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O 11
CHAPTER
R L L I N G , TO R Q U E ,
AND ANGULAR
MOMENTUM
11-1
W H AT I S P H YS I C S ?
As we discussed in Chapter 10, physics includes the study of rotation.
Arguably, the most important application of that physics is in the rolling motion
of wheels and wheel-like objects. This applied physics has long been used. For example, when the prehistoric people of Easter Island moved their gigantic stone
statues from the quarry and across the island, they dragged them over logs acting
as rollers. Much later, when settlers moved westward across America in the 1800s,
they rolled their possessions rst by wagon and then later by train. Today, like it
or not, the world is lled with cars, trucks, motorcycles, bicycles, and other rolling
vehicles.
The physics and engineering of rolling have been around for so long that you
might think no fresh ideas remain to be developed. However, skateboards and inline skates were invented and engineered fairly recently, to become huge nancial successes. Street luge is now catching on, and the self-righting Segway (Fig.
11-1) may change the way people move around in large cities. Applying the
physics of rolling can still lead to surprises and rewards. Our starting point in
exploring that physics is to simplify rolling motion.
11-2 Rolling as Translation and Rotation Combined
Here we consider only objects that roll smoothly along a surface; that is, the objects
roll without slipping or bouncing on the surface. Figure 11-2 shows how complicated
smooth rolling motion can be: Although the center of the object moves in a straight
line parallel to the surface, a point on the rim certainly does not. However, we can
study this motion by treating it as a combination of translation of the center of mass
and rotation of the rest of the object around that center.
The self-righting Segway
Human Transporter. (Justin Sullivan/Getty
Images News and Sport Services)
Fig. 11-1
A time-exposure photograph of a rolling disk. Small lights have been attached to the disk, one at its center and one at its edge. The latter traces out a curve called
a cycloid. (Richard Megna/Fundamental Photographs)
Fig. 11-2
275
275
276
CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
O
vcom
vcom
O
s
P
P
s
Fig. 11-3 The center of mass O of a
rolling wheel moves a distance s at velocv
ity :com while the wheel rotates through
angle u. The point P at which the wheel
makes contact with the surface over which
the wheel rolls also moves a distance s.
To see how we do this, pretend you are standing on a sidewalk watching the
bicycle wheel of Fig. 11-3 as it rolls along a street. As shown, you see the center of
mass O of the wheel move forward at constant speed vcom. The point P on the
street where the wheel makes contact with the street surface also moves forward
at speed vcom, so that P always remains directly below O.
During a time interval t, you see both O and P move forward by a distance s.The
bicycle rider sees the wheel rotate through an angle u about the center of the wheel,
with the point of the wheel that was touching the street at the beginning of t moving
through arc length s. Equation 10-17 relates the arc length s to the rotation angle u :
uR,
s
(11-1)
where R is the radius of the wheel. The linear speed vcom of the center of the
wheel (the center of mass of this uniform wheel) is ds/dt. The angular speed v of
the wheel about its center is du/dt. Thus, differentiating Eq. 11-1 with respect to
time (with R held constant) gives us
vcom
( a ) Pure rotation
v = vcom
+
T
vR
(11-2)
(smooth rolling motion).
( b ) Pure translation
=
( c ) Rolling motion
v = 2vcom
v = vcom
T
T
vcom
O
P
v = vcom
O
P
v = vcom
vcom
O
P
v = vcom + vcom = 0
Rolling motion of a wheel as a combination of purely rotational motion and
purely translational motion. (a) The purely rotational motion:All points on the wheel move
with the same angular speed v. Points on the outside edge of the wheel all move with the
same linear speed v vcom.The linear velocities : of two such points, at top (T ) and bottom
v
(P) of the wheel, are shown. (b) The purely translational motion:All points on the wheel
move to the right with the same linear velocity :com. (c) The rolling motion of the wheel is the
v
combination of (a) and (b).
Fig. 11-4
A photograph of a rolling bicycle wheel. The spokes near the wheels
top are more blurred than those near the
bottom because the top ones are moving
faster, as Fig. 11-4c shows. (Courtesy Alice
Halliday)
Figure 11-4 shows that the rolling motion of a wheel is a combination of
purely translational and purely rotational motions. Figure 11-4a shows the purely
rotational motion (as if the rotation axis through the center were stationary):
Every point on the wheel rotates about the center with angular speed v. (This is
the type of motion we considered in Chapter 10.) Every point on the outside
edge of the wheel has linear speed vcom given by Eq. 11-2. Figure 11-4b shows the
purely translational motion (as if the wheel did not rotate at all): Every point on
the wheel moves to the right with speed vcom.
The combination of Figs. 11-4a and 11-4b yields the actual rolling motion of
the wheel, Fig. 11-4c. Note that in this combination of motions, the portion of the
wheel at the bottom (at point P) is stationary and the portion at the top (at point
T ) is moving at speed 2vcom, faster than any other portion of the wheel. These results are demonstrated in Fig. 11-5, which is a time exposure of a rolling bicycle
wheel. You can tell that the wheel is moving faster near its top than near its bottom because the spokes are more blurred at the top than at the bottom.
The motion of any round body rolling smoothly over a surface can be separated
into purely rotational and purely translational motions, as in Figs. 11-4a and 11-4b.
Fig. 11-5
Rolling as Pure Rotation
Figure 11-6 suggests another way to look at the rolling motion of a wheel
namely, as pure rotation about an axis that always extends through the point
PA R T 1
11-3 THE KINETIC ENERGY OF ROLLING
where the wheel contacts the street as the wheel moves. We consider the rolling
motion to be pure rotation about an axis passing through point P in Fig. 11-4c and
perpendicular to the plane of the gure. The vectors in Fig. 11-6 then represent
the instantaneous velocities of points on the rolling wheel.
T
O
Question: What angular speed about this new axis will a stationary observer assign to a rolling bicycle wheel?
Answer: The same v that the rider assigns to the wheel as she or he observes it
in pure rotation about an axis through its center of mass.
Rotation axis at P
Rolling can be viewed
as pure rotation, with angular speed v,
about an axis that always extends through
P. The vectors show the instantaneous linear velocities of selected points on the
rolling wheel. You can obtain the vectors by
combining the translational and rotational
motions as in Fig. 11-4.
Fig. 11-6
To verify this answer, let us use it to calculate the linear speed of the top of the
rolling wheel from the point of view of a stationary observer. If we call the
wheels radius R, the top is a distance 2R from the axis through P in Fig. 11-6, so
the linear speed at the top should be (using Eq. 11-2)
(v)(2R)
vtop
2(vR)
2vcom,
in exact agreement with Fig. 11-4c. You can similarly verify the linear speeds
shown for the portions of the wheel at points O and P in Fig. 11-4c.
CHECKPOINT 1
The rear wheel on a clowns bicycle has twice the radius of the front wheel. (a) When
the bicycle is moving, is the linear speed at the very top of the rear wheel greater than,
less than, or the same as that of the very top of the front wheel? (b) Is the angular speed
of the rear wheel greater than, less than, or the same as that of the front wheel?
11-3 The Kinetic Energy of Rolling
Let us now calculate the kinetic energy of the rolling wheel as measured by the
stationary observer. If we view the rolling as pure rotation about an axis through
P in Fig. 11-6, then from Eq. 10-34 we have
1
2 IP
K
2
(11-3)
,
in which v is the angular speed of the wheel and IP is the rotational inertia of the
wheel about the axis through P. From the parallel-axis theorem of Eq. 10-36
(I Icom Mh2), we have
IP
Icom
MR2,
(11-4)
in which M is the mass of the wheel, Icom is its rotational inertia about an axis
through its center of mass, and R (the wheels radius) is the perpendicular
distance h. Substituting Eq. 11-4 into Eq. 11-3, we obtain
K
and using the relation vcom
1
2 Icom
2
1
2
R2
2
,
vR (Eq. 11-2) yields
K
1
2 Icom
2
1
2
v2 .
com
(11-5)
1
We can interpret the term 2 Icom 2 as the kinetic energy associated with the
rotation of the wheel about an axis through its center of mass (Fig. 11-4a), and the
term 1Mv 2 as the kinetic energy associated with the translational motion of the
com
2
wheels center of mass (Fig. 11-4b). Thus, we have the following rule:
A rolling object has two types of kinetic energy: a rotational kinetic energy(1 Icom
2
due to its rotation about its center of mass and a translational kinetic energy
1
2
(2Mvcom) due to translation of its center of mass..
2
)
277
278
CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
11-4 The Forces of Rolling
Friction and Rolling
acom
P
fs
A wheel rolls horizontally
without sliding while accelerating with linear acceleration :com. A static frictional
a
:
force f s acts on the wheel at P, opposing its
tendency to slide.
Fig. 11-7
If a wheel rolls at constant speed, as in Fig. 11-3, it has no tendency to slide at the
point of contact P, and thus no frictional force acts there. However, if a net force
acts on the rolling wheel to speed it up or to slow it, then that net force causes acceleration :com of the center of mass along the direction of travel. It also causes
a
the wheel to rotate faster or slower, which means it causes an angular
acceleration a. These accelerations tend to make the wheel slide at P. Thus, a frictional force must act on the wheel at P to oppose that tendency.
:
If the wheel does not slide, the force is a static frictional force f s and the
motion is smooth rolling. We can then relate the magnitudes of the linear acceleration :com and the angular acceleration a by differentiating Eq. 11-2 with respect
a
to time (with R held constant). On the left side, dvcom/dt is acom, and on the right
side dv/dt is a. So, for smooth rolling we have
aR
acom
(smooth rolling motion).
(11-6)
If the wheel does slide when the net force acts on it, the frictional force that
:
acts at P in Fig. 11-3 is a kinetic frictional force f k . The motion then is not smooth
rolling, and Eq. 11-6 does not apply to the motion. In this chapter we discuss only
smooth rolling motion.
Figure 11-7 shows an example in which a wheel is being made to rotate faster
while rolling to the right along a at surface, as on a bicycle at the start of a race.
The faster rotation tends to make the bottom of the wheel slide to the left at
point P. A frictional force at P, directed to the right, opposes this tendency to
:
slide. If the wheel does not slide, that frictional force is a static frictional force f s
(as shown), the motion is smooth rolling, and Eq. 11-6 applies to the motion.
(Without friction, bicycle races would be stationary and very boring.)
If the wheel in Fig. 11-7 were made to rotate slower, as on a slowing bicycle, we would change the figure in two ways: The directions of the center-of:
mass acceleration :com and the frictional force f s at point P would now be to
a
the left.
Rolling Down a Ramp
Figure 11-8 shows a round uniform body of mass M and radius R rolling smoothly
down a ramp at angle u, along an x axis.We want to nd an expression for the bodys
acceleration acom, x down the ramp. We do this by using Newtons second law in both
its linear version (Fnet Ma) and its angular version (tnet Ia).
We start by drawing the forces on the body as shown in Fig. 11-8:
:
1. The gravitational force Fg on the body is directed downward. The tail of the
vector is placed at the center of mass of the body. The component along the
ramp is Fg sin u, which is equal to Mg sin u.
:
2. A normal force FN is perpendicular to the ramp. It acts at the point of
contact P, but in Fig. 11-8 the vector has been shifted along its direction until
its tail is at the bodys center of mass.
:
3. A static frictional force f s acts at the point of contact P and is directed up
the ramp. (Do you see why? If the body were to slide at P, it would slide down the
ramp.Thus, the frictional force opposing the sliding must be up the ramp.)
We can write Newtons second law for components along the x axis in Fig. 11-8
(Fnet,x max) as
fs
Mg sin u
Macom, x.
(11-7)
PA R T 1
11-4 THE FORCES OF ROLLING
FN
Forces Fg sin and fs
determine the linear
acceleration down
the ramp.
Forces FN and Fg cos
merely balance.
Fg sin
R
fs
x
P
The torque due to fs
determines the
angular acceleration
around the com.
Fg cos
Fg
Fig. 11-8 A round uniform body of radius R rolls down a ramp. The forces that act on it
:
:
:
are the gravitational force Fg , a normal force FN , and a frictional force f s pointing up the
:
ramp. (For clarity, vector FN has been shifted in the direction it points until its tail is at the
center of the body.)
This equation contains two unknowns, f s and acom, x. (We should not assume that f s
is at its maximum value f s,max. All we know is that the value of f s is just right for
the body to roll smoothly down the ramp, without sliding.)
We now wish to apply Newtons second law in angular form to the bodys rotation about its center of mass. First, we shall use Eq. 10-41 (
r F ) to write the
:
torques on the body about that point. The frictional force f s has moment arm R
and thus produces a torque Rfs, which is positive because it tends to rotate the
:
:
body counterclockwise in Fig. 11-8. Forces Fg and FN have zero moment arms
about the center of mass and thus produce zero torques. So we can write the angular form of Newtons second law (tnet Ia) about an axis through the bodys
center of mass as
Icoma.
Rf s
(11-8)
This equation contains two unknowns, f s and a.
Because the body is rolling smoothly, we can use Eq. 11-6 (acom aR) to relate
the unknowns acom,x and a. But we must be cautious because here acom,x is negative
(in the negative direction of the x axis) and a is positive (counterclockwise). Thus
we substitute acom, x/R for a in Eq. 11-8.Then, solving for f s, we obtain
a
fs
Icom com, x .
(11-9)
R2
Substituting the right side of Eq. 11-9 for fs in Eq. 11-7, we then nd
acom, x
1
g sin
.
