotates about a fixed axis, every particle in the body moves in acircular path. The circle lies in a plane perpendicular to the axis and is centered onthe axis. The speed of a particle is directly proportional to the body’s angular veloc-ity; the faster the body rotates, the greater the speed of each particle. In Fig. 9.9,point P is a constant distance r from the axis of rotation, so it moves in a circle ofradius r. At any time, the angle (in radians) and the arc length s are related by
We take the time derivative of this, noting that r is constant for any specific parti-cle, and take the absolute value of both sides:
Now is the absolute value of the rate of change of arc length, which isequal to the instantaneous linear speed of the particle. Analogously, the absolute value of the rate of change of the angle, is the instantaneous angularspeed —that is, the magnitude of the instantaneous angular velocity in Thus
(relationship between linear and angular speeds) (9.13)
The farther a point is from the axis, the greater its linear speed. The direction ofthe linear velocity vector is tangent to its circular path at each point (Fig. 9.9).
CAUTION Speed vs. velocity Keep in mind the distinction between the linear andangular speeds and which appear in Eq. (9.13), and the linear and angular velocities
and The quantities without subscripts, and are never negative; they are the mag-nitudes of the vectors and respectively, and their values tell you only how fast a par-ticle is moving or how fast a body is rotating The corresponding quantities withsubscripts, and can be either positive or negative; their signs tell you the direction ofthe motion. ❚
Linear Acceleration in Rigid-Body RotationWe can represent the acceleration of a particle moving in a circle in terms of itscentripetal and tangential components, and (Fig. 9.10), as we did in Sec-tion 3.4. It would be a good idea to review that section now. We found that thetangential component of acceleration the component parallel to the instan-taneous velocity, acts to change the magnitude of the particle’s velocity (i.e., thespeed) and is equal to the rate of change of speed. Taking the derivative ofEq. (9.13), we find
(9.14)(tangential acceleration ofa point on a rotating body)atan 5
dvdt
5 r
dv
dt5 ra
atan ,
atanarad
vz ,vx
1v 2 .1v 2 vS ,vS
v,vvz .vx
v,v
v 5 rv
rad/s.v
0 du/dt 0 ,v0 ds/dt 0
P ds
dt P 5 r P du
dt P
s 5 ru
u
vK 5 12 mv2
Table 9.1 Comparison of Linear and Angular Motion with Constant Acceleration
Straight-Line Motion with Fixed-Axis Rotation with Constant Linear Acceleration Constant Angular Acceleration
u 2 u0 51
2 1vz 1 v0z 2 tx 2 x0 5
1
2 1vx 1 v0x 2 t
vz
2 5 v0z
2 1 2az 1 u 2 u0 2vx
2 5 v0x
2 1 2ax 1 x 2 x0 2u 5 u0 1 v0z t 1
1
2 az t 2x 5 x0 1 v0x t 1
1
2 ax t 2
vz 5 v0z 1 az tvx 5 v0x 1 ax t
az 5 constantax 5 constant
Directionof rotation
y
xQP
9.8 A line PQ on a rotating DVD at t 5 0.
Example 9.3 Rotation with constant angular acceleration
You have just finished watching a movie on DVD and the disc isslowing to a stop. The angular velocity of the disc at is
and its angular acceleration is a constant A line PQ on the surface of the disc lies along the at (Fig. 9.8). (a) What is the disc’s angular velocity at (b) What angle does the line PQ make with the at thistime?
SOLUTION
IDENTIFY: The angular acceleration of the disc is constant, so wecan use any of the equations derived in this section. Our targetvariables are the angular velocity and the angular displacement at
SET UP: We are given the initial angular velocitythe initial angle between the line PQ and the theangular acceleration and the time t 5 0.300 s.az 5 210.0 rad/s2,
1x-axis,u0 5 027.5 rad/s,v0z 5
t 5 0.300 s.
1x-axist 5 0.300 s?
t 5 01x-axis210.0 rad/s2.27.5 rad/s
t 5 0With this information it’s easiest to use Eqs. (9.7) and (9.11) to findthe target variables and respectively.
EXECUTE: (a) From Eq. (9.7), at we have
(b) From Eq. (9.11),
The DVD has turned through one complete revolution plus anadditional 0.24 revolution—that is, through an additional angle of
Hence the line PQ is at an angle ofwith the
EVALUATE: Our answer to part (a) tells us that the angularvelocity has decreased. This is as it should be, since is nega-tive. We can also use our answer for in part (a) to check ourresult for in part (b). To do so, we solve Eq. (9.12),
for the angle
which agrees with the result we found earlier.
5 0 1124.5 rad/s 2 2 2 127.5 rad/s 2 2
2 1210.0 rad/s2 2 5 7.80 rad
u 5 u0 1 1vz
2 2 v0z
2
2az2u:2az 1 u 2 u0 2 , vz
2 5 v0z
2 1uvz
az
1x-axis.87°10.24 rev 2 1 360°/rev 2 5 87°.
5 7.80 rad 5 7.80 rad 1 1 rev
2p rad 2 5 1.24 rev
5 0 1 127.5 rad/s 2 10.300 s 2 11
2 1210.0 rad/s2 2 10.300 s 2 2
u 5 u0 1 v0z t 11
2 az t 2
5 24.5 rad/s vz 5 v0z 1 az t 5 27.5 rad/s 1 1210.0 rad/s2 2 10.300 s 2
t 5 0.300 s
u,vz
Test Your Understanding of Section 9.2 Suppose the DVD in Example9.3 was initially spinning at twice the rate rather than andslowed down at twice the rate rather than (a) Com-pared to the situation in Example 9.3, how long would it take the DVD to come to a stop?(i) the same amount of time; (ii) twice as much time; (iii) 4 times as much time; (iv) asmuch time; (v) as much time. (b) Compared to the situation in Example 9.3, throughhow many revolutions would the DVD rotate before coming to a stop? (i) the same num-ber of revolutions; (ii) twice as many revolutions; (iii) 4 times as many revolutions;(iv) as many revolutions; (v) as many revolutions.
❚14
12
14
12
210.0 rad/s2 2 .1220.0 rad/s227.5 rad/s 2155.0 rad/s
Linear speed of point P(angular speed v is in rad/s)
Distance through which point P onthe body moves (angle u is in radians)
Circle followedby point P
s 5 ru
v 5 rv
r
P
u
v
v
y
Ox
9.9 A rigid body rotating about a fixedaxis through point O.
Radial and tangential acceleration components:• arad 5 v2r is point P’s centripetal acceleration.• atan 5 ra means that P’s rotation is speeding up (the body has angular acceleration).
atan 5 ra
arad 5 v2r
aS
Linearaccelerationof point P
s
v 5 rv
r
P
u
v
v
y
Ox
9.10 A rigid body whose rotation isspeeding up. The acceleration of point Phas a component toward the rotationaxis (perpendicular to and a component
along the circle that point P follows(parallel to vS).atan
vS)arad
9 .3 Relating Linear and Angular Kinematics 295294 C HAPTE R 9 Rotation of Rigid Bodies
This component of a particle’s acceleration is always tangent to the circular pathof the particle.
The quantity in Eq. (9.14) is the rate of change of the angularspeed. It is not quite the same as which is the rate of change of theangular velocity. For example, consider a body rotating so that its angular veloc-ity vector points in the (Fig. 9.5b). If the body is gaining angularspeed at a rate of per second, then But is negative andbecoming more negative as the rotation gains speed, so Therule for rotation about a fixed axis is that is equal to if is positive butequal to if is negative.
The component of the particle’s acceleration directed toward the rotationaxis, the centripetal component of acceleration is associated with thechange of direction of the particle’s velocity. In Section 3.4 we worked out therelationship We can express this in terms of by using Eq. (9.13):
(9.15)
This is true at each instant, even when and are not constant. The centripetalcomponent always points toward the axis of rotation.
The vector sum of the centripetal and tangential components of acceleration ofa particle in a rotating body is the linear acceleration (Fig. 9.10).
CAUTION Use angles in radians in all equations It’s important to remember thatEq. (9.1), is valid only when is measured in radians. The same is true of anyequation derived from this, including Eqs. (9.13), (9.14), and (9.15). When you use theseequations, you must express the angular quantities in radians, not revolutions or degrees(Fig. 9.11). ❚
Equations (9.1), (9.13), and (9.14) also apply to any particle that has the sametangential velocity as a point in a rotating rigid body. For example, when a ropewound around a circular cylinder unwraps without stretching or slipping, itsspeed and acceleration at any instant are equal to the speed and tangentialacceleration of the point at which it is tangent to the cylinder. The same princi-ple holds for situations such as bicycle chains and sprockets, belts and pulleysthat turn without slipping, and so on. We will have several opportunities to usethese relationships later in this chapter and in Chapter 10. Note that Eq. (9.15)for the centripetal component is applicable to the rope or chain only atpoints that are in contact with the cylinder or sprocket. Other points do not havethe same acceleration toward the center of the circle that points on the cylinderor sprocket have.
arad
us 5 ru,
aS
vv
(centripetal acceleration ofa point on a rotating body)arad 5
v2
r5 v2r
varad 5 v2/r.
arad ,
vz2az
vzazaaz 5 210 rad/s2.
vza 5 10 rad/s2.10 rad/s2z-direction
az 5 dvz/dt,a 5 dv/dt
In any equation that relates linear quantitiesto angular quantities, the angles MUST beexpressed in radians ...
... never in degrees or revolutions.
s 5 60r
s 5 (p/3)rRIGHT!
WRONG
y
xu 5 60° 5 p/3 rad
s 5 rur
O
9.11 Always use radians when relatinglinear and angular quantities.
Example 9.4 Throwing a discus
A discus thrower moves the discus in a circle of radius 80.0 cm. Ata certain instant, the thrower is spinning at an angular speed of
and the angular speed is increasing at Atthis instant, find the tangential and centripetal components of theacceleration of the discus and the magnitude of the acceleration.
SOLUTION
IDENTIFY: We model the discus as a particle traveling on a circularpath (Fig. 9.12a), so we can use the ideas developed in this section.
SET UP: We are given the radius the angular speedand the rate of change of angular speed
(Fig. 9.12b).The first two target variables are the accel-50.0 rad/s2.a 5v 5 10.0 rad/s,
r 5 0.800 m,
50.0 rad/s2.10.0 rad/s
eration components and which we’ll find with Eqs. (9.14)and (9.15), respectively. Given these components of the accelera-tion vector, we’ll find its magnitude (the third target variable)using the Pythagorean theorem.
EXECUTE: From Eqs. (9.14) and (9.15),
The magnitude of the acceleration vector is
a 5 "atan
2 1 arad
2 5 89.4 m/s2
arad 5 v2r 5 110.0 rad/s 2 2 10.800 m 2 5 80.0 m/s2
atan 5 ra 5 10.800 m 2 150.0 rad/s2 2 5 40.0 m/s2
a
arad ,atan
(a) (b)
r
aarad
atan
9.12 (a) Whirling a discus in a circle. (b) Our sketch showing the acceleration components for the discus.
? EVALUATE: Note that we dropped the unit “radian” from ourresults for and a. We can do this because “radian” is adimensionless quantity.
The magnitude a is about nine times g, the acceleration due togravity. Can you show that if the angular speed doubles to
arad ,atan ,while remains the same, the acceleration magnitude a
increases to or almost 33g?322 m/s2,a20.0 rad/s
rvplane 5 75.0 m/s
2400 rev/minvtan 5 rv
(a) (b)
9.13 (a) A propeller-driven airplane in flight. (b) Our sketch showing the velocity components for the propeller tip.
Example 9.5 Designing a propeller
You are asked to design an airplane propeller to turn at 2400 rpm.The forward airspeed of the plane is to be orabout and the speed of the tips of the propeller bladesthrough the air must not exceed (Fig. 9.13a). (This isabout 0.80 times the speed of sound in air. If the propeller tipswere to move too close to the speed of sound, they would producea tremendous amount of noise.) (a) What is the maximum radiusthe propeller can have? (b) With this radius, what is the accelera-tion of the propeller tip?
SOLUTION
IDENTIFY: The object of interest in this example is a particle atthe tip of the propeller; our target variables are the particle’s dis-tance from the axis and its acceleration. Note that the speed ofthis particle through the air (which cannot exceed isdue to both the propeller’s rotation and the forward motion of theairplane.
270 m/s)
270 m/s168 mi/h 2 , 1270 km/h,75.0 m/s
SET UP: As Fig. 9.13b shows, the velocity of a particle at thepropeller tip is the vector sum of its tangential velocity due to thepropeller’s rotation (magnitude given by Eq. (9.13)) andthe forward velocity of the airplane (magnitudeThe rotation plane of the propeller is perpendicular to the directionof flight, so these two vectors are perpendicular and we can usethe Pythagorean theorem to relate and to We willthen set and solve for the radius r. Note that theangular speed of the propeller is constant, so the acceleration ofthe propeller tip has only a radial component; we’ll find it usingEq. (9.15).
