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25510-7 CALCU LATI NG TH E ROTATIONAL I N E RTIA
PART 1
Table 10-2
Some Rotational Inertias
Axis 
Hoop about 
central axis 
Axis 
Annular cylinder 
(or ring) about 
central axis 
R 
I = MR 2 (b) (a) I = M(R 1
2 + R 2
2) 
R 2 
R 1 
Thin rod about 
axis through center 
perpendicular to 
length 
(e) I = ML 2 
L 
Axis 
Axis Axis 
Hoop about any 
diameter 
Slab about 
perpendicular 
axis through 
center 
 
(i) (h) 
I = MR 2 I = M(a 2 + b 2) 
R 
b 
a 
Axis 
Solid cylinder 
(or disk) about 
central axis 
(c) 
I = MR 2 
R 
L 
Axis 
Solid cylinder 
(or disk) about 
central diameter 
(d) I = MR 2 + ML 2 
R 
L 
Axis 
Thin 
spherical shell 
about any 
diameter 
(g) I = MR 2 
2R 
Solid sphere 
about any 
diameter 
(f) 
I = MR 2 
2R 
Axis 
1 __
 2 1 __
 2 
2 __
 5 
1 __
 4 
2 __
 3 
1 __
 2 
1 __
 12 
1 __
 12 
1 __
 12 
perpendicular to the plane of the figure, and another axis through point P parallel
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)
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.
I " ! (x2 ! y2) dm # 2a ! x dm # 2b ! y dm ! ! (a2 ! b2) dm.
I " ! r2 dm " ! [(x # a)2 ! (y # b)2] dm,
Fig. 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 in-
ertia 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. Both axes are perpendicular to the
plane of the figure.
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.
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)
halliday_c10_241-274hr2.qxd 29-09-2009 13:12 Page 255
264 CHAPTE R 10 ROTATION
Sample Problem
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
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.
 " !12 (6.0 N)(0.20 m)(!24 rad/s
2)(2.5 s)2
 K " W " !TR($f ! $i) " !TR(
1
2%t
2) " !12TR%t
2
$f ! $i " &it #
1
2%t
2 " 0 # 12%t
2 " 12%t
2
T
:
Work, rotational kinetic energy, torque, disk
Let the disk in Fig. 10-18 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?
We can find K with Eq. 10-34 We already know
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)
We can also get this answer by finding the disk’s kinetic 
energy from the work done on the disk.
 " 90 J.
 " 14 (2.5 kg)[(0.20 m)(!24 rad/s
2)(2.5 s)]2
 K " 12 I&
2 " 12(
1
2MR
2)(%t)2 " 14M(R%t)
2
I " 12MR
2
I " 12MR
2
(K " 12 I&
2).
KEY I DEA
KEY I DEA
Additional examples, video, and practice available at WileyPLUS
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)
$ "
s
r
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.
& "
d$
dt
.
&avg "
'$
't
.
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265R EVI EW & S U M MARY
PART 1
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,
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
(radian measure). (10-23)
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
kinetic 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)I " ! miri2
K " 12I%
2
T "
2&r
v
"
2&
%
ar "v2
r
" %2r
a:
a:
v:
' # '0 " %t #
1
2(t
2.
' # '0 "
1
2 (%0 $ %)t,
%2 " %0
2 $ 2((' # '0),
' # '0 " %0t $
1
2(t
2,
( "
d%
dt
.
(avg "
%2 # %1
t2 # t1
"
!%
!t
.
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
rotation axis due to a force . If is exerted at a point given by
the position 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
angle between and . The quantity is the perpendicular dis-
tance 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.
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 I%f
2 # 12*%i
2 " W.
P "
dW
dt
" +%.
W " "'f
'i
 
+ d'
F
:
r!
F
:
F
:
r!F
:
r:
r:F
:
+ " rFt " r!F " rF sin ,,
r:
F
:
F
:
I " " r 2 dm
halliday_c10_241-274hr.qxd 17-09-2009 12:50 Page 265
295R EVI EW & S U M MARY
PART 1
Fig. 11-22 (a) A nonspinning gyro-
scope falls by rotating in an xz plane be-
cause of torque . (b) A rapidly spinning
gyroscope, with angular momentum pre-
cesses around the z axis. Its precessional
motion is in the xy plane. (c) The change
in angular momentum leads to a ro-
tation of about O.L
:
dL
:
/dt
L
:
,
!:
x
y
z
τ
O
x
y
z
O
L
ω
= dL___
dt
(a)
(b)
x
y
z
O
L(t)
dL___
dt
dφ L(t + dt)
Circular path
taken by head
of L vector
(c)
τ
r
r
Mg
Mg
Support
simplify the situation, we assume the spin rate is so rapid that the angular
momentum due to precession is negligible relative to .We also assume the shaft
is horizontal when precession begins, as in Fig. 11-22b. The magnitude of is
given by Eq. 11-31:
L " Iv, (11-43)
where I is the rotational moment of the gyroscope about its shaft and v is the an-
gular speed at which the wheel spins about the shaft. The vector points along
the shaft, as in Fig. 11-22b. Since is parallel to torque must be
perpendicular to .