Icom /MR2
(11-10)
We can use this equation to nd the linear acceleration acom,x of any body rolling
along an incline of angle u with the horizontal.
CHECKPOINT 2
Disks A and B are identical and roll across a oor with equal speeds. Then disk A rolls
up an incline, reaching a maximum height h, and disk B moves up an incline that is identical except that it is frictionless. Is the maximum height reached by disk B greater than,
less than, or equal to h?
279
280
CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
Sample Problem
Ball rolling down a ramp
A uniform ball, of mass M 6.00 kg and radius R, rolls
smoothly from rest down a ramp at angle u 30.0 (Fig. 11-8).
(a) The ball descends a vertical height h 1.20 m to reach the
bottom of the ramp.What is its speed at the bottom?
Doing so, substituting 2MR2 for Icom (from Table 10-2f), and
5
then solving for vcom give us
vcom
2(10)gh
7
2(10)(9.8 m/s2)(1.20 m)
7
4.10 m/s.
(Answer)
Note that the answer does not depend on M or R.
KEY IDEAS
The mechanical energy E of the ball Earth system is conserved as the ball rolls down the ramp. The reason is that the
only force doing work on the ball is the gravitational force, a
conservative force. The normal force on the ball from the
ramp does zero work because it is perpendicular to the
balls path. The frictional force on the ball from the ramp
does not transfer any energy to thermal energy because the
ball does not slide (it rolls smoothly).
Therefore, we can write the conservation of mechanical
energy (Ef Ei) as
Kf
Uf
Ki
Ui,
Calculations: Substituting into Eq. 11-11 gives us
2
1
2
2
vcom)
0
0
KEY IDEA
Because the ball rolls smoothly, Eq. 11-9 gives the frictional
force on the ball.
Calculations: Before we can use Eq. 11-9, we need the
balls acceleration acom, x from Eq. 11-10:
acom, x
(11-11)
where subscripts f and i refer to the nal values (at the bottom)
and initial values (at rest), respectively.The gravitational potential energy is initially Ui Mgh (where M is the balls mass)
and nally Uf 0. The kinetic energy is initially Ki 0. For the
nal kinetic energy Kf, we need an additional idea: Because the
ball rolls, the kinetic energy involves both translation and rotation, so we include them both by using the right side of Eq. 11-5.
(1 Icom
2
(b) What are the magnitude and direction of the frictional
force on the ball as it rolls down the ramp?
Mgh,
(11-12)
where Icom is the balls rotational inertia about an axis
through its center of mass, vcom is the requested speed at the
bottom, and v is the angular speed there.
Because the ball rolls smoothly, we can use Eq. 11-2 to
substitute vcom/R for v to reduce the unknowns in Eq. 11-12.
1
g sin
Icom /MR2
g sin
1
(9.8 m/s2) sin 30.0
12
5
2
2
2
5 MR /MR
3.50 m/s2.
Note that we needed neither mass M nor radius R to nd
acom,x. Thus, any size ball with any uniform mass would have
this acceleration down a 30.0 ramp, provided the ball rolls
smoothly.
We can now solve Eq. 11-9 as
a
acom, x
2
2
2
Icom com, x
fs
5 MR
5 Macom, x
R2
R2
2
5 (6.00
kg)( 3.50 m/s2)
8.40 N.
(Answer)
Note that we needed mass M but not radius R. Thus, the
frictional force on any 6.00 kg ball rolling smoothly down
a 30.0 ramp would be 8.40 N regardless of the balls radius but would be larger for a larger mass.
Additional examples, video, and practice available at WileyPLUS
11-5 The Yo-Yo
A yo-yo is a physics lab that you can t in your pocket. If a yo-yo rolls down its
string for a distance h, it loses potential energy in amount mgh but gains kinetic
energy in both translational (1Mv2 ) and rotational (1 Icom 2) forms. As it climbs
com
2
2
back up, it loses kinetic energy and regains potential energy.
In a modern yo-yo, the string is not tied to the axle but is looped around it.
When the yo-yo hits the bottom of its string, an upward force on the axle from
the string stops the descent. The yo-yo then spins, axle inside loop, with only
rotational kinetic energy. The yo-yo keeps spinning (sleeping) until you wake
it by jerking on the string, causing the string to catch on the axle and the yo-yo to
climb back up. The rotational kinetic energy of the yo-yo at the bottom of its
PA R T 1
11-6 TORQUE REVISITED
281
string (and thus the sleeping time) can be considerably increased by throwing the
yo-yo downward so that it starts down the string with initial speeds vcom and v instead of rolling down from rest.
To nd an expression for the linear acceleration acom of a yo-yo rolling down
a string, we could use Newtons second law just as we did for the body rolling
down a ramp in Fig. 11-8. The analysis is the same except for the following:
1. Instead of rolling down a ramp at angle u with the horizontal, the yo-yo rolls
down a string at angle u 90 with the horizontal.
2. Instead of rolling on its outer surface at radius R, the yo-yo rolls on an axle of
radius R0 (Fig. 11-9a).
:
3. Instead of being slowed by frictional force f s, the yo-yo is slowed by the force
:
T on it from the string (Fig. 11-9b).
T
R
R0
The analysis would again lead us to Eq. 11-10. Therefore, let us just change the
notation in Eq. 11-10 and set u 90 to write the linear acceleration as
acom
1
g
,
Icom /MR2
0
R0
(11-13)
Fg
where Icom is the yo-yos rotational inertia about its center and M is its mass. A yoyo has the same downward acceleration when it is climbing back up.
11-6 Torque Revisited
In Chapter 10 we dened torque t for a rigid body that can rotate around a xed
axis, with each particle in the body forced to move in a path that is a circle centered on that axis. We now expand the denition of torque to apply it to an individual particle that moves along any path relative to a xed point (rather than a
xed axis). The path need no longer be a circle, and we must write the torque as a
vector : that may have any direction.
Figure 11-10a shows such a particle at point A in an xy plane. A single force
:
F in that plane acts on the particle, and the particles position relative to the origin O is given by position vector :. The torque : acting on the particle relative to
r
the xed point O is a vector quantity dened as
:
:
:
r
F
(a)
(b)
Fig. 11-9 (a) A yo-yo, shown in cross
section. The string, of assumed negligible
thickness, is wound around an axle of radius R0. (b) A free-body diagram for the
falling yo-yo. Only the axle is shown.
(11-14)
(torque dened).
We can evaluate the vector (or cross) product in this denition of : by using
t
the rules for such products given in Section 3-8. To nd the direction of :, we slide
t
z
z
z
Cross r into F.
Torque is in the
positive z direction.
(= r F)
y
O
O
F (redrawn, with
tail at origin)
F
r
r
A
F
A
y
y
O
r
r
F
A
Line of action of F
x
x
x
(a)
(b)
:
Fig. 11-10 Dening torque. (a) A force F , lying in an xy plane, acts on a particle at
:
point A. (b) This force produces a torque : ( :
r
F ) on the particle with respect to
the origin O. By the right-hand rule for vector (cross) products, the torque vector points
in the positive direction of z. Its magnitude is given by rF in (b) and by r F in (c).
(c)
F
CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
282
:
the vector F (without changing its direction) until its tail is at the origin O, so that
the two vectors in the vector product are tail to tail as in Fig. 11-10b. We
then use the right-hand rule for vector products in Fig. 3-19a, sweeping the ngers
:
of the right hand from : (the rst vector in the product) into F (the second vector).
r
The outstretched right thumb then gives the direction of :. In Fig. 11-10b, the direct
tion of : is in the positive direction of the z axis.
t
To determine the magnitude of :, we apply the general result of Eq. 3-27
t
(c ab sin f), nding
t rF sin f,
(11-15)
:
r
where f is the smaller angle between the directions of : and F when the vectors
are tail to tail. From Fig. 11-10b, we see that Eq. 11-15 can be rewritten as
(11-16)
rF ,
Sample Problem
Torque on a particle due to a force
In Fig. 11-11a, three forces, each of magnitude 2.0 N, act on a
particle. The particle is in the xz plane at point A given by
:
position vector :, where r 3.0 m and u 30. Force F1 is
r
:
parallel to the x axis, force F2 is parallel to the z axis, and
:
force F3 is parallel to the y axis. What is the torque, about the
origin O, due to each force?
vector (or cross) products, with magnitudes given by Eq. 11-15
(t rF sin f) and directions given by the right-hand rule for
vector products.
Calculations: Because we want the torques with respect to
the origin O, the vector : required for each cross product is
r
the given position vector. To determine the angle f between
the direction of : and the direction of each force, we shift
r
the force vectors of Fig. 11-11a, each in turn, so that their
tails are at the origin. Figures 11-11b, c, and d, which are di:
rect views of the xz plane, show the shifted force vectors F1,
KEY IDEA
Because the three force vectors do not lie in a plane, we cannot
evaluate their torques as in Chapter 10. Instead, we must use
z
A
z
r
x
F3
A
z
r
1 = 150
= 30
r
x
O
(b)
F1
z
x
F1
F1
F1
r
(a)
F2
O
y
Cross r into F1.
Torque 1 is into the
figure (negative y).
z
z
r
x
(a) A particle at point A is
acted on by three forces, each parallel to a
coordinate axis.The angle f (used in nd:
ing torque) is shown (b) for F1 and (c) for
:
:
F 2. (d ) Torque t 3 is perpendicular to both
:
:
:
r and F 3 (force F 3 is directed into the
plane of the gure). (e) The torques (relative to the origin O) acting on the particle.
Fig. 11-11
x
= 30
2 = 120
z
r
O
F2
r
x
x
F2
(c)
Cross r into F2.
Torque 2 is out of
the figure (positive y).
F2
PA R T 1
11-6 TORQUE REVISITED
283
:
where F ( F sin ) is the component of F perpendicular to :. From Fig. 11-10c,
r
we see that Eq. 11-15 can also be rewritten as
(11-17)
r F,
:
where r ( r sin ) is the moment arm of F (the perpendicular distance
:
between O and the line of action of F ).
CHECKPOINT 3
r
The position vector : of a particle points along the positive direction of a z axis. If
the torque on the particle is (a) zero, (b) in the negative direction of x, and (c) in the
negative direction of y, in what direction is the force causing the torque?
:
:
F2, and F3, respectively. (Note how much easier the angles
between the force vectors and the position vector are to
see.) In Fig. 11-11d, the angle between the directions of :
r
:
:
and F3 is 90 and the symbol means F3 is directed into the
page. If it were directed out of the page, it would be represented with the symbol .
Now, applying Eq. 11-15 for each force, we nd the magnitudes of the torques to be
t1 rF1 sin f1 (3.0 m)(2.0 N)(sin 150) 3.0 N m,
rF2 sin f2
t2
(3.0 m)(2.0 N)(sin 120)
5.2 N m,
t3
and
(3.0 m)(2.0 N)(sin 90)
6.0 N m.
(Answer)
To nd the directions of these torques, we use the righthand rule, placing the ngers of the right hand so as to
:
rotate : into F through the smaller of the two angles
r
between their directions. The thumb points in the direction of
the torque. Thus :1 is directed into the page in Fig. 11-11b; :2
t
t
is directed out of the page in Fig. 11-11c; and :3 is
t
directed as shown in Fig. 11-11d. All three torque vectors are
shown in Fig. 11-11e.
z
z
r
x
rF3 sin f3
z
r
F3
= 30
3
r
x
x
3
3
(d)
z
O
1
2
(e)
x
y
Cross r into F3.
Torque 3 is
in the xz plane.
These are the three torques
acting on the particle, each
measured about the origin O.
3
Additional examples, video, and practice available at WileyPLUS
284
CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
11-7 Angular Momentum
z
(= r p)
p (redrawn, with
tail at origin)
O
y
p
r
A
p
x
:
Recall that the concept of linear momentum p and the principle of conservation
of linear momentum are extremely powerful tools. They allow us to predict
the outcome of, say, a collision of two cars without knowing the details of the col:
lision. Here we begin a discussion of the angular counterpart of p, winding up in
Section 11-11 with the angular counterpart of the conservation principle.
:
:
Figure 11-12 shows a particle of mass m with linear momentum p ( m v ) as
:
it passes through point A in an xy plane. The angular momentum of this particle with respect to the origin O is a vector quantity dened as
(a)
:
z
y
O
r
r
A
p
Extension of p
x
(b)
Dening angular momentum. A particle passing through point A has
:
:
linear momentum p ( m v ) , with the vec:
tor p lying in an xy plane. The particle has
:
:
angular momentum ( : p ) with rer
spect to the origin O. By the right-hand
rule, the angular momentum vector points
in the positive direction of z. (a) The mag:
nitude of is given by
rmv .
rp
:
(b) The magnitude of is also given by
r p r mv.
Fig. 11-12
:
r
:
p
m( :
r
:
v)
(angular momentum dened),
(11-18)
r
where : is the position vector of the particle with respect to O. As the particle
:
v
moves relative to O in the direction of its momentum p ( m :), position vector
:
r rotates around O. Note carefully that to have angular momentum about O, the
particle does not itself have to rotate around O. Comparison of Eqs. 11-14 and 11-18
shows that angular momentum bears the same relation to linear momentum that
torque does to force. The SI unit of angular momentum is the kilogrammeter-squared per second (kg m2/s), equivalent to the joule-second (J s).