EXECUTE: We first convert to (see Fig. 9.11):
5 251 rad/s
v 5 2400 rpm 5 12400 rev
min 2 12p rad
1 rev 2 11 min
60 s 2rad/sv
vtip 5 270 m/svtip .vplanevtan
vplane 5 75.0 m/s).vtan ,
vStip
Continued
9 .4 Energy in Rotational Motion 297
same plane, so we specify that is the perpendicular distance from the axis tothe ith particle.
When a rigid body rotates about a fixed axis, the speed of the ith particle isgiven by Eq. (9.13), where is the body’s angular speed. Different par-ticles have different values of r, but is the same for all (otherwise, the bodywouldn’t be rigid). The kinetic energy of the ith particle can be expressed as
The total kinetic energy of the body is the sum of the kinetic energies of all itsparticles:
Taking the common factor out of this expression, we get
The quantity in parentheses, obtained by multiplying the mass of each particle bythe square of its distance from the axis of rotation and adding these products, isdenoted by I and is called the moment of inertia of the body for this rotation axis:
(9.16)
The word “moment” means that I depends on how the body’s mass is distributedin space; it has nothing to do with a “moment” of time. For a body with a givenrotation axis and a given total mass, the greater the distance from the axis to theparticles that make up the body, the greater the moment of inertia. In a rigid body,the distances are all constant and I is independent of how the body rotatesaround the given axis. The SI unit of moment of inertia is the
In terms of moment of inertia I, the rotational kinetic energy K of a rigidbody is
(rotational kinetic energy of a rigid body) (9.17)
The kinetic energy given by Eq. (9.17) is not a new form of energy; it’s simplythe sum of the kinetic energies of the individual particles that make up the rotat-ing rigid body. To use Eq. (9.17), must be measured in radians per second, notrevolutions or degrees per second, to give K in joules. That’s because we used
in our derivation.Equation (9.17) gives a simple physical interpretation of moment of inertia:
The greater the moment of inertia, the greater the kinetic energy of a rigid bodyrotating with a given angular speed We learned in Chapter 6 that the kineticenergy of a body equals the amount of work done to accelerate that body fromrest. So the greater a body’s moment of inertia, the harder it is to start the bodyrotating if it’s at rest and the harder it is to stop its rotation if it’s already rotating(Fig. 9.15). For this reason, I is also called the rotational inertia.
The next example shows how changing the rotation axis can affect the valueof I.
v.
vi 5 ri v
v
K 51
2 Iv2
1kg # m2 2 . kilogram-meter2ri
(definition ofmoment of inertia)
I 5 m1 r1
2 1 m2 r2
2 1 c5 ai
mi ri
2
K 51
2 1m1
r1
2 1 m2
r2
2 1 c2v2 51
2 1a
i
mi
ri
2 2v2
v2/2
K 51
2 m1 r1
2 v2 1
1
2 m2 r2
2v2 1 c5 ai
1
2 mi ri
2 v2
1
2 mi vi
2 51
2 mi ri
2 v2
vvvi 5 ri v,
vi
ri
296 C HAPTE R 9 Rotation of Rigid Bodies
9.4 Energy in Rotational MotionA rotating rigid body consists of mass in motion, so it has kinetic energy. As wewill see, we can express this kinetic energy in terms of the body’s angular speedand a new quantity, called moment of inertia, that depends on the body’s massand how the mass is distributed.
To begin, we think of a body as being made up of a large number of particles,with masses at distances from the axis of rotation. Welabel the particles with the index i: The mass of the ith particle is and its dis-tance from the axis of rotation is The particles don’t necessarily all lie in theri .
mi
r2 , cr1 ,m1 , m2 , c
Conceptual Example 9.6 Bicycle gears
How are the angular speeds of the two bicycle sprockets inFig. 9.14 related to the number of teeth on each sprocket?
SOLUTION
The chain does not slip or stretch, so it moves at the same tangen-tial speed on both sprockets. From Eq. (9.13),
The angular speed is inversely proportional to the radius. This rela-tionship also holds for pulleys connected by a belt, provided thebelt doesn’t slip. For chain sprockets the teeth must be equallyspaced on the circumferences of both sprockets for the chain tomesh properly with both. Let and be the numbers of teeth;the condition that the tooth spacing is the same on both sprockets is
Combining this with the other equation, we get
vrear
vfront5
Nfront
Nrear
2prfront
Nfront5
2prrear
Nrear or
rfront
rrear5
Nfront
Nrear
NrearNfront
v 5 rfront vfront 5 rrear vrear so vrear
vfront5
rfront
rrear
v
The angular speed of each sprocket is inversely proportional to thenumber of teeth. On a multispeed bike, you get the highest angularspeed of the rear wheel for a given pedaling rate whenthe ratio is maximum; this means using the largest-radius front sprocket (largest and the smallest-radius rearsprocket (smallest Nrear).
Nfront)Nfront/Nrear
vfrontvrear
Rearsprocket
Front sprocket
v
v
vrear
vfront
rrear
rfront
9.14 The sprockets and chain of a bicycle.
Test Your Understanding of Section 9.3 Information is stored on a CD orDVD (see Fig. 9.8) in a coded pattern of tiny pits. The pits are arranged in a trackthat spirals outward toward the rim of the disc. As the disc spins inside a player, thetrack is scanned at a constant linear speed. How must the rotation speed of the disc changeas the player’s scanning head moves over the track? (i) The rotation speed must increase.(ii) The rotation speed must decrease. (iii) The rotation speed must stay the same.
❚
(a) From Fig. 9.13b and Eq. (9.13), the velocity magnitude is given by
If the propeller radius is
r 5"1270 m/s 2 2 2 175.0 m/s 2 2
251 rad/s5 1.03 m
vtip 5 270 m/s,
r 2 5vtip
2 2 vplane
2
v2 and r 5"vtip
2 2 vplane
2
v
vtip
2 5 vplane
2 1 vtan
2 5 vplane
2 1 r 2v2 so
vtotal (b) The centripetal acceleration is
The tangential acceleration is zero because the angular speed isconstant.
EVALUATE: From the propeller must exert a force ofon each kilogram of material at its tip! This is why
propellers are made out of tough material, usually aluminum alloy.6.5 3 104 N
gFS
5 maS,
5 1251 rad/s 2 2 11.03 m 2 5 6.5 3 104 m/s2
arad 5 v2r
Rotation axis
Rotation axis
• Mass farther from axis• Greater moment of inertia• Harder to start apparatus rotating
• Mass close to axis• Small moment of inertia• Easy to start apparatus rotating
9.15 An apparatus free to rotate around avertical axis. To vary the moment of iner-tia, the two equal-mass cylinders can belocked into different positions on the hori-zontal shaft.
7.7 Rotational Inertia
O N L I N E
9 .4 Energy in Rotational Motion 299298 C HAPTE R 9 Rotation of Rigid Bodies
CAUTION Moment of inertia depends on the choice of axis The results of parts (a)and (b) of Example 9.7 show that the moment of inertia of a body depends on the locationand orientation of the axis. It’s not enough to just say, “The moment of inertia of this bodyis We have to be specific and say, “The moment of inertia of this bodyabout the axis through B and C is ” ❚
In Example 9.7 we represented the body as several point masses, and we eval-uated the sum in Eq. (9.16) directly. When the body is a continuous distributionof matter, such as a solid cylinder or plate, the sum becomes an integral, and weneed to use calculus to calculate the moment of inertia. We will give severalexamples of such calculations in Section 9.6; meanwhile, Table 9.2 givesmoments of inertia for several familiar shapes in terms of their masses anddimensions. Each body shown in Table 9.2 is uniform; that is, the density has thesame value at all points within the solid parts of the body.
CAUTION Computing the moment of inertia You may be tempted to try to com-pute the moment of inertia of a body by assuming that all the mass is concentrated at thecenter of mass and multiplying the total mass by the square of the distance from the centerof mass to the axis. Resist that temptation; it doesn’t work! For example, when a uniformthin rod of length L and mass M is pivoted about an axis through one end, perpendicular tothe rod, the moment of inertia is [case (b) in Table 9.2]. If we took the mass asconcentrated at the center, a distance from the axis, we would obtain the incorrectresult ❚
Now that we know how to calculate the kinetic energy of a rotating rigid body,we can apply the energy principles of Chapter 7 to rotational motion. Here aresome points of strategy and some examples.
I 5 M 1L/2 2 2 5 ML2/4.L/2
I 5 ML2/3
0.048 kg # m2.0.048 kg # m2.”
Example 9.7 Moments of inertia for different rotation axes
An engineer is designing a machine part consisting of three heavydisks linked by lightweight struts (Fig. 9.16). (a) What is themoment of inertia of this body about an axis through the center ofdisk A, perpendicular to the plane of the diagram? (b) What is themoment of inertia about an axis through the centers of disks B andC? (c) If the body rotates about an axis through A perpendicular tothe plane of the diagram, with angular speed whatis its kinetic energy?
SOLUTION
IDENTIFY: We’ll consider the disks as massive particles and thelightweight struts as massless rods. Then we can use the ideas of
v 5 4.0 rad/s,
this section to calculate the moment of inertia of this collection ofthree particles.
SET UP: In parts (a) and (b), we’ll use Eq. (9.16) to find themoments of inertia for each of the two axes. Given the moment ofinertia for axis A, we’ll use Eq. (9.17) in part (c) to find the rota-tional kinetic energy.
EXECUTE: (a) The particle at point A lies on the axis. Its distance rfrom the axis is zero, so it contributes nothing to the moment ofinertia. Equation (9.16) gives
(b) The particles at B and C both lie on the axis, so for themand neither contributes to the moment of inertia. Only A
contributes, and we have
(c) From Eq. (9.17),
EVALUATE: Our results show that the moment of inertia for theaxis through A is greater than that for the axis through B and C.Hence, of the two axes, it’s easier to make the machine part rotateabout the axis through B and C.
K 51
2 Iv2 5
1
2 10.057 kg # m2 2 14.0 rad/s 2 2 5 0.46 J
I 5 ami ri
2 5 10.30 kg 2 1 0.40 m 2 2 5 0.048 kg # m2
r 5 0
5 0.057 kg # m2
I 5 ami ri
2 5 10.10 kg 2 10.50 m 2 2 1 10.20 kg 2 10.40 m 2 2
Axis passingthrough disk A
0.50 m
mB 5 0.10 kg
mC 5 0.20 kg
mA 5 0.30 kg
0.30 m
0.40 mA
B
C
Axis passing through disks B and C
9.16 An oddly shaped machine part.
L
(a) Slender rod, axis through center
(b) Slender rod, axis through one end
(c) Rectangular plate, axis through center
(d) Thin rectangular plate, axis along edge
I 5 ML2
Lb
a
b
a
R1
R2R R R R
I 5 MR2
(e) Hollow cylinder (f) Solid cylinder (g) Thin-walled hollow cylinder
(i) Thin-walled hollow sphere
(h) Solid sphere
112
I 5 M 1a2 � b2 2112
I 5 M 1R12 1 R2
2 212
I 5 MR212
I 5 MR225
I 5 MR223
I 5 ML213
I 5 Ma213
Table 9.2 Moments of Inertia of Various Bodies
Problem-Solving Strategy 9.1 Rotational Energy
IDENTIFY the relevant concepts: You can use work–energy rela-tionships and conservation of energy to find relationships involv-ing position and motion of a rigid body rotating around a fixedaxis. As we saw in Chapter 7, the energy method is usually nothelpful for problems that involve elapsed time. In Chapter 10 we’llsee how to approach rotational problems of this kind.
SET UP the problem using the same steps as in Problem-SolvingStrategy 7.1 (Section 7.1), with the following addition:5. Many problems involve a rope or cable wrapped around a
rotating rigid body, which functions as a pulley. In these situa-tions, remember that the point on the pulley that contacts therope has the same linear speed as the rope, provided the ropedoesn’t slip on the pulley. You can then take advantage ofEqs. (9.13) and (9.14), which relate the linear speed and tan-gential acceleration of a point on a rigid body to the angularvelocity and angular acceleration of the body. Examples 9.8and 9.9 illustrate this point.
EXECUTE the solution: As in Chapter 7, write expressions for theinitial and final kinetic and potential energies and and the nonconservative work (if any). The new feature isrotational kinetic energy, which is expressed in terms of the body’smoment of inertia I for the given axis and its angular speed
instead of its mass m and speed Substitute theseexpressions into (if nonconservativework is done) or (if only conservative work isdone) and solve for the target variable(s). As in Chapter 7, it’shelpful to draw bar graphs showing the initial and final values ofK, U, and
EVALUATE your answer: As always, check whether your answermakes physical sense.
E 5 K 1 U.
