Fundamentals of Physics "ROTATION"

E X T E N D E D
FUNDAMENTALS OF PHYSICS
T E N T H E D I T I O N

257
C H A P T E R 1 0
Rotation
10-1ROTATIONAL VARIABLES
After reading this module, you should be able to . . .
10.01 Identify that if all parts of a body rotate around a fixed
axis locked together, the body is a rigid body. (This chapter
is about the motion of such bodies.)
10.02 Identify that the angular position of a rotating rigid body
is the angle that an internal reference line makes with a
fixed, external reference line.
10.03 Apply the relationship between angular displacement
and the initial and final angular positions.
10.04 Apply the relationship between average angular veloc-
ity, angular displacement, and the time interval for that dis-
placement.
10.05 Apply the relationship between average angular accel-
eration, change in angular velocity, and the time interval for
that change.
10.06 Identify that counterclockwise motion is in the positive
direction and clockwise motion is in the negative direction.
10.07 Given angular position as a function of time, calculate the
instantaneous angular velocity at any particular time and the
average angular velocity between any two particular times.
10.08 Given a graph of angular position versus time, deter-
mine the instantaneous angular velocity at a particular time
and the average angular velocity between any two particu-
lar times.
10.09 Identify instantaneous angular speed as the magnitude
of the instantaneous angular velocity.
10.10 Given angular velocity as a function of time, calculate
the instantaneous angular acceleration at any particular
time and the average angular acceleration between any
two particular times.
10.11 Given a graph of angular velocity versus time, deter-
mine the instantaneous angular acceleration at any partic-
ular time and the average angular acceleration between
any two particular times.
10.12 Calculate a body’s change in angular velocity by
integrating its angular acceleration function with respect
to time.
10.13 Calculate a body’s change in angular position by inte-
grating its angular velocity function with respect to time.
● To describe the rotation of a rigid body about a fixed axis,
called the rotation axis, we assume a reference line is fixed in
the body, perpendicular to that axis and rotating with the
body. We measure the angular position u of this line
relative to a fixed direction. When u is measured in radians,
(radian measure),
where s is the arc length of a circular path of radius r and
angle u.
● Radian measure is related to angle measure in revolutions
and degrees by
1 rev ! 360# ! 2p rad.
● A body that rotates about a rotation axis, changing its angu-
lar position from u1 to u2, undergoes an angular displacement
"u ! u2 % u1,
where "u is positive for counterclockwise rotation and nega-
tive for clockwise rotation.
● If a body rotates through an angular displacement "u in a
time interval "t, its average angular velocity vavg is
u !
s
r
The (instantaneous) angular velocity v of the body is
Both vavg and v are vectors, with directions given by a
right-hand rule. They are positive for counterclockwise rota-
tion and negative for clockwise rotation. The magnitude of the
body’s angular velocity is the angular speed.
● If the angular velocity of a body changes from v1 to v2 in a
time interval "t ! t2 % t1, the average angular acceleration
aavg of the body is
The (instantaneous) angular acceleration a of the body is
Both aavg and a are vectors.
a !
dv
dt
.
aavg !
v2 % v1
t2 % t1
!
"v
"t
.
v !
du
dt
.
vavg !
"u
"t
.
Key Ideas
Learning Objectives

258 CHAPTER 10 ROTATION
What Is Physics?
As we have discussed, one focus of physics is motion. However, so far we
have examined only the motion of translation, in which an object moves along
a straight or curved line, as in Fig. 10-1a. We now turn to the motion of rotation,
in which an object turns about an axis, as in Fig. 10-1b.
You see rotation in nearly every machine, you use it every time you open a
beverage can with a pull tab, and you pay to experience it every time you go to an
amusement park. Rotation is the key to many fun activities, such as hitting a long
drive in golf (the ball needs to rotate in order for the air to keep it aloft longer)
and throwing a curveball in baseball (the ball needs to rotate in order for the air
to push it left or right). Rotation is also the key to more serious matters, such as
metal failure in aging airplanes.
We begin our discussion of rotation by defining the variables for the
motion, just as we did for translation in Chapter 2. As we shall see, the vari-
ables for rotation are analogous to those for one-dimensional motion and, as
in Chapter 2, an important special situation is where the acceleration (here the
rotational acceleration) is constant. We shall also see that Newton’s second
law can be written for rotational motion, but we must use a new quantity
called torque instead of just force. Work and the work–kinetic energy
theorem can also be applied to rotational motion, but we must use a new quan-
tity called rotational inertia instead of just mass. In short, much of what we
have discussed so far can be applied to rotational motion with, perhaps, a few
changes.
Caution: In spite of this repetition of physics ideas, many students find this
and the next chapter very challenging. Instructors have a variety of reasons as
to why, but two reasons stand out: (1) There are a lot of symbols (with Greek
Figure 10-1 Figure skater Sasha Cohen in motion of (a) pure translation in a fixed
direction and (b) pure rotation about a vertical axis.
(b)
(a)
Mike
Segar/Reuters/Landov
LLC
Elsa/Getty
Images,
Inc.

259
10-1 ROTATIONAL VARIABLES
letters) to sort out. (2) Although you are very familiar with linear motion (you
can get across the room and down the road just fine), you are probably very
unfamiliar with rotation (and that is one reason why you are willing to pay so
much for amusement park rides). If a homework problem looks like a foreign
language to you, see if translating it into the one-dimensional linear motion of
Chapter 2 helps. For example, if you are to find, say, an angular distance, tem-
porarily delete the word angular and see if you can work the problem with the
Chapter 2 notation and ideas.
Rotational Variables
We wish to examine the rotation of a rigid body about a fixed axis.A rigid body is
a body that can rotate with all its parts locked together and without any change in
its shape. A fixed axis means that the rotation occurs about an axis that does not
move.Thus, we shall not examine an object like the Sun, because the parts of the
Sun (a ball of gas) are not locked together. We also shall not examine an object
like a bowling ball rolling along a lane, because the ball rotates about a moving
axis (the ball’s motion is a mixture of rotation and translation).
Figure 10-2 shows a rigid body of arbitrary shape in rotation about a fixed
axis, called the axis of rotation or the rotation axis. In pure rotation (angular
motion), every point of the body moves in a circle whose center lies on the axis of
rotation, and every point moves through the same angle during a particular time
interval. In pure translation (linear motion), every point of the body moves in a
straight line, and every point moves through the same linear distance during a
particular time interval.
We deal now—one at a time—with the angular equivalents of the linear
quantities position, displacement, velocity, and acceleration.
Angular Position
Figure 10-2 shows a reference line, fixed in the body, perpendicular to the rotation
axis and rotating with the body. The angular position of this line is the angle of
the line relative to a fixed direction, which we take as the zero angular position.
In Fig. 10-3, the angular position u is measured relative to the positive direction of
the x axis. From geometry, we know that u is given by
(radian measure). (10-1)
Here s is the length of a circular arc that extends from the x axis (the zero angular
position) to the reference line, and r is the radius of the circle.
u !
s
r
Figure 10-2 A rigid body of arbitrary shape in pure rotation about the z axis of a coordinate
system.The position of the reference line with respect to the rigid body is arbitrary, but it is
perpendicular to the rotation axis. It is fixed in the body and rotates with the body.
z
O
Reference line
Rotation
axis
x
y
Body This reference line is part of the body
and perpendicular to the rotation axis.
We use it to measure the rotation of the
body relative to a fixed direction.
Figure 10-3 The rotating rigid body of
Fig. 10-2 in cross section, viewed from
above. The plane of the cross section is
perpendicular to the rotation axis, which
now extends out of the page, toward you.
In this position of the body, the reference
line makes an angle u with the x axis.
x
y
Reference
line
θ
r
s
Rotation
axis
The body has rotated
counterclockwise
by angle . This is the
positive direction.
θ
This dot means that
the rotation axis is
out toward you.

An angle defined in this way is measured in radians (rad) rather than in
revolutions (rev) or degrees. The radian, being the ratio of two lengths, is a
pure number and thus has no dimension. Because the circumference of a circle of
radius r is 2pr, there are 2p radians in a complete circle:
(10-2)
and thus 1 rad ! 57.3# ! 0.159 rev. (10-3)
We do not reset u to zero with each complete rotation of the reference line about
the rotation axis. If the reference line completes two revolutions from the zero
angular position, then the angular position u of the line is u ! 4p rad.
For pure translation along an x axis, we can know all there is to know
about a moving body if we know x(t), its position as a function of time.
Similarly, for pure rotation, we can know all there is to know about a rotating
body if we know u(t), the angular position of the body’s reference line as a
function of time.
Angular Displacement
If the body of Fig. 10-3 rotates about the rotation axis as in Fig. 10-4, changing the
angular position of the reference line from u1 to u2, the body undergoes an
angular displacement "u given by
"u ! u2 % u1. (10-4)
This definition of angular displacement holds not only for the rigid body as a
whole but also for every particle within that body.
Clocks Are Negative. If a body is in translational motion along an x axis, its
displacement "x is either positive or negative, depending on whether the body is
moving in the positive or negative direction of the axis. Similarly, the angular dis-
placement "u of a rotating body is either positive or negative, according to the
following rule:
1 rev ! 360# !
2pr
r
! 2p rad,
An angular displacement in the counterclockwise direction is positive, and one in
the clockwise direction is negative.
Checkpoint 1
A disk can rotate about its central axis like a merry-go-round.Which of the following
pairs of values for its initial and final angular positions, respectively, give a negative
angular displacement: (a) %3 rad, &5 rad, (b) %3 rad, %7 rad, (c) 7 rad, %3 rad?
The phrase “clocks are negative” can help you remember this rule (they certainly
are negative when their alarms sound off early in the morning).
Angular Velocity
Suppose that our rotating body is at angular position u1 at time t1 and at
angular position u2 at time t2 as in Fig. 10-4.We define the average angular velocity
of the body in the time interval "t from t1 to t2 to be
(10-5)
where "u is the angular displacement during "t (v is the lowercase omega).
vavg !
u2 % u1
t2 % t1
!
"u
"t
,

261
Figure 10-4 The reference line of the rigid body of Figs. 10-2 and 10-3 is at angular position
u1 at time t1 and at angular position u2 at a later time t2.The quantity "u (! u2 % u1) is the
angular displacement that occurs during the interval "t (! t2 % t1).The body itself is not
shown.
x
y
Rotation axis
O
θ1
θ2
∆θ
At t2
At t1
Reference line
This change in the angle of the reference line
(which is part of the body) is equal to the angular
displacement of the body itself during this
time interval.
The (instantaneous) angular velocity v, with which we shall be most con-
cerned, is the limit of the ratio in Eq. 10-5 as "t approaches zero.Thus,
(10-6)
If we know u(t), we can find the angular velocity v by differentiation.
Equations 10-5 and 10-6 hold not only for the rotating rigid body as a whole
but also for every particle of that body because the particles are all locked
together. The unit of angular velocity is commonly the radian per second (rad/s)
or the revolution per second (rev/s). Another measure of angular velocity was
used during at least the first three decades of rock: Music was produced by vinyl
(phonograph) records that were played on turntables at “ ” or “45 rpm,”
meaning at or 45 rev/min.
If a particle moves in translation along an x axis, its linear velocity v is either
positive or negative, depending on its direction along the axis. Similarly, the angu-
lar velocity v of a rotating rigid body is either positive or negative, depending on
whether the body is rotating counterclockwise (positive) or clockwise (negative).
(“Clocks are negative” still works.) The magnitude of an angular velocity is called
the angular speed, which is also represented with v.
Angular Acceleration
If the angular velocity of a rotating body is not constant, then the body has an an-
gular acceleration. Let v2 and v1 be its angular velocities at times t2 and t1,
respectively.The average angular acceleration of the rotating body in the interval
from t1 to t2 is defined as
(10-7)
in which "v is the change in the angular velocity that occurs during the time
interval "t. The (instantaneous) angular acceleration a, with which we shall be
most concerned, is the limit of this quantity as "t approaches zero.Thus,
(10-8)
As the name suggests, this is the angular acceleration of the body at a given in-
stant. Equations 10-7 and 10-8 also hold for every particle of that body. The unit of
angular acceleration is commonly the radian per second-squared (rad/s2
) or the
revolution per second-squared (rev/s2
).
a ! lim
"t:0
"v
"t
!
dv
dt
.
aavg !
v2 % v1
t2 % t1
!
"v
"t
,
331
3 rev/min
331
3 rpm
v ! lim
"t:0
"u
"t
!
du
dt
.

Calculations: To sketch the disk and its reference line at a
particular time, we need to determine u for that time. To do
so, we substitute the time into Eq. 10-9. For t ! %2.0 s, we get
This means that at t ! %2.0 s the reference line on the disk
is rotated counterclockwise from the zero position by angle
1.2 rad ! 69# (counterclockwise because u is positive). Sketch
1 in Fig.10-5b shows this position of the reference line.
Similarly, for t ! 0, we find u ! %1.00 rad ! %57#,
which means that the reference line is rotated clockwise
from the zero angular position by 1.0 rad, or 57#, as shown
in sketch 3. For t ! 4.0 s, we find u ! 0.60 rad ! 34#
(sketch 5). Drawing sketches for when the curve crosses
the t axis is easy, because then u ! 0 and the reference line
is momentarily aligned with the zero angular position
(sketches 2 and 4).
(b) At what time tmin does u(t) reach the minimum
value shown in Fig. 10-5b? What is that minimum value?
! 1.2 rad ! 1.2 rad
360#
20 rad
! 69#.
u ! %1.00 % (0.600)(%2.0) & (0.250)(%2.0)2
Sample Problem 10.01 Angular velocity derived from angular position
The disk in Fig. 10-5a is rotating about its central axis like a
merry-go-round. The angular position u(t) of a reference
line on the disk is given by
u ! %1.00 % 0.600t & 0.250t2
, (10-9)
with t in seconds, u in radians, and the zero angular position
as indicated in the figure. (If you like, you can translate all
this into Chapter 2 notation by momentarily dropping the
word “angular” from “angular position” and replacing the
symbol u with the symbol x.What you then have is an equa-
tion that gives the position as a function of time, for the one-
dimensional motion of Chapter 2.)
(a) Graph the angular position of the disk versus time
from t ! %3.0 s to t ! 5.4 s. Sketch the disk and its angular
position reference line at t ! %2.0 s, 0 s, and 4.0 s, and
when the curve crosses the t axis.
KEY IDEA
The angular position of the disk is the angular position
u(t) of its reference line,which is given by Eq.10-9 as a function
of time t.So we graph Eq.10-9;the result is shown in Fig.10-5b.
A
Zero
angular
position
Reference
line
Rotation axis
(a)
(b)
2
0
–2
0 2 4 6
(rad)
(1) (2) (3) (4) (5)
t (s)
θ
–2
The angular position
of the disk is the angle
between these two lines.
Now, the disk is
at a zero angle.
θ
At t = −2 s, the disk
is at a positive
(counterclockwise)
angle. So, a positive
value is plotted.
This is a plot of the angle
of the disk versus time.
Now, it is at a
negative (clockwise)
angle. So, a negative
value is plotted.
θ
It has reversed
its rotation and
is again at a
zero angle.
Now, it is
back at a
positive
angle.
Figure 10-5 (a) A rotating disk. (b) A plot of the disk’s angular position u(t). Five sketches indicate the angular position of the refer-
ence line on the disk for five points on the curve. (c) A plot of the disk’s angular velocity v(t). Positive values of v correspond to
counterclockwise rotation, and negative values to clockwise rotation.