:
To nd the direction of the angular momentum vector in Fig. 11-12, we
:
slide the vector p until its tail is at the origin O. Then we use the right-hand rule
:
for vector products, sweeping the ngers from : into p . The outstretched thumb
r
:
then shows that the direction of is in the positive direction of the z axis in Fig. 11-12.
This positive direction is consistent with the counterclockwise rotation of position
:
vector : about the z axis, as the particle moves. (A negative direction of would be
r
consistent with a clockwise rotation of : about the z axis.)
r
:
To nd the magnitude of , we use the general result of Eq. 3-27 to write
(11-19)
rmv sin ,
:
:
where f is the smaller angle between r and p when these two vectors are tail
to tail. From Fig. 11-12a, we see that Eq. 11-19 can be rewritten as
rp
(11-20)
rmv ,
:
:
where p is the component of p perpendicular to r and v is the component
of : perpendicular to :. From Fig. 11-12b, we see that Eq. 11-19 can also be
r
v
rewritten as
(11-21)
r p r mv,
:
where r is the perpendicular distance between O and the extension of p .
Note two features here: (1) angular momentum has meaning only with respect to a specied origin and (2) its direction is always perpendicular to the
:
plane formed by the position and linear momentum vectors : and p .
r
CHECKPOINT 4
1
3
In part a of the gure, particles
1 and 2 move around point O
O
O
in circles with radii 2 m and 4
5
m. In part b, particles 3 and
4 travel along straight lines at
4
2
perpendicular distances of 4 m
and 2 m from point O. Particle
(a)
(b)
5 moves directly away from O.
All ve particles have the same mass and the same constant speed. (a) Rank the particles according to the magnitudes of their angular momentum about point O, greatest
rst. (b) Which particles have negative angular momentum about point O?
PA R T 1
11-8 NEWTONS SECOND LAW IN ANGULAR FORM
285
Sample Problem
Angular momentum of a two-particle system
Figure 11-13 shows an overhead view of two particles moving
at constant momentum along horizontal paths. Particle 1, with
momentum magnitude p1 5.0 kg m/s, has position vector :1
r
and will pass 2.0 m from point O. Particle 2, with momentum
magnitude p2 2.0 kg m/s, has position vector :2 and will pass
r
4.0 m from point O. What are the magnitude and direction of
:
the net angular momentum L about point O of the twoparticle system?
p1
Two particles
pass near point O.
Fig. 11-13
1
:
Calculations: For particle 1, Eq. 11-21 yields
1
r 1 p1
Similarly, the magnitude of
r 2 p2
2
To nd the direction of vector
we use Eq. 11-18 and the
:
right-hand rule for vector products. For :
r1 p1, the vector
product is out of the page, perpendicular to the plane of Fig.
11-13. This is the positive direction, consistent with the
counterclockwise rotation of the particles position vector
10 kg m2/s.
:
2
is
(4.0 m)(2.0 kg m /s)
:
and the vector product :
r2 p2 is into the page, which is the
negative direction, consistent with the clockwise rotation of
:
r 2 around O as particle 2 moves. Thus, the angular momentum vector for particle 2 is
2
8.0 kg m2/s.
L
1
2
10 kg m2/s
( 8.0 kg m2/s)
2.0 kg m2/s.
11-8 Newtons Second Law in Angular Form
Newtons second law written in the form
:
dp
dt
(single particle)
(11-22)
expresses the close relation between force and linear momentum for a single particle. We have seen enough of the parallelism between linear and angular quantities to be pretty sure that there is also a close relation between torque and angular momentum. Guided by Eq. 11-22, we can even guess that it must be
:
:
net
d
dt
(single particle).
(Answer)
The plus sign means that the systems net angular momentum about point O is out of the page.
Additional examples, video, and practice available at WileyPLUS
:
O
The net angular momentum for the two-particle system is
:
1,
F net
r1
8.0 kg m2/s,
(2.0 m)(5.0 kg m /s)
10 kg m2/s.
r2
r1
:
r 1 around O as particle 1 moves. Thus, the angular momentum vector for particle 1 is
KEY IDEA
To nd L, we can rst nd the individual angular momenta
:
:
1 and
2 and then add them. To evaluate their magnitudes,
we can use any one of Eqs. 11-18 through 11-21. However,
Eq. 11-21 is easiest, because we are given the perpendicular
distances r1 ( 2.0 m) and r2 ( 4.0 m) and the momentum magnitudes p1 and p2.
r2
p2
(11-23)
Equation 11-23 is indeed an angular form of Newtons second law for a single
particle:
The (vector) sum of all the torques acting on a particle is equal to the time rate of
change of the angular momentum of that particle.
286
CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
Equation 11-23 has no meaning unless the torques : and the angular momentum
:
are dened with respect to the same point, usually the origin of the coordinate
system being used.
Proof of Equation 11-23
We start with Eq. 11-18, the denition of the angular momentum of a particle:
:
m( :
r
:
v ),
:
where r is the position vector of the particle and : is the velocity of the particle.
v
Differentiating* each side with respect to time t yields
:
d
dt
m:
r
d:
v
dt
d:
r
dt
:
v.
(11-24)
:
However, dv /dt is the acceleration : of the particle, and d :/dt is its velocity :.
a
r
v
Thus, we can rewrite Eq. 11-24 as
:
d
a
v
v
m( : : : :).
r
dt
Now : : 0 (the vector product of any vector with itself is zero because the
v
v
angle between the two vectors is necessarily zero). Thus, the last term of this expression is eliminated and we then have
:
d
m( : :) : m:.
r
a
r
a
dt
:
:
We now use Newtons second law ( Fnet m:) to replace ma with its equal, the
a
vector sum of the forces that act on the particle, obtaining
:
d
:
:
:
(11-25)
Fnet
(:
r
F ).
r
dt
:
Here the symbol indicates that we must sum the vector products :
F for all
r
the forces. However, from Eq. 11-14, we know that each one of those vector products is the torque associated with one of the forces. Therefore, Eq. 11-25 tells us
that
:
d
:
.
net
dt
This is Eq. 11-23, the relation that we set out to prove.
CHECKPOINT 5
r
The gure shows the position vector : of a particle at a certain instant, and four choices
for the direction of a force that is to accelerate the particle. All four choices lie in the xy
:
plane. (a) Rank the choices according to the magnitude of the time rate of change (d /dt)
they produce in the angular momentum of the particle about point O, greatest rst. (b)
Which choice results in a negative rate of change about O?
F2
y
F3
r
F1
F4
O
x
*In differentiating a vector product, be sure not to change the order of the two quantities (here : and
r
:
v ) that form that product. (See Eq. 3-28.)
PA R T 1
11-8 NEWTONS SECOND LAW IN ANGULAR FORM
287
Sample Problem
Torque, time derivative of angular momentum, penguin fall
In Fig. 11-14, a penguin of mass m falls from rest at point A,
a horizontal distance D from the origin O of an xyz coordinate system. (The positive direction of the z axis is directly
outward from the plane of the gure.)
(a) What is the angular momentum
about O?
:
y
D
O
A
x
,
of the falling penguin
r
KEY IDEA
We can treat the penguin as a particle, and thus its angular
:
:
:
:
momentum is given by Eq. 11-18 (
r
p ), where :
r
is the penguins position vector (extending from O to the
:
penguin) and p is the penguins linear momentum. (The
penguin has angular momentum about O even though it
moves in a straight line, because vector : rotates about O
r
as the penguin falls.)
Fg or p
:
Calculations: To nd the magnitude of , we can use any
one of the scalar equations derived from Eq. 11-18
namely, Eqs. 11-19 through 11-21. However, Eq. 11-21
(
r mv) is easiest because the perpendicular distance r
:
between O and an extension of vector p is the given distance D. The speed of an object that has fallen from rest for
a time t is v gt. We can now write Eq. 11-21 in terms of
given quantities as
r mv
(Answer)
Dmgt.
:
To find the direction of , we use the right-hand rule
for the vector product : : in Eq. 11-18. Mentally shift
r
p
:
p until its tail is at the origin, and then use the fingers of
:
your right hand to rotate : into p through the smaller anr
gle between the two vectors. Your outstretched thumb
then points into the plane of the figure, indicating that the
:
product : : and thus also
are directed into that
r
p
plane, in the negative direction of the z axis. We represent
:
:
with an encircled cross
at O. The vector changes
with time in magnitude only; its direction remains unchanged.
(b) About the origin O, what is the torque
:
guin due to the gravitational force Fg?
:
on the pen-
KEY IDEAS
:
(1) The torque is given by Eq. 11-14 (: : F ), where
r
:
:
now the force is Fg. (2) Force Fg causes a torque on the penguin, even though the penguin moves in a straight line,
because : rotates about O as the penguin moves.
r
A penguin falls vertically from point A. The torque
:
and the angular momentum of the falling penguin with respect
to the origin O are directed into the plane of the gure at O.
Fig. 11-14
:
Calculations: To nd the magnitude of :, we can use any
one of the scalar equations derived from Eq. 11-14
namely, Eqs. 11-15 through 11-17. However, Eq. 11-17
(
r F ) is easiest because the perpendicular distance r
:
between O and the line of action of Fg is the given distance D.
:
So, substituting D and using mg for the magnitude of Fg, we
can write Eq. 11-17 as
t
DFg
Dmg.
(Answer)
:
Using the right-hand rule for the vector product : F in
r
Eq. 11-14, we nd that the direction of : is the negative
:
direction of the z axis, the same as .
The results we obtained in parts (a) and (b) must be
consistent with Newtons second law in the angular form of
:
Eq. 11-23 ( :net d /dt).To check the magnitudes we got, we
write Eq. 11-23 in component form for the z axis and then
substitute our result
Dmgt.We nd
d
dt
d(Dmgt)
dt
Dmg,
which is the magnitude we found for :. To check the
:
directions, we note that Eq. 11-23 tells us that : and d /dt
:
:
must have the same direction. So and must also have
the same direction, which is what we found.
Additional examples, video, and practice available at WileyPLUS
288
CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
11-9 The Angular Momentum of a System of Particles
Now we turn our attention to the angular momentum of a system of particles with
:
respect to an origin. The total angular momentum L of the system is the (vector)
:
sum of the angular momenta of the individual particles (here with label i):
:
L
:
1
:
2
:
3
n
:
n
i
:
i.
(11-26)
1
With time, the angular momenta of individual particles may change because
of interactions between the particles or with the outside. We can nd the resulting
:
change in L by taking the time derivative of Eq. 11-26. Thus,
:
:
n
dL
dt
i
1
di
.
dt
(11-27)
:
From Eq. 11-23, we see that d i /dt is equal to the net torque
particle. We can rewrite Eq. 11-27 as
:
n
dL
dt
:
:
.
on the ith
(11-28)
net,i
i
net,i
1
:
That is, the rate of change of the systems angular momentum L is equal to the
vector sum of the torques on its individual particles. Those torques include internal torques (due to forces between the particles) and external torques (due to
forces on the particles from bodies external to the system). However, the forces
between the particles always come in third-law force pairs so their torques sum to
:
zero. Thus, the only torques that can change the total angular momentum L of
the system are the external torques acting on the system.
Let :net represent the net external torque, the vector sum of all external
torques on all particles in the system. Then we can write Eq. 11-28 as
:
:
net
dL
dt
(system of particles),
(11-29)
which is Newtons second law in angular form. It says:
The net external torque :net acting on a system of particles is equal to the time rate of
:
change of the systems total angular momentum L .
:
:
Equation 11-29 is analogous to Fnet d P/dt (Eq. 9-27) but requires extra
caution: Torques and the systems angular momentum must be measured relative
to the same origin. If the center of mass of the system is not accelerating relative
to an inertial frame, that origin can be any point. However, if it is accelerating,
then it must be the origin. For example, consider a wheel as the system of particles. If it is rotating about an axis that is xed relative to the ground, then the origin for applying Eq. 11-29 can be any point that is stationary relative to the
ground. However, if it is rotating about an axis that is accelerating (such as when
it rolls down a ramp), then the origin can be only at its center of mass.
11-10 The Angular Momentum of a Rigid Body
Rotating About a Fixed Axis
We next evaluate the angular momentum of a system of particles that form a rigid
body that rotates about a xed axis. Figure 11-15a shows such a body. The xed axis
of rotation is a z axis, and the body rotates about it with constant angular speed v.
We wish to nd the angular momentum of the body about that axis.
PA R T 1
11-10 THE ANGULAR MOMENTUM OF A RIGID BODY ROTATING ABOUT A FIXED AXIS
We can nd the angular momentum by summing the z components of the angular momenta of the mass elements in the body. In Fig. 11-15a, a typical mass element, of mass mi, moves around the z axis in a circular path. The position of
r
the mass element is located relative to the origin O by position vector :i. The
radius of the mass elements circular path is r i, the perpendicular distance
between the element and the z axis.
:
The magnitude of the angular momentum i of this mass element, with
respect to O, is given by Eq. 11-19:
(r i)( pi)(sin 90 )
i
z
r
i
sin
(r i)( mi vi),
(r i sin )( mi vi)
r
i
i
mi
pi
ri
where pi and vi are the linear momentum and linear speed of the mass element,
:
:
and 90 is the angle between :i and pi. The angular momentum vector i for the
r
mass element in Fig. 11-15a is shown in Fig. 11-15b; its direction must be perpen:
dicular to those of :i and pi.
r
:
We are interested in the component of i that is parallel to the rotation axis,
here the z axis. That z component is
iz
n
n
x
(a)
z
mi vi.
i
1
i
1
i
i
1
x
(11-30)
1
We can remove v from the summation here because it has the same value for all
points of the rotating rigid body.