K1 1 U1 5 K2 1 U2
K1 1 U1 1 Wother 5 K2 1 U2
v.v AK 5 12 Iv2B
Wother
U2 2U1 ,K2 ,1K1 ,
9 .5 Parallel-Axis Theorem 301300 C HAPTE R 9 Rotation of Rigid Bodies
Gravitational Potential Energy for an Extended BodyIn Example 9.9 the cable was of negligible mass, so we could ignore its kineticenergy as well as the gravitational potential energy associated with it. If the massis not negligible, we need to know how to calculate the gravitational potentialenergy associated with such an extended body. If the acceleration of gravity g isthe same at all points on the body, the gravitational potential energy is the sameas though all the mass were concentrated at the center of mass of the body. Sup-pose we take the y-axis vertically upward. Then for a body with total mass M, thegravitational potential energy U is simply
(gravitational potential energy for an extended body) (9.18)
where is the y-coordinate of the center of mass. This expression applies to anyextended body, whether it is rigid or not (Fig. 9.19).
To prove Eq. (9.18), we again represent the body as a collection of mass elementsThe potential energy for element is so the total potential energy is
But from Eq. (8.28), which defines the coordinates of the center of mass,
where is the total mass. Combining this with the aboveexpression for U, we find in agreement with Eq. (9.18).
We leave the application of Eq. (9.18) to the problems. We’ll make use of thisrelationship in Chapter 10 in the analysis of rigid-body problems in which theaxis of rotation moves.
U 5 Mgycm
M 5 m1 1 m2 1 c
m1 y1 1 m2 y2 1 c5 1m1 1 m2 1 c2 ycm 5 Mycm
U 5 m1 gy1 1 m2 gy2 1 c5 1m1 y1 1 m2 y2 1 c2gmi gyi ,mimi .
ycm
U 5 Mgycm
(a) (b)
9.18 Our sketches for this problem.
Example 9.9 An unwinding cable II
We wrap a light, flexible cable around a solid cylinder with massM and radius R. The cylinder rotates with negligible friction abouta stationary horizontal axis. We tie the free end of the cable to ablock of mass m and release the object with no initial velocity at adistance h above the floor. As the block falls, the cable unwindswithout stretching or slipping, turning the cylinder. Find the speedof the falling block and the angular speed of the cylinder just as theblock strikes the floor.
SOLUTION
IDENTIFY: As in Example 9.8, the cable doesn’t slip and frictiondoes no work. The cable does no net work; at its upper end theforce and displacement are in the same direction, and at its lowerend they are in opposite directions. Thus the total work done by thetwo ends of the cable is zero. Hence only gravity does work, andso mechanical energy is conserved.
SET UP: Figure 9.18a shows the situation just before the blockbegins to fall. At this point the system has no kinetic energy, so
0.120 m50 kg
9.0 N2.0 m
9.17 A cable unwinds from a cylinder (side view).
We take the potential energy to be zero when the block isat floor level; then and (We can ignore the grav-itational potential energy for the rotating cylinder, since its heightdoesn’t change.) Just before the block hits the floor (Fig. 9.18b),both the block and the cylinder have kinetic energy. The totalkinetic energy at that instant is
From Table 9.2 the moment of inertia of the cylinder is Also, and are related by since the speed of the fallingblock must be equal to the tangential speed at the outer surface ofthe cylinder. We’ll use these relationships to solve for the targetvariables and shown in Fig. 9.18b.vv
v 5 Rv,vvI 5 1
2 MR2.
K2 51
2 mv2 1
1
2 Iv2
K2
U2 5 0.U1 5 mghK1 5 0. EXECUTE: We use our expressions for and and the
relationship in the energy-conservation equation We then solve for
The final angular speed of the cylinder is
EVALUATE: Let’s check some particular cases. When M is muchlarger than m, is very small, as we would expect. When M ismuch smaller than m, is nearly equal to which is thespeed of a body that falls freely from an initial height h. Does itsurprise you that doesn’t depend on the radius of the cylinder?v
!2gh ,vv
v 5 v/R.
v 5 Å2gh
1 1 M/2m
0 1 mgh 51
2 mv2 1
1
2 112 MR2 2 1 vR 2 2
1 0 51
2 1m 1
1
2 M 2v2
v:U1 5 K2 1 U2 .K1 1v 5 v/R
U2K2 ,U1 ,K1 ,
Test Your Understanding of Section 9.4 Suppose the cylinder and blockin Example 9.9 have the same mass, so Just before the block strikes thefloor, which statement is correct about the relationship between the kinetic energyof the falling block and the rotational kinetic energy of the cylinder? (i) The block hasmore kinetic energy than the cylinder. (ii) The block has less kinetic energy than thecylinder. (iii) The block and the cylinder have equal amounts of kinetic energy.
❚
m 5 M.
9.5 Parallel-Axis TheoremWe pointed out in Section 9.4 that a body doesn’t have just one moment ofinertia. In fact, it has infinitely many, because there are infinitely many axesabout which it might rotate. But there is a simple relationship between themoment of inertia of a body of mass M about an axis through its center ofIcm
cm
9.19 In a technique called the “Fosburyflop” after its innovator, this athlete archeshis body as he passes over the bar in thehigh jump. As a result, his center of massactually passes under the bar. This tech-nique requires a smaller increase in gravi-tational potential energy [Eq. (9.18)] thanthe older method of straddling the bar.
Example 9.8 An unwinding cable I
A light, flexible, nonstretching cable is wrapped several timesaround a winch drum, a solid cylinder of mass 50 kg and diameter0.120 m, which rotates about a stationary horizontal axis held byfrictionless bearings (Fig. 9.17). The free end of the cable is pulledwith a constant 9.0-N force for a distance of 2.0 m. It unwindswithout slipping and turns the cylinder. If the cylinder is initially atrest, find its final angular speed and the final speed of the cable.
SOLUTION
IDENTIFY: We will solve this problem using energy methods.Point 1 is when the cylinder first begins to move, and point 2 iswhen the cable has moved 2.0 m. We’ll assume that the light cableis massless, so that only the cylinder has kinetic energy. The cylin-der doesn’t move vertically, so there are no changes in gravita-tional potential energy. There is friction between the cable and thecylinder, which is what makes the cylinder rotate when the cable ispulled. But because the cable doesn’t slip, there is no sliding of thecable relative to the cylinder and no mechanical energy is lost infriction. Because the cable is massless, the force that the cableexerts on the cylinder rim is equal to the applied force F.
SET UP: The cylinder starts at rest, so the initial kinetic energy isBetween points 1 and 2 the force F does work on the
cylinder over a distance As a result, the kinetic energyat point 2 is One of our target variables is the otheris the speed of the cable at point 2, which is equal to the tangentialspeed of the cylinder at that point. We’ll find from by usingEq. (9.13).
EXECUTE: The work done on the cylinder is From Table 9.2 the moment of inertia is
(The radius R is half the diameter of the cylinder.) The relationshipthen gives
The final tangential speed of the cylinder, and hence the final speedof the cable, is
EVALUATE: If the mass of the cable can’t be neglected, then someof the work done would go into the kinetic energy of the cable.Hence the cylinder would end up with less kinetic energy and asmaller angular speed than we calculated here.
v 5 Rv 5 10.060 m 2 120 rad/s 2 5 1.2 m/s
5 20 rad/s v 5 Å
2Wother
I5 Å
2 118 J 20.090 kg # m2
0 1 0 1 Wother 51
2 Iv2 1 0
K1 1 U1 1 Wother 5 K2 1 U2
I 51
2 mR2 5
1
2 150 kg 2 10.060 m 2 2 5 0.090 kg # m2
19.0 N 2 1 2.0 m 2 5 18 J.Wother 5 Fs 5
vvv
v;K2 5 12 Iv2.
s 5 2.0 m.K1 5 0.
7.12 Woman and Flywheel Elevator—Energy Approach
7.13 Rotoride—Energy Approach
O N L I N E
9 .6 Moment-of-Inertia Calculations 303302 C HAPTE R 9 Rotation of Rigid Bodies
mass and the moment of inertia about any other axis parallel to the originalone but displaced from it by a distance d. This relationship, called the parallel-axis theorem, states that
(parallel-axis theorem) (9.19)
To prove this theorem, we consider two axes, both parallel to the z-axis, onethrough the center of mass and the other through a point P (Fig. 9.20). First we takea very thin slice of the body, parallel to the xy-plane and perpendicular to the z-axis.We take the origin of our coordinate system to be at the center of mass of the body;the coordinates of the center of mass are then The axisthrough the center of mass passes through this thin slice at point O, and the parallelaxis passes through point P, whose x- and y-coordinates are The distanceof this axis from the axis through the center of mass is d, where
We can write an expression for the moment of inertia about the axisthrough point P. Let be a mass element in our slice, with coordinates
Then the moment of inertia of the slice about the axis through the centerof mass (at O) is
The moment of inertia of the slice about the axis through P is
These expressions don’t involve the coordinates measured perpendicular to theslices, so we can extend the sums to include all particles in all slices. Then becomes the moment of inertia of the entire body for an axis through P. We thenexpand the squared terms and regroup, and obtain
The first sum is From Eq. (8.28), the definition of the center of mass, the sec-ond and third sums are proportional to and these are zero because wehave taken our origin to be the center of mass. The final term is multiplied bythe total mass, or This completes our proof that
As Eq. (9.19) shows, a rigid body has a lower moment of inertia about an axisthrough its center of mass than about any other parallel axis. Thus it’s easier tostart a body rotating if the rotation axis passes through the center of mass. Thissuggests that it’s somehow most natural for a rotating body to rotate about an axisthrough its center of mass; we’ll make this idea more quantitative in Chapter 10.
IP 5 Icm 1 Md 2.Md 2.d 2
ycm ;xcm
Icm .
IP 5 ai
mi 1 xi
2 1 yi
2 2 2 2aai
mi xi 2 2bai
mi yi 1 1a2 1 b2 2 ai
mi
IP
zi
IP 5 ai
mi 3 1 xi 2 a 2 2 1 1 yi 2 b 2 2 4
Icm 5 ai
mi 1 xi
2 1 yi
2 2Icmzi 2 .
yi ,1 xi ,mi
IP
d 2 5 a2 1 b2.1a, b 2 .
xcm 5 ycm 5 zcm 5 0.
IP 5 Icm 1 Md 2
IP
*9.6 Moment-of-Inertia CalculationsNOTE: This optional section is for students who are familiar with integralcalculus.If a rigid body is a continuous distribution of mass—like a solid cylinder or asolid sphere—it cannot be represented by a few point masses. In this case thesum of masses and distances that defines the moment of inertia [Eq. (9.16)]becomes an integral. Imagine dividing the body into elements of mass dm that arevery small, so that all points in a particular element are at essentially the sameperpendicular distance from the axis of rotation. We call this distance r, as before.Then the moment of inertia is
(9.20)
To evaluate the integral, we have to represent r and dm in terms of the sameintegration variable. When the object is effectively one-dimensional, such as theslender rods (a) and (b) in Table 9.2, we can use a coordinate x along the length andrelate dm to an increment dx. For a three-dimensional object it is usually easiest toexpress dm in terms of an element of volume dV and the density of the body.Density is mass per unit volume, so we may also write Eq. (9.20) as
This expression tells us that a body’s moment of inertia depends on how its den-sity varies within its volume (Fig. 9.22). If the body is uniform in density, thenwe may take outside the integral:
(9.21)
To use this equation, we have to express the volume element dV in terms of thedifferentials of the integration variables, such as The element dVmust always be chosen so that all points within it are at very nearly the same dis-tance from the axis of rotation. The limits on the integral are determined by theshape and dimensions of the body. For regularly shaped bodies, this integration isoften easy to do.
dV 5 dx dy dz.
I 5 r3r 2 dV
r
I 5 3r 2r dV
r 5 dm/dV,r
I 5 3r 2 dm
Mass element mi
Axis of rotation passing through cm andperpendicular to the plane of the figure
Second axis of rotationparallel to the one through the cm
y
yi � b
xix
miyi
xi � a
b
a
dP
cmO
Slice of a body of mass M
9.20 The mass element has coordinateswith respect to an axis of rotation
through the center of mass (cm) and coor-dinates with respect to theparallel axis through point P.
yi 2 b 21 xi 2 a,
1 xi, yi 2 mi
Axis throughcenter of mass
Axis through P
cm
0.15 m
P
9.21 Calculating from a measurement of IP .Icm
Example 9.10 Using the parallel-axis theorem
A part of a mechanical linkage (Fig. 9.21) has a mass of 3.6 kg. Wemeasure its moment of inertia about an axis 0.15 m from its centerof mass to be What is the moment of inertia about a parallel axis through the center of mass?
IcmIP 5 0.132 kg # m2.
SOLUTION
IDENTIFY: The parallel-axis theorem allows us to relate themoments of inertia and through the two parallel axes.
SET UP: We’ll use Eq. (9.19) to determine the target variable
EXECUTE: Rearranging the equation and substituting the values,
EVALUATE: Our result shows that is less than This is as itshould be: As we saw earlier, the moment of inertia for an axisthrough the center of mass is lower than for any other parallel axis.