263
t ! %3.0 s to t ! 6.0 s. Sketch the disk and indicate the direc-
tion of turning and the sign of v at t ! %2.0 s,4.0 s,and tmin.
KEY IDEA
From Eq. 10-6, the angular velocity v is equal to du/dt as
given in Eq. 10-10. So, we have
v ! %0.600 & 0.500t. (10-11)
The graph of this function v(t) is shown in Fig. 10-5c.
Because the function is linear, the plot is a straight line. The
slope is 0.500 rad/s2
and the intercept with the vertical axis
(not shown) is %0.600 rad/s.
Calculations: To sketch the disk at t ! %2.0 s, we substitute
that value into Eq. 10-11, obtaining
v ! %1.6 rad/s. (Answer)
The minus sign here tells us that at t ! %2.0 s, the disk is
turning clockwise (as indicated by the left-hand sketch in
Fig. 10-5c).
Substituting t ! 4.0 s into Eq. 10-11 gives us
v ! 1.4 rad/s. (Answer)
The implied plus sign tells us that now the disk is turning
counterclockwise (the right-hand sketch in Fig. 10-5c).
For tmin, we already know that du/dt ! 0. So, we must
also have v ! 0. That is, the disk momentarily stops when
the reference line reaches the minimum value of u in
Fig. 10-5b, as suggested by the center sketch in Fig. 10-5c. On
the graph of v versus t in Fig. 10-5c, this momentary stop is
the zero point where the plot changes from the negative
clockwise motion to the positive counterclockwise motion.
(d) Use the results in parts (a) through (c) to describe the
motion of the disk from t ! %3.0 s to t ! 6.0 s.
Description: When we first observe the disk at t ! %3.0 s, it
has a positive angular position and is turning clockwise but
slowing. It stops at angular position u ! %1.36 rad and then
begins to turn counterclockwise, with its angular position
eventually becoming positive again.
KEY IDEA
To find the extreme value (here the minimum) of a function,
we take the first derivative of the function and set the result
to zero.
Calculations: The first derivative of u(t) is
(10-10)
Setting this to zero and solving for t give us the time at
which u(t) is minimum:
tmin ! 1.20 s. (Answer)
To get the minimum value of u, we next substitute tmin into
Eq. 10-9, finding
u ! %1.36 rad % %77.9#. (Answer)
This minimum of u(t) (the bottom of the curve in Fig. 10-5b)
corresponds to the maximum clockwise rotation of the disk
from the zero angular position, somewhat more than is
shown in sketch 3.
(c) Graph the angular velocity v of the disk versus time from
du
dt
! %0.600 & 0.500t.
(c)
2
0
–2
–2 0 2 4 6
(rad/s)
ω
t (s)
negative ω zero ω positive ω
This is a plot of the angular
velocity of the disk versus time.
The angular velocity is
initially negative and slowing,
then momentarily zero during
reversal, and then positive and
increasing.
Additional examples, video, and
practice available at WileyPLUS

Are Angular Quantities Vectors?
We can describe the position, velocity, and acceleration of a single particle by
means of vectors. If the particle is confined to a straight line, however, we do not
really need vector notation. Such a particle has only two directions available to it,
and we can indicate these directions with plus and minus signs.
In the same way, a rigid body rotating about a fixed axis can rotate only
clockwise or counterclockwise as seen along the axis, and again we can select
between the two directions by means of plus and minus signs.The question arises:
“Can we treat the angular displacement, velocity, and acceleration of a rotating
body as vectors?” The answer is a qualified “yes” (see the caution below, in con-
nection with angular displacements).
Angular Velocities. Consider the angular velocity. Figure 10-6a shows a
vinyl record rotating on a turntable. The record has a constant angular speed
in the clockwise direction. We can represent its angular ve-
locity as a vector pointing along the axis of rotation, as in Fig. 10-6b. Here’s
how: We choose the length of this vector according to some convenient scale,
for example, with 1 cm corresponding to 10 rev/min. Then we establish a direc-
tion for the vector by using a right-hand rule, as Fig. 10-6c shows: Curl your
right hand about the rotating record, your fingers pointing in the direction of
rotation. Your extended thumb will then point in the direction of the angular
velocity vector. If the record were to rotate in the opposite sense, the right-
v
:
v
:
v (! 331
3 rev/min)
To evaluate the constant of integration C, we note that v !
5 rad/s at t ! 0. Substituting these values in our expression
for v yields
,
so C ! 5 rad/s.Then
. (Answer)
(b) Obtain an expression for the angular position u(t) of the
top.
KEY IDEA
By definition, v(t) is the derivative of u(t) with respect to
time. Therefore, we can find u(t) by integrating v(t) with
respect to time.
Calculations: Since Eq. 10-6 tells us that
du ! v dt,
we can write
(Answer)
where C/ has been evaluated by noting that u ! 2 rad at t! 0.
! 1
4 t5
% 2
3 t3
& 5t & 2,
! 1
4 t5
% 2
3 t3
& 5t & C/
u ! "v dt ! "(5
4 t4
% 2t2
& 5) dt
v ! 5
4 t4
% 2t2
& 5
5 rad/s ! 0 % 0 & C
Sample Problem 10.02 Angular velocity derived from angular acceleration
A child’s top is spun with angular acceleration
,
with t in seconds and a in radians per second-squared. At
t ! 0, the top has angular velocity 5 rad/s, and a reference
line on it is at angular position u ! 2 rad.
(a) Obtain an expression for the angular velocity v(t) of the
top.That is,find an expression that explicitly indicates how the
angular velocity depends on time. (We can tell that there is
such a dependence because the top is undergoing an angular
acceleration,which means that its angular velocity is changing.)
KEY IDEA
By definition,a(t) is the derivative of v(t) with respect to time.
Thus, we can find v(t) by integrating a(t) with respect to time.
Calculations: Equation 10-8 tells us
,
so .
From this we find
.
v ! "(5t3
% 4t) dt ! 5
4t4
% 4
2t2
& C
"dv ! "a dt
dv ! a dt
a ! 5t3
% 4t
Additional examples, video, and practice available at WileyPLUS

265
hand rule would tell you that the angular velocity vector then points in the op-
posite direction.
It is not easy to get used to representing angular quantities as vectors.We in-
stinctively expect that something should be moving along the direction of a vec-
tor. That is not the case here. Instead, something (the rigid body) is rotating
around the direction of the vector. In the world of pure rotation, a vector defines
an axis of rotation, not a direction in which something moves. Nonetheless, the
vector also defines the motion. Furthermore, it obeys all the rules for vector
manipulation discussed in Chapter 3. The angular acceleration is another
vector, and it too obeys those rules.
In this chapter we consider only rotations that are about a fixed axis. For such
situations, we need not consider vectors—we can represent angular velocity with
v and angular acceleration with a, and we can indicate direction with an implied
plus sign for counterclockwise or an explicit minus sign for clockwise.
Angular Displacements. Now for the caution: Angular displacements
(unless they are very small) cannot be treated as vectors.Why not? We can cer-
tainly give them both magnitude and direction, as we did for the angular veloc-
ity vector in Fig. 10-6. However, to be represented as a vector, a quantity must
also obey the rules of vector addition, one of which says that if you add two
vectors, the order in which you add them does not matter. Angular displace-
ments fail this test.
Figure 10-7 gives an example. An initially horizontal book is given two
90# angular displacements, first in the order of Fig. 10-7a and then in the order
of Fig. 10-7b.Although the two angular displacements are identical, their order
is not, and the book ends up with different orientations. Here’s another exam-
ple. Hold your right arm downward, palm toward your thigh. Keeping your
wrist rigid, (1) lift the arm forward until it is horizontal, (2) move it horizon-
tally until it points toward the right, and (3) then bring it down to your side.
Your palm faces forward. If you start over, but reverse the steps, which way
does your palm end up facing? From either example, we must conclude that
the addition of two angular displacements depends on their order and they
cannot be vectors.
a
:
Figure 10-6 (a) A record rotating about a vertical axis that coincides with the axis of the
spindle. (b) The angular velocity of the rotating record can be represented by the vector
, lying along the axis and pointing down, as shown. (c) We establish the direction of the
angular velocity vector as downward by using a right-hand rule. When the fingers of the
right hand curl around the record and point the way it is moving, the extended thumb
points in the direction of .
v
:
v
:
z z z
(a) (b) (c)
Axis Axis Axis
ω
Spindle
ω
This right-hand rule
establishes the
direction of the
angular velocity
vector.
Figure 10-7 (a) From its initial position, at
the top, the book is given two successive
90# rotations, first about the (horizontal)
x axis and then about the (vertical) y axis.
(b) The book is given the same rotations,
but in the reverse order.
PHYSICS
P
H
Y
S
I
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S
PHYSICS
P
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Y
S
I
C
S
P
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Y
S
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C
S
PHYSICS
P
H
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(a) (b)
PHYSICS
y
x
z
y
x
z
y
x
z z
y
x
z
y
x
y
x
z The order of the
rotations makes
a big difference
in the result.

Rotation with Constant Angular Acceleration
In pure translation, motion with a constant linear acceleration (for example, that
of a falling body) is an important special case. In Table 2-1, we displayed a series
of equations that hold for such motion.
In pure rotation, the case of constant angular acceleration is also important,
and a parallel set of equations holds for this case also. We shall not derive them
here, but simply write them from the corresponding linear equations, substituting
equivalent angular quantities for the linear ones.This is done in Table 10-1, which
lists both sets of equations (Eqs. 2-11 and 2-15 to 2-18; 10-12 to 10-16).
Recall that Eqs. 2-11 and 2-15 are basic equations for constant linear
acceleration—the other equations in the Linear list can be derived from them.
Similarly, Eqs. 10-12 and 10-13 are the basic equations for constant angular
acceleration, and the other equations in the Angular list can be derived from
them.To solve a simple problem involving constant angular acceleration, you can
usually use an equation from the Angular list (if you have the list). Choose
an equation for which the only unknown variable will be the variable requested
in the problem.A better plan is to remember only Eqs. 10-12 and 10-13, and then
solve them as simultaneous equations whenever needed.
10-2 ROTATION WITH CONSTANT ANGULAR ACCELERATION
10.14 For constant angular acceleration, apply the relation-
ships between angular position, angular displacement,
Key Idea
● Constant angular acceleration (a ! constant) is an important special case of rotational motion. The appropriate kinematic
equations are
v ! v0 & at,
u % u0 ! vt % 1
2 at2
.
u % u0 ! 1
2 (v0 & v)t,
v2
! v0
2
& 2a(u % u0),
u % u0 ! v0t & 1
2at2
,
Learning Objective
angular velocity, angular acceleration, and elapsed time
(Table 10-1).
Table 10-1 Equations of Motion for Constant Linear Acceleration and for Constant Angular Acceleration
Equation Linear Missing Angular Equation
Number Equation Variable Equation Number
(2-11) v ! v0 & at x % x0 u % u0 v ! v0 & at (10-12)
(2-15) v v (10-13)
(2-16) t t (10-14)
(2-17) a a (10-15)
(2-18) v0 v0 (10-16)
u % u0 ! vt % 1
2at2
x % x0 ! vt % 1
2at2
u % u0 ! 1
2(v0 & v)t
x % x0 ! 1
2(v0 & v)t
v2
! v0
2
& 2a(u % u0)
v2
! v0
2
& 2a(x % x0)
u % u0 ! v0t & 1
2at2
x % x0 ! v0 t & 1
2 at2
Checkpoint 2
In four situations, a rotating body has angular position u(t) given by (a) u ! 3t % 4,
(b) u ! %5t3
& 4t2
& 6, (c) u ! 2/t2
% 4/t, and (d) u ! 5t2
% 3.To which situations do
the angular equations of Table 10-1 apply?

267
10-2 ROTATION WITH CONSTANT ANGULAR ACCELERATION
(We converted 5.0 rev to 10p rad to keep the units consis-
tent.) Solving this quadratic equation for t,we find
t ! 32 s. (Answer)
Now notice something a bit strange. We first see the wheel
when it is rotating in the negative direction and through the
u ! 0 orientation.Yet, we just found out that 32 s later it is at
the positive orientation of u ! 5.0 rev. What happened in
that time interval so that it could be at a positive orientation?
(b) Describe the grindstone’s rotation between t ! 0 and
t ! 32 s.
Description: The wheel is initially rotating in the negative
(clockwise) direction with angular velocity v0 ! %4.6 rad/s,
but its angular acceleration a is positive.This initial opposi-
tion of the signs of angular velocity and angular accelera-
tion means that the wheel slows in its rotation in the nega-
tive direction, stops, and then reverses to rotate in the
positive direction. After the reference line comes back
through its initial orientation of u ! 0, the wheel turns an
additional 5.0 rev by time t ! 32 s.
(c) At what time t does the grindstone momentarily stop?
Calculation: We again go to the table of equations for con-
stant angular acceleration, and again we need an equation
that contains only the desired unknown variable t. However,
now the equation must also contain the variable v, so that we
can set it to 0 and then solve for the corresponding time t. We
choose Eq.10-12,which yields
(Answer)
t !
v % v0
a
!
0 % (%4.6 rad/s)
0.35 rad/s2
! 13 s.
Sample Problem 10.03 Constant angular acceleration, grindstone
A grindstone (Fig. 10-8) rotates at constant angular acceler-
ation a ! 0.35 rad/s2
. At time t ! 0, it has an angular velocity
of v0 ! %4.6 rad/s and a reference line on it is horizontal, at
the angular position u0 ! 0.
(a) At what time after t ! 0 is the reference line at the
angular position u ! 5.0 rev?
KEY IDEA
The angular acceleration is constant, so we can use the rota-
tion equations of Table 10-1.We choose Eq. 10-13,
,
because the only unknown variable it contains is the desired
time t.
Calculations: Substituting known values and setting u0 ! 0
and u ! 5.0 rev ! 10p rad give us
.
10p rad ! (%4.6 rad/s)t & 1
2 (0.35 rad/s2
)t2
u % u0 ! v0t & 1
2 at2
Figure 10-8 A grindstone. At t ! 0 the reference line (which we
imagine to be marked on the stone) is horizontal.
Axis
Reference
line
Zero angular
position
We measure rotation by using
this reference line.
Clockwise = negative
Counterclockwise = positive
rad/s, the angular displacement is u % u0 ! 20.0 rev, and the
angular velocity at the end of that displacement is v ! 2.00
rad/s. In addition to the angular acceleration a that we want,
both basic equations also contain time t, which we do not
necessarily want.
To eliminate the unknown t, we use Eq. 10-12 to write
which we then substitute into Eq. 10-13 to write
Solving for a, substituting known data, and converting
20 rev to 125.7 rad, we find
(Answer)
! %0.0301 rad/s2
.
a !
v2
% v0
2
2(u % u0)
!
(2.00 rad/s)2
% (3.40 rad/s)2
2(125.7 rad)
u % u0 ! v0#v % v0
a $& 1
2 a#v % v0
a $
2
.
t !
v % v0
a
,
Sample Problem 10.04 Constant angular acceleration, riding a Rotor
While you are operating a Rotor (a large, vertical, rotating
cylinder found in amusement parks), you spot a passenger in
acute distress and decrease the angular velocity of the cylin-
der from 3.40 rad/s to 2.00 rad/s in 20.0 rev, at constant angu-
lar acceleration.(The passenger is obviously more of a“trans-
lation person”than a“rotation person.”)
(a) What is the constant angular acceleration during this
decrease in angular speed?
KEY IDEA
Because the cylinder’s angular acceleration is constant, we
can relate it to the angular velocity and angular displacement
via the basic equations for constant angular acceleration
(Eqs.10-12 and 10-13).
Calculations: Let’s first do a quick check to see if we can solve
the basic equations. The initial angular velocity is v0 ! 3.40

Relating the Linear and Angular Variables
In Module 4-5,we discussed uniform circular motion,in which a particle travels at con-
stant linear speed v along a circle and around an axis of rotation.When a rigid body,
such as a merry-go-round,rotates around an axis,each particle in the body moves in its
own circle around that axis. Since the body is rigid, all the particles make one revolu-
tion in the same amount of time;that is,they all have the same angular speed v.
However, the farther a particle is from the axis, the greater the circumference
of its circle is, and so the faster its linear speed v must be.You can notice this on a
merry-go-round. You turn with the same angular speed v regardless of your dis-
tance from the center, but your linear speed v increases noticeably if you move to
the outside edge of the merry-go-round.
We often need to relate the linear variables s, v, and a for a particular point in
a rotating body to the angular variables u, v, and a for that body. The two sets of
variables are related by r, the perpendicular distance of the point from the
rotation axis. This perpendicular distance is the distance between the point and
the rotation axis, measured along a perpendicular to the axis. It is also the radius r
of the circle traveled by the point around the axis of rotation.
(b) How much time did the speed decrease take?
Calculation: Now that we know a, we can use Eq. 10-12 to
solve for t: (Answer)
! 46.5 s.
t !
v % v0
a
!
2.00 rad/s % 3.40 rad/s
%0.0301 rad/s2
10-3 RELATING THE LINEAR AND ANGULAR VARIABLES
10.15 For a rigid body rotating about a fixed axis, relate the angular
variables of the body (angular position, angular velocity, and an-
gular acceleration) and the linear variables of a particle on the
body (position, velocity, and acceleration) at any given radius.
10.16 Distinguish between tangential acceleration and radial
acceleration, and draw a vector for each in a sketch of a
particle on a body rotating about an axis, for both an in-
crease in angular speed and a decrease.
● A point in a rigid rotating body, at a perpendicular distance
r from the rotation axis, moves in a circle with radius r. If the
body rotates through an angle u, the point moves along an
arc with length s given by
s ! ur (radian measure),
where u is in radians.
● The linear velocity of the point is tangent to the circle; the
point’s linear speed v is given by
v ! vr (radian measure),
where v is the angular speed (in radians per second) of the body,
and thus also the point.
v
:
● The linear acceleration of the point has both tangential
and radial components. The tangential component is
at ! ar (radian measure),
where a is the magnitude of the angular acceleration (in radi-
ans per second-squared) of the body. The radial component
of is
(radian measure).
● If the point moves in uniform circular motion, the period T of
the motion for the point and the body is
(radian measure).
T !
2pr
v
!
2p
v
ar !
v2
r
! v2
r
a
:
a
:
Learning Objectives
Key Ideas