The quantity
mi r 2 i in Eq. 11-30 is the rotational inertia I of the body
about the xed axis (see Eq. 10-33). Thus Eq. 11-30 reduces to
Iv
(11-31)
(rigid body, xed axis).
We have dropped the subscript z, but you must remember that the angular
momentum dened by Eq. 11-31 is the angular momentum about the rotation
axis. Also, I in that equation is the rotational inertia about that same axis.
Table 11-1, which supplements Table 10-3, extends our list of corresponding
linear and angular relations.
Table 11-1
More Corresponding Variables and Relations for Translational
and Rotational Motiona
Translational
Force
Linear momentum
Linear momentumb
Linear momentumb
Newtons second lawb
Conservation lawd
a
:
F
:
p
:
:
P(
p i)
:
:
P M vcom
:
dP
:
Fnet
dt
:
P a constant
Rotational
Torque
Angular momentum
Angular momentumb
Angular momentumc
Newtons second lawb
Conservation lawd
See also Table 10-3.
For systems of particles, including rigid bodies.
c
For a rigid body about a xed axis, with L being the component along that axis.
d
For a closed, isolated system.
b
y
O
mi r 2 i .
L
iz
mi( r i)r
i
n
i
i
n
mi vir
iz
y
O
The z component of the angular momentum for the rotating rigid body as a
whole is found by adding up the contributions of all the mass elements that make
up the body. Thus, because v
r , we may write
Lz
289
:
:
( : F)
r
( : :)
r
p
:
:
L(
)
i
L Iv
:
dL
:
net
dt
:
L a constant
:
(b)
(a) A rigid body rotates
about a z axis with angular speed v. A mass
element of mass mi within the body
moves about the z axis in a circle with radius r i . The mass element has linear momentum : i, and it is located relative to the
p
origin O by position vector :i. Here the
r
mass element is shown when r i is parallel to
:
the x axis. (b) The angular momentum i,
with respect to O, of the mass element in (a).
The z component iz is also shown.
Fig. 11-15
290
CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
CHECKPOINT 6
Hoop
Sphere
In the gure, a disk, a Disk
hoop, and a solid sphere
are made to spin about
F
F
F
xed central axes (like a
top) by means of strings wrapped around them, with the strings producing the same
:
constant tangential force F on all three objects. The three objects have the same mass
and radius, and they are initially stationary. Rank the objects according to (a) their angular momentum about their central axes and (b) their angular speed, greatest rst,
when the strings have been pulled for a certain time t.
11-11 Conservation of Angular Momentum
So far we have discussed two powerful conservation laws, the conservation of
energy and the conservation of linear momentum. Now we meet a third law of
this type, involving the conservation of angular momentum. We start from
:
Eq. 11-29 ( :net dL /dt), which is Newtons second law in angular form. If no
:
net external torque acts on the system, this equation becomes dL/dt 0, or
:
L
a constant
(isolated system).
(11-32)
This result, called the law of conservation of angular momentum, can also be
written as
net angular momentum
net angular momentum
,
at some initial time t i
at some later time t f
or
:
Li
:
Lf
(isolated system).
(11-33)
Equations 11-32 and 11-33 tell us:
:
If the net external torque acting on a system is zero, the angular momentum L of the
system remains constant, no matter what changes take place within the system.
Equations 11-32 and 11-33 are vector equations; as such, they are equivalent
to three component equations corresponding to the conservation of angular momentum in three mutually perpendicular directions. Depending on the torques
acting on a system, the angular momentum of the system might be conserved in
only one or two directions but not in all directions:
If the component of the net external torque on a system along a certain axis is zero,
then the component of the angular momentum of the system along that axis cannot
change, no matter what changes take place within the system.
We can apply this law to the isolated body in Fig. 11-15, which rotates around
the z axis. Suppose that the initially rigid body somehow redistributes its mass
relative to that rotation axis, changing its rotational inertia about that axis.
Equations 11-32 and 11-33 state that the angular momentum of the body cannot
change. Substituting Eq. 11-31 (for the angular momentum along the rotational
axis) into Eq. 11-33, we write this conservation law as
Iivi
If vf.
(11-34)
Here the subscripts refer to the values of the rotational inertia I and angular
speed v before and after the redistribution of mass.
Like the other two conservation laws that we have discussed, Eqs. 11-32 and
11-33 hold beyond the limitations of Newtonian mechanics. They hold for parti-
PA R T 1
11-11 CONSERVATION OF ANGULAR MOMENTUM
cles whose speeds approach that of light (where the theory of special relativity
reigns), and they remain true in the world of subatomic particles (where quantum
physics reigns). No exceptions to the law of conservation of angular momentum
have ever been found.
We now discuss four examples involving this law.
1. The spinning volunteer Figure 11-16 shows a student seated on a stool that
can rotate freely about a vertical axis. The student, who has been set into
rotation at a modest initial angular speed vi, holds two dumbbells in his
:
outstretched hands. His angular momentum vector L lies along the vertical rotation axis, pointing upward.
The instructor now asks the student to pull in his arms; this action reduces
his rotational inertia from its initial value Ii to a smaller value If because he
moves mass closer to the rotation axis. His rate of rotation increases markedly,
from vi to vf.The student can then slow down by extending his arms once more,
moving the dumbbells outward.
No net external torque acts on the system consisting of the student, stool,
and dumbbells. Thus, the angular momentum of that system about the rotation
axis must remain constant, no matter how the student maneuvers the dumbbells. In Fig. 11-16a, the students angular speed vi is relatively low and his rotational inertia Ii is relatively high. According to Eq. 11-34, his angular speed
in Fig. 11-16b must be greater to compensate for the decreased If.
2. The springboard diver Figure 11-17 shows a diver doing a forward one-anda-half-somersault dive. As you should expect, her center of mass follows a par:
abolic path. She leaves the springboard with a denite angular momentum L
about an axis through her center of mass, represented by a vector pointing
into the plane of Fig. 11-17, perpendicular to the page. When she is in the air,
no net external torque acts on her about her center of mass, so her angular
momentum about her center of mass cannot change. By pulling her arms and
legs into the closed tuck position, she can considerably reduce her rotational
inertia about the same axis and thus, according to Eq. 11-34, considerably
increase her angular speed. Pulling out of the tuck position (into the open layout position) at the end of the dive increases her rotational inertia and thus
slows her rotation rate so she can enter the water with little splash. Even in a
more complicated dive involving both twisting and somersaulting, the angular
momentum of the diver must be conserved, in both magnitude and direction,
throughout the dive.
3. Long jump When an athlete takes off from the ground in a running long
jump, the forces on the launching foot give the athlete an angular momentum
with a forward rotation around a horizontal axis. Such rotation would not allow
L
The divers angular
:
momentum L is constant throughout the dive,
being represented by the tail of an arrow that
is perpendicular to the plane of the gure. Note
also that her center of mass (see the dots) follows a parabolic path.
Fig. 11-17
Her angular momentum
is fixed but she can still
control her spin rate.
L
291
L
i
Ii
Rotation axis
(a)
L
f
If
(b)
(a) The student has a relatively large rotational inertia about the rotation axis and a relatively small angular
speed. (b) By decreasing his rotational inertia, the student automatically increases
his angular speed. The angular momentum
:
L of the rotating system remains unchanged.
Fig. 11-16
292
CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
Fig. 11-18 Windmill motion of the arms during a long jump helps maintain body orientation for a proper landing.
the jumper to land properly: In the landing, the legs should be together and extended forward at an angle so that the heels mark the sand at the greatest distance. Once airborne, the angular momentum cannot change (it is conserved)
because no external torque acts to change it. However, the jumper can shift
most of the angular momentum to the arms by rotating them in windmill fashion (Fig. 11-18). Then the body remains upright and in the proper orientation
for landing.
4. Tour jet In a tour jet, a ballet performer leaps with a small twisting motion
on the oor with one foot while holding the other leg perpendicular to the body
(Fig. 11-19a). The angular speed is so small that it may not be perceptible to the
audience. As the performer ascends, the outstretched leg is brought down and
the other leg is brought up, with both ending up at angle u to the body (Fig.
11-19b). The motion is graceful, but it also serves to increase the rotation because bringing in the initially outstretched leg decreases the performers rotational inertia. Since no external torque acts on the airborne performer, the angular momentum cannot change. Thus, with a decrease in rotational inertia, the
angular speed must increase. When the jump is well executed, the performer
seems to suddenly begin to spin and rotates 180 before the initial leg orientations are reversed in preparation for the landing. Once a leg is again outstretched, the rotation seems to vanish.
(a)
CHECKPOINT 7
(b )
(a) Initial phase of a tour
jet: large rotational inertia and small angular speed. (b) Later phase: smaller rotational inertia and larger angular speed.
Fig. 11-19
A rhinoceros beetle rides the rim of a small disk that rotates like a merry-go-round. If
the beetle crawls toward the center of the disk, do the following (each relative to the
central axis) increase, decrease, or remain the same for the beetle disk system: (a)
rotational inertia, (b) angular momentum, and (c) angular speed?
Sample Problem
Conservation of angular momentum, rotating wheel demo
Figure 11-20a shows a student, again sitting on a stool that
can rotate freely about a vertical axis. The student, initially
at rest, is holding a bicycle wheel whose rim is loaded with
lead and whose rotational inertia Iwh about its central axis is
1.2 kg m2. (The rim contains lead in order to make the
value of Iwh substantial.) The wheel is rotating at an angular
speed vwh of 3.9 rev/s; as seen from overhead, the rotation is
counterclockwise. The axis of the wheel is vertical, and the
:
angular momentum Lwh of the wheel points vertically upward. The student now inverts the wheel (Fig. 11-20b) so
that, as seen from overhead, it is rotating clockwise. Its angu:
lar momentum is now Lwh. The inversion results in the student, the stool, and the wheels center rotating together as a
composite rigid body about the stools rotation axis, with rotational inertia Ib 6.8 kg m2. (The fact that the wheel is
also rotating about its center does not affect the mass distribution of this composite body; thus, Ib has the same value
whether or not the wheel rotates.) With what angular speed
vb and in what direction does the composite body rotate after the inversion of the wheel?
PA R T 1
11-11 CONSERVATION OF ANGULAR MOMENTUM
:
Lb
Lwh
Lwh
wh
wh
b
293
:
3. The vector addition of Lb and Lwh gives the total angular
:
momentum Ltot of the system of the student, stool, and
wheel.
4. As the wheel is inverted, no net external torque acts on
:
that system to change Ltot about any vertical axis.
(Torques due to forces between the student and the
wheel as the student inverts the wheel are internal to the
system.) So, the systems total angular momentum is conserved about any vertical axis.
:
Calculations: The conservation of Ltot is represented with
vectors in Fig. 11-20c. We can also write this conservation in
terms of components along a vertical axis as
(a)
(b)
=
+
Lwh
Lb
Initial
Lwh
Lb,f
The student now has
angular momentum,
and the net of these
two vectors equals
the initial vector.
Final
(c)
(a) A student holds a bicycle wheel rotating around
a vertical axis. (b) The student inverts the wheel, setting himself
into rotation. (c) The net angular momentum of the system must
remain the same in spite of the inversion.
Lb,f
Lwh,i,
(11-35)
2Lwh,i.
Using Eq. 11-31, we next substitute Ibvb for Lb,f and Iwhvwh
for Lwh,i and solve for vb, nding
2Iwh
Ib
b
1. The angular speed vb we seek is related to the nal angu:
lar momentum Lb of the composite body about the stools
rotation axis by Eq. 11-31 (L Iv).
2. The initial angular speed vwh of the wheel is related to the
:
angular momentum Lwh of the wheels rotation about its
center by the same equation.
Lb,i
where i and f refer to the initial state (before inversion of
the wheel) and the nal state (after inversion). Because
inversion of the wheel inverted the angular momentum
vector of the wheels rotation, we substitute Lwh,i for Lwh,f.
Then, if we set Lb,i 0 (because the student, the stool, and
the wheels center were initially at rest), Eq. 11-35 yields
Fig. 11-20
KEY IDEAS
Lwh,f
wh
(2)(1.2 kg m2)(3.9 rev/s)
6.8 kg m2
1.4 rev/s.
(Answer)
This positive result tells us that the student rotates counterclockwise about the stool axis as seen from overhead. If the
student wishes stop to rotating, he has only to invert the
wheel once more.
Sample Problem
Conservation of angular momentum, cockroach on disk
In Fig. 11-21, a cockroach with mass m rides on a disk of mass
6.00m and radius R. The disk rotates like a merry-go-round
around its central axis at angular speed i 1.50 rad/s. The
cockroach is initially at radius r 0.800R, but then it crawls
out to the rim of the disk. Treat the cockroach as a particle.
What then is the angular speed?
KEY IDEAS
(1) The cockroachs crawl changes the mass distribution (and
thus the rotational inertia) of the cockroachdisk system.
(2) The angular momentum of the system does not change
because there is no external torque to change it. (The forces
i
R
r
Rotation axis
Fig. 11-21 A cockroach rides at radius r on a disk rotating like
a merry-go-round.
and torques due to the cockroachs crawl are internal to the
system.) (3) The magnitude of the angular momentum of a
rigid body or a particle is given by Eq. 11-31 (L I ).
294
CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
Calculations: We want to nd the nal angular speed. Our
key is to equate the nal angular momentum Lf to the initial
angular momentum Li, because both involve angular speed.