IP .Icm
5 0.051 kg # m2
Icm 5 IP 2 Md 2 5 0.132 kg # m2 2 13.6 kg 2 1 0.15 m 2 2
Icm .
IPIcm
Test Your Understanding of Section 9.5 A pool cue is a wooden rod with a uni-form composition and tapered with a larger diameter at one end than at the other end. Usethe parallel-axis theorem to decide whether a pool cue has a larger moment of inertia(i) for an axis through the thicker end of the rod and perpendicular to the length of therod, or (ii) for an axis through the thinner end of the rod and perpendicular to the lengthof the rod.
❚
9.22 By measuring small variations in theorbits of satellites, geophysicists can meas-ure the earth’s moment of inertia. This tellsus how our planet’s mass is distributedwithin its interior. The data show that theearth is far denser at the core than in itsouter layers.
Example 9.11 Uniform thin rod, axis perpendicular to length
Figure 9.23 shows a slender uniform rod with mass M and lengthL. It might be a baton held by a twirler in a marching band (less therubber end caps). Compute its moment of inertia about an axisthrough O, at an arbitrary distance h from one end.
SOLUTION
IDENTIFY: The rod is a continuous distribution of mass, so wemust use integration to find the moment of inertia. We choose as anelement of mass a short section of rod with length dx at a distancex from point O.
Mass element: rodsegment of length dx
xxM
O
hL � hL
Axis
dx
9.23 Finding the moment of inertia of a thin rod about an axisthrough O.
9 .6 Moment-of-Inertia Calculations 305
Example 9.12 Hollow or solid cylinder, rotating about axis of symmetry
Figure 9.24 shows a hollow, uniform cylinder with length L, innerradius and outer radius It might be a steel cylinder in aprinting press or a sheet-steel rolling mill. Find the moment ofinertia about the axis of symmetry of the cylinder.
SOLUTION
IDENTIFY: Again we must use integration to find the moment ofinertia, but now we choose as a volume element a thin cylindricalshell of radius r, thickness dr, and length L. All parts of this ele-ment are at very nearly the same distance from the axis.
SET UP: The volume of the element is very nearly equal to that ofa flat sheet with thickness dr, length L, and width (the circum-ference of the shell). Then
We will use this expression in Eq. (9.20) and integrate from to
EXECUTE: The moment of inertia is given by
It is usually more convenient to express the moment of inertia interms of the total mass M of the body, which is its density multi-plied by the total volume V. The volume is
so the total mass M is
M 5 rV 5 pLr 1R2
2 2 R1
2 2V 5 pL 1R2
2 2 R1
2 2r
5prL
2 1R2
2 2 R1
2 2 1R2
2 1 R1
2 2 5
2prL
4 1R2
4 2 R1
4 2 5 2prL 3
R2
R1
r 3 dr
I 5 3r 2 dm 5 3R2
R1
r 2r 12prL dr 2
r 5 R2 .r 5 R1
dm 5 r dV 5 r 12prL dr 22pr
R2 .R1 ,
Hence the moment of inertia is
EVALUATE: This agrees with Table 9.2, case (e). If the cylinder issolid, such as a lawn roller, Calling the outer radius simply R, we find that the moment of inertia of a solid cylinder ofradius R is
If the cylinder has a very thin wall (like a pipe), and are verynearly equal; if R represents this common radius,
We could have predicted this last result; in a thin-walled cylinder,all the mass is the same distance from the axis, so ∫r 2 dm 5 R2 ∫ dm 5 MR2.
I 5r 5 R
I 5 MR2
R2R1
I 51
2 MR2
R2R1 5 0.
I 51
2 M 1R1
2 1 R2
2 2
304 C HAPTE R 9 Rotation of Rigid Bodies
SET UP: The ratio of the mass dm of an element to the total massM is equal to the ratio of its length dx to the total length L:
We’ll determine I from Eq. (9.20) with r replaced by x (seeFig. 9.23).
EXECUTE: Figure 9.23 shows that the integration limits on x arefrom to Hence we obtain
5 SML 1 x3
3 2 TL2h
2h
51
3 M 1L2 2 3Lh 1 3h2 2
I 5 3x2 dm 5M
L 3
L2h
2h
x2 dx
1L 2 h 2 .2h
dm
M5
dx
L so dm 5
M
L dx
Mass element:cylindrical shellwith radius r andthickness dr
Axis
drr
L
R1
R2
9.24 Finding the moment of inertia of a hollow cylinder about itssymmetry axis.
EVALUATE: From this general expression we can find the momentof inertia about an axis through any point on the rod. For example,if the axis is at the left end, and
If the axis is at the right end, we should get the same result. Puttingwe again get
If the axis passes through the center, the usual place for a twirledbaton, then and
These results agree with the expressions in Table 9.2.
I 51
12 ML2
h 5 L/2
I 51
3 ML2
h 5 L,
I 51
3 ML2
h 5 0
Example 9.13 Uniform sphere with radius R, axis through center
Find the moment of inertia of a solid, uniform sphere (like a bil-liard ball or ball bearing) about an axis through its center.
SOLUTION
IDENTIFY: To calculate the moment of inertia we divide thesphere into thin disks of thickness dx (Fig. 9.25), whose moment ofinertia we know from Example 9.12. We’ll integrate over these tofind the total moment of inertia. The only tricky point is that theradius and mass of a disk depend on its distance x from the centerof the sphere.
SET UP: The radius r of the disk shown in Fig. 9.25 is
Its volume is
dV 5 pr 2 dx 5 p 1R2 2 x2 2 dx
r 5 "R2 2 x2
and its mass is
EXECUTE: From Example 9.12, the moment of inertia of a disk ofradius r and mass dm is
Integrating this expression from to gives the momentof inertia of the right hemisphere. The total I for the entire sphere,including both hemispheres, is just twice this:
Carrying out the integration, we obtain
The mass M of the sphere of volume is
By comparing the expressions for I and M, we find
EVALUATE: This result agrees with the expression in Table 9.2,case (h). Note that the moment of inertia of a solid sphere of massM and radius R is less than the moment of inertia of a solidcylinder of the same mass and radius, The reason is thatmore of the sphere’s mass is located close to the axis.
I 5 12 MR2.
I 52
5 MR2
M 5 rV 54prR3
3
V 5 4pR3/3
I 58pr
15 R5
I 5 12 2
pr
2 3
R
0
1R2 2 x2 2 2 dx
x 5 Rx 5 0
5pr
2 1R2 2 x2 2 2 dx
dI 51
2 r 2 dm 5
1
2 A"R2 2 x2B 2 3pr 1R2 2 x2 2 dx 4
dm 5 r dV 5 pr 1R2 2 x2 2 dx
Test Your Understanding of Section 9.6 Two hollow cylinders have the sameinner and outer radii and the same mass, but they have different lengths. One is made oflow-density wood and the other of high-density lead. Which cylinder has the greatermoment of inertia around its axis of symmetry? (i) the wood cylinder; (ii) the lead cylin-der; (iii) the two moments of inertia are equal.
❚
Mass element: disk ofradius r and thickness dxlocated a distance x fromthe center of the sphere
Axis
rR
x
dx
9.25 Finding the moment of inertia of a sphere about an axisthrough its center.
Discussion Questions 307
306
CHAPTER 9 SUMMARY
Rotational kinematics: When a rigid body rotatesabout a stationary axis (usually called the z-axis), itsposition is described by an angular coordinate Theangular velocity is the time derivative of and theangular acceleration is the time derivative of orthe second derivative of (See Examples 9.1 and 9.2.)If the angular acceleration is constant, then and are related by simple kinematic equations analogous tothose for straight-line motion with constant linear accel-eration. (See Example 9.3.)
azvz ,u,u.
vzaz
u,vz
u.
(9.3)
(9.5)
(9.11)
(constant only)
(9.10)
(constant only)
(9.7)(constant only)
(9.12)(constant only)az
vz
2 5 v0z
2 1 2az 1 u 2 u0 2az
vz 5 v0z 1 az t
az
u 2 u0 51
2 1vz 1 v0z 2 taz
u 5 u0 1 v0z t 11
2 az t 2
az 5 limDtS0
Dvz
Dt5
dvz
dt5
d 2u
dt 2
vz 5 limDtS0
Du
Dt5
du
dt
Relating linear and angular kinematics: The angularspeed of a rigid body is the magnitude of its angularvelocity. The rate of change of is For aparticle in the body a distance r from the rotation axis,the speed and the components of the acceleration are related to and (See Examples 9.4–9.6.)a.v
aSv
a 5 dv/dt.vv
(9.13)
(9.14)
(9.15)arad 5v2
r5 v2r
atan 5dvdt
5 r
dv
dt5 ra
v 5 rv
Duu2
u1O
x
At t2 At t1
yvz 5dudt az 5
dvz
dt
aSLinearaccelerationof point P s
v 5 rv
r
P
u
y
Ox
vatan 5 ra
arad 5 v2r
Axis ofrotation
v
12K 5 Iv2
miri2I 5
iS
r1
r2
r3
m1
m2
m3
Moment of inertia and rotational kinetic energy: Themoment of inertia I of a body about a given axis is ameasure of its rotational inertia: The greater the value ofI, the more difficult it is to change the state of the body’srotation. The moment of inertia can be expressed as asum over the particles that make up the body, each ofwhich is at its own perpendicular distance from theaxis. The rotational kinetic energy of a rigid body rotat-ing about a fixed axis depends on the angular speed and the moment of inertia I for that rotation axis. (SeeExamples 9.7–9.9.)
v
ri
mi
(9.16)
(9.17)K 51
2 Iv2
5ai
mi ri
2
I 5 m1 r1
2 1 m2 r2
2 1 c
cm
PMass MIcm
IP 5 Icm 1 Md2
dCalculating the moment of inertia: The parallel-axistheorem relates the moments of inertia of a rigid body ofmass M about two parallel axes: an axis through thecenter of mass (moment of inertia and a parallelaxis a distance d from the first axis (moment of inertia
(See Example 9.10.) If the body has a continuousmass distribution, the moment of inertia can be calcu-lated by integration. (See Examples 9.11–9.13.)
IP).
Icm)
(9.19)IP 5 Icm 1 Md 2
Key Termsrigid body, 285radian, 286average angular velocity, 286angular displacement, 286instantaneous angular velocity, 287
average angular acceleration, 289instantaneous angular acceleration, 289angular speed, 293tangential component of acceleration, 293centripetal component of acceleration, 294
Answer to Chapter Opening Question ?Both segments of the rigid blade have the same angular speed From Eqs. (9.13) and (9.15), doubling the distance r for the same
doubles the linear speed and doubles the radial accelera-tion
Answers to Test Your Understanding Questions9.1 Answers: (a) (i) and (iii), (b) (ii) The rotation is speeding upwhen the angular velocity and angular acceleration have the samesign, and slowing down when they have opposite signs. Hence it isspeeding up for and are both positive) and for
and are both negative), but is slowing downfor is positive and is negative). Note that thebody is rotating in one direction for is positive) and inthe opposite direction for is negative).9.2 Answers: (a) (i), (b) (ii) When the DVD comes to rest,From Eq. (9.7), the time when this occurs is
(this is a positive time because is negative). If wedouble the initial angular velocity and also double the angu-lar acceleration their ratio is unchanged and the rotation stopsin the same amount of time. The angle through which the DVD rotates is given by Eq. (9.10): u 2 u0 5 1
2 1v0z 1 vz 2 t 5 12 v0z t
az ,v0z
az2v0z/az
t 5 1vz 2 v0z 2 /az 5vz 5 0.
(vzt . 4 s(vzt , 4 s
az(vz2 s , t , 4 saz(vz4 s , t , 6 s
az(vz0 , t , 2 s
arad 5 v2r.v 5 rvv
v.
(since the final angular velocity is The initial angularvelocity has been doubled but the time t is the same, so theangular displacement (and hence the number of revolu-tions) has doubled. You can also come to the same conclusionusing Eq. (9.12).9.3 Answer: (ii) From Eq. (9.13), To maintain a con-stant linear speed the angular speed must decrease as the scan-ning head moves outward (greater r).9.4 Answer: (i) The kinetic energy in the falling block is and the kinetic energy in the rotating cylinder is
Hence the total kinetic energy of the systemis of which two-thirds is in the block and one-third is in thecylinder.9.5 Answer: (ii) More of the mass of the pool cue is concen-trated at the thicker end, so the center of mass is closer to that end.The moment of inertia through a point P at either end is
the thinner end is farther from the center of mass, sothe distance d and the moment of inertia are greater for the thin-ner end.9.6 Answer: (iii) Our result from Example 9.12 does not dependon the cylinder length L. The moment of inertia depends only onthe radial distribution of mass, not on its distribution along theaxis.
IP
Icm 1 Md 2 ;
IP 5
34 mv2,
12 A12 mR2B AvRB 2 5 1
4 mv2.