269
10-3 RELATING THE LINEAR AND ANGULAR VARIABLES
The Position
If a reference line on a rigid body rotates through an angle u, a point within the
body at a position r from the rotation axis moves a distance s along a circular arc,
where s is given by Eq. 10-1:
s ! ur (radian measure). (10-17)
This is the first of our linear–angular relations. Caution: The angle u here must be
measured in radians because Eq. 10-17 is itself the definition of angular measure
in radians.
The Speed
Differentiating Eq. 10-17 with respect to time—with r held constant—leads to
However, ds/dt is the linear speed (the magnitude of the linear velocity) of the
point in question, and du/dt is the angular speed v of the rotating body. So
v ! vr (radian measure). (10-18)
Caution: The angular speed v must be expressed in radian measure.
Equation 10-18 tells us that since all points within the rigid body have the
same angular speed v, points with greater radius r have greater linear speed v.
Figure 10-9a reminds us that the linear velocity is always tangent to the circular
path of the point in question.
If the angular speed v of the rigid body is constant, then Eq. 10-18 tells
us that the linear speed v of any point within it is also constant. Thus, each point
within the body undergoes uniform circular motion. The period of revolution T
for the motion of each point and for the rigid body itself is given by Eq. 4-35:
. (10-19)
This equation tells us that the time for one revolution is the distance 2pr traveled
in one revolution divided by the speed at which that distance is traveled.
Substituting for v from Eq. 10-18 and canceling r, we find also that
This equivalent equation says that the time for one revolution is the angular dis-
tance 2p rad traveled in one revolution divided by the angular speed (or rate) at
which that angle is traveled.
The Acceleration
Differentiating Eq. 10-18 with respect to time—again with r held constant—
leads to
(10-21)
Here we run up against a complication. In Eq. 10-21, dv/dt represents only the
part of the linear acceleration that is responsible for changes in the magnitude v
of the linear velocity . Like , that part of the linear acceleration is tangent to
the path of the point in question.We call it the tangential component at of the lin-
ear acceleration of the point, and we write
at ! ar (radian measure), (10-22)
v
:
v
:
dv
dt
!
dv
dt
r.
T !
2p
v
T !
2pr
v
ds
dt
!
du
dt
r.
Figure 10-9 The rotating rigid body of Fig. 10-2,
shown in cross section viewed from above.
Every point of the body (such as P) moves
in a circle around the rotation axis. (a) The
linear velocity of every point is tangent to
the circle in which the point moves. (b) The
linear acceleration of the point has (in
general) two components: tangential at and
radial ar.
a
:
v
:
x
y
r
Rotation
axis
P
Circle
traveled by P
(a)
v
The velocity vector is
always tangent to this
circle around the
rotation axis.
x
y
ar
P
(b)
at
Rotation
axis
The acceleration always
has a radial (centripetal)
component and may have
a tangential component.

where a ! dv/dt. Caution: The angular acceleration a in Eq. 10-22 must be
expressed in radian measure.
In addition, as Eq. 4-34 tells us, a particle (or point) moving in a circular path
has a radial component of linear acceleration, ar ! v2
/r (directed radially inward),
that is responsible for changes in the direction of the linear velocity . By substi-
tuting for v from Eq. 10-18, we can write this component as
Thus, as Fig. 10-9b shows, the linear acceleration of a point on a rotating rigid
body has, in general, two components. The radially inward component ar (given
by Eq. 10-23) is present whenever the angular velocity of the body is not zero.
The tangential component at (given by Eq. 10-22) is present whenever the angu-
lar acceleration is not zero.
ar !
v2
r
! v2
r
v
:
Checkpoint 3
A cockroach rides the rim of a rotating merry-go-round. If the angular speed of this
system (merry-go-round & cockroach) is constant, does the cockroach have (a) radial
acceleration and (b) tangential acceleration? If v is decreasing, does the cockroach
have (c) radial acceleration and (d) tangential acceleration?
and radial accelerations are the (perpendicular) compo-
nents of the (full) acceleration .
Calculations: Let’s go through the steps. We first find the
angular velocity by taking the time derivative of the given
angular position function and then substituting the given
time of t ! 2.20 s:
v ! (ct3
) ! 3ct2
(10-25)
! 3(6.39 $ 10%2
rad/s3
)(2.20 s)2
! 0.928 rad/s. (Answer)
From Eq. 10-18, the linear speed just then is
v ! vr ! 3ct2
r (10-26)
! 3(6.39 $ 10%2
rad/s3
)(2.20 s)2
(33.1 m)
! 30.7 m/s. (Answer)
du
dt
!
d
dt
a
:
Sample Problem 10.05 Designing The Giant Ring, a large-scale amusement park ride
We are given the job of designing a large horizontal ring
that will rotate around a vertical axis and that will have a ra-
dius of r ! 33.1 m (matching that of Beijing’s The Great
Observation Wheel, the largest Ferris wheel in the world).
Passengers will enter through a door in the outer wall of the
ring and then stand next to that wall (Fig. 10-10a).We decide
that for the time interval t ! 0 to t ! 2.30 s, the angular posi-
tion u(t) of a reference line on the ring will be given by
u ! ct3
, (10-24)
with c ! 6.39 $ 10%2
rad/s3
. After t ! 2.30 s, the angular
speed will be held constant until the end of the ride. Once
the ring begins to rotate, the floor of the ring will drop away
from the riders but the riders will not fall—indeed, they feel
as though they are pinned to the wall. For the time t ! 2.20 s,
let’s determine a rider’s angular speed v, linear speed v, an-
gular acceleration a, tangential acceleration at, radial accel-
eration ar, and acceleration .
KEY IDEAS
(1) The angular speed v is given by Eq. 10-6 (v ! du/dt).
(2) The linear speed v (along the circular path) is related to
the angular speed (around the rotation axis) by Eq. 10-18
(v ! vr). (3) The angular acceleration a is given by Eq. 10-8
(a ! dv/dt). (4) The tangential acceleration at (along the cir-
cular path) is related to the angular acceleration (around
the rotation axis) by Eq. 10-22 (at ! ar). (5) The radial accel-
eration ar is given Eq. 10-23 (ar ! v2
r). (6) The tangential
a
:
u
a
ar
at
(b)
(a)
Figure 10-10 (a) Overhead view of
a passenger ready to ride The
Giant Ring. (b) The radial and
tangential acceleration compo-
nents of the (full) acceleration.

271
10-4 KINETIC ENERGY OF ROTATION
The radial and tangential accelerations are perpendicu-
lar to each other and form the components of the rider’s
acceleration (Fig. 10-10b). The magnitude of is given by
a ! (10-29)
39.9 m/s2
, (Answer)
or 4.1g (which is really exciting!). All these values are
acceptable.
To find the orientation of , we can calculate the angle u
shown in Fig. 10-10b:
tan u !
However, instead of substituting our numerical results, let’s
use the algebraic results from Eqs. 10-27 and 10-28:
u ! tan%1
. (10-30)
The big advantage of solving for the angle algebraically is that
we can then see that the angle (1) does not depend on the
ring’s radius and (2) decreases as t goes from 0 to 2.20 s.That
is, the acceleration vector swings toward being radially in-
ward because the radial acceleration (which depends on t4
)
quickly dominates over the tangential acceleration (which
depends on only t).At our given time t ! 2.20 s,we have
u #. (Answer)
! tan%1
2
3(6.39 $ 10%2
rad/s3
)(2.20 s)3
! 44.4
a
:
# 6ctr
9c2
t4
r $! tan%1
# 2
3ct3 $
at
ar
.
a
:
%
! 2(28.49 m/s2
)2
& (27.91 m/s2
)2
2a2
r & a2
t
a
:
a
:
Although this is fast (111 km/h or 68.7 mi/h), such speeds are
common in amusement parks and not alarming because (as
mentioned in Chapter 2) your body reacts to accelerations but
not to velocities. (It is an accelerometer, not a speedometer.)
From Eq.10-26 we see that the linear speed is increasing as the
square of the time (but this increase will cut off at t ! 2.30 s).
Next, let’s tackle the angular acceleration by taking the
time derivative of Eq. 10-25:
a ! (3ct2
) ! 6ct
! 6(6.39 $ 10%2
rad/s3
)(2.20 s) ! 0.843 rad/s2
. (Answer)
The tangential acceleration then follows from Eq. 10-22:
at ! ar ! 6ctr (10-27)
! 6(6.39 $ 10%2
rad/s3
)(2.20 s)(33.1 m)
! 27.91 m/s2
27.9 m/s2
, (Answer)
or 2.8g (which is reasonable and a bit exciting). Equation
10-27 tells us that the tangential acceleration is increasing
with time (but it will cut off at t ! 2.30 s). From Eq. 10-23,
we write the radial acceleration as
ar ! v2
r.
Substituting from Eq. 10-25 leads us to
ar ! (3ct2
)2
r ! 9c2
t4
r (10-28)
! 9(6.39 $ 10%2
rad/s3
)2
(2.20 s)4
(33.1 m)
! 28.49 m/s2
28.5 m/s2
, (Answer)
or 2.9g (which is also reasonable and a bit exciting).
%
%
dv
dt
!
d
dt
10-4 KINETIC ENERGY OF ROTATION
10.17 Find the rotational inertia of a particle about a point.
10.18 Find the total rotational inertia of many particles moving
around the same fixed axis.
10.19 Calculate the rotational kinetic energy of a
body in terms of its rotational inertia and its angular
speed.
● The kinetic energy K of a rigid body rotating about a fixed
axis is given by
(radian measure),
K ! 1
2Iv2
in which I is the rotational inertia of the body, defined as
for a system of discrete particles.
I ! ' miri
2
Learning Objectives
Key Idea
Kinetic Energy of Rotation
The rapidly rotating blade of a table saw certainly has kinetic energy due to that
rotation. How can we express the energy? We cannot apply the familiar formula
to the saw as a whole because that would give us the kinetic energy
only of the saw’s center of mass, which is zero.
K ! 1
2 mv2

Figure 10-11 A long rod is much easier to
rotate about (a) its central (longitudinal)
axis than about (b) an axis through its
center and perpendicular to its length. The
reason for the difference is that the mass
is distributed closer to the rotation axis in
(a) than in (b).
Rotation
axis
(a)
(b)
Rod is easy to rotate
this way.
Harder this way.
Instead, we shall treat the table saw (and any other rotating rigid body) as a
collection of particles with different speeds. We can then add up the kinetic
energies of all the particles to find the kinetic energy of the body as a whole.
In this way we obtain, for the kinetic energy of a rotating body,
(10-31)
in which mi is the mass of the ith particle and vi is its speed.The sum is taken over
all the particles in the body.
The problem with Eq. 10-31 is that vi is not the same for all particles.We solve
this problem by substituting for v from Eq. 10-18 (v ! vr), so that we have
(10-32)
in which v is the same for all particles.
The quantity in parentheses on the right side of Eq. 10-32 tells us how
the mass of the rotating body is distributed about its axis of rotation. We call
that quantity the rotational inertia (or moment of inertia) I of the body with
respect to the axis of rotation. It is a constant for a particular rigid body and
a particular rotation axis. (Caution: That axis must always be specified if the
value of I is to be meaningful.)
We may now write
(rotational inertia) (10-33)
and substitute into Eq. 10-32, obtaining
(radian measure) (10-34)
as the expression we seek. Because we have used the relation v ! vr in deriving
Eq. 10-34, v must be expressed in radian measure. The SI unit for I is the
kilogram–square meter (kg)m2
).
The Plan. If we have a few particles and a specified rotation axis, we find mr2
for each particle and then add the results as in Eq.10-33 to get the total rotational in-
ertia I. If we want the total rotational kinetic energy, we can then substitute that I
into Eq. 10-34.That is the plan for a few particles, but suppose we have a huge num-
ber of particles such as in a rod. In the next module we shall see how to handle such
continuous bodies and do the calculation in only a few minutes.
Equation 10-34, which gives the kinetic energy of a rigid body in pure rotation,
is the angular equivalent of the formula ,which gives the kinetic energy
K ! 1
2 Mvcom
2
K ! 1
2 I12
I ! ' miri
2
K ! ' 1
2 mi(vri)2
! 1
2 #' miri
2
$v2
,
! ' 1
2mivi
2
,
K ! 1
2 m1v2
1 & 1
2 m2v2
2
& 1
2 m3v2
3 & ) ) )
of a rigid body in pure translation. In both formulas there is a factor of . Where
mass M appears in one equation, I (which involves both mass and its distribution)
appears in the other. Finally, each equation contains as a factor the square of a
speed—translational or rotational as appropriate. The kinetic energies of transla-
tion and of rotation are not different kinds of energy.They are both kinetic energy,
expressed in ways that are appropriate to the motion at hand.
We noted previously that the rotational inertia of a rotating body involves
not only its mass but also how that mass is distributed. Here is an example that
you can literally feel. Rotate a long, fairly heavy rod (a pole, a length of lumber,
or something similar), first around its central (longitudinal) axis (Fig. 10-11a)
and then around an axis perpendicular to the rod and through the center
(Fig. 10-11b). Both rotations involve the very same mass, but the first rotation is
much easier than the second. The reason is that the mass is distributed much
closer to the rotation axis in the first rotation. As a result, the rotational inertia
of the rod is much smaller in Fig. 10-11a than in Fig. 10-11b. In general, smaller
rotational inertia means easier rotation.
1
2

273
10-5 CALCULATING THE ROTATIONAL INERTIA
Checkpoint 4
The figure shows three small spheres that rotate
about a vertical axis.The perpendicular distance
between the axis and the center of each sphere is
given. Rank the three spheres according to their
rotational inertia about that axis, greatest first.
Rotation
axis
4 kg
3 m
2 m
1 m
9 kg
36 kg
10.20 Determine the rotational inertia of a body if it is given in
Table 10-2.
10.21 Calculate the rotational inertia of a body by integration
over the mass elements of the body.
10.22 Apply the parallel-axis theorem for a rotation axis that is
displaced from a parallel axis through the center of mass of
a body.
● I is the rotational inertia of the body, defined as
for a system of discrete particles and defined as
for a body with continuously distributed mass. The r and ri in
these expressions represent the perpendicular distance from
the axis of rotation to each mass element in the body, and the
integration is carried out over the entire body so as to include
every mass element.
I ! "r 2
dm
I ! ' miri
2
● The parallel-axis theorem relates the rotational inertia I of a
body about any axis to that of the same body about a parallel
axis through the center of mass:
I ! Icom & Mh2
.
Here h is the perpendicular distance between the two axes,
and Icom is the rotational inertia of the body about the axis
through the com. We can describe h as being the distance
the actual rotation axis has been shifted from the rotation axis
through the com.
Learning Objectives
Key Ideas
Calculating the Rotational Inertia
If a rigid body consists of a few particles, we can calculate its rotational inertia
about a given rotation axis with Eq. 10-33 ; that is, we can find the
product mr2
for each particle and then sum the products. (Recall that r is the per-
pendicular distance a particle is from the given rotation axis.)
If a rigid body consists of a great many adjacent particles (it is continuous, like
a Frisbee), using Eq. 10-33 would require a computer.Thus, instead, we replace the
sum in Eq.10-33 with an integral and define the rotational inertia of the body as
(rotational inertia, continuous body). (10-35)
Table 10-2 gives the results of such integration for nine common body shapes and
the indicated axes of rotation.
Parallel-Axis Theorem
Suppose we want to find the rotational inertia I of a body of mass M about a
given axis. In principle, we can always find I with the integration of Eq. 10-35.
However, there is a neat shortcut if we happen to already know the rotational in-
ertia Icom of the body about a parallel axis that extends through the body’s center
of mass. Let h be the perpendicular distance between the given axis and the axis
I ! "r2
dm
(I ! ' miri
2
)

Table 10-2 Some Rotational Inertias
Axis
Hoop about
central axis
Axis
Annular cylinder
(or ring) about
central axis
R
I = MR2 (b)
(a) I = M(R1
2 + R2
2)
R2
R1
Thin rod about
axis through center
perpendicular to
length
(e)
I = ML2
L
Axis
Axis
Axis
Hoop about any
diameter
Slab about
perpendicular
axis through
center
(i)
(h)
I = MR2
I = M(a2
+ b2
)
R
b
a
Axis
Solid cylinder
(or disk) about
central axis
(c)
I = MR2
R
L
Axis
Solid cylinder
(or disk) about
central diameter
(d)
I = MR2
+ ML2
R
L
Axis
Thin
spherical shell
about any
diameter
(g)
I = MR2
2R
Solid sphere
about any
diameter
(f)
I = MR2
2R
Axis
1
_
_
2 1
_
_
2
2
_
_
5
1
_
_
4
2
_
_
3
1
_
_
2
1
__
12
1
__
12
1
__
12
Figure 10-12 A rigid body in cross section,
with its center of mass at O.The parallel-
axis theorem (Eq. 10-36) relates the
rotational inertia of the body about an axis
through O to that about a parallel axis
through a point such as P, a distance h
from the body’s center of mass.
dm
r
P
h
a
b
x – a
y – b
com
O
Rotation axis
through
center of mass
Rotation axis
through P
y
x
We need to relate the rotational inertia
around the axis at P to that around the
axis at the com.
through the center of mass (remember these two axes must be parallel).Then the
rotational inertia I about the given axis is
I ! Icom & Mh2
(parallel-axis theorem). (10-36)
Think of the distance h as being the distance we have shifted the rotation axis
from being through the com.This equation is known as the parallel-axis theorem.
We shall now prove it.
Proof of the Parallel-Axis Theorem
Let O be the center of mass of the arbitrarily shaped body shown in cross section
in Fig. 10-12. Place the origin of the coordinates at O. Consider an axis through O
perpendicular to the plane of the figure, and another axis through point P paral-
lel to the first axis. Let the x and y coordinates of P be a and b.
Let dm be a mass element with the general coordinates x and y. The rota-
tional inertia of the body about the axis through P is then, from Eq. 10-35,
which we can rearrange as
(10-37)
From the definition of the center of mass (Eq. 9-9), the middle two integrals of
Eq. 10-37 give the coordinates of the center of mass (multiplied by a constant)
I ! "(x2
& y2
) dm % 2a "x dm % 2b "y dm & "(a2
& b2
) dm.
I ! "r2
dm ! "[(x % a)2
& (y % b)2
] dm,

275
and thus must each be zero. Because x2
& y2
is equal to R2
, where R is the dis-
tance from O to dm, the first integral is simply Icom, the rotational inertia of the
body about an axis through its center of mass. Inspection of Fig. 10-12 shows that
the last term in Eq. 10-37 is Mh2
, where M is the body’s total mass. Thus,
Eq. 10-37 reduces to Eq. 10-36, which is the relation that we set out to prove.
Checkpoint 5
The figure shows a book-like object (one side is
longer than the other) and four choices of rotation
axes, all perpendicular to the face of the object.
Rank the choices according to the rotational inertia
of the object about the axis, greatest first.
(1) (2) (3) (4)
left and L for the particle on the right. Now Eq. 10-33
gives us
I ! m(0)2
& mL2
! mL2
. (Answer)
Second technique: Because we already know Icom about an
axis through the center of mass and because the axis here is
parallel to that “com axis,” we can apply the parallel-axis
theorem (Eq. 10-36).We find
(Answer)
! mL2
.
I ! Icom & Mh2
! 1
2 mL2
& (2m)(1
2 L)2
Sample Problem 10.06 Rotational inertia of a two-particle system
Figure 10-13a shows a rigid body consisting of two particles of
mass m connected by a rod of length L and negligible mass.
(a) What is the rotational inertia Icom about an axis through the
centerofmass,perpendiculartotherodasshown?
KEY IDEA
Because we have only two particles with mass, we can find
the body’s rotational inertia Icom by using Eq. 10-33 rather
than by integration. That is, we find the rotational inertia of
each particle and then just add the results.
Calculations: For the two particles, each at perpendicular
distance from the rotation axis, we have
(Answer)
(b) What is the rotational inertia I of the body about an axis
through the left end of the rod and parallel to the first axis
(Fig. 10-13b)?
KEY IDEAS
This situation is simple enough that we can find I using
either of two techniques. The first is similar to the one used
in part (a). The other, more powerful one is to apply the
parallel-axis theorem.
First technique: We calculate I as in part (a), except here
the perpendicular distance ri is zero for the particle on the
! 1
2 mL2
.
I ! ' miri
2
! (m)(1
2 L)2
& (m)(1
2 L)2
1
2 L
m m
(a)
L
L
com
Rotation axis
through
center of mass
m m
(b)
L
com
Rotation axis through
end of rod
1
_
_
2
1
_
_
2
Here the rotation axis is through the com.
Here it has been shifted from the com
without changing the orientation. We
can use the parallel-axis theorem.
Figure 10-13 A rigid body consisting of two particles of mass m
joined by a rod of negligible mass.