They also involve rotational inertia I. So, lets start by nding
the rotational inertia of the system of cockroach and disk
before and after the crawl.
The rotational inertia of a disk rotating about its central
axis is given by Table 10-2c as 1MR2. Substituting 6.00m for
2
the mass M, our disk here has rotational inertia
Id
3.00mR2.
(11-36)
(We dont have values for m and R, but we shall continue
with physics courage.)
From Eq. 10-33, we know that the rotational inertia of
the cockroach (a particle) is equal to mr 2. Substituting the
cockroachs initial radius (r 0.800R) and nal radius
(r R), we nd that its initial rotational inertia about the
rotation axis is
Ici
0.64mR2
(11-37)
and its nal rotational inertia about the rotation axis is
mR2.
Icf
(11-38)
So, the cockroachdisk system initially has the rotational
inertia
(11-39)
Ii Id Ici 3.64mR2,
and nally has the rotational inertia
If
Id
4.00mR2.
Icf
(11-40)
Next, we use Eq. 11-31 (L I ) to write the fact that
the systems nal angular momentum Lf is equal to the systems initial angular momentum Li:
If
or
4.00mR
2
Ii
f
i
2
3.64mR (1.50 rad/s).
f
After canceling the unknowns m and R, we come to
f
1.37 rad/s.
(Answer)
Note that the angular speed decreased because part of the
mass moved outward from the rotation axis, thus increasing
the rotational inertia of the system.
Additional examples, video, and practice available at WileyPLUS
11-12 Precession of a Gyroscope
A simple gyroscope consists of a wheel xed to a shaft and free to spin about the
axis of the shaft. If one end of the shaft of a nonspinning gyroscope is placed on a
support as in Fig. 11-22a and the gyroscope is released, the gyroscope falls by rotating downward about the tip of the support. Since the fall involves rotation, it is
governed by Newtons second law in angular form, which is given by Eq. 11-29:
:
dL
(11-41)
.
dt
This equation tells us that the torque causing the downward rotation (the fall)
:
changes the angular momentum L of the gyroscope from its initial value of zero.
:
:
The torque is due to the gravitational force Mg acting at the gyroscopes center
of mass, which we take to be at the center of the wheel.The moment arm relative to
the support tip, located at O in Fig. 11-22a, is :.The magnitude of : is
r
:
t
Mgr sin 90
:
:
Mgr
(11-42)
(because the angle between Mg and r is 90), and its direction is as shown in Fig.
11-22a.
A rapidly spinning gyroscope behaves differently. Assume it is released with
the shaft angled slightly upward. It rst rotates slightly downward but then, while
it is still spinning about its shaft, it begins to rotate horizontally about a vertical
axis through support point O in a motion called precession.
Why does the spinning gyroscope stay aloft instead of falling over like the nonspinning gyroscope? The clue is that when the spinning gyroscope is released, the
:
torque due to Mg must change not an initial angular momentum of zero but rather
some already existing nonzero angular momentum due to the spin.
To see how this nonzero initial angular momentum leads to precession, we
:
rst consider the angular momentum L of the gyroscope due to its spin. To
PA R T 1
REVIEW & SUMMARY
simplify the situation, we assume the spin rate is so rapid that the angular
:
momentum due to precession is negligible relative to L . We also assume the shaft
:
is horizontal when precession begins, as in Fig. 11-22b. The magnitude of L is
given by Eq. 11-31:
L Iv,
(11-43)
z
dL
:
dt.
x
Support
Mg
(a)
z
(11-44)
However, for a rapidly spinning gyroscope, the magnitude of L is xed by Eq.
:
11-43. Thus the torque can change only the direction of L, not its magnitude.
:
From Eq. 11-44 we see that the direction of dL is in the direction of :, per:
:
pendicular to L . The only way that L can be changed in the direction of :
:
without the magnitude L being changed is for L to rotate around the z axis as
:
:
shown in Fig. 11-22c. L maintains its magnitude, the head of the L vector follows
:
:
a circular path, and
is always tangent to that path. Since L must always
point along the shaft, the shaft must rotate about the z axis in the direction of :.
Thus we have precession. Because the spinning gyroscope must obey Newtons
law in angular form in response to any change in its initial angular momentum, it
must precess instead of merely toppling over.
We can nd the precession rate by rst using Eqs. 11-44 and 11-42 to get
:
the magnitude of dL :
dL t dt Mgr dt.
(11-45)
y
dL
= ___
dt
x
Mg
(b)
Circular path
taken by head
of L vector
L(t)
:
z
O
d
y
L(t + dt)
dL
___
dt
x
(c)
(a) A nonspinning gyroscope falls by rotating in an xz plane because of torque :. (b) A rapidly spinning
:
gyroscope, with angular momentum L, precesses around the z axis. Its precessional
motion is in the xy plane. (c) The change
:
dL/dt in angular momentum leads to a ro:
tation of L about O.
Fig. 11-22
df/dt, we obtain
(precession rate).
O
r
L
As L changes by an incremental amount in an incremental time interval dt, the
:
shaft and L precess around the z axis through incremental angle df. (In Fig.
11-22c, angle df is exaggerated for clarity.) With the aid of Eqs. 11-43 and 11-45,
we nd that df is given by
dL
Mgr dt
d
.
L
I
Mgr
I
y
:
Dividing this expression by dt and setting the rate
O
r
where I is the rotational moment of the gyroscope about its shaft and v is the an:
gular speed at which the wheel spins about the shaft. The vector L points along
:
r
the shaft, as in Fig. 11-22b. Since L is parallel to :, torque : must be
:
perpendicular to L .
:
According to Eq. 11-41, torque : causes an incremental change dL in the angular momentum of the gyroscope in an incremental time interval dt; that is,
:
295
(11-46)
This result is valid under the assumption that the spin rate v is rapid. Note that
decreases as v is increased. Note also that there would be no precession if the
:
gravitational force Mg did not act on the gyroscope, but because I is a function of
M, mass cancels from Eq. 11-46; thus is independent of the mass.
Equation 11-46 also applies if the shaft of a spinning gyroscope is at an angle
to the horizontal. It holds as well for a spinning top, which is essentially a spinning
gyroscope at an angle to the horizontal.
Rolling Bodies For a wheel of radius R rolling smoothly,
vcom
vR,
(11-2)
where vcom is the linear speed of the wheels center of mass and v is
the angular speed of the wheel about its center. The wheel may
also be viewed as rotating instantaneously about the point P of the
road that is in contact with the wheel. The angular speed of the
wheel about this point is the same as the angular speed of the
wheel about its center. The rolling wheel has kinetic energy
K
1
2 Icom
2
1
2
v2 ,
com
(11-5)
where Icom is the rotational moment of the wheel about its center
of mass and M is the mass of the wheel. If the wheel is being accel-
CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
296
erated but is still rolling smoothly, the acceleration of the center of
mass :com is related to the angular acceleration a about the center
a
with
acom aR.
(11-6)
If the wheel rolls smoothly down a ramp of angle u, its acceleration
along an x axis extending up the ramp is
g sin
(11-10)
acom, x
.
1 Icom /MR2
:
is a vector
quantity dened relative to a xed point (usually an origin); it is
law for a particle can be written in angular form as
:
d
,
dt
where :net is the net torque acting on the particle and
gular momentum of the particle.
:
net
:
r
:
(11-14)
F,
:
where F is a force applied to a particle and : is a position vector
r
locating the particle relative to the xed point. The magnitude of :
is given by
rF sin
rF
(11-15, 11-16, 11-17)
r F,
:
:
:
where f is the angle between F and r , F is the component of F
:
perpendicular to :, and r is the moment arm of F . The direction
r
:
of is given by the right-hand rule.
:
lar momentum L of a system of particles is the vector sum of the
angular momenta of the individual particles:
:
L
:
of a particle with linear momentum p, mass m, and linear velocity : is a vector quantity defined relative to a xed point (usually
v
an origin) as
:
The magnitude of
:
:
r
:
p
m( :
r
:
is given by
:
where f is the angle between : and p, p and v are the compor
:
nents of p and : perpendicular to :, and r is the perpendicular
v
r
:
distance between the xed point and the extension of p. The direc:
tion of is given by the right-hand rule for cross products.
Newtons Second Law in Angular Form Newtons second
1 Figure 11-23 shows three particles of the same mass and the same
constant speed moving as indicated by the velocity vectors.
Points a, b, c, and d form a square,
with point e at the center. Rank the
points according to the magnitude
of the net angular momentum of
the three-particle system when
measured about the points, greatest rst.
a
b
e
d
Fig. 11-23
c
Question 1.
2 Figure 11-24 shows two particles A and B at xyz coordinates
(1 m, 1 m, 0) and (1 m, 0, 1 m). Acting on each particle are three
n
:
2
:
(11-26)
i.
n
1
Angular Momentum of a Rigid Body For a rigid body
rotating about a xed axis, the component of its angular
momentum parallel to the rotation axis is
Iv
L
(11-31)
(rigid body, xed axis).
Conservation of Angular Momentum The angular mo:
mentum L of a system remains constant if the net external torque
acting on the system is zero:
:
L
or
(11-19)
(11-20)
(11-21)
rmv sin
rp
rmv
r p r mv,
:
1
The time rate of change of this angular momentum is equal to the
net external torque on the system (the vector sum of the torques
due to interactions of the particles of the system with particles external to the system):
:
dL
:
(system of particles).
(11-29)
net
dt
(11-18)
v ).
:
i
Angular Momentum of a Particle The angular momentum
:
is the an-
Angular Momentum of a System of Particles The angu-
Torque as a Vector In three dimensions, torque
:
(11-23)
:
a constant
:
Li
:
Lf
(11-32)
(isolated system)
(11-33)
(isolated system).
This is the law of conservation of angular momentum.
Precession of a Gyroscope A spinning gyroscope can precess about a vertical axis through its support at the rate
Mgr
,
I
(11-46)
where M is the gyroscopes mass, r is the moment arm, I is the rotational inertia, and v is the spin rate.
numbered forces, all of the same
magnitude and each directed parallel to an axis. (a) Which of the
forces produce a torque about the
origin that is directed parallel
to y? (b) Rank the forces according to the magnitudes of the
torques they produce on the particles about the origin, greatest
first.
y
A
3
1
2
4
x
5
z
B
6
Fig. 11-24 Question 2.
3 What happens to the initially
stationary yo-yo in Fig. 11-25 if
:
you pull it via its string with (a) force F2 (the line of action passes
PA R T 1
PROBLEMS
through the point of contact on the
:
table, as indicated), (b) force F1 (the
line of action passes above the point
:
of contact), and (c) force F3 (the line
of action passes to the right of the
point of contact)?
4 The position vector : of a particle
r
relative to a certain point has a mag:
nitude of 3 m, and the force F on the
particle has a magnitude of 4 N. What
is the angle between the directions of
:
:
r and F if the magnitude of the associated torque equals (a) zero and (b)
12 N m?
5 In Fig. 11-26, three forces of the
same magnitude are applied to a par:
ticle at the origin ( F1 acts directly into
the plane of the gure). Rank the
forces according to the magnitudes of
the torques they create about (a)
point P1, (b) point P2, and (c) point
P3, greatest rst.
of the disk? (d) What are your answers if the beetle walks in the direction opposite the rotation?
F3
F2
1
8 Figure 11-27 shows an overhead
O
view of a rectangular slab that can
2
7
spin like a merry-go-round about its
3
center at O. Also shown are seven
4
5
6
paths along which wads of bubble
gum can be thrown (all with the Fig. 11-27 Question 8.
same speed and mass) to stick onto
the stationary slab. (a) Rank the paths according to the angular
speed that the slab (and gum) will have after the gum sticks, greatest rst. (b) For which paths will the angular momentum of the slab
(and gum) about O be negative from the view of Fig. 11-27?
F1
Fig. 11-25
Question 3.
y
P3
F2
P2
P1
F3
x
F1
Fig. 11-26
297
Question 5.
6 The angular momenta (t) of a particle in four situations are
6t 2; (3)
2; (4)
4/t. In which situa(1)
3t 4; (2)
tion is the net torque on the particle (a) zero, (b) positive and constant, (c) negative and increasing in magnitude (t 0), and (d)
negative and decreasing in magnitude (t 0)?
9 Figure 11-28 gives the angular mo- L
mentum magnitude L of a wheel versus
time t. Rank the four lettered time intervals according to the magnitude of the
torque acting on the wheel, greatest rst.
t
A
B
C
D
10 Figure 11-29 shows a particle
Fig. 11-28
v
moving at constant velocity : and ve
Question 9.
points with their xy coordinates. Rank
the points according to the magnitude of the angular momentum
of the particle measured about them, greatest rst.
y
c (1, 3)
7 A rhinoceros beetle rides the rim of a horizontal disk rotating
counterclockwise like a merry-go-round. If the beetle then walks
along the rim in the direction of the rotation, will the magnitudes
of the following quantities (each measured about the rotation axis)
increase, decrease, or remain the same (the disk is still rotating in
the counterclockwise direction): (a) the angular momentum of the
beetle disk system, (b) the angular momentum and angular velocity of the beetle, and (c) the angular momentum and angular velocity
(3, 1)
v
a
e
(9, 1)
x
(1, 2)
d (4, 1)
b
Fig. 11-29
Question 10.