12 Iv2 5
12 mv2,
vv,v 5 rv.
u 2 u0
v0z
vz 5 0).
PROBLEMS For instructor-assigned homework, go to www.masteringphysics.com
Discussion QuestionsQ9.1. Which of the following formulas is valid if the angular accel-eration of an object is not constant? Explain your reasoning in eachcase. (a) (b) (c) (d)(e)Q9.2. A diatomic molecule can be modeled as two point masses,
and slightly separated (Fig. 9.26). If the molecule is ori-ented along the y-axis, it has kinetic energy K when it spins about
m2 ,m1
K 5 12 Iv2.
atan 5 rv2;v 5 v0 1 at;atan 5 ra;v 5 rv;
the x-axis. What will its kinetic energy (in terms of K) be if it spinsat the same angular speed about (a) the z-axis and (b) the y-axis?Q9.3. What is the difference between tangential and radial acceler-ation for a point on a rotating body?Q9.4. In Fig. 9.14, all points on the chain have the same linearspeed. Is the magnitude of the linear acceleration also the same forall points on the chain? How are the angular accelerations of thetwo sprockets related? Explain.Q9.5. In Fig. 9.14, how are the radial accelerations of points at theteeth of the two sprockets related? Explain the reasoning behindyour answer.Q9.6. A flywheel rotates with constant angular velocity. Does apoint on its rim have a tangential acceleration? A radial accelera-tion? Are these accelerations constant in magnitude? In direction?In each case give the reasoning behind your answer.Q9.7. What is the purpose of the spin cycle of a washing machine?Explain in terms of acceleration components.Q9.8. Although angular velocity and angular acceleration can betreated as vectors, the angular displacement despite having amagnitude and a direction, cannot. This is because does not fol-low the commutative law of vector addition (Eq. 1.3). Prove this toyourself in the following way: Lay your physics textbook flat on
uu,
moment of inertia, 297rotational kinetic energy, 297parallel-axis theorem, 302
m1
m2
z
x
y
O
Figure 9.26 Question Q9.2.
308 C HAPTE R 9 Rotation of Rigid Bodies Exercises 309
the desk in front of you with the cover side up so you can read thewriting on it. Rotate it through 90° about a horizontal axis so thatthe farthest edge comes toward you. Call this angular displacement
Then rotate it by about a vertical axis so that the left edgecomes toward you. Call this angular displacement The spine ofthe book should now face you, with the writing on it oriented sothat you can read it. Now start over again but carry out the tworotations in the reverse order. Do you get a different result? That is,does equal Now repeat this experiment but thistime with an angle of rather than Do you think that theinfinitesimal displacement obeys the commutative law of addi-tion and hence qualifies as a vector? If so, how is the direction of
related to the direction of Q9.9. Can you think of a body that has the same moment of iner-tia for all possible axes? If so, give an example, and if not,explain why this is not possible. Can you think of a body that hasthe same moment of inertia for all axes passing through a certainpoint? If so, give an example and indicate where the point islocated.Q9.10. To maximize the moment of inertia of a flywheel whileminimizing its weight, what shape and distribution of mass shouldit have? Explain.Q9.11. How might you determine experimentally the moment ofinertia of an irregularly shaped body about a given axis?Q9.12. A cylindrical body has mass M and radius R. Can the massbe distributed within the body in such a way that its moment ofinertia about its axis of symmetry is greater than Explain.Q9.13. Describe how you could use part (b) of Table 9.2 to derivethe result in part (d).Q9.14. A hollow spherical shell of radius R that is rotating about anaxis through its center has rotational kinetic energy K. If you wantto modify this sphere so that it has three times as much kineticenergy at the same angular speed while keeping the same mass,what should be its radius in terms of R?Q9.15. For the equations for I given in parts (a) and (b) of Table 9.2to be valid, must the rod have a circular cross section? Is there anyrestriction on the size of the cross section for these equations toapply? Explain.Q9.16. In part (d) of Table 9.2, the thickness of the plate must bemuch less than a for the expression given for I to apply. But inpart (c), the expression given for I applies no matter how thick theplate is. Explain.Q9.17. Two identical balls, A and B, are each attached to very lightstring, and each string is wrapped around the rim of a frictionlesspulley of mass M. The only difference is that the pulley for ball Ais a solid disk, while the one for ball B is a hollow disk, likepart (e) in Table 9.2. If both balls are released from rest and fall thesame distance, which one will have more kinetic energy, or willthey have the same kinetic energy? Explain your reasoning.Q9.18. An elaborate pulley consistsof four identical balls at the ends ofspokes extending out from a rotatingdrum (Fig. 9.27). A box is connectedto a light thin rope wound around therim of the drum. When it is releasedfrom rest, the box acquires a speed Vafter having fallen a distance d. Nowthe four balls are moved inwardcloser to the drum, and the box isagain released from rest. After it hasfallen a distance d, will its speed beequal to V, greater than V, or lessthan V ? Show or explain why.
MR2 ?
vS ?du
S
duS
90°.1°u2 1 u1 ?u1 1 u2
u2 .90°u1 .
Q9.19. You can use any angular measure—radians, degrees, orrevolutions—in some of the equations in Chapter 9, but you canuse only radian measure in others. Identify those for which usingradians is necessary and those for which it is not, and in each casegive the reasoning behind your answer.Q9.20. When calculating the moment of inertia of an object, canwe treat all its mass as if it were concentrated at the center of massof the object? Justify your answer.Q9.21. A wheel is rotating about an axis perpendicular to the planeof the wheel and passing through the center of the wheel. Theangular speed of the wheel is increasing at a constant rate. Point Ais on the rim of the wheel and point B is midway between the rimand center of the wheel. For each of the following quantities, is itsmagnitude larger at point A, at point B, or is it the same at bothpoints? (a) angular speed; (b) tangential speed; (c) angular acceler-ation; (d) tangential acceleration; (e) radial acceleration. Justifyeach of your answers.
ExercisesSection 9.1 Angular Velocity and Acceleration9.1. (a) What angle in radians is subtended by an arc 1.50 m longon the circumference of a circle of radius 2.50 m? What is thisangle in degrees? (b) An arc 14.0 cm long on the circumference ofa circle subtends an angle of What is the radius of the circle?(c) The angle between two radii of a circle with radius 1.50 m is0.700 rad. What length of arc is intercepted on the circumferenceof the circle by the two radii?9.2. An airplane propeller is rotating at 1900 rpm (a) Compute the propeller’s angular velocity in (b) Howmany seconds does it take for the propeller to turn through 9.3. The angular velocity of a flywheel obeys the equation
where t is in seconds and A and B are constantshaving numerical values 2.75 (for A) and 1.50 (for B). (a) Whatare the units of A and B if is in (b) What is the angularacceleration of the wheel at (i) and (ii) (c) Through what angle does the flywheel turn during the first2.00 s? (Hint: See Section 2.6.)9.4. A fan blade rotates with angular velocity given by
where and (a) Calcu-late the angular acceleration as a function of time. (b) Calculate theinstantaneous angular acceleration at and the averageangular acceleration for the time interval to How do these two quantities compare? If they are different, whyare they different?9.5. A child is pushing a merry-go-round. The angle throughwhich the merry-go-round has turned varies with time according to
where and (a) Calculate the angular velocity of the merry-go-round as a func-tion of time. (b) What is the initial value of the angular velocity?(c) Calculate the instantaneous value of the angular velocity at
and the average angular velocity for the time inter-val to Show that is not equal to the averageof the instantaneous angular velocities at and andexplain why it is not.9.6. At the current to a dc electric motor is reversed, result-ing in an angular displacement of the motor shaft given by
(a) Atwhat time is the angular velocity of the motor shaft zero? (b) Cal-culate the angular acceleration at the instant that the motor shafthas zero angular velocity. (c) How many revolutions does themotor shaft turn through between the time when the current is
u 1 t 2 5 1250 rad/s 2 t 2 120.0 rad/s2 2 t 2 2 11.50 rad/s3 2 t 3.
t 5 0
t 5 5.00 s,t 5 0vav-zt 5 5.00 s.t 5 0
vav-zt 5 5.00 svz
b 5 0.0120 rad/s3.g 5 0.400 rad/su 1 t 2 5 gt 1 bt 3,
t 5 3.00 s.t 5 0aav-z
t 5 3.00 saz
b 5 0.800 rad/s3.g 5 5.00 rad/sg 2 bt 2,vz 1 t 2 5
t 5 5.00 s?t 5 0.00rad/s?v
vz 1 t 2 5 A 1 Bt 2,
35°?rad/s.
1 rev/min 2 .
128°.
Drum
Box
Figure 9.27Question 9.18.
reversed and the instant when the angular velocity is zero? (d) Howfast was the motor shaft rotating at when the current wasreversed? (e) Calculate the average angular velocity for the timeperiod from to the time calculated in part (a).9.7. The angle through which a disk drive turns is given by
where a, b, and c are constants, t is in sec-onds, and is in radians. When and the angularvelocity is and when the angular accelera-tion is (a) Find a, b, and c, including their units.(b) What is the angular acceleration when (c) Whatare and the angular velocity when the angular acceleration is
9.8. A wheel is rotating about an axis that is in the z-direction. Theangular velocity is at increases linearly withtime, and is at We have taken counterclock-wise rotation to be positive. (a) Is the angular acceleration duringthis time interval positive or negative? (b) During what time inter-val is the speed of the wheel increasing? Decreasing? (c) What isthe angular displacement of the wheel at
Section 9.2 Rotation with Constant Angular Acceleration9.9. A bicycle wheel has an initial angular velocity of (a) If its angular acceleration is constant and equal to what is its angular velocity at (b) Through what anglehas the wheel turned between and 9.10. An electric fan is turned off, and its angular velocitydecreases uniformly from to in 4.00 s.(a) Find the angular acceleration in and the number of revo-lutions made by the motor in the 4.00-s interval. (b) How manymore seconds are required for the fan to come to rest if the angularacceleration remains constant at the value calculated in part (a)?9.11. The rotating blade of a blender turns with constant angularacceleration (a) How much time does it take to reachan angular velocity of starting from rest? (b) Throughhow many revolutions does the blade turn in this time interval?9.12. (a) Derive Eq. (9.12) by combining Eqs. (9.7) and (9.11) toeliminate t. (b) The angular velocity of an airplane propellerincreases from to while turning through7.00 rad. What is the angular acceleration in 9.13. A turntable rotates with a constant angular accel-eration. After 4.00 s it has rotated through an angle of 60.0 rad.What was the angular velocity of the wheel at the beginning of the4.00-s interval?9.14. A circular saw blade 0.200 m in diameter starts from rest. In6.00 s it accelerates with constant angular acceleration to an angu-lar velocity of Find the angular acceleration and theangle through which the blade has turned.9.15. A high-speed flywheel in a motor is spinning at 500 rpmwhen a power failure suddenly occurs. The flywheel has mass40.0 kg and diameter 75.0 cm. The power is off for 30.0 s, and dur-ing this time the flywheel slows due to friction in its axle bearings.During the time the power is off, the flywheel makes 200 completerevolutions. (a) At what rate is the flywheel spinning when thepower comes back on? (b) How long after the beginning of thepower failure would it have taken the flywheel to stop if the powerhad not come back on, and how many revolutions would the wheelhave made during this time?9.16. A computer disk drive is turned on starting from rest and hasconstant angular acceleration. If it took 0.750 s for the drive tomake its second complete revolution, (a) how long did it take tomake the first complete revolution, and (b) what is its angularacceleration, in rad/s2?
140 rad/s.
2.25 rad/s2rad/s2?
16.0 rad/s12.0 rad/s
36.0 rad/s,1.50 rad/s2.
rev/s2200 rev/min500 rev/min
t 5 2.50 s?t 5 0t 5 2.50 s?
0.300 rad/s2,1.50 rad/s.
t 5 7.00 s?
t 5 7.00 s.18.00 m/st 5 0,26.00 rad/svz
3.50 rad/s2?u
u 5 p/4 rad?1.25 rad/s2.
t 5 1.50 s,2.00 rad/s,u 5 p/4 radt 5 0,u
u 1 t 2 5 a 1 bt 2 ct 3,u
t 5 0
t 5 0,9.17. A safety device brings the blade of a power mower from aninitial angular speed of to rest in 1.00 revolution. At the sameconstant acceleration, how many revolutions would it take theblade to come to rest from an initial angular speed that wasthree times as great, 9.18. A straight piece of reflecting tape extends from the center ofa wheel to its rim. You darken the room and use a camera andstrobe unit that flashes once every 0.050 s to take pictures of thewheel as it rotates counterclockwise. You trigger the strobe so thatthe first flash occurs when the tape is horizontal to theright at an angular displacement of zero. For the following situa-tions draw a sketch of the photo you will get for the time exposureover five flashes (at 0.050 s, 0.100 s, 0.150 s, and 0.200 s),and graph versus t and versus t for to (a) The angular velocity is constant at (b) The wheelstarts from rest with a constant angular acceleration of (c) The wheel is rotating at at and changes angularvelocity at a constant rate of 9.19. At a grinding wheel has an angular velocity of
It has a constant angular acceleration of until a circuit breaker trips at From then on, it turnsthrough 432 rad as it coasts to a stop at constant angular accelera-tion. (a) Through what total angle did the wheel turn betweenand the time it stopped? (b) At what time did it stop? (c) What wasits acceleration as it slowed down?