Sample Problem 10.07 Rotational inertia of a uniform rod, integration
Figure 10-14 shows a thin, uniform rod of mass M and length
L,on an x axis with the origin at the rod’s center.
(a) What is the rotational inertia of the rod about the
perpendicular rotation axis through the center?
KEY IDEAS
(1) The rod consists of a huge number of particles at a great
many different distances from the rotation axis. We certainly
don’t want to sum their rotational inertias individually. So, we
first write a general expression for the rotational inertia of a
mass element dm at distance r from the rotation axis: r2
dm.
(2) Then we sum all such rotational inertias by integrating the
expression (rather than adding them up one by one). From
Eq.10-35,we write
(10-38)
(3) Because the rod is uniform and the rotation axis is at the
center, we are actually calculating the rotational inertia Icom
about the center of mass.
Calculations: We want to integrate with respect to coordinate
x (not mass m as indicated in the integral), so we must relate
the mass dm of an element of the rod to its length dx along the
rod. (Such an element is shown in Fig. 10-14.) Because the rod
is uniform,the ratio of mass to length is the same for all the el-
ements and for the rod as a whole.Thus,we can write
or dm !
M
L
dx.
element’s mass dm
element’s length dx
!
rod’s mass M
rod’s length L
I ! "r2
dm.
Figure 10-14 A uniform rod of length L
and mass M. An element of mass dm
and length dx is represented.
A
We can now substitute this result for dm and x for r in
Eq.10-38.Then we integrate from end to end of the rod (from
x ! %L/2 to x ! L/2) to include all the elements.We find
(Answer)
(b) What is the rod’s rotational inertia I about a new rotation
axis that is perpendicular to the rod and through the left end?
KEY IDEAS
We can find I by shifting the origin of the x axis to the left end
of the rod and then integrating from to .However,
here we shall use a more powerful (and easier) technique by
applying the parallel-axis theorem (Eq. 10-36), in which we
shift the rotation axis without changing its orientation.
Calculations: If we place the axis at the rod’s end so that it
is parallel to the axis through the center of mass, then we
can use the parallel-axis theorem (Eq. 10-36). We know
from part (a) that Icom is . From Fig. 10-14, the perpen-
dicular distance h between the new rotation axis and the
center of mass is . Equation 10-36 then gives us
(Answer)
Actually, this result holds for any axis through the left
or right end that is perpendicular to the rod.
! 1
3 ML2
.
I ! Icom & Mh2
! 1
12 ML2
& (M)(1
2 L)2
1
2 L
1
12 ML2
x ! L
x ! 0
! 1
12 ML2
.
!
M
3L (x3
)%L/2
&L/2
!
M
3L (#L
2 $
3
% #%
L
2 $
3
)
I ! "x!&L/2
x!%L/2
x2
#M
L $dx
x
Rotation
axis
L
__
2
L
__
2
com M
We want the
rotational inertia.
x
Rotation
axis
x dm
dx
First, pick any tiny element
and write its rotational
inertia as x2 dm.
x
x = −
Rotation
axis
Leftmost Rightmost
L
__
2
x = L
__
2
Then, using integration, add up
the rotational inertias for all of
the elements, from leftmost to
rightmost.

277
10-6 TORQUE
KEY IDEA
The released energy was equal to the rotational kinetic en-
ergy K of the rotor just as it reached the angular speed of
14 000 rev/min.
Calculations: We can find K with Eq. 10-34 , but
first we need an expression for the rotational inertia I.Because
the rotor was a disk that rotated like a merry-go-round, I is
given in Table 10-2c .Thus,
The angular speed of the rotor was
Then, with Eq. 10-34, we find the (huge) energy release:
(Answer)
! 2.1 $ 107
J.
K ! 1
2 Iv2
! 1
2(19.64 kg)m2
)(1.466 $ 103
rad/s)2
! 1.466 $ 103
rad/s.
v ! (14 000 rev/min)(2p rad/rev)#1 min
60 s $
I ! 1
2 MR2
! 1
2 (272 kg)(0.38 m)2
! 19.64 kg)m2
.
(I ! 1
2 MR2
)
(K ! 1
2 Iv2
)
Sample Problem 10.08 Rotational kinetic energy, spin test explosion
Large machine components that undergo prolonged, high-
speed rotation are first examined for the possibility of fail-
ure in a spin test system. In this system, a component is spun
up (brought up to high speed) while inside a cylindrical
arrangement of lead bricks and containment liner, all within
a steel shell that is closed by a lid clamped into place. If the
rotation causes the component to shatter, the soft lead
bricks are supposed to catch the pieces for later analysis.
In 1985,Test Devices, Inc. (www.testdevices.com) was spin
testing a sample of a solid steel rotor (a disk) of mass M !
272 kg and radius R ! 38.0 cm. When the sample reached
an angular speed v of 14 000 rev/min, the test engineers
heard a dull thump from the test system, which was
located one floor down and one room over from them.
Investigating, they found that lead bricks had been thrown
out in the hallway leading to the test room, a door to the
room had been hurled into the adjacent parking lot, one
lead brick had shot from the test site through the wall of a
neighbor’s kitchen, the structural beams of the test build-
ing had been damaged, the concrete floor beneath the
spin chamber had been shoved downward by about 0.5
cm, and the 900 kg lid had been blown upward through
the ceiling and had then crashed back onto the test equip-
ment (Fig. 10-15). The exploding pieces had not pene-
trated the room of the test engineers only by luck.
How much energy was released in the explosion of the
rotor?
Figure 10-15 Some of the
destruction caused by
the explosion of a rap-
idly rotating steel disk.
Courtesy
Test
Devices,
Inc.
10-6 TORQUE
10.23 Identify that a torque on a body involves a force and a
position vector, which extends from a rotation axis to the
point where the force is applied.
10.24 Calculate the torque by using (a) the angle between
the position vector and the force vector, (b) the line of ac-
tion and the moment arm of the force, and (c) the force
component perpendicular to the position vector.
10.25 Identify that a rotation axis must always be specified to
calculate a torque.
10.26 Identify that a torque is assigned a positive or negative
sign depending on the direction it tends to make the body
rotate about a specified rotation axis: “clocks are negative.”
10.27 When more than one torque acts on a body about a
rotation axis, calculate the net torque.
Learning Objectives
● Torque is a turning or twisting action on a body about a
rotation axis due to a force . If is exerted at a point given
by the position vector relative to the axis, then the magni-
tude of the torque is
where Ft is the component of perpendicular to and
f is the angle between and . The quantity is the
r"
F
:
r
:
r
:
F
:
t ! rFt ! r"F ! rF sin f,
r
:
F
:
F
:
perpendicular distance between the rotation axis and
an extended line running through the vector. This line
is called the line of action of , and is called the
moment arm of . Similarly, r is the moment arm of Ft.
● The SI unit of torque is the newton-meter (N)m). A
torque t is positive if it tends to rotate a body at rest
counterclockwise and negative if it tends to rotate the
body clockwise.
F
:
r"
F
:
F
:
Key Ideas

Checkpoint 6
The figure shows an overhead view of a meter stick that can pivot about the dot at the position
marked 20 (for 20 cm).All five forces on the stick are horizontal and have the same magnitude.
Rank the forces according to the magnitude of the torque they produce, greatest first.
0 20 40
Pivot point
100
F1
F2
F3
F4
F5
Figure 10-16 (a) A force acts on a rigid
body, with a rotation axis perpendicular to
the page. The torque can be found with
(a) angle f, (b) tangential force compo-
nent Ft, or (c) moment arm .
r"
F
:
(a)
(b)
(c)
O
P
φ Fr
Ft
Rotation
axis
F
r
O
P
φ
Rotation
axis
φ
Line of
action of F
r
Moment arm
of F
F
r
O
P
φ
Rotation
axis
F
r
The torque due to this force
causes rotation around this axis
(which extends out toward you).
You calculate the same torque by
using this moment arm distance
and the full force magnitude.
But actually only the tangential
component of the force causes
the rotation.
magnitude Ft ! F sin f.This component does cause rotation.
Calculating Torques. The ability of to rotate the body depends not only
on the magnitude of its tangential component Ft, but also on just how far from O
the force is applied. To include both these factors, we define a quantity called
torque t as the product of the two factors and write it as
t ! (r)(F sin f). (10-39)
Two equivalent ways of computing the torque are
t ! (r)(F sin f) ! rFt (10-40)
and (10-41)
where is the perpendicular distance between the rotation axis at O and an extended
r"
t ! (r sin f)(F) ! r"F,
F
:
Torque
A doorknob is located as far as possible from the door’s hinge line for a good rea-
son. If you want to open a heavy door, you must certainly apply a force, but that
is not enough.Where you apply that force and in what direction you push are also
important. If you apply your force nearer to the hinge line than the knob, or at
any angle other than 90# to the plane of the door, you must use a greater force
than if you apply the force at the knob and perpendicular to the door’s plane.
Figure 10-16a shows a cross section of a body that is free to rotate about an
axis passing through O and perpendicular to the cross section. A force is
applied at point P, whose position relative to O is defined by a position vector .
The directions of vectors and make an angle f with each other. (For simplic-
ity, we consider only forces that have no component parallel to the rotation axis;
thus, is in the plane of the page.)
To determine how results in a rotation of the body around the rotation
axis, we resolve into two components (Fig. 10-16b). One component, called the
radial component Fr, points along . This component does not cause rotation,
because it acts along a line that extends through O. (If you pull on a door par-
allel to the plane of the door, you do not rotate the door.) The other compo-
nent of , called the tangential component Ft, is perpendicular to and has
r
:
F
:
r
:
F
:
F
:
F
:
r
:
F
:
r
:
F
:
line running through the vector (Fig. 10-16c).This extended line is called the line
of action of , and is called the moment arm of . Figure 10-16b shows that we
can describe r,the magnitude of ,as being the moment arm of the force componentFt.
Torque, which comes from the Latin word meaning “to twist,” may be loosely
identified as the turning or twisting action of the force .When you apply a force
to an object—such as a screwdriver or torque wrench—with the purpose of turn-
ing that object, you are applying a torque. The SI unit of torque is the newton-
meter (N)m). Caution: The newton-meter is also the unit of work. Torque and
work, however, are quite different quantities and must not be confused. Work is
often expressed in joules (1 J ! 1 N)m),but torque never is.
Clocks Are Negative. In Chapter 11 we shall use vector notation for torques,
but here, with rotation around a single axis, we use only an algebraic sign. If a
torque would cause counterclockwise rotation, it is positive. If it would cause
clockwise rotation, it is negative. (The phrase “clocks are negative” from Module
10-1 still works.)
Torques obey the superposition principle that we discussed in Chapter 5 for
forces:When several torques act on a body, the net torque (or resultant torque) is
the sum of the individual torques.The symbol for net torque is tnet.
F
:
r
:
F
:
r"
F
:
F
:

279
10-7 NEWTON’S SECOND LAW FOR ROTATION
10.28 Apply Newton’s second law for rotation to relate the
net torque on a body to the body’s rotational inertia and
rotational acceleration, all calculated relative to a specified
rotation axis.
● The rotational analog of Newton’s second law is
tnet ! Ia,
where tnet is the net torque acting on a particle or rigid body,
I is the rotational inertia of the particle or body about the
rotation axis, and a is the resulting angular acceleration about
that axis.
Learning Objective
Key Idea
Newton’s Second Law for Rotation
A torque can cause rotation of a rigid body, as when you use a torque to rotate
a door. Here we want to relate the net torque tnet on a rigid body to the angular
acceleration a that torque causes about a rotation axis.We do so by analogy with
Newton’s second law (Fnet ! ma) for the acceleration a of a body of mass m due
to a net force Fnet along a coordinate axis.We replace Fnet with tnet, m with I, and a
with a in radian measure, writing
tnet ! Ia (Newton’s second law for rotation). (10-42)
Proof of Equation 10-42
We prove Eq. 10-42 by first considering the simple situation shown in Fig. 10-17.
The rigid body there consists of a particle of mass m on one end of a massless rod
of length r.The rod can move only by rotating about its other end, around a rota-
tion axis (an axle) that is perpendicular to the plane of the page.Thus, the particle
can move only in a circular path that has the rotation axis at its center.
A force acts on the particle. However, because the particle can move
only along the circular path, only the tangential component Ft of the force (the
component that is tangent to the circular path) can accelerate the particle along
the path. We can relate Ft to the particle’s tangential acceleration at along the
path with Newton’s second law, writing
Ft ! mat.
The torque acting on the particle is, from Eq. 10-40,
t ! Ftr ! matr.
From Eq. 10-22 (at ! ar) we can write this as
t ! m(ar)r ! (mr2
)a. (10-43)
The quantity in parentheses on the right is the rotational inertia of the particle
about the rotation axis (see Eq. 10-33, but here we have only a single particle).
Thus, using I for the rotational inertia, Eq. 10-43 reduces to
t ! Ia (radian measure). (10-44)
If more than one force is applied to the particle, Eq. 10-44 becomes
tnet ! Ia (radian measure), (10-45)
which we set out to prove.We can extend this equation to any rigid body rotating
about a fixed axis, because any such body can always be analyzed as an assembly
of single particles.
F
:
Figure 10-17 A simple rigid body, free to
rotate about an axis through O, consists of
a particle of mass m fastened to the end of
a rod of length r and negligible mass. An
applied force causes the body to rotate.
F
:
O
x
y
Rod
θ
Rotation axis
r
m
Fr
Ft
φ
F
The torque due to the tangential
component of the force causes
an angular acceleration around
the rotation axis.