Tutoring problem available (at instructors discretion) in WileyPLUS and WebAssign
SSM
Worked-out solution available in Student Solutions Manual
WWW Worked-out solution is at
Number of dots indicates level of problem difculty
ILW
Interactive solution is at
http://www.wiley.com/college/halliday
Additional information available in The Flying Circus of Physics and at yingcircusofphysics.com
sec. 11-2 Rolling as Translation and Rotation Combined
1 A car travels at 80 km/h on a level road in the positive direction
of an x axis. Each tire has a diameter of 66 cm. Relative to a woman
riding in the car and in unit-vector notation, what are the velocity :
v
at the (a) center, (b) top, and (c) bottom of the tire and the magnitude a of the acceleration at the (d) center, (e) top, and (f) bottom
of each tire? Relative to a hitchhiker sitting next to the road and in
v
unit-vector notation, what are the velocity : at the (g) center,
(h) top, and (i) bottom of the tire and the magnitude a of the
acceleration at the (j) center, (k) top, and (l) bottom of each tire?
2 An automobile traveling at 80.0 km/h has tires of 75.0 cm diameter. (a) What is the angular speed of the tires about their axles?
(b) If the car is brought to a stop uniformly in 30.0 complete turns
of the tires (without skidding), what is the magnitude of the angular acceleration of the wheels? (c) How far does the car move during the braking?
sec. 11-4 The Forces of Rolling
3 SSM A 140 kg hoop rolls along a horizontal oor so that the
hoops center of mass has a speed of 0.150 m/s. How much work
must be done on the hoop to stop it?
4 A uniform solid sphere rolls down an incline. (a) What must be
the incline angle if the linear acceleration of the center of the
sphere is to have a magnitude of 0.10g? (b) If a frictionless block
were to slide down the incline at that angle, would its acceleration
magnitude be more than, less than, or equal to 0.10g? Why?
298
CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
5 ILW A 1000 kg car has four 10 kg wheels. When the car is moving, what fraction of its total kinetic energy is due to rotation of the
wheels about their axles? Assume that the wheels have the same
rotational inertia as uniform disks of the same mass and size. Why
do you not need to know the radius of the wheels?
7 ILW In Fig. 11-31, a solid
cylinder of radius 10 cm and
mass 12 kg starts from rest and
rolls without slipping a distance
L 6.0 m down a roof that is inclined at the angle u 30.
(a) What is the angular speed of
the cylinder about its center as it
leaves the roof? (b) The roofs
edge is at height H 5.0 m. How
far horizontally from the roofs
edge does the cylinder hit the
level ground?
vs
v (m/s)
6 Figure 11-30 gives the
speed v versus time t for a
0.500 kg object of radius 6.00
cm that rolls smoothly down
a 30 ramp. The scale on the velocity axis is set by vs 4.0 m/s.
What is the rotational inertia of
the object?
0
0.2
0.4 0.6
t (s)
0.8
1
Problem 6.
Fig. 11-30
L
H
8 Figure 11-32 shows the potential energy U(x) of a solid ball
Fig. 11-31 Problem 7.
that can roll along an x axis. The
scale on the U axis is set by Us
100 J. The ball is uniform, rolls U (J)
smoothly, and has a mass of 0.400 Us
kg. It is released at x 7.0 m
headed in the negative direction
of the x axis with a mechanical
energy of 75 J. (a) If the ball can
reach x 0 m, what is its speed
there, and if it cannot, what is its
2 4 6 8 10 12 14
0
turning point? Suppose, instead,
x (m)
it is headed in the positive direcFig. 11-32 Problem 8.
tion of the x axis when it is released at x 7.0 m with 75 J. (b)
If the ball can reach x 13 m,
what is its speed there, and if it
cannot, what is its turning point?
9
In Fig. 11-33, a solid
ball rolls smoothly from rest
(starting at height H 6.0 m)
until it leaves the horizontal section at the end of the track, at
height h 2.0 m. How far horizontally from point A does the
ball hit the oor?
position? When the sphere has moved 1.0 m up the incline from its
initial position, what are (c) its total kinetic energy and (d) the speed
of its center of mass?
11 In Fig. 11-34, a constant hor:
izontal force Fapp of magnitude 10
N is applied to a wheel of mass 10
Fapp
kg and radius 0.30 m. The wheel
rolls smoothly on the horizontal
surface, and the acceleration of its
x
center of mass has magnitude
Fig. 11-34 Problem 11.
0.60 m/s2. (a) In unit-vector notation, what is the frictional force on
the wheel? (b) What is the rotational inertia of the wheel about the
rotation axis through its center of mass?
12
In Fig. 11-35, a solid brass ball of mass 0.280 g will roll
smoothly along a loop-the-loop track when released from rest
along the straight section. The circular loop has radius R 14.0 cm,
and the ball has radius r R. (a)
What is h if the ball is on the verge
of leaving the track when it h
R
Q
reaches the top of the loop? If the
ball is released at height h
6.00R, what are the (b) magnitude
and (c) direction of the horizontal
Fig. 11-35 Problem 12.
force component acting on the
ball at point Q?
13 Nonuniform ball. In Fig.
11-36, a ball of mass M and radius
h
R rolls smoothly from rest down a
ramp and onto a circular loop of
radius 0.48 m. The initial height of
the ball is h 0.36 m. At the loop Fig. 11-36 Problem 13.
bottom, the magnitude of the normal force on the ball is 2.00Mg. The ball consists of an outer spherical shell (of a certain uniform density) that is glued to a central
sphere (of a different uniform density). The rotational inertia of
the ball can be expressed in the general form I bMR2, but b is
not 0.4 as it is for a ball of uniform density. Determine b.
14
In Fig. 11-37, a small, solid, uniform ball is to be shot
from point P so that it rolls smoothly along a horizontal path, up
along a ramp, and onto a plateau. Then it leaves the plateau horizontally to land on a game board, at a horizontal distance d from
the right edge of the plateau. The vertical heights are h1 5.00 cm
and h2 1.60 cm. With what speed must the ball be shot at point P
for it to land at d 6.00 cm?
H
h2
h
A
Fig. 11-33
h1
Ball
Problem 9.
10 A hollow sphere of radius 0.15 m, with rotational inertia
I 0.040 kg m2 about a line through its center of mass, rolls
without slipping up a surface inclined at 30 to the horizontal. At
a certain initial position, the spheres total kinetic energy is 20 J.
(a) How much of this initial kinetic energy is rotational? (b)
What is the speed of the center of mass of the sphere at the initial
d
P
Fig. 11-37
Problem 14.
15
A bowler throws a bowling ball of radius R 11 cm
along a lane. The ball (Fig. 11-38) slides on the lane with initial
speed vcom,0 8.5 m/s and initial angular speed v0 0. The coefcient of kinetic friction between the ball and the lane is 0.21. The
PA R T 1
PROBLEMS
:
kinetic frictional force f k acting on
the ball causes a linear acceleration
vcom
of the ball while producing a
fk
x
torque that causes an angular acceleration of the ball. When speed Fig. 11-38 Problem 15.
vcom has decreased enough and angular speed v has increased enough, the ball stops sliding and then
rolls smoothly. (a) What then is vcom in terms of v? During the sliding, what are the balls (b) linear acceleration and (c) angular acceleration? (d) How long does the ball slide? (e) How far does the
ball slide? (f) What is the linear speed of the ball when smooth
rolling begins?
16 Nonuniform cylindrical object. In Fig. 11-39, a cylindrical
object of mass M and radius R rolls smoothly from rest down a
ramp and onto a horizontal section. From there it rolls off the ramp
and onto the oor, landing a horizontal distance d 0.506 m from
the end of the ramp. The initial height of the object is H 0.90 m;
the end of the ramp is at height h 0.10 m. The object consists of
an outer cylindrical shell (of a certain uniform density) that is
glued to a central cylinder (of a different uniform density). The rotational inertia of the object can be expressed in the general form
I bMR2, but b is not 0.5 as it is for a cylinder of uniform density.
Determine b.
H
h
299
vector notation, what is the torque about the origin on the plum if
:
that torque is due to a force F whose only component is (a) Fx
6.0 N, (b) Fx
6.0 N, (c) Fz 6.0 N, and (d) Fz
6.0 N?
21 In unit-vector notation, what is the torque about the origin on a
particle located at coordinates (0, 4.0 m, 3.0 m) if that torque is due
:
to (a) force F1 with components F1x 2.0 N, F1y F1z 0, and (b)
:
force F2 with components F2x 0, F2y 2.0 N, F2z 4.0 N?
22 A particle moves through an xyz coordinate system while
a force acts on the particle. When the particle has the position
r
vector : (2.00 m)i (3.00 m)j (2.00 m)k , the force is given by
:
F F xi (7.00 N)j (6.00 N)k and the corresponding torque
about the origin is : (4.00 N m)i (2.00 N m)j (1.00 N m)k.
Determine Fx.
:
23 Force F (2.0 N)i (3.0 N)k acts on a pebble with posi:
tion vector r
(0.50 m)j (2.0 m)k relative to the origin. In unitvector notation, what is the resulting torque on the pebble about
(a) the origin and (b) the point (2.0 m, 0, 3.0 m)?
24 In unit-vector notation, what is the torque about the origin
on a jar of jalapeo peppers located at coordinates (3.0 m, 2.0 m,
:
(4.0 N)j (5.0 N)k , (b)
4.0 m) due to (a) force F1 (3.0 N)i
:
(4.0 N)j (5.0 N)k , and (c) the vector sum
force F2 ( 3.0 N)i
:
:
of F1 and F 2? (d) Repeat part (c) for the torque about the point
with coordinates (3.0 m, 2.0 m, 4.0 m).
:
25 SSM Force F ( 8.0 N)i (6.0 N)j acts on a particle with
:
(3.0 m)i (4.0 m)j. What are (a) the torque
position vector r
on the particle about the origin, in unit-vector notation, and (b) the
:
r
angle between the directions of : and F ?
d
Fig. 11-39
Problem 16.
sec. 11-5 The Yo-Yo
17 SSM
A yo-yo has a rotational inertia of 950 g cm2
and a mass of 120 g. Its axle radius is 3.2 mm, and its string is
120 cm long. The yo-yo rolls from rest down to the end of the
string. (a) What is the magnitude of its linear acceleration? (b)
How long does it take to reach the end of the string? As it reaches
the end of the string, what are its (c) linear speed, (d) translational
kinetic energy, (e) rotational kinetic energy, and (f) angular
speed?
18
In 1980, over San Francisco Bay, a large yo-yo was
released from a crane. The 116 kg yo-yo consisted of two uniform
disks of radius 32 cm connected by an axle of radius 3.2 cm. What
was the magnitude of the acceleration of the yo-yo during (a) its
fall and (b) its rise? (c) What was the tension in the cord on which
it rolled? (d) Was that tension near the cords limit of 52 kN?
Suppose you build a scaled-up version of the yo-yo (same shape
and materials but larger). (e) Will the magnitude of your yo-yos
acceleration as it falls be greater than, less than, or the same as that
of the San Francisco yo-yo? (f) How about the tension in the
cord?
sec. 11-6 Torque Revisited
19 In unit-vector notation, what is the net torque about the origin on a ea located at coordinates (0, 4.0 m, 5.0 m) when forces
:
:
F1 (3.0 N)k and F2 ( 2.0 N)j act on the ea?
20 A plum is located at coordinates ( 2.0 m, 0, 4.0 m). In unit-
sec. 11-7 Angular Momentum
y
26 At the instant of Fig. 11-40, a 2.0 kg
:
particle P has a position vector r of magniF
tude 3.0 m and angle u1 45 and a veloc3
ity vector : of magnitude 4.0 m/s and angle
v
:
u2 30. Force F, of magnitude 2.0 N and
P
v
angle u3 30, acts on P. All three vectors
2 r
lie in the xy plane. About the origin, what
1
x
are the (a) magnitude and (b) direction of
O
the angular momentum of P and the (c)
Fig. 11-40
magnitude and (d) direction of the torque
Problem 26.
acting on P?
:
27 SSM At one instant, force F 4.0j N acts on a 0.25 kg object
that has position vector : (2.0i 2.0k ) m and velocity vector
r
:
v ( 5.0i 5.0k ) m/s. About the origin and in unit-vector notation, what are (a) the objects angular momentum and (b) the
torque acting on the object?
28 A 2.0 kg particle-like object moves in a plane with velocity
components vx 30 m/s and vy 60 m/s as it passes through the
point with (x, y) coordinates of (3.0, 4.0) m. Just then, in unitvector notation, what is its angular momentum relative to (a) the
origin and (b) the point located at
P1
v1
( 2.0, 2.0) m?
29 ILW In the instant of Fig. 11-41,
two particles move in an xy plane.
Particle P1 has mass 6.5 kg and
speed v1 2.2 m/s, and it is at distance d1 1.5 m from point O.
Particle P2 has mass 3.1 kg and speed
y
d1
x
O
Fig. 11-41
v2
d2
P2
Problem 29.
300
CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
v2 3.6 m/s, and it is at distance d2 2.8 m from point O. What are
the (a) magnitude and (b) direction of the net angular momentum of
the two particles about O?
same density (mass per unit volume) and thickness. What is the ratio of the magnitude of the angular momentum of disk C to that of
disk B?
30 At the instant the displacement of a 2.00 kg object relative
:
to the origin is d (2.00 m)i (4.00 m) j (3.00 m)k , its veloc:
ity is v
(6.00 m/s)i (3.00 m/s)j (3.00 m/s)k and it is subject
:
to a force F (6.00 N)i (8.00 N)j (4.00 N)k. Find (a) the acceleration of the object, (b) the angular momentum of the object about
the origin, (c) the torque about the origin acting on the object, and (d)
the angle between the velocity of the oby
ject and the force acting on the object.