Section 9.3 Relating Linear and Angular Kinematics9.20. In a charming 19th-century hotel, an old-style ele-vator is connected to a counter-weight by a cable that passesover a rotating disk 2.50 m indiameter (Fig. 9.28). The eleva-tor is raised and lowered byturning the disk, and the cabledoes not slip on the rim of thedisk but turns with it. (a) At howmany rpm must the disk turn toraise the elevator at (b) To start the elevator moving,it must be accelerated at What must be the angular accel-eration of the disk, in (c) Through what angle (in radi-ans and degrees) has the diskturned when it has raised theelevator 3.25 m between floors?9.21. Using astronomical data from Appendix F, along with thefact that the earth spins on its axis once per day, calculate (a) theearth’s orbital angular speed (in due to its motion aroundthe sun, (b) its angular speed (in due to its axial spin, (c) thetangential speed of the earth around the sun (assuming a circularorbit), (d) the tangential speed of a point on the earth’s equator dueto the planet’s axial spin, and (e) the radial and tangential accelera-tion components of the point in part (d).9.22. Compact Disc. A compact disc (CD) stores music in acoded pattern of tiny pits deep. The pits are arranged in atrack that spirals outward toward the rim of the disc; the inner andouter radii of this spiral are 25.0 mm and 58.0 mm, respectively.As the disc spins inside a CD player, the track is scanned at aconstant linear speed of (a) What is the angular speedof the CD when the innermost part of the track is scanned? The
1.25 m/s.
1027 m
rad/s)rad/s)
rad/s2?
18 g.
25.0 cm/s?
t 5 0
t 5 2.00 s.30.0 rad/s224.0 rad/s.
t 5 0250.0 rev/s2.
t 5 010.0 rev/s25.0 rev/s2.
10.0 rev/s.t 5 0.200 s.t 5 0vu
t 5 0,
1 t 5 0 2
v3 5 3v1 ?v3
v1
Disk
ElevatorCounterweight
Figure 9.28 Exercise 9.20.
Exercises 311310 C HAPTE R 9 Rotation of Rigid Bodies
outermost part of the track? (b) The maximum playing time of aCD is 74.0 min. What would be the length of the track on such amaximum-duration CD if it were stretched out in a straight line?(c) What is the average angular acceleration of a maximum-durationCD during its 74.0-min playing time? Take the direction of rota-tion of the disc to be positive.9.23. A wheel of diameter 40.0 cm starts from rest and rotates witha constant angular acceleration of At the instant thewheel has computed its second revolution, compute the radialacceleration of a point on the rim in two ways: (a) using the rela-tionship and (b) from the relationship 9.24. Ultracentrifuge. Find the required angular speed (in
of an ultracentrifuge for the radial acceleration of a point2.50 cm from the axis to equal 400,000g (that is, 400,000 times theacceleration due to gravity).9.25. A flywheel with a radius of 0.300 m starts from rest andaccelerates with a constant angular acceleration of Compute the magnitude of the tangential acceleration, the radialacceleration, and the resultant acceleration of a point on its rim(a) at the start; (b) after it has turned through (c) after it hasturned through 9.26. An electric turntable 0.750 m in diameter is rotating about afixed axis with an initial angular velocity of and a con-stant angular acceleration of (a) Compute the angularvelocity of the turntable after 0.200 s. (b) Through how many rev-olutions has the turntable spun in this time interval? (c) What isthe tangential speed of a point on the rim of the turntable at
(d) What is the magnitude of the resultant accelera-tion of a point on the rim at 9.27. Centrifuge. An advertisement claims that a centrifugetakes up only 0.127 m of bench space but can produce a radialacceleration of 3000g at Calculate the requiredradius of the centrifuge. Is the claim realistic?9.28. (a) Derive an equation for the radial acceleration that includes
and but not r. (b) You are designing a merry-go-round forwhich a point on the rim will have a radial acceleration of
when the tangential velocity of that point has magnitudeWhat angular velocity is required to achieve these values?
9.29. Electric Drill. According to the shop manual, when drill-ing a 12.7-mm-diameter hole in wood, plastic, or aluminum, a drillshould have a speed of For a 12.7-mm-diameterdrill bit turning at a constant find (a) the maximumlinear speed of any part of the bit and (b) the maximum radialacceleration of any part of the bit.9.30. At a point on the rim of a 0.200-m-radius wheelhas a tangential speed of as the wheel slows down with atangential acceleration of constant magnitude (a) Calcu-late the wheel’s constant angular acceleration. (b) Calculate theangular velocities at and (c) Through what angledid the wheel turn between and (d) At what timewill the radial acceleration equal g?9.31. The spin cycles of a washing machine have two angularspeeds, and The internal diameter of thedrum is 0.470 m. (a) What is the ratio of the maximum radial forceon the laundry for the higher angular speed to that for the lowerspeed? (b) What is the ratio of the maximum tangential speed of thelaundry for the higher angular speed to that for the lower speed?(c) Find the laundry’s maximum tangential speed and the maxi-mum radial acceleration, in terms of g.9.32. You are to design a rotating cylindrical axle to lift 800-Nbuckets of cement from the ground to a rooftop 78.0 m above theground. The buckets will be attached to a hook on the free end of acable that wraps around the rim of the axle; as the axle turns, the
640 rev/min.423 rev/min
t 5 3.00 s?t 5 0t 5 0.t 5 3.00 s
10.0 m/s2.50.0 m/s
t 5 3.00 s
1250 rev/min,1250 rev/min.
2.00 m/s.0.500 m/s2
v,v
5000 rev/min.
t 5 0.200 s?t 5 0.200 s?
0.900 rev/s2.0.250 rev/s
120.0°.60.0°;
0.600 rad/s2.
rev/min 2arad 5 v2/r.arad 5 v2r
3.00 rad/s2.
buckets will rise. (a) What should the diameter of the axle be inorder to raise the buckets at a steady when it is turningat 7.5 rpm? (b) If instead the axle must give the buckets an upwardacceleration of what should the angular acceleration ofthe axle be?9.33. While riding a multispeed bicycle, the rider can select theradius of the rear sprocket that is fixed to the rear axle. The frontsprocket of a bicycle has radius 12.0 cm. If the angular speed ofthe front sprocket is what is the radius of the rearsprocket for which the tangential speed of a point on the rim of therear wheel will be The rear wheel has radius 0.330 m.
Section 9.4 Energy in Rotational Motion9.34. Four small spheres, each ofwhich you can regard as a point ofmass 0.200 kg, are arranged in asquare 0.400 m on a side andconnectedbyextremelylightrods(Fig. 9.29). Find the moment ofinertia of the system about anaxis (a) through the center of the square, perpendicular to itsplane (an axis through point O in the figure); (b) bisecting twoopposite sides of the square (an axis along the line AB in the fig-ure); (c) that passes through the centers of the upper left and lowerright spheres and through point O.9.35. Calculate the moment of inertia of each of the following uni-form objects about the axes indicated. Consult Table 9.2 as needed.(a) A thin 2.50-kg rod of length 75.0 cm, about an axis perpendicu-lar to it and passing through (i) one end and (ii) its center, and(iii) about an axis parallel to the rod and passing through it. (b) A3.00-kg sphere 38.0 cm in diameter, about an axis through its cen-ter, if the sphere is (i) solid and (ii) a thin-walled hollow shell.(c) An 8.00-kg cylinder, of length 19.5 cm and diameter 12.0 cm,about the central axis of the cylinder, if the cylinder is (i) thin-walled and hollow, and (ii) solid.9.36. Small blocks, each with mass m, are clamped at the ends andat the center of a rod of length L and negligible mass. Compute themoment of inertia of the system about an axis perpendicular to therod and passing through (a) the center of the rod and (b) a pointone-fourth of the length from one end.9.37. A uniform bar has two small balls glued to its ends. The bar is2.00 m long and has mass 4.00 kg, while the balls each have mass0.500 kg and can be treated as point masses. Find the moment ofinertia of this combination about each of the following axes: (a) anaxis perpendicular to the bar through its center; (b) an axis perpendi-cular to the bar through one of the balls; (c) an axis parallel to the barthrough both balls; (d) an axis parallel to the bar and 0.500 m from it.9.38. A twirler’s baton is made of aslender metal cylinder of mass Mand length L. Each end has a rubbercap of mass m, and you can accu-rately treat each cap as a particle inthis problem. Find the total momentof inertia of the baton about theusual twirling axis (perpendicular tothe baton through its center).9.39. A wagon wheel is constructedas shown in Fig. 9.30. The radius ofthe wheel is 0.300 m, and the rimhas mass 1.40 kg. Each of the eightspokes that lie along a diameter and
5.00 m/s?
0.600 rev/s,
0.400 m/s2,
2.00 cm/s
0.400 m
BO
0.200 kg
A
Figure 9.29 Exercise 9.34.
are 0.300 m long has mass 0.280 kg. What is the moment of inertiaof the wheel about an axis through its center and perpendicular tothe plane of the wheel? (Use the formulas given in Table 9.2.)9.40. A uniform disk of radius R is cut inhalf so that the remaining half has mass M(Fig. 9.31a). (a) What is the moment of iner-tia of this half about an axis perpendicular toits plane through point A? (b) Why did youranswer in part (a) come out the same as if thiswere a complete disk of mass M? (c) Whatwould be the moment of inertia of a quarterdisk of mass M and radius R about an axisperpendicular to its plane passing throughpoint B (Fig. 9.31b)?9.41. A compound disk of outside diameter140.0 cm is made up of a uniform solid diskof radius 50.0 cm and area densitysurrounded by a concentric ring of innerradius 50.0 cm, outer radius 70.0 cm, andarea density Find the moment ofinertia of this object about an axis perpendi-cular to the plane of the object and passingthrough its center.9.42. An airplane propeller is 2.08 m inlength (from tip to tip) with mass 117 kg andis rotating at 2400 rpm about anaxis through its center. You can model the propeller as a slender rod.(a) What is its rotational kinetic energy? (b) Suppose that, due toweight constraints, you had to reduce the propeller’s mass to 75.0%of its original mass, but you still needed to keep the same size andkinetic energy. What would its angular speed have to be, in rpm?9.43. Energy from the Moon? Suppose that some time in thefuture we decide to tap the moon’s rotational energy for use onearth. In additional to the astronomical data in Appendix F, youmay need to know that the moon spins on its axis once every27.3 days. Assume that the moon is uniform throughout. (a) Howmuch total energy could we get from the moon’s rotation? (b) Theworld presently uses about of energy per year. If inthe future the world uses five times as much energy yearly, for howmany years would the moon’s rotation provide us energy? In lightof your answer, does this seem like a cost-effective energy sourcein which to invest?9.44. You need to design an industrial turntable that is 60.0 cm indiameter and has a kinetic energy of 0.250 J when turning at45.0 rpm (a) What must be the moment of inertia of theturntable about the rotation axis? (b) If your workshop makes thisturntable in the shape of a uniform solid disk, what must be its mass?9.45. The flywheel of a gasoline engine is required to give up500 J of kinetic energy while its angular velocity decreases from
to What moment of inertia is required?9.46. A light, flexible rope is wrapped several times around ahollow cylinder, with a weight of 40.0 N and a radius of 0.25 m,that rotates without friction about a fixed horizontal axis. Thecylinder is attached to the axle by spokes of a negligible momentof inertia. The cylinder is initially at rest. The free end of the ropeis pulled with a constant force P for a distance of 5.00 m, at whichpoint the end of the rope is moving at If the rope doesnot slip on the cylinder, what is the value of P?9.47. Energy is to be stored in a 70.0-kg flywheel in the shape of auniform solid disk with radius To prevent structuralfailure of the flywheel, the maximum allowed radial accelerationof a point on its rim is What is the maximum kineticenergy that can be stored in the flywheel?
3500 m/s2.
R 5 1.20 m.
6.00 m/s.
520 rev/min.650 rev/min
1 rev/min 2 .
4.0 3 1020 J
1 rev/min 2
2.00 g/cm2.