KEY IDEA
Because the moment arm for is no longer zero, the torque
F
:
g
Checkpoint 7
The figure shows an overhead view of a meter stick that can pivot about the point indicated,which is
to the left of the stick’s midpoint.Two horizontal forces, and ,are applied to the stick.Only is
shown.Force is perpendicular to the stick and is applied at the right end.If the stick is not to turn,
(a) what should be the direction of ,and (b) should F2 be greater than,less than,or equal to F1?
F
:
2
F
:
2
F
:
1
F
:
2
F
:
1
F1
Pivot point
Sample Problem 10.09 Using Newton’s second law for rotation in a basic judo hip throw
To throw an 80 kg opponent with a basic judo hip throw, you
intend to pull his uniform with a force and a moment arm
d1 ! 0.30 m from a pivot point (rotation axis) on your right
hip (Fig. 10-18). You wish to rotate him about the pivot
point with an angular acceleration a of %6.0 rad/s2
—that is,
with an angular acceleration that is clockwise in the figure.
Assume that his rotational inertia I relative to the pivot
point is 15 kg)m2
.
(a) What must the magnitude of be if, before you throw
him, you bend your opponent forward to bring his center of
mass to your hip (Fig. 10-18a)?
KEY IDEA
We can relate your pull on your opponent to the given an-
gular acceleration a via Newton’s second law for rotation
(tnet ! Ia).
Calculations: As his feet leave the floor, we can assume that
only three forces act on him: your pull , a force on him
from you at the pivot point (this force is not indicated in Fig.
10-18), and the gravitational force .To use tnet ! Ia, we need
the corresponding three torques,each about the pivot point.
From Eq. 10-41 (t ! F), the torque due to your pull F
:
r"
F
:
g
N
:
F
:
F
:
F
:
F
:
Figure 10-18 A judo hip throw (a) correctly executed and (b) incor-
rectly executed.
Opponent's
center of
mass
Moment arm d1
of your pull
Pivot
on hip
Moment arm d2
of gravitational
force on
opponent
Moment
arm d1
of your pull
Fg
Fg
(a) (b)
F
F
is equal to % F, where is the moment arm and the
sign indicates the clockwise rotation this torque tends to
cause. The torque due to is zero, because acts at the
N
:
N
:
r"
d1
d1
pivot point and thus has moment arm ! 0.
To evaluate the torque due to , we can assume that
acts at your opponent’s center of mass. With the center of
mass at the pivot point, has moment arm ! 0 and thus
r"
F
:
g
F
:
g
F
:
g
r"
ponent is due to your pull , and we can write tnet ! Ia as
%d1F ! Ia.
We then find
! 300 N. (Answer)
(b) What must the magnitude of be if your opponent
remains upright before you throw him, so that has a mo-
ment arm d2 ! 0.12 m (Fig. 10-18b)?
F
:
g
F
:
F !
%Ia
d1
!
%(15 kg)m2
)(%6.0 rad/s2
)
0.30 m
F
:
the torque due to is zero. So, the only torque on your op-
F
:
g
due to is now equal to d2mg and is positive because the
torque attempts counterclockwise rotation.
Calculations: Now we write tnet ! Ia as
%d1F & d2mg ! Ia,
which gives
From (a), we know that the first term on the right is equal to
300 N. Substituting this and the given data, we have
! 613.6 N 610 N. (Answer)
The results indicate that you will have to pull much harder if
you do not initially bend your opponent to bring his center
of mass to your hip. A good judo fighter knows this lesson
from physics. Indeed, physics is the basis of most of the mar-
tial arts, figured out after countless hours of trial and error
over the centuries.
%
F ! 300 N &
(0.12 m)(80 kg)(9.8 m/s2
)
0.30 m
F ! %
Ia
d1
&
d2mg
d1
.
F
:
g

281
Sample Problem 10.10 Newton’s second law, rotation, torque, disk
Figure 10-19a shows a uniform disk, with mass M ! 2.5 kg
and radius R ! 20 cm, mounted on a fixed horizontal axle.
A block with mass m ! 1.2 kg hangs from a massless cord that
is wrapped around the rim of the disk.Find the acceleration of
the falling block, the angular acceleration of the disk, and the
tension in the cord.The cord does not slip, and there is no fric-
tion at the axle.
KEY IDEAS
(1) Taking the block as a system, we can relate its accelera-
tion a to the forces acting on it with Newton’s second law
( ). (2) Taking the disk as a system, we can relate
its angular acceleration a to the torque acting on it with
Newton’s second law for rotation (tnet ! Ia). (3) To combine
the motions of block and disk, we use the fact that the linear
acceleration a of the block and the (tangential) linear accel-
eration of the disk rim are equal. (To avoid confusion
about signs, let’s work with acceleration magnitudes and
explicit algebraic signs.)
Forces on block: The forces are shown in the block’s free-
body diagram in Fig. 10-19b: The force from the cord is ,
and the gravitational force is , of magnitude mg. We can
now write Newton’s second law for components along a ver-
tical y axis (Fnet,y ! may) as
T % mg ! m(%a), (10-46)
where a is the magnitude of the acceleration (down the y
axis). However, we cannot solve this equation for a because
it also contains the unknown T.
Torque on disk: Previously, when we got stuck on the y axis,
we switched to the x axis. Here, we switch to the rotation of
the disk and use Newton’s second law in angular form. To
calculate the torques and the rotational inertia I, we take
the rotation axis to be perpendicular to the disk and through
its center, at point O in Fig. 10-19c.
The torques are then given by Eq. 10-40 (t ! rFt). The
gravitational force on the disk and the force on the disk from
the axle both act at the center of the disk and thus at distance
r ! 0,so their torques are zero.The force on the disk due to
the cord acts at distance r ! R and is tangent to the rim of the
disk. Therefore, its torque is %RT, negative because the
torque rotates the disk clockwise from rest. Let a be the mag-
nitude of the negative (clockwise) angular acceleration. From
Table 10-2c, the rotational inertia I of the disk is . Thus
we can write the general equation tnet ! Ia as
(10-47)
%RT ! 1
2 MR2
(%a).
1
2MR2
T
:
F
:
g
T
:
at
F
:
net ! m:
a
This equation seems useless because it has two
unknowns, a and T, neither of which is the desired a.
However, mustering physics courage, we can make it useful
with this fact: Because the cord does not slip, the magnitude
a of the block’s linear acceleration and the magnitude at of
the (tangential) linear acceleration of the rim of the disk are
equal. Then, by Eq. 10-22 (at ! ar) we see that here a !
a/R. Substituting this in Eq. 10-47 yields
(10-48)
Combining results: Combining Eqs. 10-46 and 10-48 leads to
. (Answer)
We then use Eq. 10-48 to find T:
(Answer)
As we should expect, acceleration a of the falling block is less
than g, and tension T in the cord (! 6.0 N) is less than the
gravitational force on the hanging block (! mg ! 11.8 N).
We see also that a and T depend on the mass of the disk but
not on its radius.
As a check, we note that the formulas derived above
predict a ! g and T ! 0 for the case of a massless disk (M !
0). This is what we would expect; the block simply falls as a
free body. From Eq. 10-22, the magnitude of the angular ac-
celeration of the disk is
(Answer)
a !
a
R
!
4.8 m/s2
0.20 m
! 24 rad/s2
.
! 6.0 N.
T ! 1
2 Ma ! 1
2(2.5 kg)(4.8 m/s2
)
! 4.8 m/s2
a ! g
2m
M & 2m
! (9.8 m/s2
)
(2)(1.2 kg)
2.5 kg & (2)(1.2 kg)
T ! 1
2 Ma.
m
M
M
R
O
Fg
(b)
(a)
(c)
m
T
T
The torque due to the
cord's pull on the rim
causes an angular
acceleration of the disk.
These two forces
determine the block's
(linear) acceleration.
We need to relate
those two
accelerations.
y
Figure 10-19 (a) The falling block causes the disk to rotate. (b) A
free-body diagram for the block. (c) An incomplete free-body
diagram for the disk.

Work and Rotational Kinetic Energy
As we discussed in Chapter 7, when a force F causes a rigid body of mass m to ac-
celerate along a coordinate axis, the force does work W on the body. Thus, the
body’s kinetic energy can change. Suppose it is the only energy of the
(K ! 1
2 mv2
)
10-8 WORK AND ROTATIONAL KINETIC ENERGY
10.29 Calculate the work done by a torque acting on a rotat-
ing body by integrating the torque with respect to the an-
gle of rotation.
10.30 Apply the work–kinetic energy theorem to relate the
work done by a torque to the resulting change in the rota-
tional kinetic energy of the body.
10.31 Calculate the work done by a constant torque by relat-
ing the work to the angle through which the body rotates.
10.32 Calculate the power of a torque by finding the rate at
which work is done.
10.33 Calculate the power of a torque at any given instant by
relating it to the torque and the angular velocity at that instant.
● The equations used for calculating work and power in rota-
tional motion correspond to equations used for translational
motion and are
and P !
dW
dt
! tv.
W ! "uf
ui
t du
● When t is constant, the integral reduces to
W ! t(uf % ui).
● The form of the work–kinetic energy theorem used for
rotating bodies is
"K ! Kf % Ki ! 1
2 Ivf
2
% 1
22vi
2
! W.
Learning Objectives
Key Ideas
body that changes.Then we relate the change "K in kinetic energy to the work W
with the work–kinetic energy theorem (Eq. 7-10), writing
(work–kinetic energy theorem). (10-49)
For motion confined to an x axis, we can calculate the work with Eq. 7-32,
(work, one-dimensional motion). (10-50)
This reduces to W ! Fd when F is constant and the body’s displacement is d.
The rate at which the work is done is the power, which we can find with Eqs. 7-43
and 7-48,
(power, one-dimensional motion). (10-51)
Now let us consider a rotational situation that is similar. When a torque
accelerates a rigid body in rotation about a fixed axis, the torque does work W
on the body. Therefore, the body’s rotational kinetic energy can
change. Suppose that it is the only energy of the body that changes. Then we
can still relate the change "K in kinetic energy to the work W with the
work–kinetic energy theorem, except now the kinetic energy is a rotational
kinetic energy:
(work–kinetic energy theorem). (10-52)
Here, I is the rotational inertia of the body about the fixed axis and vi and vf are
the angular speeds of the body before and after the work is done.
"K ! Kf % Ki ! 1
2 Ivf
2
% 1
22vi
2
! W
(K ! 1
2 I12
)
P !
dW
dt
! Fv
W ! "xf
xi
F dx
"K ! Kf % Ki ! 1
2 mvf
2
% 1
2 mvi
2
! W

283
10-8 WORK AND ROTATIONAL KINETIC ENERGY
Also, we can calculate the work with a rotational equivalent of Eq. 10-50,
(work, rotation about fixed axis), (10-53)
where t is the torque doing the work W, and ui and uf are the body’s angular
positions before and after the work is done, respectively. When t is constant,
Eq. 10-53 reduces to
W ! t(uf % ui) (work, constant torque). (10-54)
The rate at which the work is done is the power, which we can find with the rota-
tional equivalent of Eq. 10-51,
(power, rotation about fixed axis). (10-55)
Table 10-3 summarizes the equations that apply to the rotation of a rigid body
about a fixed axis and the corresponding equations for translational motion.
Proof of Eqs. 10-52 through 10-55
Let us again consider the situation of Fig. 10-17, in which force rotates a rigid
body consisting of a single particle of mass m fastened to the end of a massless
rod. During the rotation, force does work on the body. Let us assume that the
only energy of the body that is changed by is the kinetic energy. Then we can
apply the work–kinetic energy theorem of Eq. 10-49:
"K ! Kf % Ki ! W. (10-56)
Using and Eq. 10-18 (v ! vr), we can rewrite Eq. 10-56 as
(10-57)
From Eq. 10-33, the rotational inertia for this one-particle body is I ! mr2
.
Substituting this into Eq. 10-57 yields
which is Eq. 10-52.We derived it for a rigid body with one particle, but it holds for
any rigid body rotated about a fixed axis.
We next relate the work W done on the body in Fig. 10-17 to the torque t
on the body due to force . When the particle moves a distance ds along its
F
:
"K ! 1
2 Ivf
2
% 1
2 2vi
2
! W,
"K ! 1
2 mr2
vf
2
% 1
2 mr2
vi
2
! W.
K ! 1
2 mv2
F
:
F
:
F
:
P !
dW
dt
! tv
W ! "uf
ui
t du
Table 10-3 Some Corresponding Relations for Translational and Rotational Motion
Pure Translation (Fixed Direction) Pure Rotation (Fixed Axis)
Position x Angular position u
Velocity v ! dx/dt Angular velocity v ! du/dt
Acceleration a ! dv/dt Angular acceleration a ! dv/dt
Mass m Rotational inertia I
Newton’s second law Fnet ! ma Newton’s second law tnet ! Ia
Work W ! * F dx Work W ! * t du
Kinetic energy Kinetic energy K ! 1
2 Iv2
K ! 1
2 mv2
Power (constant force) P ! Fv Power (constant torque) P ! tv
Work–kinetic energy theorem W ! "K Work–kinetic energy theorem W ! "K

Sample Problem 10.11 Work, rotational kinetic energy, torque, disk
Let the disk in Fig. 10-19 start from rest at time t ! 0 and
also let the tension in the massless cord be 6.0 N and the an-
gular acceleration of the disk be %24 rad/s2
.What is its rota-
tional kinetic energy K at t ! 2.5 s?
KEY IDEA
We can find K with Eq. 10-34 We already know
(K ! 1
2 Iv2
).
Calculations: First, we relate the change in the kinetic
energy of the disk to the net work W done on the disk, using
the work–kinetic energy theorem of Eq. 10-52 (Kf % Ki ! W).
With K substituted for Kf and 0 for Ki,we get
K ! Ki & W ! 0 & W ! W. (10-60)
Next we want to find the work W. We can relate W to
the torques acting on the disk with Eq. 10-53 or 10-54. The
only torque causing angular acceleration and doing work is
the torque due to force on the disk from the cord, which is
T
:
that , but we do not yet know v at t ! 2.5 s.
However, because the angular acceleration a has the con-
stant value of %24 rad/s2
, we can apply the equations for
constant angular acceleration in Table 10-1.
Calculations: Because we want v and know a and v0 (! 0),
we use Eq.10-12:
v ! v0 & at ! 0 & at ! at.
Substituting v ! at and into Eq.10-34,we find
(Answer)
KEY IDEA
We can also get this answer by finding the disk’s kinetic
energy from the work done on the disk.
! 90 J.
! 1
4 (2.5 kg)[(0.20 m)(%24 rad/s2
)(2.5 s)]2
K ! 1
2 Iv2
! 1
2(1
2MR2
)(at)2
! 1
4M(Rat)2
I ! 1
2 MR2
I ! 1
2 MR2
circular path, only the tangential component Ft of the force accelerates the parti-
cle along the path. Therefore, only Ft does work on the particle. We write that
work dW as Ft ds. However, we can replace ds with r du, where du is the angle
through which the particle moves.Thus we have
dW ! Ftr du. (10-58)
From Eq. 10-40, we see that the product Ftr is equal to the torque t, so we can
rewrite Eq. 10-58 as
dW ! t du. (10-59)
The work done during a finite angular displacement from ui to uf is then
which is Eq. 10-53. It holds for any rigid body rotating about a fixed axis.
Equation 10-54 comes directly from Eq. 10-53.
We can find the power P for rotational motion from Eq. 10-59:
which is Eq. 10-55.
P !
dW
dt
! t
du
dt
! tv,
W ! "uf
ui
t du,
equal to %TR. Because a is constant, this torque also must
be constant.Thus, we can use Eq. 10-54 to write
W ! t(uf % ui) ! %TR(uf % ui). (10-61)
Because a is constant, we can use Eq. 10-13 to find
uf % ui.With vi ! 0, we have
.
Now we substitute this into Eq. 10-61 and then substitute the
result into Eq. 10-60. Inserting the given values T ! 6.0 N
and a ! %24 rad/s2
, we have
(Answer)
! 90 J.
! %1
2 (6.0 N)(0.20 m)(%24 rad/s2
)(2.5 s)2
K ! W ! %TR(uf % ui) ! %TR(1
2at2
) ! %1
2TRat2
uf % ui ! vit & 1
2at2
! 0 & 1
2at2
! 1
2at2

285
REVIEW & SUMMARY
Angular Position To describe the rotation of a rigid body about
a fixed axis, called the rotation axis, we assume a reference line is
fixed in the body, perpendicular to that axis and rotating with the
body.We measure the angular position u of this line relative to a fixed
direction.When u is measured in radians,
(radian measure), (10-1)
where s is the arc length of a circular path of radius r and angle u.
Radian measure is related to angle measure in revolutions and de-
grees by
1 rev ! 360# ! 2p rad. (10-2)
Angular Displacement A body that rotates about a rotation
axis, changing its angular position from u1 to u2, undergoes an angu-
lar displacement
"u ! u2 % u1, (10-4)
where "u is positive for counterclockwise rotation and negative for
clockwise rotation.
Angular Velocity and Speed If a body rotates through an
angular displacement "u in a time interval "t, its average angular
velocity vavg is
(10-5)
The (instantaneous) angular velocity v of the body is
(10-6)
Both vavg and v are vectors, with directions given by the right-hand
rule of Fig. 10-6. They are positive for counterclockwise rotation
and negative for clockwise rotation. The magnitude of the body’s
angular velocity is the angular speed.
Angular Acceleration If the angular velocity of a body
changes from v1 to v2 in a time interval "t ! t2 % t1, the average
angular acceleration aavg of the body is
(10-7)
The (instantaneous) angular acceleration a of the body is
(10-8)
Both aavg and a are vectors.
The Kinematic Equations for Constant Angular Accel-
eration Constant angular acceleration (a ! constant) is an im-
portant special case of rotational motion. The appropriate kine-
matic equations, given in Table 10-1, are
v ! v0 & at, (10-12)
(10-13)
(10-14)
(10-15)
(10-16)
Linear and Angular Variables Related A point in a rigid
rotating body, at a perpendicular distance r from the rotation axis,
u % u0 ! vt % 1
2 at2
.
u % u0 ! 1
2 (v0 & v)t,
v2
! v0
2
& 2a(u % u0),
u % u0 ! v0t & 1
2at2
,
a !
dv
dt
.
aavg !
v2 % v1
t2 % t1
!
"v
"t
.
v !
du
dt
.
vavg !
"u
"t
.
u !
s
r
Review & Summary
moves in a circle with radius r. If the body rotates through an angle u,
the point moves along an arc with length s given by
s ! ur (radian measure), (10-17)
where u is in radians.
The linear velocity of the point is tangent to the circle; the
point’s linear speed v is given by
v ! vr (radian measure), (10-18)
where v is the angular speed (in radians per second) of the body.
The linear acceleration of the point has both tangential and
radial components.The tangential component is
at ! ar (radian measure), (10-22)
where a is the magnitude of the angular acceleration (in radians
per second-squared) of the body.The radial component of is
If the point moves in uniform circular motion, the period T of
the motion for the point and the body is
(radian measure). (10-19, 10-20)
Rotational Kinetic Energy and Rotational Inertia The ki-
netic energy K of a rigid body rotating about a fixed axis is given by
(radian measure), (10-34)
in which I is the rotational inertia of the body, defined as
(10-33)
for a system of discrete particles and defined as
(10-35)
for a body with continuously distributed mass. The r and ri in these
expressions represent the perpendicular distance from the axis of
rotation to each mass element in the body, and the integration is car-
ried out over the entire body so as to include every mass element.
The Parallel-Axis Theorem The parallel-axis theorem relates
the rotational inertia I of a body about any axis to that of the same
body about a parallel axis through the center of mass:
I ! Icom & Mh2
. (10-36)
Here h is the perpendicular distance between the two axes, and
Icom is the rotational inertia of the body about the axis through the
com. We can describe h as being the distance the actual rotation
axis has been shifted from the rotation axis through the com.
Torque Torque is a turning or twisting action on a body about a ro-
tation axis due to a force . If is exerted at a point given by the po-
sition vector relative to the axis, then the magnitude of the torque is
(10-40, 10-41, 10-39)
where Ft is the component of perpendicular to and f is the an-
gle between and .The quantity is the perpendicular distance
between the rotation axis and an extended line running through
the vector. This line is called the line of action of , and is
called the moment arm of . Similarly, r is the moment arm of Ft.
F
:
r"
F
:
F
:
r"
F
:
r
:
r
:
F
:
t ! rFt ! r"F ! rF sin f,
r
:
F
:
F
:
I ! "r2
dm
I ! ' miri
2
K ! 1
2Iv2
T !
2pr
v
!
2p
v
ar !
v2
r
! v2
r
a
:
a
:
v
:

8 Figure 10-25b shows an overhead view of a horizontal bar that
is rotated about the pivot point by two horizontal forces, and ,
with at angle f to the bar. Rank the following values of f accord-
ing to the magnitude of the angular acceleration of the bar, greatest
first:90#,70#,and 110#.
9 Figure 10-26 shows a uniform metal plate
that had been square before 25% of it was
snipped off. Three lettered points are indicated.
Rank them according to the rotational inertia of
the plate around a perpendicular axis through
them, greatest first.
F
:
2
F
:
2
F
:
1
angles during the rotation, which is
counterclockwise and at a constant
rate. However, we are to decrease the
angle u of without changing the
magnitude of . (a) To keep the an-
gular speed constant, should we in-
crease, decrease, or maintain the mag-
nitude of ? Do forces (b) and (c)
tend to rotate the disk clockwise or
counterclockwise?
6 In the overhead view of Fig. 10-24,
five forces of the same magnitude act
on a strange merry-go-round; it is a
square that can rotate about point P, at
midlength along one of the edges.
Rank the forces according to the mag-
nitude of the torque they create about
point P,greatest first.
7 Figure 10-25a is an overhead view
of a horizontal bar that can pivot; two horizontal forces act on the
bar, but it is stationary. If the angle between the bar and is now
F
:
2
F
:
2
F
:
1
F
:
2
F
:
1
F
:
1
1 Figure 10-20 is a graph of the an-
gular velocity versus time for a disk
rotating like a merry-go-round. For a
point on the disk rim, rank the in-
stants a, b, c, and d according to the
magnitude of the (a) tangential and
(b) radial acceleration, greatest first.
2 Figure 10-21 shows plots of angu-
lar position u versus time t for three
cases in which a disk is rotated like a
merry-go-round. In each case, the ro-
tation direction changes at a certain
angular position uchange. (a) For each
case, determine whether uchange is
clockwise or counterclockwise from
u ! 0, or whether it is at u ! 0. For
each case, determine (b) whether
v is zero before, after, or at t ! 0
and (c) whether a is positive,negative,or zero.
3 A force is applied to the rim of a disk that can rotate like
a merry-go-round, so as to change its angular velocity. Its initial
and final angular velocities, respectively, for four situations are:
(a) %2 rad/s, 5 rad/s; (b) 2 rad/s, 5 rad/s; (c) %2 rad/s, %5 rad/s; and
(d) 2 rad/s, %5 rad/s. Rank the situations according to the work
done by the torque due to the force, greatest first.
4 Figure 10-22b is a graph of the angular position of the rotating
disk of Fig. 10-22a. Is the angular velocity of the disk positive, nega-
tive, or zero at (a) t ! 1 s, (b) t ! 2 s, and (c) t ! 3 s? (d) Is the an-
gular acceleration positive or negative?
Questions
ω
t
a b c d
Figure 10-20 Question 1.
1
2
3
0
–90°
90°
θ
t
Rotation axis
t (s)
(rad)
θ
1 2 3
(a) (b)
F1
θ
F2
F5
F4
F3
F2
F1
P
Pivot point
F1 F2
Pivot point
(a) (b)
φ
F1
F2
a
b
c
Figure 10-26
Question 9.
Figure 10-25 Questions 7 and 8.
The SI unit of torque is the newton-meter (N)m). A torque t
is positive if it tends to rotate a body at rest counterclockwise and
negative if it tends to rotate the body clockwise.
Newton’s Second Law in Angular Form The rotational
analog of Newton’s second law is
tnet ! Ia, (10-45)
where tnet is the net torque acting on a particle or rigid body,I is the ro-
tational inertia of the particle or body about the rotation axis, and a is
the resulting angular acceleration about that axis.
Work and Rotational Kinetic Energy The equations used
for calculating work and power in rotational motion correspond to
equations used for translational motion and are
(10-53)
and (10-55)
When t is constant, Eq. 10-53 reduces to
W ! t(uf % ui). (10-54)
The form of the work–kinetic energy theorem used for rotating
bodies is
(10-52)
"K ! Kf % Ki ! 1
2 Ivf
2
% 1
22vi
2
! W.
P !
dW
dt
! tv.
W ! "uf
ui
t du
5 In Fig. 10-23, two forces and act on a disk that turns about
its center like a merry-go-round.The forces maintain the indicated
F
:
2
F
:
1
decreased from 90# and the bar is still not to turn, should F2 be
made larger, made smaller, or left the same?

287
PROBLEMS
10 Figure 10-27 shows three flat disks (of the same radius) that
can rotate about their centers like merry-go-rounds. Each disk con-
sists of the same two materials, one denser than the other (density is
mass per unit volume). In disks 1 and 3, the denser material forms
the outer half of the disk area. In disk 2, it forms the inner half of the
disk area. Forces with identical magnitudes are applied tangentially
to the disk, either at the outer edge or at the interface of the two ma-
terials, as shown. Rank the disks according to (a) the torque about
the disk center, (b) the rotational inertia about the disk center, and
(c) the angular acceleration of the disk,greatest first.
F
Denser
Disk 1
Denser
Disk 3
F F
Lighter
Disk 2
11 Figure 10-28a shows a meter stick, half wood and half steel,
that is pivoted at the wood end at O.A force is applied to the steel
end at a. In Fig. 10-28b, the stick is reversed and pivoted at the steel
end at O/, and the same force is applied at the wood end at a/. Is the
resulting angular acceleration of Fig. 10-28a greater than, less than, or
the same as that of Fig.10-28b?
F
:
R:
M:
1 m
26 kg
(a)
2 m
7 kg
(b)
3 m
3 kg
(c)
Module 10-1 Rotational Variables
•1 A good baseball pitcher can throw a baseball toward home
plate at 85 mi/h with a spin of 1800 rev/min. How many revolutions
does the baseball make on its way to home plate? For simplicity,
assume that the 60 ft path is a straight line.
•2 What is the angular speed of (a) the second hand, (b) the
minute hand, and (c) the hour hand of a smoothly running analog
watch? Answer in radians per second.
••3 When a slice of buttered toast is accidentally pushed
over the edge of a counter, it rotates as it falls. If the distance to the
floor is 76 cm and for rotation less than 1 rev, what are the (a)
smallest and (b) largest angular speeds that cause the toast to hit
and then topple to be butter-side down?
••4 The angular position of a point on a rotating wheel is given
by u ! 2.0 & 4.0t2
& 2.0t3
, where u is in radians and t is in seconds.At
t ! 0, what are (a) the point’s angular position and (b) its angular ve-
locity? (c)What is its angular velocity at t ! 4.0 s? (d) Calculate its an-
gular acceleration at t ! 2.0 s. (e) Is its angular acceleration constant?
••5 A diver makes 2.5 revolutions on the way from a 10-m-high
platform to the water. Assuming zero initial vertical velocity, find
the average angular velocity during the dive.
••6 The angular position of a point on the rim of a rotating wheel is
given by u ! 4.0t % 3.0t2
& t3
, where u is in radians and t is in seconds.
What are the angular velocities at (a) t ! 2.0 s and (b) t ! 4.0 s?
(c) What is the average angular acceleration for the time interval
that begins at t ! 2.0 s and ends at t ! 4.0 s? What are the instanta-
neous angular accelerations at (d) the beginning and (e) the end of
this time interval?
•••7 The wheel in Fig. 10-30 has eight equally spaced spokes and
a radius of 30 cm. It is mounted on a fixed axle and is spinning at 2.5
rev/s.You want to shoot a 20-cm-long arrow parallel to this axle and
ILW
through the wheel without hitting any
of the spokes. Assume that the arrow
and the spokes are very thin. (a) What
minimum speed must the arrow have?
(b) Does it matter where between the
axle and rim of the wheel you aim? If
so,what is the best location?
•••8 The angular acceleration of a
wheel is a ! 6.0t4
% 4.0t2
, with a in ra-
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign
SSM Worked-out solution available in Student Solutions Manual
• – ••• Number of dots indicates level of problem difficulty
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
WWW Worked-out solution is at
ILW Interactive solution is at
http://www.wiley.com/college/halliday
Problems
Figure 10-30 Problem 7.
(b)
(a)
O O´ a´
a
F F
Figure 10-28
Question 11.
12 Figure 10-29 shows three disks,
each with a uniform distribution of
mass. The radii R and masses M are
indicated. Each disk can rotate
around its central axis (perpendicular
to the disk face and through the cen-
ter). Rank the disks according to their
rotational inertias calculated about
their central axes, greatest first.
dians per second-squared and t in seconds.At time t ! 0, the wheel
has an angular velocity of &2.0 rad/s and an angular position of
&1.0 rad.Write expressions for (a) the angular velocity (rad/s) and
(b) the angular position (rad) as functions of time (s).
Module 10-2 Rotation with Constant Angular Acceleration
•9 A drum rotates around its central axis at an angular velocity
of 12.60 rad/s. If the drum then slows at a constant rate of 4.20
rad/s2
, (a) how much time does it take and (b) through what angle
does it rotate in coming to rest?
•10 Starting from rest, a disk rotates about its central axis with
constant angular acceleration. In 5.0 s, it rotates 25 rad. During that
time, what are the magnitudes of (a) the angular acceleration and
(b) the average angular velocity? (c) What is the instantaneous an-
gular velocity of the disk at the end of the 5.0 s? (d) With the angu-
lar acceleration unchanged, through what additional angle will the
disk turn during the next 5.0 s?
•11 A disk, initially rotating at 120 rad/s, is slowed down
with a constant angular acceleration of magnitude 4.0 rad/s2
. (a) How
much time does the disk take to stop? (b) Through what angle does
the disk rotate during that time?
•12 The angular speed of an automobile engine is increased at a
constant rate from 1200 rev/min to 3000 rev/min in 12 s. (a) What is

a point on Earth’s surface at latitude 40# N? (Earth rotates about
that axis.) (b) What is the linear speed v of the point? What are
(c) v and (d) v for a point at the equator?
••26 The flywheel of a steam engine runs with a constant angular ve-
locity of 150 rev/min.When steam is shut off,the friction of the bearings
and of the air stops the wheel in 2.2 h. (a) What is the constant angular
acceleration, in revolutions per minute-squared, of the wheel during
the slowdown? (b) How many revolutions does the wheel make before
stopping? (c) At the instant the flywheel is turning at 75 rev/min, what
is the tangential component of the linear acceleration of a flywheel par-
ticle that is 50 cm from the axis of rotation? (d) What is the magnitude
of the net linear acceleration of the particle in (c)?
••27 A seed is on a turntable rotating at rev/min, 6.0 cm
from the rotation axis.What are (a) the seed’s acceleration and (b)
the least coefficient of static friction to avoid slippage? (c) If the
turntable had undergone constant angular acceleration from rest
in 0.25 s, what is the least coefficient to avoid slippage?
••28 In Fig. 10-31, wheel A of radius
rA ! 10 cm is coupled by belt B to
wheel C of radius rC ! 25 cm.The an-
gular speed of wheel A is increased
from rest at a constant rate of
1.6 rad/s2
. Find the time needed for
wheel C to reach an angular speed of
100 rev/min, assuming the belt does
not slip. (Hint: If the belt does not slip, the linear speeds at the two
rims must be equal.)
••29 Figure 10-32 shows an early method of measuring the
speed of light that makes use of a rotating slotted wheel.A beam of
331
3
celeration (in revolutions per minute-squared) will increase the
wheel’s angular speed to 1000 rev/min in 60.0 s? (d) How many
revolutions does the wheel make during that 60.0 s?
•24 A vinyl record is played by rotating the record so that an ap-
proximately circular groove in the vinyl slides under a stylus.
Bumps in the groove run into the stylus, causing it to oscillate. The
equipment converts those oscillations to electrical signals and then
to sound. Suppose that a record turns at the rate of , the
groove being played is at a radius of 10.0 cm, and the bumps in the
groove are uniformly separated by 1.75 mm.At what rate (hits per
second) do the bumps hit the stylus?
••25 (a) What is the angular speed v about the polar axis of
SSM
331
3 rev/min
its angular acceleration in revolutions per minute-squared? (b) How
many revolutions does the engine make during this 12 s interval?
••13 A flywheel turns through 40 rev as it slows from an
angular speed of 1.5 rad/s to a stop. (a) Assuming a constant angu-
lar acceleration, find the time for it to come to rest. (b) What is its
angular acceleration? (c) How much time is required for it to com-
plete the first 20 of the 40 revolutions?
••14 A disk rotates about its central axis starting from rest and
accelerates with constant angular acceleration.At one time it is ro-
tating at 10 rev/s; 60 revolutions later, its angular speed is 15 rev/s.
Calculate (a) the angular acceleration, (b) the time required to
complete the 60 revolutions, (c) the time required to reach the
10 rev/s angular speed, and (d) the number of revolutions from rest
until the time the disk reaches the 10 rev/s angular speed.
••15 Starting from rest, a wheel has constant a = 3.0 rad/s2
.
During a certain 4.0 s interval, it turns through 120 rad. How much
time did it take to reach that 4.0 s interval?
••16 A merry-go-round rotates from rest with an angular accel-
eration of 1.50 rad/s2
. How long does it take to rotate through
(a) the first 2.00 rev and (b) the next 2.00 rev?
••17 At t ! 0, a flywheel has an angular velocity of 4.7 rad/s, a
constant angular acceleration of %0.25 rad/s2
, and a reference line
at u0 ! 0. (a) Through what maximum angle umax will the reference
line turn in the positive direction? What are the (b) first and
(c) second times the reference line will be at ? At what
(d) negative time and (e) positive time will the reference line be
at 10.5 rad? (f) Graph u versus t, and indicate your answers.
•••18 A pulsar is a rapidly rotating neutron star that emits a radio
beam the way a lighthouse emits a light beam. We receive a radio
pulse for each rotation of the star.The period T of rotation is found
by measuring the time between pulses.The pulsar in the Crab neb-
ula has a period of rotation of T ! 0.033 s that is increasing at the
rate of 1.26 $ 10%5
s/y. (a) What is the pulsar’s angular acceleration
a? (b) If a is constant, how many years from now will the pulsar
stop rotating? (c) The pulsar originated in a supernova explosion
seen in the year 1054.Assuming constant a, find the initial T.
Module 10-3 Relating the Linear and Angular Variables
•19 What are the magnitudes of (a) the angular velocity, (b) the ra-
dial acceleration, and (c) the tangential acceleration of a spaceship
taking a circular turn of radius 3220 km at a speed of 29 000 km/h?
•20 An object rotates about a fixed axis, and the angular posi-
tion of a reference line on the object is given by u ! 0.40e2t
, where
u is in radians and t is in seconds. Consider a point on the object
that is 4.0 cm from the axis of rotation.At t ! 0, what are the mag-
nitudes of the point’s (a) tangential component of acceleration
and (b) radial component of acceleration?
•21 Between 1911 and 1990, the top of the leaning bell
tower at Pisa, Italy, moved toward the south at an average rate of
1.2 mm/y. The tower is 55 m tall. In radians per second, what is the
average angular speed of the tower’s top about its base?
•22 An astronaut is tested in a centrifuge with radius 10 m and
rotating according to u ! 0.30t2
. At t ! 5.0 s, what are the magni-
tudes of the (a) angular velocity, (b) linear velocity, (c) tangential
acceleration, and (d) radial acceleration?
•23 A flywheel with a diameter of 1.20 m is rotating
at an angular speed of 200 rev/min. (a) What is the angular speed
of the flywheel in radians per second? (b) What is the linear speed
of a point on the rim of the flywheel? (c) What constant angular ac-
WWW
SSM
u !
u ! 1
2umax
SSM
ILW
B
C
rA
A
rC
Light
beam
Light
source
Rotating
slotted wheel
Mirror
perpendicular
to light beam
L