31 In Fig. 11-42, a 0.400 kg ball is
shot directly upward at initial speed
Ball
P
x
40.0 m/s. What is its angular momentum about P, 2.00 m horizontally from Fig. 11-42 Problem 31.
the launch point, when the ball is (a) at
maximum height and (b) halfway back to the ground? What is the
torque on the ball about P due to the gravitational force when the
ball is (c) at maximum height and (d) halfway back to the ground?
37
In Fig. 11-44, three parti
m
cles of mass m 23 g are fastened
d
to three rods of length d 12 cm
m
d
and negligible mass. The rigid assembly rotates around point O at
d
the angular speed v 0.85 rad/s. O
About O, what are (a) the rota- Fig. 11-44 Problem 37.
tional inertia of the assembly, (b)
the magnitude of the angular momentum of the middle particle,
and (c) the magnitude of the angular momentum of the asssembly?
38 A sanding disk with rotational inertia 1.2 10 3 kg m2 is attached to an electric drill whose motor delivers a torque
of magnitude 16 N m about the central axis of the disk.
About that axis and with the torque applied for 33 ms, what is the
magnitude of the (a) angular momentum and (b) angular velocity
of the disk?
sec. 11-8 Newtons Second Law in Angular Form
32 A particle is acted on by two torques about the origin: :1 has
a magnitude of 2.0 N m and is directed in the positive direction of
the x axis, and :2 has a magnitude of 4.0 N m and is
directed in the negative direction of the y axis. In unit-vector nota:
:
tion, nd d /dt, where is the angular momentum of the particle
about the origin.
39 S SM The angular momentum of a ywheel having a rotational inertia of 0.140 kg m2 about its central axis decreases from
3.00 to 0.800 kg m2/s in 1.50 s. (a) What is the magnitude of the average torque acting on the ywheel about its central axis during
this period? (b) Assuming a constant angular acceleration, through
what angle does the ywheel turn? (c) How much work is done on
the wheel? (d) What is the average power of the ywheel?
33 SSM ILW WWW At time t 0, a 3.0 kg particle with velocity
:
v (5.0 m/s)i (6.0 m/s)j is at x 3.0 m, y 8.0 m. It is pulled by
a 7.0 N force in the negative x direction.About the origin, what are (a)
the particles angular momentum, (b) the torque acting on the particle, and (c) the rate at which the angular momentum is changing?
40 A disk with a rotational inertia of 7.00 kg m2 rotates like a
merry-go-round while undergoing a variable torque given by
(5.00 2.00t) N m. At time t 1.00 s, its angular momentum
is 5.00 kg m2/s. What is its angular momentum at t 3.00 s?
34 A particle is to move in an xy plane, clockwise around the
origin as seen from the positive side of the z axis. In unit-vector notation, what torque acts on the particle if the magnitude of its angular momentum about the origin is (a) 4.0 kg m2/s, (b) 4.0t 2
kg m2/s, (c) 4.0 1t kg m2/s, and (d) 4.0/t2 kg m2/s?
r
35 At time t, the vector : 4.0t2 i (2.0t 6.0t2)j gives the
position of a 3.0 kg particle relative to the origin of an xy coordinate
r
system (: is in meters and t is in seconds). (a) Find an expression for
the torque acting on the particle relative to the origin. (b) Is the
magnitude of the particles angular momentum relative to the origin
increasing, decreasing, or unchanging?
sec. 11-10 The Angular Momentum of a Rigid Body
Rotating About a Fixed Axis
36 Figure 11-43 shows three rotating, uniform disks that are
coupled by belts. One belt runs around the rims of disks A and C.
Another belt runs around a central hub on disk A and the rim of
disk B. The belts move smoothly without slippage on the rims and
hub. Disk A has radius R; its hub has radius 0.5000R; disk B has radius 0.2500R; and disk C has radius 2.000R. Disks B and C have the
m
41
Figure 11-45 shows a rigid
structure consisting of a circular hoop
of radius R and mass m, and a square
made of four thin bars, each of length
R and mass m. The rigid structure rotates at a constant speed about a vertical axis, with a period of rotation of
2.5 s. Assuming R 0.50 m and m
2.0 kg, calculate (a) the structures rotational inertia about the axis of rotation and (b) its angular momentum
about that axis.
Belt
Fig. 11-45
s
4
8
12
C
Fig. 11-43
Problem 36.
Problem 41.
(N m)
B
A
2R
R
42 Figure 11-46 gives the torque t that acts on an initially stationary disk that can rotate about its center like a merry-go-round.
0
Belt
Rotation axis
Fig. 11-46
Problem 42.
16
t (s)
20
PA R T 1
PROBLEMS
The scale on the t axis is set by ts 4.0 N m. What is the angular
momentum of the disk about the rotation axis at times (a) t 7.0 s
and (b) t 20 s?
sec. 11-11 Conservation of Angular Momentum
43 In Fig. 11-47, two skaters, each
of mass 50 kg, approach each other
along parallel paths separated by 3.0
m. They have opposite velocities of
1.4 m/s each. One skater carries one
end of a long pole of negligible mass,
and the other skater grabs the other
Fig. 11-47 Problem 43.
end as she passes. The skaters then
rotate around the center of the pole. Assume that the friction
between skates and ice is negligible. What are (a) the radius of the
circle, (b) the angular speed of the skaters, and (c) the kinetic energy
of the two-skater system? Next, the skaters pull along the pole until
they are separated by 1.0 m. What then are (d) their angular speed
and (e) the kinetic energy of the system? (f) What provided the energy for the increased kinetic energy?
44 A Texas cockroach of mass 0.17 kg runs counterclockwise
around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has radius 15 cm, rotational inertia 5.0 10 3 kg m2,
and frictionless bearings. The cockroachs speed (relative to the
ground) is 2.0 m/s, and the lazy Susan turns clockwise with angular
speed v0 2.8 rad/s. The cockroach nds a bread crumb on the rim
and, of course, stops. (a) What is the angular speed of the lazy
Susan after the cockroach stops? (b) Is mechanical energy conserved as it stops?
45 SSM WWW A man stands on a platform that is rotating
(without friction) with an angular speed of 1.2 rev/s; his arms are
outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the man, bricks, and platform about
the central vertical axis of the platform is 6.0 kg m2. If by moving
the bricks the man decreases the rotational inertia of the system to
2.0 kg m2, what are (a) the resulting angular speed of the platform
and (b) the ratio of the new kinetic energy of the system to the
original kinetic energy? (c) What source provided the added
kinetic energy?
46 The rotational inertia of a collapsing spinning star drops to 1
3
its initial value. What is the ratio of the new rotational kinetic energy to the initial rotational kinetic energy?
(rad/s)
47 SSM A track is mounted on a
large wheel that is free to turn with
negligible friction about a vertical
axis (Fig. 11-48). A toy train of mass
m is placed on the track and, with Fig. 11-48 Problem 47.
the system initially at rest, the
trains electrical power is turned on. The train reaches speed 0.15
m/s with respect to the track.What is the angular speed of the wheel
if its mass is 1.1m and its radius is
0.43 m? (Treat the wheel as a hoop,
b
and neglect the mass of the spokes
and hub.)
a
48 A Texas cockroach rst rides
at the center of a circular disk that
0
R
rotates freely like a merry-goRadial distance
round without external torques.
The cockroach then walks out to Fig. 11-49 Problem 48.
301
the edge of the disk, at radius R. Figure 11-49 gives the angular
speed v of the cockroach disk system during the walk. The scale
on the v axis is set by va 5.0 rad/s and vb 6.0 rad/s. When the
cockroach is on the edge at radius R, what is the ratio of the bugs
rotational inertia to that of the disk, both calculated about the rotation axis?
49 Two disks are mounted (like a merry-go-round) on low-friction bearings on the same axle and can be brought together so that
they couple and rotate as one unit. The rst disk, with rotational inertia 3.30 kg m2 about its central axis, is set spinning counterclockwise at 450 rev/min. The second disk, with rotational inertia 6.60
kg m2 about its central axis, is set spinning counterclockwise at 900
rev/min. They then couple together. (a) What is their angular speed
after coupling? If instead the second disk is set spinning clockwise
at 900 rev/min, what are their (b) angular speed and (c) direction
of rotation after they couple together?
50 The rotor of an electric motor has rotational inertia Im
2.0 10 3 kg m2 about its central axis. The motor is used to
change the orientation of the space probe in which it is mounted.
The motor axis is mounted along the central axis of the probe; the
probe has rotational inertia Ip 12 kg m2 about this axis.
Calculate the number of revolutions of the rotor required to turn
the probe through 30 about its central axis.
51 SSM ILW A wheel is rotating freely at angular speed 800
rev/min on a shaft whose rotational inertia is negligible. A second
wheel, initially at rest and with twice the rotational inertia of the
rst, is suddenly coupled to the same shaft. (a) What is the angular
speed of the resultant combination of the shaft and two wheels?
(b) What fraction of the original rotational kinetic energy is lost?
52
A cockroach of mass m lies on the rim of a uniform disk
of mass 4.00m that can rotate freely about its center like a merrygo-round. Initially the cockroach and disk rotate together with an
angular velocity of 0.260 rad/s. Then the cockroach walks halfway
to the center of the disk. (a) What then is the angular velocity of
the cockroach disk system? (b) What is the ratio K /K0 of the new
kinetic energy of the system to its initial kinetic energy? (c) What
accounts for the change in the kiAxis
netic energy?
53
A uniform thin rod of
length 0.500 m and mass 4.00 kg can
rotate in a horizontal plane about a
vertical axis through its center. The
rod is at rest when a 3.00 g bullet
traveling in the rotation plane is red Fig. 11-50 Problem 53.
into one end of the rod. As viewed
from above, the bullets path makes angle u 60.0 with the rod (Fig.
11-50). If the bullet lodges in the rod and the angular velocity of the
rod is 10 rad/s immediately after the collision, what is the bullets
speed just before impact?
54
Figure 11-51 shows an
overhead view of a ring that can rotate about its center like a merrygo-round. Its outer radius R2 is 0.800
m, its inner radius R1 is R2/2.00, its
mass M is 8.00 kg, and the mass of
the crossbars at its center is negligible. It initially rotates at an angular speed of 8.00 rad/s with a cat of
R1
R2
Fig. 11-51
Problem 54.
302
CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
mass m M/4.00 on its outer edge, at radius R2. By how much does
the cat increase the kinetic energy of the cat ring system if the cat
crawls to the inner edge, at radius R1?
the rotational inertia of the block rod bullet system about point
A? (b) If the angular speed of the system about A just after impact
is 4.5 rad/s, what is the bullets speed just before impact?
55 A horizontal vinyl record of mass 0.10 kg and radius 0.10 m
rotates freely about a vertical axis through its center with an angular speed of 4.7 rad/s. The rotational inertia of the record about its
axis of rotation is 5.0 10 4 kg m2. A wad of wet putty of mass
0.020 kg drops vertically onto the record from above and sticks to
the edge of the record. What is the angular speed of the record immediately after the putty sticks to it?
61 The uniform rod (length 0.60
m, mass 1.0 kg) in Fig. 11-54 rotates
in the plane of the gure about an
axis through one end, with a rotational inertia of 0.12 kg m2. As the
rod swings through its lowest position, it collides with a 0.20 kg putty
wad that sticks to the end of the rod.
If the rods angular speed just before
collision is 2.4 rad/s, what is the angular speed of the rod putty system
immediately after collision?
56
In a long jump, an athlete leaves the ground with an
initial angular momentum that tends to rotate her body forward,
threatening to ruin her landing. To counter this tendency, she rotates her outstretched arms to take up the angular momentum
(Fig. 11-18). In 0.700 s, one arm sweeps through 0.500 rev and the
other arm sweeps through 1.000 rev. Treat each arm as a thin rod of
mass 4.0 kg and length 0.60 m, rotating around one end. In the athletes reference frame, what is the magnitude of the total angular
momentum of the arms around the common rotation axis through
the shoulders?
57 A uniform disk of mass 10m and radius 3.0r can rotate freely
about its xed center like a merry-go-round. A smaller uniform
disk of mass m and radius r lies on top of the larger disk, concentric
with it. Initially the two disks rotate together with an angular velocity of 20 rad/s. Then a slight disturbance causes the smaller disk
to slide outward across the larger disk, until the outer edge of the
smaller disk catches on the outer edge of the larger disk. Afterward,
the two disks again rotate together (without further sliding). (a) What
then is their angular velocity about the center of the larger disk? (b)
What is the ratio K /K0 of the new kinetic energy of the two-disk system to the systems initial kinetic energy?
58 A horizontal platform in the shape of a circular disk rotates
on a frictionless bearing about a vertical axle through the center of
the disk. The platform has a mass of 150 kg, a radius of 2.0 m, and a
rotational inertia of 300 kg m2 about the axis of rotation. A 60 kg
student walks slowly from the rim of the platform toward the center. If the angular speed of the system is 1.5 rad/s when the student
starts at the rim, what is the angular speed when she is 0.50 m from
the center?