3.00 g/cm2
9.48. Suppose the solid cylinder in the apparatus described inExample 9.9 (Section 9.4) is replaced by a thin-walled, hollowcylinder with the same mass M and radius R. The cylinder isattached to the axle by spokes of a negligible moment of inertia.(a) Find the speed of the hanging mass m just as it strikes the floor.(b) Use energy concepts to explain why the answer to part (a) isdifferent from the speed found in Example 9.9.9.49. A frictionless pulley hasthe shape of a uniform solid diskof mass 2.50 kg and radius20.0 cm. A 1.50-kg stone isattached to a very light wire thatis wrapped around the rim of thepulley (Fig. 9.32), and the systemis released from rest. (a) How farmust the stone fall so that the pul-ley has 4.50 J of kinetic energy?(b) What percent of the totalkinetic energy does the pulleyhave?9.50. A bucket of mass m is tiedto a massless cable that is wrapped around the outer rim of a fric-tionless uniform pulley of radius R, similar to the system shown inFig. 9.32. In terms of the stated variables, what must be themoment of inertia of the pulley so that it always has half as muchkinetic energy as the bucket?9.51. How I Scales. If we multiply all the design dimensions ofan object by a scaling factor its volume and mass will be multi-plied by (a) By what factor will its moment of inertia be multi-plied? (b) If a model has a rotational kinetic energy of2.5 J, what will be the kinetic energy for the full-scale object of thesame material rotating at the same angular velocity?9.52. A uniform 2.00-m ladder of mass 9.00 kg is leaning against avertical wall while making an angle of 53.0° with the floor. Aworker pushes the ladder up against the wall until it is vertical.How much work did this person do against gravity?9.53. A uniform 3.00-kg rope 24.0 m long lies on the ground at thetop of a vertical cliff. A mountain climber at the top lets down halfof it to help his partner climb up the cliff. What was the change inpotential energy of the rope during this maneuver?
Section 9.5 Parallel-Axis Theorem9.54. Find the moment of inertia of a hoop (a thin-walled, hollowring) with mass M and radius R about an axis perpendicular to thehoop’s plane at an edge.9.55. About what axis will a uniform, balsa-wood sphere have thesame moment of inertia as does a thin-walled, hollow, lead sphereof the same mass and radius, with the axis along a diameter?9.56. Use the parallel-axis theorem to show that the moments ofinertia given in parts (a) and (b) of Table 9.2 are consistent.9.57. A thin, rectangular sheet of metal has mass M and sides oflength a and b. Use the parallel-axis theorem to calculate themoment of inertia of the sheet for an axis that is perpendicular tothe plane of the sheet and that passes through one corner of thesheet.9.58. (a) For the thin rectangular plate shown in part (d) ofTable 9.2, find the moment of inertia about an axis that lies in theplane of the plate, passes through the center of the plate, and is par-allel to the axis shown in the figure. (b) Find the moment of inertiaof the plate for an axis that lies in the plane of the plate, passesthrough the center of the plate, and is perpendicular to the axis inpart (a).
148-scale
f 3.f,
A
B
R
M
R
M
(a)
(b)
Figure 9.31Exercise 9.40.
0.600 m
Figure 9.30 Exercise9.39.
2.50-kgpulley
1.50-kgstone
Figure 9.32 Exercise 9.49.
Problems 313312 C HAPTE R 9 Rotation of Rigid Bodies
9.59. A thin uniform rod of mass M and length L is bent at its cen-ter so that the two segments are now perpendicular to each other.Find its moment of inertia about an axis perpendicular to its planeand passing through (a) the point where the two segments meetand (b) the midpoint of the line connecting its two ends.
*Section 9.6 Moment-of-Inertia Calculations*9.60. Using the information in Table 9.2 and the parallel-axis the-orem, find the moment of inertia of the slender rod with mass Mand length L shown in Fig. 9.23 about an axis through O, at anarbitrary distance h from one end. Compare your result to thatfound by integration in Example 9.11 (Section 9.6).*9.61. Use Eq. (9.20) to calculate the moment of inertia of a uni-form, solid disk with mass M and radius R for an axis perpendicu-lar to the plane of the disk and passing through its center.*9.62. Use Eq. (9.20) to calculate the moment of inertia of a slen-der, uniform rod with mass M and length L about an axis at oneend, perpendicular to the rod.*9.63. A slender rod with length L has a mass per unit length thatvaries with distance from the left end, where according to
where has units of (a) Calculate the totalmass of the rod in terms of and L. (b) Use Eq. (9.20) to calculatethe moment of inertia of the rod for an axis at the left end, perpen-dicular to the rod. Use the expression you derived in part (a) toexpress I in terms of M and L. How does your result compare tothat for a uniform rod? Explain this comparison. (c) Repeat part (b)for an axis at the right end of the rod. How do the results for parts(b) and (c) compare? Explain this result.
Problems9.64. Sketch a wheel lying in the plane of your paper and rotatingcounterclockwise. Choose a point on the rim and draw a vector from the center of the wheel to that point. (a) What is the directionof (b) Show that the velocity of the point is (c) Show that the radial acceleration of the point is
(see Exercise 9.28).9.65. Trip to Mars. You are working on a project with NASA tolaunch a rocket to Mars, with the rocket blasting off from earthwhen earth and Mars are aligned along a straight line from the sun.If Mars is now ahead of earth in its orbit around the sun, whenshould you launch the rocket? (Note: All the planets orbit the sunin the same direction, 1 year on Mars is 1.9 earth-years, andassume circular orbits for both planets.)9.66. A roller in a printing press turns through an anglegiven by where and
(a) Calculate the angular velocity of the roller as afunction of time. (b) Calculate the angular acceleration of the rolleras a function of time. (c) What is the maximum positive angularvelocity, and at what value of t does it occur?*9.67. A disk of radius 25.0 cm is free to turn about an axle perpen-dicular to it through its center. It has very thin but strong stringwrapped around its rim, and the string is attached to a ball that ispulled tangentially away from the rim of the disk (Fig. 9.33). Thepull increases in magnitude and produces an acceleration of the ballthat obeys the equation where t is in seconds and A is aconstant. The cylinder starts from rest, and at the end of the thirdsecond, the ball’s acceleration is (a) Find A. (b) Expressthe angular acceleration of the disk as a function of time. (c) Howmuch time after the disk has begun to turn does it reach an angularspeed of (d) Through what angle has the disk turnedjust as it reaches (Hint: See Section 2.6.)15.0 rad/s?
15.0 rad/s?
1.80 m/s2.
a 1 t 2 5 At,
0.500 rad/s3.b 5g 5 3.20 rad/s2u 1 t 2 5 gt 2 2 bt 3,u 1 t 2
60°
vS
3 vS 5 vS
3 1vS 3 rS 2 aSrad 5vS 5 v
S
3 rS.vS ?
rS
gkg/m2.gdm/dx 5 gx,
x 5 0,
Ball
Disk
Pull
Figure 9.33 Problem 9.67. is the speed of a point on the belt? (b) What is the angular velocityof the wheel, in 9.72. The motor of a table saw is rotating at A pul-ley attached to the motor shaft drives a second pulley of half thediameter by means of a V-belt. A circular saw blade of diameter0.208 m is mounted on the same rotating shaft as the second pulley.(a) The operator is careless and the blade catches and throws back asmall piece of wood. This piece of wood moves with linear speedequal to the tangential speed of the rim of the blade. What is thisspeed? (b) Calculate the radial acceleration of points on the outeredge of the blade to see why sawdust doesn’t stick to its teeth.9.73. A wheel changes its angular velocity with a constant angularacceleration while rotating about a fixed axis through its center.(a) Show that the change in the magnitude of the radial accelera-tion during any time interval of a point on the wheel is twice theproduct of the angular acceleration, the angular displacement, andthe perpendicular distance of the point from the axis. (b) The radialacceleration of a point on the wheel that is 0.250 m from the axischanges from to as the wheel rotates through15.0 rad. Calculate the tangential acceleration of this point.(c) Show that the change in the wheel’s kinetic energy during anytime interval is the product of the moment of inertia about the axis,the angular acceleration, and the angular displacement. (d) Duringthe 15.0-rad angular displacement of part (b), the kinetic energyof the wheel increases from 20.0 J to 45.0 J. What is the momentof inertia of the wheel about the rotation axis?9.74. A sphere consists of a solid wooden ball of uniform density
and radius 0.20 m and is covered with a thin coating oflead foil with area density Calculate the moment of iner-tia of this sphere about an axis passing through its center.9.75. Estimate your own moment of inertia about a vertical axisthrough the center of the top of your head when you are standingup straight with your arms outstretched. Make reasonable approxi-mations and measure or estimate necessary quantities.9.76. A thin uniform rod 50.0 cm long with mass 0.320 kg is bentat its center into a V shape, with a angle at its vertex. Findthe moment of inertia of this V-shaped object about an axis perpen-dicular to the plane of the V at its vertex.9.77. It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energyand store it until it is needed during peak load times, such as themiddle of the day. One suggestion has been to store the energy inlarge flywheels spinning on nearly frictionless ball bearings. Con-sider a flywheel made of iron (density in the shape ofa 10.0-cm-thick uniform disk. (a) What would the diameter of sucha disk need to be if it is to store 10.0 megajoules of kinetic energywhen spinning at 90.0 rpm about an axis perpendicular to the diskat its center? (b) What would be the centripetal acceleration of apoint on its rim when spinning at this rate?9.78. While redesigning a rocket engine, you want to reduce itsweight by replacing a solid spherical part with a hollow sphericalshell of the same size. The parts rotate about an axis through theircenter. You need to make sure that the new part always has thesame rotational kinetic energy as the original part had at any givenrate of rotation. If the original part had mass M, what must be themass of the new part?9.79. The earth, which is not a uniform sphere, has a moment ofinertia of about an axis through its north and southpoles. It takes the earth 86,164 s to spin once about this axis. UseAppendix F to calculate (a) the earth’s kinetic energy due to itsrotation about this axis and (b) the earth’s kinetic energy due to itsorbital motion around the sun. (c) Explain how the value of the
0.3308MR2
7800 kg/m3)
70.0°
20 kg/m2.800 kg/m3
85.0 m/s225.0 m/s2
3450 rev/min.rad/s?
earth’s moment of inertia tells us that the mass of the earth is con-centrated toward the planet’s center.9.80. A uniform, solid disk with mass m and radius R is pivotedabout a horizontal axis through its center. A small object of thesame mass m is glued to the rim of the disk. If the disk is releasedfrom rest with the small object at the end of a horizontal radius, findthe angular speed when the small object is directly below the axis.9.81. A metal sign for a car dealership is a thin, uniform right trian-gle with base length b and height h. The sign has mass M. (a) Whatis the moment of inertia of the sign for rotation about the side oflength h? (b) If and whatis the kinetic energy of the sign when it is rotating about an axisalong the 1.20-m side at 9.82. Measuring I. As an intern with an engineering firm, youare asked to measure the moment of inertia of a large wheel, forrotation about an axis through its center. Since you were a goodphysics student, you know what to do. You measure the diameterof the wheel to be 0.740 m and find that it weighs 280 N. Youmount the wheel, using frictionless bearings, on a horizontal axisthrough the wheel’s center. You wrap a light rope around the wheeland hang a 8.00-kg mass from the free end of the rope, as shown inFig. 9.18. You release the mass from rest; the mass descends andthe wheel turns as the rope unwinds. You find that the mass hasspeed after it has descended 2.00 m. (a) What is themoment of inertia of the wheel for an axis perpendicular to thewheel at its center? (b) Your boss tells you that a larger I is needed.He asks you to design a wheel of the same mass and radius that has
How do you reply?9.83. A meter stick with a mass of 0.160 kg is pivoted about oneend so it can rotate without friction about a horizontal axis. Themeter stick is held in a horizontal position and released. As itswings through the vertical, calculate (a) the change in gravita-tional potential energy that has occurred; (b) the angular speed ofthe stick; (c) the linear speed of the end of the stick opposite theaxis. (d) Compare the answer in part (c) to the speed of a particlethat has fallen 1.00 m, starting from rest.9.84. Exactly one turn of a flexible rope with mass m is wrappedaround a uniform cylinder with mass M and radius R. The cylinderrotates without friction about a horizontal axle along the cylinderaxis. One end of the rope is attached to the cylinder. The cylin-der starts with angular speed After one revolution of thecylinder the rope has unwrapped and, at this instant, hangs verticallydown, tangent to the cylinder. Find the angular speed of the cylinderand the linear speed of the lower end of the rope at this time. You canignore the thickness of the rope. [Hint: Use Eq. (9.18).]9.85. The pulley in Fig. 9.35 has radius R and a moment of inertiaI. The rope does not slip over the pulley, and the pulley spins on africtionless axle. The coefficient of kinetic friction between blockA and the tabletop is The system is released from rest, andblock B descends. Block A has mass and block B has mass Use energy methods to calculate the speed of block B as a functionof the distance d that it has descended.
mB .mA
mk .
v0 .
I 5 19.0 kg # m2.