289
PROBLEMS
••41 In Fig. 10-37, two particles,
each with mass m 0.85 kg, are fas-
tened to each other, and to a rotation
axis at O, by two thin rods, each with
length d ! 5.6 cm and mass M !
1.2 kg. The combination rotates
around the rotation axis with the an-
gular speed v ! 0.30 rad/s. Measured
about O, what are the combination’s
(a) rotational inertia and (b) kinetic energy?
••42 The masses and coordinates of four particles are as
follows: 50 g, x ! 2.0 cm, y ! 2.0 cm; 25 g, x ! 0, y ! 4.0 cm; 25 g,
x ! %3.0 cm, y ! %3.0 cm; 30 g, x ! %2.0 cm, y ! 4.0 cm. What
are the rotational inertias of this collection about the (a) x, (b) y,
and (c) z axes? (d) Suppose that we symbolize the answers to (a)
and (b) as A and B, respectively. Then what is the answer to (c)
in terms of A and B?
!
L
O
light passes through one of the slots at the outside edge of the
wheel, travels to a distant mirror, and returns to the wheel just in
time to pass through the next slot in the wheel. One such slotted
wheel has a radius of 5.0 cm and 500 slots around its edge.
Measurements taken when the mirror is L ! 500 m from the
wheel indicate a speed of light of 3.0 $ 105
km/s. (a) What is the
(constant) angular speed of the wheel? (b) What is the linear
speed of a point on the edge of the wheel?
••30 A gyroscope flywheel of radius 2.83 cm is accelerated from
rest at 14.2 rad/s2
until its angular speed is 2760 rev/min. (a) What is
the tangential acceleration of a point on the rim of the flywheel during
this spin-up process? (b) What is the radial acceleration of this point
when the flywheel is spinning at full speed? (c)Through what distance
does a point on the rim move during the spin-up?
••31 A disk, with a radius of 0.25 m, is to be rotated like a merry-
go-round through 800 rad, starting from rest, gaining angular speed
at the constant rate a1 through the first 400 rad and then losing an-
gular speed at the constant rate %a1 until it is again at rest.The mag-
nitude of the centripetal acceleration of any portion of the disk is
not to exceed 400 m/s2
. (a) What is the least time required for the ro-
tation? (b)What is the corresponding value of a1?
••32 A car starts from rest and moves around a circular track of
radius 30.0 m. Its speed increases at the constant rate of 0.500 m/s2
.
(a) What is the magnitude of its net linear acceleration 15.0 s later?
(b) What angle does this net acceleration vector make with the
car’s velocity at this time?
Module 10-4 Kinetic Energy of Rotation
•33 Calculate the rota-
tional inertia of a wheel that has
a kinetic energy of 24 400 J when
rotating at 602 rev/min.
•34 Figure 10-33 gives angu-
lar speed versus time for a thin
rod that rotates around one
end. The scale on the v axis is
set by (a) What
is the magnitude of the rod’s an-
gular acceleration? (b) At t
4.0 s, the rod has a rotational ki-
netic energy of 1.60 J.What is its kinetic energy at t ! 0?
Module 10-5 Calculating the Rotational Inertia
•35 Two uniform solid cylinders, each rotating about its cen-
tral (longitudinal) axis at 235 rad/s, have the same mass of 1.25 kg but
differ in radius.What is the rotational kinetic energy of (a) the smaller
cylinder,of radius 0.25 m,and (b) the larger cylinder,of radius 0.75 m?
•36 Figure 10-34a shows a disk that can rotate about an axis at
SSM
!
vs ! 6.0 rad/s.
SSM
a radial distance h from the center of the disk. Figure 10-34b gives
the rotational inertia I of the disk about the axis as a function of that
distance h, from the center out to the edge of the disk. The scale on
the I axis is set by and What is
the mass of the disk?
•37 Calculate the rotational inertia of a meter stick, with
mass 0.56 kg, about an axis perpendicular to the stick and located
at the 20 cm mark. (Treat the stick as a thin rod.)
•38 Figure 10-35 shows three
0.0100 kg particles that have been
glued to a rod of length L ! 6.00 cm
and negligible mass. The assembly
can rotate around a perpendicular
axis through point O at the left end.
If we remove one particle (that is,
33% of the mass), by what percent-
age does the rotational inertia of the assembly around the rotation
axis decrease when that removed particle is (a) the innermost one
and (b) the outermost one?
••39 Trucks can be run on energy stored in a rotating flywheel,
with an electric motor getting the flywheel up to its top speed of
200p rad/s. Suppose that one such flywheel is a solid, uniform
cylinder with a mass of 500 kg and a radius of 1.0 m. (a) What is the
kinetic energy of the flywheel after charging? (b) If the truck uses
an average power of 8.0 kW, for how many minutes can it operate
between chargings?
••40 Figure 10-36 shows an arrangement of 15 identical disks that
have been glued together in a rod-like shape of length L ! 1.0000 m
and (total) mass M ! 100.0 mg.The disks are uniform, and the disk
arrangement can rotate about a perpendicular axis through its cen-
tral disk at point O. (a) What is the rotational inertia of the
arrangement about that axis? (b) If we approximated the arrange-
ment as being a uniform rod of mass M and length L, what percent-
age error would we make in using the formula in Table 10-2e to cal-
culate the rotational inertia?
SSM
IB ! 0.150 kg)m2
.
IA ! 0.050 kg)m2
0
0 1 2 3 4 5 6
t (s)
◊ (rad/s)
s
ω
I
(kg
•
m
2
)
IB
IA
0 0.1
h (m)
0.2
(b)
(a)
Axis
h
Figure 10-35 Problems
38 and 62.
Axis
L
m
O
d d d
m m
Rotation axis
m
d
m
d
M
M
O
ω

••53 Figure 10-43 shows a uniform
disk that can rotate around its center like a
merry-go-round. The disk has a radius of
2.00 cm and a mass of 20.0 grams and is ini-
tially at rest. Starting at time t ! 0, two
forces are to be applied tangentially to the
rim as indicated, so that at time t ! 1.25 s
the disk has an angular velocity of 250
rad/s counterclockwise. Force
has a magnitude of 0.100 N. What
is magnitude F2?
••54 In a judo foot-sweep
move, you sweep your opponent’s
left foot out from under him while
pulling on his gi (uniform) toward
that side. As a result, your oppo-
nent rotates around his right foot
and onto the mat. Figure 10-44
shows a simplified diagram of
your opponent as you face him,
with his left foot swept out. The
rotational axis is through point O.
The gravitational force on him
effectively acts at his center of
mass, which is a horizontal dis-
tance d ! 28 cm from point O. His
mass is 70 kg, and his rotational in-
ertia about point O is 65 kg)m2
.What is the magnitude of his initial
angular acceleration about point O if your pull on his gi is (a) neg-
ligible and (b) horizontal with a magnitude of 300 N and applied at
height h ! 1.4 m?
••55 In Fig. 10-45a, an irregularly shaped plastic plate with
uniform thickness and density (mass per unit volume) is to be
rotated around an axle that is perpendicular to the plate face
and through point O. The rotational inertia of the plate about
F
:
a
F
:
g
F
:
1
••43 The uniform solid
block in Fig. 10-38 has mass 0.172 kg
and edge lengths a ! 3.5 cm, b ! 8.4
cm, and c ! 1.4 cm. Calculate its rota-
tional inertia about an axis through
one corner and perpendicular to the
large faces.
••44 Four identical particles of
mass 0.50 kg each are placed at the
vertices of a 2.0 m $ 2.0 m square
and held there by four massless rods, which form the sides of the
square. What is the rotational inertia of this rigid body about an
axis that (a) passes through the midpoints of opposite sides and
lies in the plane of the square, (b) passes through the midpoint of
one of the sides and is perpendicular to the plane of the square,
and (c) lies in the plane of the square and passes through two di-
agonally opposite particles?
WWW
SSM rest, block 2 falls 75.0 cm in 5.00 s without the cord slipping on
the pulley. (a) What is the magnitude of the acceleration of the
blocks? What are (b) tension and (c) tension ? (d) What is
the magnitude of the pulley’s angular acceleration? (e) What is
its rotational inertia?
••52 In Fig. 10-42, a cylinder having a mass of 2.0 kg can rotate
about its central axis through point O. Forces are applied as shown:
F1 ! 6.0 N, F2 ! 4.0 N, F3 ! 2.0 N, and F4 ! 5.0 N. Also, r ! 5.0 cm
and R ! 12 cm. Find the (a) magnitude and (b) direction of the an-
gular acceleration of the cylinder. (During the rotation, the forces
maintain their same angles relative to the cylinder.)
T1
T2
b
a
c
Rotation
axis
F2
F1
Figure 10-43
Problem 53.
F1
F4 R
r
O
Rotation
axis
F2
F3
O
r1
θ1
F1
r2
θ2
F2
Module 10-6 Torque
•45 The body in
Fig. 10-39 is pivoted at O, and
two forces act on it as shown. If
r1 ! 1.30 m, r2 ! 2.15 m, F1 !
4.20 N, F2 ! 4.90 N, u1 ! 75.0#,
and u2 ! 60.0#, what is the net
torque about the pivot?
•46 The body in Fig. 10-40 is
pivoted at O. Three forces act
on it: FA ! 10 N at point A, 8.0
m from O; FB ! 16 N at B, 4.0
m from O; and FC ! 19 N at C,
3.0 m from O. What is the net
torque about O?
•47 A small ball of mass
0.75 kg is attached to one end
of a 1.25-m-long massless rod,
SSM
ILW
SSM
and the other end of the rod is hung from a pivot.When the resulting
pendulum is 30# from the vertical, what is the magnitude of the gravi-
tational torque calculated about the pivot?
•48 The length of a bicycle pedal arm is 0.152 m, and a down-
ward force of 111 N is applied to the pedal by the rider.What is the
magnitude of the torque about the pedal arm’s pivot when the arm
is at angle (a) 30#, (b) 90#, and (c) 180# with the vertical?
Module 10-7 Newton’s Second Law for Rotation
•49 During the launch from a board, a diver’s angular
speed about her center of mass changes from zero to 6.20 rad/s in 220
ms. Her rotational inertia about her center of mass is 12.0 kg)m2
.
During the launch, what are the magnitudes of (a) her average angu-
lar acceleration and (b) the average external
torque on her from the board?
•50 If a 32.0 N)m torque on a wheel
causes angular acceleration 25.0 rad/s2
,
what is the wheel’s rotational inertia?
••51 In Fig. 10-41, block 1 has mass
, block 2 has mass ,
and the pulley, which is mounted on a hor-
izontal axle with negligible friction, has
radius . When released from
R ! 5.00 cm
m2 ! 500 g
m1 ! 460 g
ILW
SSM
O
d
com
Fg
Fa
h
FC
C
FB
B
160°
90°
O
FA
A
135°
m1
T1 T2
m2
R
Figure 10-41
Problems 51 and 83.

291
PROBLEMS
that axle is measured with the following
method. A circular disk of mass 0.500 kg and
radius 2.00 cm is glued to the plate, with its
center aligned with point O (Fig. 10-45b). A
string is wrapped around the edge of the disk
the way a string is wrapped around a top.
Then the string is pulled for 5.00 s. As a re-
sult, the disk and plate are rotated by a con-
stant force of 0.400 N that is applied by the
string tangentially to the edge of the disk.
The resulting angular speed is 114 rad/s.
What is the rotational inertia of the plate
about the axle?
••56 Figure 10-46 shows
particles 1 and 2, each of mass
m, fixed to the ends of a rigid
massless rod of length L1 &
L2, with L1 ! 20 cm and L2 !
80 cm. The rod is held hori-
zontally on the fulcrum and then released. What are the magni-
tudes of the initial accelerations of (a) particle 1 and (b) particle 2?
•••57 A pulley,with a rotational inertia of 1.0 $ 10%3
kg)m2
about
its axle and a radius of 10 cm,is acted on by a force applied tangentially
at its rim.The force magnitude varies in time as F ! 0.50t & 0.30t2
,with
F in newtons and t in seconds.The pulley is initially at rest.At t ! 3.0 s
what are its (a) angular acceleration and (b) angular speed?
Module 10-8 Work and Rotational Kinetic Energy
•58 (a) If R ! 12 cm, M ! 400 g, and m ! 50 g in Fig. 10-19, find
the speed of the block after it has descended 50 cm starting from
rest. Solve the problem using energy conservation principles.
(b) Repeat (a) with R ! 5.0 cm.
•59 An automobile crankshaft transfers energy from the engine
to the axle at the rate of 100 hp (! 74.6 kW) when rotating at a
speed of 1800 rev/min. What torque (in newton-meters) does the
crankshaft deliver?
•60 A thin rod of length 0.75 m and mass 0.42 kg is suspended
freely from one end.It is pulled to one side and then allowed to swing
like a pendulum, passing through its lowest position with angular
speed 4.0 rad/s. Neglecting friction and air resistance, find (a) the
rod’s kinetic energy at its lowest position and (b) how far above that
position the center of mass rises.
•61 A 32.0 kg wheel, essentially a thin hoop with radius 1.20 m, is
rotating at 280 rev/min. It must be brought to a stop in 15.0 s.
(a) How much work must be done to stop it? (b) What is the
required average power?
••62 In Fig. 10-35, three 0.0100 kg particles have been glued to a
rod of length L ! 6.00 cm and negligible mass and can rotate
around a perpendicular axis through point O at one end. How
much work is required to change the rotational rate (a) from 0 to
20.0 rad/s, (b) from 20.0 rad/s to 40.0 rad/s, and (c) from 40.0 rad/s to
60.0 rad/s? (d) What is the slope of a plot of the assembly’s kinetic
energy (in joules) versus the square of its rotation rate (in radians-
squared per second-squared)?
••63 A meter stick is held vertically with one end on
the floor and is then allowed to fall. Find the speed of the other end
just before it hits the floor, assuming that the end on the floor does
not slip. (Hint: Consider the stick to be a thin rod and use the con-
servation of energy principle.)
ILW
SSM
••64 A uniform cylinder of radius 10 cm and mass 20 kg is
mounted so as to rotate freely about a horizontal axis that is paral-
lel to and 5.0 cm from the central longitudinal axis of the cylinder.
(a) What is the rotational inertia of the cylinder about the axis of
rotation? (b) If the cylinder is released from rest with its central
longitudinal axis at the same height as the axis about which the
cylinder rotates, what is the angular speed of the cylinder as it
passes through its lowest position?
•••65 A tall, cylindrical chimney falls over when its base
is ruptured.Treat the chimney as a thin rod of length 55.0 m.At the
instant it makes an angle of 35.0# with the vertical as it falls, what
are (a) the radial acceleration of the top, and (b) the tangential ac-
celeration of the top. (Hint: Use energy considerations, not a torque.)
(c)At what angle u is the tangential acceleration equal to g?
•••66 A uniform spherical shell of mass M ! 4.5 kg and radius
(b)
(a)
O
Axle
Plate
Disk
String
Figure 10-45
Problem 55.
1 2
L1 L2
R ! 8.5 cm can rotate about a vertical axis on frictionless bearings
(Fig. 10-47).A massless cord passes around the equator of the shell,
over a pulley of rotational inertia I ! 3.0 $ 10%3
kg)m2
and radius
r ! 5.0 cm, and is attached to a small object of mass m ! 0.60 kg.
There is no friction on the pulley’s axle; the cord does not slip on
the pulley.What is the speed of the object when it has fallen 82 cm
after being released from rest? Use energy considerations.
M, R
I, r
m
•••67 Figure 10-48 shows a rigid as-
sembly of a thin hoop (of mass m and ra-
dius R ! 0.150 m) and a thin radial rod
(of mass m and length L ! 2.00R). The
assembly is upright, but if we give it a
slight nudge, it will rotate around a hori-
zontal axis in the plane of the rod and
hoop, through the lower end of the rod.
Assuming that the energy given to the
assembly in such a nudge is negligible,
what would be the assembly’s angular speed about the rotation axis
when it passes through the upside-down (inverted) orientation?
Additional Problems
68 Two uniform solid spheres have the same mass of 1.65 kg, but
one has a radius of 0.226 m and the other has a radius of 0.854 m.
Each can rotate about an axis through its center. (a) What is the
magnitude t of the torque required to bring the smaller sphere
from rest to an angular speed of 317 rad/s in 15.5 s? (b) What is the
magnitude F of the force that must be
applied tangentially at the sphere’s
equator to give that torque? What are
the corresponding values of (c) t and
(d) F for the larger sphere?
69 In Fig. 10-49, a small disk of radius
r ! 2.00 cm has been glued to the edge of
a larger disk of radius R ! 4.00 cm so that
Rotation
axis
Hoop
Rod
O