59 Figure 11-52 is an overhead
view of a thin uniform rod of length
0.800 m and mass M rotating horizonRotation
axis
tally at angular speed 20.0 rad/s about
an axis through its center. A particle Fig. 11-52 Problem 59.
of mass M/3.00 initially attached to
one end is ejected from the rod and travels along a path that is perpendicular to the rod at the instant of ejection. If the particles speed
vp is 6.00 m/s greater than the speed of
A
the rod end just after ejection, what is
the value of vp?
60 In Fig. 11-53, a 1.0 g bullet is
Rod
red into a 0.50 kg block attached to
the end of a 0.60 m nonuniform rod of
mass 0.50 kg. The block rod bullet
system then rotates in the plane of the
gure, about a xed axis at A. The roBlock
tational inertia of the rod alone about
Bullet
2
that axis at A is 0.060 kg m . Treat the
block as a particle. (a) What then is Fig. 11-53 Problem 60.
Rotation axis
Rod
Fig. 11-54
Problem 61.
62
During a jump to his
partner, an aerialist is to make a quadruple somersault lasting a
time t 1.87 s. For the rst and last quarter-revolution, he is in the
extended orientation shown in Fig. 11-55, with rotational inertia
I1
19.9 kg m2 around his center of mass (the dot). During the
rest of the ight he is in a tight tuck, with rotational inertia I2
3.93 kg m2. What must be his angular speed v 2 around his center
of mass during the tuck?
2
Tuck
I2
Parabolic
path of
aerialist
1
1
I1
I1
Release
Catch
Fig. 11-55
Problem 62.
Ball
63
In Fig. 11-56, a 30 kg
child stands on the edge of a stav
tionary merry-go-round of radius
Child
2.0 m. The rotational inertia of the
merry-go-round about its rotation
axis is 150 kg m2. The child catches
a ball of mass 1.0 kg thrown by a
friend. Just before the ball is caught,
it has a horizontal velocity : of magv
nitude 12 m/s, at angle f 37 with
Fig. 11-56 Problem 63.
a line tangent to the outer edge of
the merry-go-round, as shown. What is the angular speed of the
merry-go-round just after the ball is caught?
64
A ballerina begins a tour jet (Fig. 11-19a) with angular speed i and a rotational inertia consisting of two parts:
Ileg 1.44 kg m2 for her leg extended outward at angle
90.0
to her body and Itrunk 0.660 kg m2 for the rest of her body (pri-
PA R T 1
PROBLEMS
marily her trunk). Near her maximum height she holds both legs
at angle
30.0 to her body and has angular speed f (Fig. 1119b). Assuming that Itrunk has not changed, what is the ratio f / i?
65 SSM WWW
Two 2.00 kg
Putty wad
balls are attached to the ends of a
thin rod of length 50.0 cm and negliRotation
gible mass. The rod is free to rotate in
axis
a vertical plane without friction
about a horizontal axis through its
center. With the rod initially horizon- Fig. 11-57 Problem 65.
tal (Fig. 11-57), a 50.0 g wad of wet
putty drops onto one of the balls, hitting it with a speed of 3.00 m/s
and then sticking to it. (a) What is the angular speed of the system
just after the putty wad hits? (b) What is the ratio of the kinetic energy of the system after the collision to that of the putty wad just
before? (c) Through what angle will the system rotate before it
momentarily stops?
O
66 In Fig. 11-58, a small 50 g
block slides down a frictionless surface through height h 20 cm and
then sticks to a uniform rod of mass
100 g and length 40 cm. The rod pivots about point O through angle u before momentarily stopping. Find u.
h
67 Figure 11-59 is an overhead
view of a thin uniform rod of length
0.600 m and mass M rotating hori- Fig. 11-58 Problem 66.
zontally at 80.0 rad/s counterclockwise about an axis through its center. A particle of mass M/3.00 and
traveling horizontally at speed 40.0 m/s hits the rod and sticks. The
particles path is perpendicular to the rod at the instant of the hit,
at a distance d from the rods center. (a) At what value of d are rod
and particle stationary after the hit? (b) In which direction do rod
and particle rotate if d is greater than this value?
Rotation axis
Particle
d
Fig. 11-59
Problem 67.
sec. 11-12 Precession of a Gyroscope
68 A top spins at 30 rev/s about an axis that makes an angle of
30 with the vertical. The mass of the top is 0.50 kg, its rotational inertia about its central axis is 5.0 10 4 kg m2, and its center of
mass is 4.0 cm from the pivot point. If the spin is clockwise from an
overhead view, what are the (a) precession rate and (b) direction of
the precession as viewed from overhead?
69 A certain gyroscope consists of a uniform disk with a 50 cm
radius mounted at the center of an axle that is 11 cm long and of
negligible mass. The axle is horizontal and supported at one end. If
the disk is spinning around the axle at 1000 rev/min, what is the
precession rate?
303
Additional Problems
70 A uniform solid ball rolls smoothly along a oor, then up a
ramp inclined at 15.0. It momentarily stops when it has rolled 1.50
m along the ramp. What was its initial speed?
71 SSM In Fig. 11-60, a constant
Fapp
:
horizontal force Fapp of magnitude
12 N is applied to a uniform solid
Fishing line
cylinder by shing line wrapped
around the cylinder. The mass of the
x
cylinder is 10 kg, its radius is 0.10 m,
and the cylinder rolls smoothly Fig. 11-60 Problem 71.
on the horizontal surface. (a) What
is the magnitude of the acceleration of the center of mass of the
cylinder? (b) What is the magnitude of the angular acceleration of
the cylinder about the center of mass? (c) In unit-vector notation,
what is the frictional force acting on the cylinder?
72 A thin-walled pipe rolls along the oor. What is the ratio of its
translational kinetic energy to its rotational kinetic energy about
the central axis parallel to its length?
73 SSM A 3.0 kg toy car moves along an x axis with a velocity
given by :
v
2.0t 3i m/s, with t in seconds. For t 0, what are (a)
:
the angular momentum L of the car and (b) the torque : on the
:
car, both calculated about the origin? What are (c) L and (d) :
:
about the point (2.0 m, 5.0 m, 0)? What are (e) L and (f) : about
the point (2.0 m, 5.0 m, 0)?
74 A wheel rotates clockwise about its central axis with an angular momentum of 600 kg m2/s. At time t 0, a torque of magnitude 50 N m is applied to the wheel to reverse the rotation. At
what time t is the angular speed zero?
75 SSM In a playground, there is a small merry-go-round of
radius 1.20 m and mass 180 kg. Its radius of gyration (see Problem
79 of Chapter 10) is 91.0 cm. A child of mass 44.0 kg runs at a speed
of 3.00 m/s along a path that is tangent to the rim of the initially
stationary merry-go-round and then jumps on. Neglect friction between the bearings and the shaft of the merry-go-round. Calculate
(a) the rotational inertia of the merry-go-round about its axis of
rotation, (b) the magnitude of the angular momentum of the running child about the axis of rotation of the merry-go-round, and (c)
the angular speed of the merry-go-round and child after the child
has jumped onto the merry-go-round.
76 A uniform block of granite in the shape of a book has face dimensions of 20 cm and 15 cm and a thickness of 1.2 cm. The density
(mass per unit volume) of granite is 2.64 g/cm3. The block rotates
around an axis that is perpendicular to its face and halfway between its center and a corner. Its angular momentum about that
axis is 0.104 kg m2/s. What is its rotational kinetic energy about
that axis?
77 SSM Two particles, each of mass 2.90 10 4 kg and speed
5.46 m/s, travel in opposite directions along parallel lines separated
by 4.20 cm. (a) What is the magnitude L of the angular momentum
of the two-particle system around a point midway between the two
lines? (b) Does the value of L change if the point about which it is
calculated is not midway between the lines? If the direction of
travel for one of the particles is reversed, what would be (c) the answer to part (a) and (d) the answer to part (b)?
78 A wheel of radius 0.250 m, which is moving initially at 43.0
m/s, rolls to a stop in 225 m. Calculate the magnitudes of (a) its lin-
304
CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
ear acceleration and (b) its angular acceleration. (c) The wheels
rotational inertia is 0.155 kg m2 about its central axis. Calculate
the magnitude of the torque about the central axis due to friction
on the wheel.
79 Wheels A and B in Fig. 11-61
A
are connected by a belt that does not
B
slip. The radius of B is 3.00 times the
radius of A. What would be the ratio
of the rotational inertias IA /IB if the
two wheels had (a) the same angular Fig. 11-61 Problem 79.
momentum about their central axes
and (b) the same rotational kinetic energy?
80 A 2.50 kg particle that is moving horizontally over a oor with
velocity ( 3.00 m/s) undergoes a completely inelastic collision
j
with a 4.00 kg particle that is moving horizontally over the oor
with velocity (4.50 m/s)i. The collision occurs at xy coordinates
( 0.500 m, 0.100 m). After the collision and in unit-vector notation, what is the angular momentum of the stuck-together particles
with respect to the origin?
81 SSM A uniform wheel of mass 10.0 kg and radius 0.400 m is
mounted rigidly on a massless axle through its center (Fig. 11-62).
The radius of the axle is 0.200 m, and the rotational inertia of the
wheel axle combination about its central axis is 0.600 kg m2. The
wheel is initially at rest at the top of a surface that is inclined at angle u 30.0 with the horizontal; the axle rests on the surface while
the wheel extends into a groove in the surface without touching
the surface. Once released, the axle rolls down along the surface
smoothly and without slipping. When the wheel axle combination
has moved down the surface by 2.00 m, what are (a) its rotational
kinetic energy and (b) its translational kinetic energy?
Wheel
Axle
Groove
Fig. 11-62
Problem 81.
82 A uniform rod rotates in a horizontal plane about a vertical
axis through one end. The rod is 6.00 m long, weighs 10.0 N, and rotates at 240 rev/min. Calculate (a) its rotational inertia about the
axis of rotation and (b) the magnitude of its angular momentum
about that axis.
83 A solid sphere of weight 36.0 N rolls up an incline at an angle
of 30.0. At the bottom of the incline the center of mass of the
sphere has a translational speed of 4.90 m/s. (a) What is the kinetic
energy of the sphere at the bottom of the incline? (b) How far does
the sphere travel up along the incline? (c) Does the answer to (b)
depend on the spheres mass?
84
Suppose that the yo-yo in Problem 17, instead of
rolling from rest, is thrown so that its initial speed down the string
is 1.3 m/s. (a) How long does the yo-yo take to reach the end of the
string? As it reaches the end of the string, what are its (b) total ki-
netic energy, (c) linear speed, (d) translational kinetic energy, (e)
angular speed, and (f) rotational kinetic energy?
85 A girl of mass M stands on the rim of a frictionless merry-goround of radius R and rotational inertia I that is not moving. She
throws a rock of mass m horizontally in a direction that is tangent
to the outer edge of the merry-go-round. The speed of the rock,
relative to the ground, is v. Afterward, what are (a) the angular
speed of the merry-go-round and (b) the linear speed of the girl?
86 At time t 0, a 2.0 kg particle has the position vector
:
r (4.0 m)i (2.0 m)j relative to the origin. Its velocity is
:
2
given by v ( 6.0t m/s)i for t 0 in seconds. About the origin,
:
what are (a) the particles angular momentum L and (b) the torque
:
acting on the particle, both in unit-vector notation and for t 0?
:
About the point ( 2.0 m, 3.0 m, 0), what are (c) L and (d) :
for t 0?
87 If Earths polar ice caps fully melted and the water returned
to the oceans, the oceans would be deeper by about 30 m. What effect would this have on Earths rotation? Make an estimate of the
resulting change in the length of the day.
88 A 1200 kg airplane is ying in a straight line at 80 m/s,
1.3 km above the ground. What is the magnitude of its angular momentum with respect to a point on the ground directly under the
path of the plane?
89 With axle and spokes of negligible mass and a thin rim, a certain bicycle wheel has a radius of 0.350 m and weighs 37.0 N; it can
turn on its axle with negligible friction. A man holds the wheel
above his head with the axle vertical while he stands on a turntable
that is free to rotate without friction; the wheel rotates clockwise, as
seen from above, with an angular speed of 57.7 rad/s, and the
turntable is initially at rest. The rotational inertia of wheel man
turntable about the common axis of rotation is 2.10 kg m2. The
mans free hand suddenly stops the rotation of the wheel (relative
to the turntable). Determine the resulting (a) angular speed and
(b) direction of rotation of the system.
90 For an 84 kg person standing at the equator, what is the magnitude of the angular momentum about Earths center due to
Earths rotation?
91 A small solid sphere with radius 0.25 cm and mass 0.56 g rolls
without slipping on the inside of a large xed hemisphere with radius 15 cm and a vertical axis of symmetry. The sphere starts at the
top from rest. (a) What is its kinetic energy at the bottom? (b)
What fraction of its kinetic energy at the bottom is associated with
rotation about an axis through its com? (c) What is the magnitude
of the normal force on the hemisphere from the sphere when the
sphere reaches the bottom?
92 An automobile has a total mass of 1700 kg. It accelerates from
rest to 40 km/h in 10 s. Assume each wheel is a uniform 32 kg disk.
Find, for the end of the 10 s interval, (a) the rotational kinetic
energy of each wheel about its axle, (b) the total kinetic energy of
each wheel, and (c) the total kinetic energy of the automobile.
93 A body of radius R and mass m is rolling smoothly with speed
v on a horizontal surface. It then rolls up a hill to a maximum
height h. (a) If h 3v2/4g, what is the bodys rotational inertia
about the rotational axis through its center of mass? (b) What
might the body be?

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