5.00 m/s
2.00 rev/s?
h 5 1.20 m,b 5 1.60 m,M 5 5.40 kg,9.68. When a toy car is rapidly scooted across the floor, it storesenergy in a flywheel. The car has mass 0.180 kg, and its flywheelhas moment of inertia The car is 15.0 cmlong. An advertisement claims that the car can travel at a scalespeed of up to The scale speed is the speedof the toy car multiplied by the ratio of the length of an actual carto the length of the toy. Assume a length of 3.0 m for a real car.(a) For a scale speed of what is the actual translationalspeed of the car? (b) If all the kinetic energy that is initially in theflywheel is converted to the translational kinetic energy of the toy,how much energy is originally stored in the flywheel? (c) What ini-tial angular velocity of the flywheel was needed to store theamount of energy calculated in part (b)?9.69. A classic 1957 Chevrolet Corvette of mass 1240 kg startsfrom rest and speeds up with a constant tangential acceleration of
on a circular test track of radius 60.0 m. Treat the car as aparticle. (a) What is its angular acceleration? (b) What is its angularspeed 6.00 s after it starts? (c) What is its radial acceleration at thistime? (d) Sketch a view from above showing the circular track, thecar, the velocity vector, and the acceleration component vectors6.00 s after the car starts. (e) What are the magnitudes of the totalacceleration and net force for the car at this time? (f) What angle dothe total acceleration and net force make with the car’s velocity atthis time?9.70. Engineers are designing asystem by which a falling mass mimparts kinetic energy to a rotatinguniform drum to which it isattached by thin, very light wirewrapped around the rim of thedrum (Fig. 9.34). There is noappreciable friction in the axle ofthe drum, and everything startsfrom rest. This system is beingtested on earth, but it is to be usedon Mars, where the accelerationdue to gravity is In theearth tests, when m is set to 15.0 kgand allowed to fall through 5.00 m, it gives 250.0 J of kineticenergy to the drum. (a) If the system is operated on Mars, throughwhat distance would the 15.0-0 mass have to fall to give the sameamount of kinetic energy to the drum? (b) How fast would the15.0-kg mass be moving on Mars just as the drum gained 250.0 Jof kinetic energy?9.71. A vacuum cleaner belt is looped over a shaft of radius0.45 cm and a wheel of radius 2.00 cm. The arrangement of thebelt, shaft, and wheel is similar to that of the chain and sprockets inFig. 9.14. The motor turns the shaft at and the movingbelt turns the wheel, which in turn is connected by another shaft tothe roller that beats the dirt out of the rug being vacuumed. Assumethat the belt doesn’t slip on either the shaft or the wheel. (a) What
60.0 rev/s
3.71 m/s2.
3.00 m/s2
700 km/h,
1440 mi/h 2 .700 km/h
4.00 3 1025 kg # m2.
m
Drum
Figure 9.34 Problem 9.70.
B
I
A
Figure 9.35 Problem 9.85.
Challenge Problems 315314 C HAPTE R 9 Rotation of Rigid Bodies
9.86. The pulley in Fig. 9.36 hasradius 0.160 m and moment ofinertia The ropedoes not slip on the pulley rim.Use energy methods to calculatethe speed of the 4.00-kg blockjust before it strikes the floor.9.87. You hang a thin hoop withradius R over a nail at the rim ofthe hoop. You displace it to theside (within the plane of the hoop)through an angle from its equi-librium position and let it go.What is its angular speed when it returns to its equilibrium posi-tion? [Hint: Use Eq. (9.18).]9.88. A passenger bus in Zurich, Switzerland, derived its motivepower from the energy stored in a large flywheel. The wheel wasbrought up to speed periodically, when the bus stopped at a station,by an electric motor, which could then be attached to the electricpower lines. The flywheel was a solid cylinder with mass 1000 kgand diameter 1.80 m; its top angular speed was (a) At this angular speed, what is the kinetic energy of the fly-wheel? (b) If the average power required to operate the bus is
how long could it operate between stops?9.89. Two metal disks, one with radius
and mass andthe other with radius andmass are welded togetherand mounted on a frictionless axis throughtheir common center (Fig. 9.37). (a) Whatis the total moment of inertia of the twodisks? (b) A light string is wrapped aroundthe edge of the smaller disk, and a 1.50-kgblock is suspended from the free end of thestring. If the block is released from rest ata distance of 2.00 m above the floor, whatis its speed just before it strikes the floor?(c) Repeat the calculation of part (b), thistime with the string wrapped around theedge of the larger disk. In which case is thefinal speed of the block the greatest? Explain why this is so.9.90. In the cylinder and mass combination described in Example9.9 (Section 9.4), suppose the falling mass m is made of ideal rub-ber, so that no mechanical energy is lost when the mass hits theground. (a) If the cylinder is originally not rotating and the mass mis released from rest at a height h above the ground, to what heightwill this mass rebound if it bounces straight back up from thefloor? (b) Explain, in terms of energy, why the answer to part (a) isless than h.9.91. In the system shown in Fig. 9.18, a 12.0-kg mass is releasedfrom rest and falls, causing the uniform 10.0-kg cylinder of diame-ter 30.0 cm to turn about a frictionless axle through its center. Howfar will the mass have to descend to give the cylinder 250 J ofkinetic energy?9.92. In Fig. 9.38, the cylinderand pulley turn without frictionabout stationary horizontal axlesthat pass through their centers. Alight rope is wrapped around thecylinder, passes over the pulley,and has a 3.00-kg box suspendedfrom its free end. There is no slip-
M2 5 1.60 kg,R2 5 5.00 cmM1 5 0.80 kgR1 5 2.50 cm
1.86 3 104 W,
3000 rev/min.
b
0.480 kg # m2.
ping between the rope and the pulley surface. The uniform cylin-der has mass 5.00 kg and radius 40.0 cm. The pulley is a uniformdisk with mass 2.00 kg and radius 20.0 cm. The box is releasedfrom rest and descends as the rope unwraps from the cylinder. Findthe speed of the box when it has fallen 1.50 m.9.93. A thin, flat, uniform disk has mass M and radius R. A circularhole of radius centered at a point from the disk’s center, isthen punched in the disk. (a) Find the moment of inertia of the diskwith the hole about an axis through the original center of the disk,perpendicular to the plane of the disk. (Hint: Find the moment ofinertia of the piece punched from the disk.) (b) Find the momentof inertia of the disk with the hole about an axis through the centerof the hole, perpendicular to the plane of the disk.9.94. A pendulum is made of a uniform solid sphere with mass Mand radius R suspended from the end of a light rod. The distancefrom the pivot at the upper end of the rod to the center of thesphere is L. The pendulum’s moment of inertia for rotation aboutthe pivot is usually approximated as (a) Use the parallel-axistheorem to show that if R is 5% of L and the mass of the rod isignored, is only 0.1% greater than (b) If the mass of the rodis 1% of M and R is much less than L, what is the ratio of for anaxis at the pivot to 9.95. Perpendicular-Axis Theorem. Consider a rigid body thatis a thin, plane sheet of arbitrary shape. Take the body to lie in thexy-plane and let the origin O of coordinates be located at any pointwithin or outside the body. Let and be the moments of inertiaabout the x- and y-axes, and let be the moment of inertia aboutan axis through O perpendicular to the plane. (a) By consideringmass elements with coordinates show that This is called the perpendicular-axis theorem. Note that point Odoes not have to be the center of mass. (b) For a thin washer withmass M and with inner and outer radii and use the perpendi-cular-axis theorem to find the moment of inertia about an axis thatis in the plane of the washer and that passes through its center. Youmay use the information in Table 9.2. (c) Use the perpendicular-axis theorem to show that for a thin, square sheet with mass M andside L, the moment of inertia about any axis in the plane of thesheet that passes through the center of the sheet is You mayuse the information in Table 9.2.9.96. A thin, uniform rod is bent into a square of side length a. Ifthe total mass is M, find the moment of inertia about an axisthrough the center and perpendicular to the plane of the square.(Hint: Use the parallel-axis theorem.)*9.97. A cylinder with radius R and mass M has density thatincreases linearly with distance r from the cylinder axis, where is a positive constant. (a) Calculate the moment of inertiaof the cylinder about a longitudinal axis through its center in termsof M and R. (b) Is your answergreater or smaller than the momentof inertia of a cylinder of the samemass and radius but of uniformdensity? Explain why this resultmakes qualitative sense.9.98. Neutron Stars and Super-nova Remnants. The Crab Neb-ula is a cloud of glowing gasabout 10 light-years across, locatedabout 6500 light years from theearth (Fig. 9.39). It is the remnantof a star that underwent a super-nova explosion, seen on earth in1054 A.D. Energy is released by the
ar 5 ar,
12 ML2.
R2 ,R1
Ix 1 Iy 5 IO .1 xi , yi 2 ,mi
IO
IyIx
ML2 ?
Irod
ML2.IP
ML2.IP
R/2R/4,
Pulley
BoxCylinder
Figure 9.38 Problem 9.92.
Crab Nebula at a rate of about about times therate at which the sun radiates energy. The Crab Nebula obtains itsenergy from the rotational kinetic energy of a rapidly spinningneutron star at its center. This object rotates once every 0.0331 s,and this period is increasing by for each second oftime that elapses. (a) If the rate at which energy is lost by the neu-tron star is equal to the rate at which energy is released by the neb-ula, find the moment of inertia of the neutron star. (b) Theories ofsupernovae predict that the neutron star in the Crab Nebula has amass about 1.4 times that of the sun. Modeling the neutron star as asolid uniform sphere, calculate its radius in kilometers. (c) What isthe linear speed of a point on the equator of the neutron star? Com-pare to the speed of light. (d) Assume that the neutron star is uni-form and calculate its density. Compare to the density of ordinaryrock and to the density of an atomic nucleus (about
Justify the statement that a neutron star is essentiallya large atomic nucleus.
Challenge Problems9.99. The moment of inertia of a sphere with uniform densityabout an axis through its center is Satelliteobservations show that the earth’s moment of inertia isGeophysical data suggest the earth consists of five main re-gions: the inner core to of average density
the outer core to ofaverage density the lower mantle to
of average density the upper mantleto of average density
and the outer crust and oceans to ofaverage density (a) Show that the moment of inertiaabout a diameter of a uniform spherical shell of inner radius outer radius and density is (Hint:Form the shell by superposition of a sphere of density and asmaller sphere of density (b) Check the given data by usingthem to calculate the mass of the earth. (c) Use the given data tocalculate the earth’s moment of inertia in terms of MR2.
2r.)r
I 5 r 18p/15 2 1R2
5 2 R1
5 2 .rR2 ,R1 ,
2400 kg/m3.r 5 6370 km 21 r 5 6350 km
3600 kg/m3,r 5 6350 km 21 r 5 5700 km4900 kg/m3,r 5 5700 km 2 1 r 5 3480 km10,900 kg/m3,
r 5 3480 km 21 r 5 1220 km12,900 kg/m3,r 5 1220 km 21 r 5 0
0.3308MR2.
25 MR2 5 0.400MR2.
1017 kg/m3).13000 kg/m3 2
4.22 3 10213 s
1055 3 1031 W, *9.100. Calculate the moment of iner-tia of a uniform solid cone about anaxis through its center (Fig. 9.40).The cone has mass M and altitude h.The radius of its circular base is R.9.101. On a compact disc (CD), musicis coded in a pattern of tiny pitsarranged in a track that spirals out-ward toward the rim of the disc. Asthe disc spins inside a CD player, thetrack is scanned at a constant linearspeed of Because theradius of the track varies as it spiralsoutward, the angular speed of thedisc must change as the CD is played. (See Exercise 9.22.) Let’ssee what angular acceleration is required to keep constant. Theequation of a spiral is where is the radius ofthe spiral at and is a constant. On a CD, is the innerradius of the spiral track. If we take the rotation direction of theCD to be positive, must be positive so that r increases as the discturns and increases. (a) When the disc rotates through a smallangle the distance scanned along the track is Usingthe above expression for integrate ds to find the total dis-tance s scanned along the track as a function of the total anglethrough which the disc has rotated. (b) Since the track is scannedat a constant linear speed the distance s found in part (a) isequal to Use this to find as a function of time. There will betwo solutions for choose the positive one, and explain why thisis the solution to choose. (c) Use your expression for to findthe angular velocity and the angular acceleration as func-tions of time. Is constant? (d) On a CD, the inner radius of thetrack is 25.0 mm, the track radius increases by per revo-lution, and the playing time is 74.0 min. Find the values of and
and find the total number of revolutions made during the play-ing time. (e) Using your results from parts (c) and (d), makegraphs of (in versus t and (in versus t between
and t 5 74.0 min.t 5 0rad/s2)azrad/s)vz
b,r0
1.55 mmaz
azvz
u 1 t 2u ;uvt.
v,
ur 1 u 2 , ds 5 r du.du,
ub
r0bu 5 0r0r 1 u 2 5 r0 1 bu,
v
v 5 1.25 m/s.
h
R
Axis
Figure 9.40 ChallengeProblem 9.100.
5.00 m
4.00 kg
2.00 kg
Figure 9.36 Problem 9.86.
R2
R1
1.50 kg
Figure 9.37Problem 9.89.
Figure 9.39 Problem 9.98.