the disks lie in the same plane.The disks can be rotated around a per-
pendicular axis through point O at the center of the larger disk. The
disks both have a uniform density (mass per unit volume) of 1.40 $
103
kg/m3
and a uniform thickness of 5.00 mm. What is the rota-
tional inertia of the two-disk assembly about the rotation axis
through O?
70 A wheel, starting from rest, rotates with a constant angular
acceleration of 2.00 rad/s2
. During a certain 3.00 s interval, it turns
through 90.0 rad. (a) What is the angular velocity of the wheel at
the start of the 3.00 s interval? (b) How long has the wheel been
turning before the start of the 3.00 s interval?
71 In Fig. 10-50, two 6.20 kg
blocks are connected by a massless
string over a pulley of radius 2.40 cm
and rotational inertia 7.40 $ 10%4
kg)m2
. The string does not slip on
the pulley; it is not known whether
there is friction between the table and
the sliding block; the pulley’s axis is
frictionless. When this system is re-
leased from rest, the pulley turns through 0.130 rad in 91.0 ms and the
acceleration of the blocks is constant.What are (a) the magnitude of
the pulley’s angular acceleration, (b) the magnitude of either block’s
acceleration,(c) string tension T1,and (d) string tension T2?
72 Attached to each end of a thin steel rod of length 1.20 m and
mass 6.40 kg is a small ball of mass 1.06 kg. The rod is constrained
to rotate in a horizontal plane about a vertical axis through its mid-
point.At a certain instant, it is rotating at 39.0 rev/s. Because of fric-
tion,it slows to a stop in 32.0 s.Assuming a constant retarding torque
due to friction, compute (a) the angular acceleration, (b) the retard-
ing torque, (c) the total energy transferred from mechanical energy
to thermal energy by friction, and (d) the number of revolutions ro-
tated during the 32.0 s. (e) Now suppose that the retarding torque is
known not to be constant.If any of the quantities (a),(b),(c),and (d)
can still be computed without additional information, give its value.
73 A uniform helicopter rotor blade is 7.80 m long, has a mass of
110 kg, and is attached to the rotor axle by a single bolt. (a) What is
the magnitude of the force on the bolt from the axle when the ro-
tor is turning at 320 rev/min? (Hint: For this calculation the blade
can be considered to be a point mass at its center of mass. Why?)
(b) Calculate the torque that must be applied to the rotor to bring
it to full speed from rest in 6.70 s. Ignore air resistance. (The blade
cannot be considered to be a point mass for this calculation. Why
not? Assume the mass distribution of a uniform thin rod.) (c) How
much work does the torque do on the blade in order for the blade
to reach a speed of 320 rev/min?
74 Racing disks. Figure 10-51 shows
two disks that can rotate about their
centers like a merry-go-round. At
time t ! 0, the reference lines of the
two disks have the same orientation.
Disk A is already rotating, with a con-
stant angular velocity of 9.5 rad/s.
Disk B has been stationary but now begins to rotate at a constant
angular acceleration of 2.2 rad/s2
. (a) At what time t will the refer-
ence lines of the two disks momentarily have the same angular dis-
placement u? (b) Will that time t be the first time since t ! 0 that
the reference lines are momentarily aligned?
75 A high-wire walker always attempts to keep his center of
mass over the wire (or rope). He normally carries a long, heavy pole
SSM
to help:If he leans,say,to his right (his com moves to the right) and is
in danger of rotating around the wire, he moves the pole to his left
(its com moves to the left) to slow the rotation and allow himself
time to adjust his balance. Assume that the walker has a mass of
70.0 kg and a rotational inertia of about the wire.What is
the magnitude of his angular acceleration about the wire if his com is
5.0 cm to the right of the wire and (a) he carries no pole and (b) the
14.0 kg pole he carries has its com 10 cm to the left of the wire?
76 Starting from rest at t ! 0, a wheel undergoes a constant an-
gular acceleration. When t ! 2.0 s, the angular velocity of the
wheel is 5.0 rad/s.The acceleration continues until t ! 20 s, when it
abruptly ceases. Through what angle does the wheel rotate in the
interval t ! 0 to t ! 40 s?
77 A record turntable rotating at slows down
and stops in 30 s after the motor is turned off. (a) Find its (con-
stant) angular acceleration in revolutions per minute-squared.
(b) How many revolutions does it make in this time?
78 A rigid body is made of three
identical thin rods, each with length
L ! 0.600 m, fastened together in the
form of a letter H (Fig. 10-52). The
body is free to rotate about a hori-
zontal axis that runs along the length
of one of the legs of the H. The body
is allowed to fall from rest from a position in which the plane of the
H is horizontal. What is the angular speed of the body when the
plane of the H is vertical?
79 (a) Show that the rotational inertia of a solid cylinder of
mass M and radius R about its central axis is equal to the rotational
inertia of a thin hoop of mass M and radius about its central
axis. (b) Show that the rotational inertia I of any given body of
mass M about any given axis is equal to the rotational inertia of an
equivalent hoop about that axis, if the hoop has the same mass M
and a radius k given by
The radius k of the equivalent hoop is called the radius of gyration
of the given body.
80 A disk rotates at constant angular acceleration, from angular
position u1 ! 10.0 rad to angular position u2 ! 70.0 rad in 6.00 s. Its
angular velocity at u2 is 15.0 rad/s. (a) What was its angular velocity
at u1? (b) What is the angular acceleration? (c) At what angular
position was the disk initially at rest? (d) Graph u versus time t and
angular speed v versus t for the disk, from the beginning of the
motion (let t ! 0 then).
81 The thin uniform rod in Fig. 10-53 has
length 2.0 m and can pivot about a horizontal,
frictionless pin through one end. It is released
from rest at angle u ! 40# above the horizontal.
Use the principle of conservation of energy to
determine the angular speed of the rod as it
passes through the horizontal position.
82 George Washington Gale Ferris, Jr., a
civil engineering graduate from Rensselaer Polytechnic Institute,
built the original Ferris wheel for the 1893 World’s Columbian
Exposition in Chicago. The wheel, an astounding engineering con-
struction at the time, carried 36 wooden cars, each holding up to 60
passengers, around a circle 76 m in diameter.The cars were loaded 6
at a time, and once all 36 cars were full, the wheel made a complete
k !
A
I
M
.
R/22
SSM
331
3 rev/min
SSM
15.0 kg)m2
T2
T1
Disk A Disk B
L
L
L
θ
Pin
Figure 10-53
Problem 81.

the rings are given in the following table. A tangential force of
magnitude 12.0 N is applied to the outer edge of the outer ring for
0.300 s.What is the change in the angular speed of the construction
during the time interval?
Ring Mass (kg) Inner Radius (m) Outer Radius (m)
1 0.120 0.0160 0.0450
2 0.240 0.0900 0.1400
87 In Fig. 10-55, a wheel of ra-
dius 0.20 m is mounted on a friction-
less horizontal axle. A massless cord
is wrapped around the wheel and at-
tached to a 2.0 kg box that slides on
a frictionless surface inclined at an-
gle u ! 20# with the horizontal. The
box accelerates down the surface at 2.0 m/s2
. What is the rota-
tional inertia of the wheel about the axle?
88 A thin spherical shell has a radius of 1.90 m.An applied torque
of 960 N)m gives the shell an angular acceleration of 6.20 rad/s2
about an axis through the center of the shell. What are (a) the rota-
tional inertia of the shell about that axis and (b) the mass of the shell?
89 A bicyclist of mass 70 kg puts all his mass on each downward-
moving pedal as he pedals up a steep road. Take the diameter of
293
PROBLEMS
rotation at constant angular speed in about 2 min. Estimate the
amount of work that was required of the machinery to rotate the
passengers alone.
83 In Fig.10-41,two blocks,of mass m1 ! 400 g and m2 ! 600 g,are
connected by a massless cord that is wrapped around a uniform disk
of mass M ! 500 g and radius R ! 12.0 cm.The disk can rotate with-
out friction about a fixed horizontal axis through its center; the cord
cannot slip on the disk.The system is released from rest. Find (a) the
magnitude of the acceleration of the blocks, (b) the tension T1 in the
cord at the left,and (c) the tension T2 in the cord at the right.
84 At 7!14 A.M. on June 30, 1908, a huge explosion
the circle in which the pedals rotate to be 0.40 m, and determine
the magnitude of the maximum torque he exerts about the rota-
tion axis of the pedals.
90 The flywheel of an engine is rotating at 25.0 rad/s. When the
engine is turned off, the flywheel slows at a constant rate and stops
in 20.0 s. Calculate (a) the angular acceleration of the flywheel,
(b) the angle through which the flywheel rotates in stopping, and
(c) the number of revolutions made by the flywheel in stopping.
91 In Fig.10-19a,a wheel of radius 0.20 m is mounted on a fric-
tionless horizontal axis. The rotational inertia of the wheel about the
axis is 0.40 kg)m2
. A massless cord wrapped around the wheel’s cir-
cumference is attached to a 6.0 kg box. The system is released from
rest.When the box has a kinetic energy of 6.0 J,what are (a) the wheel’s
rotational kinetic energy and (b) the distance the box has fallen?
92 Our Sun is 2.3 $ 104
ly (light-years) from the center of our
Milky Way galaxy and is moving in a circle around that center at a
speed of 250 km/s. (a) How long does it take the Sun to make one
revolution about the galactic center? (b) How many revolutions has
the Sun completed since it was formed about 4.5 $ 109
years ago?
93 A wheel of radius 0.20 m
is mounted on a frictionless horizon-
tal axis. The rotational inertia of the
wheel about the axis is 0.050 kg)m2
.
A massless cord wrapped around
the wheel is attached to a 2.0 kg
block that slides on a horizontal frictionless surface. If a horizontal
force of magnitude P ! 3.0 N is applied to the block as shown in
Fig. 10-56, what is the magnitude of the angular acceleration of the
wheel? Assume the cord does not slip on the wheel.
94 If an airplane propeller rotates at 2000 rev/min while the air-
plane flies at a speed of 480 km/h relative to the ground, what is the
linear speed of a point on the tip of the propeller, at radius 1.5 m, as
seen by (a) the pilot and (b) an observer on the ground? The plane’s
velocity is parallel to the propeller’s axis of rotation.
95 The rigid body shown in
Fig. 10-57 consists of three particles
connected by massless rods. It is to be
rotated about an axis perpendicular
to its plane through point P. If M !
0.40 kg, a ! 30 cm, and b ! 50 cm,
how much work is required to take
the body from rest to an angular
speed of 5.0 rad/s?
96 Beverage engineering. The pull
tab was a major advance in the engi-
neering design of beverage contain-
ers.The tab pivots on a central bolt in the can’s top.When you pull
upward on one end of the tab, the other end presses downward on
a portion of the can’s top that has been scored. If you pull upward
with a 10 N force, what force magnitude acts on the scored section?
(You will need to examine a can with a pull tab.)
97 Figure 10-58 shows a propeller blade that rotates at
2000 rev/min about a perpendicular axis at point B. Point A is at
the outer tip of the blade, at radial distance
1.50 m. (a) What is the difference in the
magnitudes a of the centripetal acceleration
of point A and of a point at radial distance
0.150 m? (b) Find the slope of a plot of a
versus radial distance along the blade.
SSM
SSM
86 Figure 10-54 shows a flat construction of
two circular rings that have a common center and
are held together by three rods of negligible mass.
The construction, which is initially at rest, can
rotate around the common center (like a merry-
go-round), where another rod of negligible mass
lies. The mass, inner radius, and outer radius of
occurred above remote central Siberia, at latitude 61# N and lon-
gitude 102# E; the fireball thus created was the brightest flash
seen by anyone before nuclear weapons. The Tunguska Event,
which according to one chance witness “covered an enormous part
of the sky,” was probably the explosion of a stony asteroid about 140
m wide. (a) Considering only Earth’s rotation, determine how much
later the asteroid would have had to arrive to put the explosion
above Helsinki at longitude 25# E. This would have obliterated the
city. (b) If the asteroid had, instead, been a metallic asteroid, it could
have reached Earth’s surface. How much later would such an
asteroid have had to arrive to put the impact in the Atlantic Ocean
at longitude 20# W? (The resulting tsunamis would have wiped out
coastal civilization on both sides of theAtlantic.)
85 A golf ball is launched at an angle of 20# to the horizontal,
with a speed of 60 m/s and a rotation rate of 90 rad/s. Neglecting air
drag, determine the number of revolutions the ball makes by the
time it reaches maximum height.
Figure 10-54
Problem 86.
P
P
b b
M
2M 2M
a a
θ
B
A
Figure 10-58
Problem 97.

98 A yo-yo-shaped device
mounted on a horizontal fric-
tionless axis is used to lift a 30 kg
box as shown in Fig. 10-59. The
outer radius R of the device is
0.50 m, and the radius r of the
hub is 0.20 m. When a constant
horizontal force of magni-
tude 140 N is applied to a rope
wrapped around the outside of
the device, the box, which is sus-
pended from a rope wrapped
around the hub, has an upward
acceleration of magnitude 0.80
m/s2
.What is the rotational iner-
tia of the device about its axis of rotation?
99 A small ball with mass 1.30 kg is mounted on one end of a rod
0.780 m long and of negligible mass.The system rotates in a horizon-
tal circle about the other end of the rod at 5010 rev/min.(a) Calculate
the rotational inertia of the system about the axis of rotation. (b)
There is an air drag of 2.30 $ 10%2
N on the ball, directed opposite its
motion.What torque must be applied to the system to keep it rotat-
ing at constant speed?
100 Two thin rods (each of mass 0.20
kg) are joined together to form a rigid
body as shown in Fig. 10-60. One of the
rods has length L1 ! 0.40 m, and the
other has length L2 ! 0.50 m. What is
the rotational inertia of this rigid body
about (a) an axis that is perpendicular
to the plane of the paper and passes
through the center of the shorter rod
and (b) an axis that is perpendicular to
the plane of the paper and passes
through the center of the longer rod?
101 In Fig. 10-61, four pul-
leys are connected by two
belts. Pulley A (radius 15 cm)
is the drive pulley, and it ro-
tates at 10 rad/s. Pulley B (ra-
dius 10 cm) is connected by
belt 1 to pulley A. Pulley B/
(radius 5 cm) is concentric with
pulley B and is rigidly attached
to it. Pulley C (radius 25 cm) is
connected by belt 2 to pulley B/.
Calculate (a) the linear speed of
a point on belt 1, (b) the angular
F
:
app
and three connecting rods, with and
. The balls may be treated as particles, and the connecting
rods have negligible mass. Determine the rotational kinetic energy
of the object if it has an angular speed of 1.2 rad/s about (a) an axis
that passes through point P and is perpendicular to the plane of the
figure and (b) an axis that passes through point P, is perpendicular
to the rod of length 2L, and lies in the plane of the figure.
103 In Fig. 10-63, a thin uniform rod
(mass 3.0 kg, length 4.0 m) rotates
freely about a horizontal axis A that is
perpendicular to the rod and passes
through a point at distance d ! 1.0 m
from the end of the rod. The kinetic
energy of the rod as it passes through
the vertical position is 20 J. (a) What is
the rotational inertia of the rod about
axis A? (b) What is the (linear) speed
of the end B of the rod as the rod
passes through the vertical position?
(c) At what angle u will the rod mo-
mentarily stop in its upward swing?
104 Four particles, each of mass,
0.20 kg, are placed at the vertices of a
square with sides of length 0.50 m.The
particles are connected by rods of neg-
ligible mass. This rigid body can rotate
in a vertical plane about a horizontal
axis A that passes through one of the
particles. The body is released from
rest with rod AB horizontal (Fig. 10-64).
(a) What is the rotational inertia of the
body about axis A? (b) What is the an-
gular speed of the body about axis A
when rod AB swings through the verti-
cal position?
105 Cheetahs running at top speed have been reported at an as-
tounding 114 km/h (about 71 mi/h) by observers driving alongside
the animals. Imagine trying to measure a cheetah’s speed by keeping
your vehicle abreast of the animal while also glancing at your
speedometer, which is registering 114 km/h. You keep the vehicle a
constant 8.0 m from the cheetah, but the noise of the vehicle causes
the cheetah to continuously veer away from you along a circular
path of radius 92 m. Thus, you travel along a circular path of radius
100 m. (a) What is the angular speed of you and the cheetah around
the circular paths? (b) What is the linear speed of the cheetah along
its path? (If you did not account for the circular motion, you would
conclude erroneously that the cheetah’s speed is 114 km/h, and that
type of error was apparently made in the published reports.)
106 A point on the rim of a 0.75-m-diameter grinding wheel
changes speed at a constant rate from 12 m/s to 25 m/s in 6.2 s.
What is the average angular acceleration of the wheel?
107 A pulley wheel that is 8.0 cm in diameter has a 5.6-m-long
cord wrapped around its periphery. Starting from rest, the wheel is
given a constant angular acceleration of 1.5 rad/s2
. (a) Through
what angle must the wheel turn for the cord to unwind com-
pletely? (b) How long will this take?
108 A vinyl record on a turntable rotates at 33 rev/min.
(a) What is its angular speed in radians per second? What is the
linear speed of a point on the record (b) 15 cm and (c) 7.4 cm from
the turntable axis?
1
3
' ! 30#
M ! 1.6 kg, L ! 0.60 m,
R
r
Rope
Yo-yo-shaped
device
Rigid mount
Hub
Fapp
A
B
Belt 1
Drive
pulley
Belt 2
B'
C
L1
L2
1
__
2
L1
1
__
2
θ
θ
2M
2M
M
2L
L
L
P
Figure 10-62
Problem 102.
θ
A
B
d
A B
Rotation
axis
speed of pulley B, (c) the angular speed of pulley B/, (d) the linear
speed of a point on belt 2,and (e) the angular speed of pulley C.(Hint:
If the belt between two pulleys does not slip, the linear speeds at the
rims of the two pulleys must be equal.)
102 The rigid object shown in Fig. 10-62 consists of three balls

Fundamentals of Physics "ROTATION"

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