According to Eq. 11-41, torque causes an incremental change in the an-
gular momentum of the gyroscope in an incremental time interval dt; that is,
(11-44)
However, for a rapidly spinning gyroscope, the magnitude of is fixed by Eq.
11-43.Thus the torque can change only the direction of not its magnitude.
From Eq. 11-44 we see that the direction of is in the direction of per-
pendicular to . The only way that can be changed in the direction of 
without the magnitude L being changed is for to rotate around the z axis as
shown in Fig. 11-22c. maintains its magnitude, the head of the vector follows
a circular path, and is always tangent to that path. Since must always 
point along the shaft, the shaft must rotate about the z axis in the direction of .
Thus we have precession. Because the spinning gyroscope must obey Newton’s
law in angular form in response to any change in its initial angular momentum, it
must precess instead of merely toppling over.
We can find the precession rate # by first using Eqs. 11-44 and 11-42 to get
the magnitude of :
dL " t dt " Mgr dt. (11-45)
As changes by an incremental amount in an incremental time interval dt, the
shaft and precess around the z axis through incremental angle df. (In Fig.
11-22c, angle df is exaggerated for clarity.) With the aid of Eqs. 11-43 and 11-45,
we find that df is given by
Dividing this expression by dt and setting the rate # " df/dt, we obtain
(precession rate). (11-46)
This result is valid under the assumption that the spin rate v is rapid. Note that #
decreases as v is increased. Note also that there would be no precession if the
gravitational force did not act on the gyroscope, but because I is a function of
M, mass cancels from Eq. 11-46; thus # is independent of the mass.
Equation 11-46 also applies if the shaft of a spinning gyroscope is at an angle
to the horizontal. It holds as well for a spinning top, which is essentially a spinning
gyroscope at an angle to the horizontal.
Mg:
# "
Mgr
I$
d% "
dL
L
"
Mgr dt
I$
.
L
:
L
:
dL
:
!:
L
:
!:
L
:
L
:
L
:
!:L
:
L
:
!:,dL
:
L
:
,
L
:
dL
:
" !: dt.
dL
:
!:
L
:
!:r:,L
:
L
:
L
:
L
:
Rolling Bodies For a wheel of radius R rolling smoothly,
vcom " vR, (11-2)
where vcom is the linear speed of the wheel’s center of mass and v is
the angular speed of the wheel about its center. The wheel may
also be viewed as rotating instantaneously about the point P of the
“road” that is in contact with the wheel. The angular speed of the
wheel about this point is the same as the angular speed of the
wheel about its center.The rolling wheel has kinetic energy
(11-5)
where Icom is the rotational moment of the wheel about its center
of mass and M is the mass of the wheel. If the wheel is being accel-
K " 12Icom$
2 & 12'v
2
com,
halliday_c11_275-304hr2.qxd 29-09-2009 13:17 Page 295
296 CHAPTE R 11 ROLLI NG, TORQU E, AN D ANG U LAR MOM E NTU M
1 Figure 11-23 shows three parti-
cles of the same mass and the same
constant speed moving as indi-
cated by the velocity vectors.
Points a, b, c, and d form a square,
with point e at the center. Rank the
points according to the magnitude
of the net angular momentum of
the three-particle system when
measured about the points, great-
est first.
2 Figure 11-24 shows two particles A and B at xyz coordinates
(1 m, 1 m, 0) and (1 m, 0, 1 m). Acting on each particle are three
numbered forces, all of the same
magnitude and each directed par-
allel to an axis. (a) Which of the
forces produce a torque about the
origin that is directed parallel
to y? (b) Rank the forces accord-
ing to the magnitudes of the
torques they produce on the par-
ticles about the origin, greatest
first.
3 What happens to the initially
stationary yo-yo in Fig. 11-25 if
you pull it via its string with (a) force (the line of action passesF
:
2
Fig. 11-23 Question 1.
cd
e
a b
Fig. 11-24 Question 2.
y
x
z
A
B 6
5
4
2
1
3
erated but is still rolling smoothly, the acceleration of the center of
mass is related to the angular acceleration a about the center
with
acom ! aR. (11-6)
If the wheel rolls smoothly down a ramp of angle u, its acceleration
along an x axis extending up the ramp is
(11-10)
Torque as a Vector In three dimensions, torque is a vector
quantity defined relative to a fixed point (usually an origin); it is
(11-14)
where is a force applied to a particle and is a position vectorlocating the particle relative to the fixed point.The magnitude of 
is given by
(11-15, 11-16, 11-17)
where f is the angle between and is the component of 
perpendicular to and is the moment arm of . The direction
of is given by the right-hand rule.
Angular Momentum of a Particle The angular momentum
of a particle with linear momentum mass m, and linear veloc-
ity is a vector quantity defined relative to a fixed point (usually
an origin) as
(11-18)
The magnitude of is given by
(11-19)
(11-20)
(11-21)
where f is the angle between and and are the compo-
nents of and perpendicular to and is the perpendicular
distance between the fixed point and the extension of .The direc-
tion of is given by the right-hand rule for cross products.
Newton’s Second Law in Angular Form Newton’s second
!
:
p:
r!r
:,v:p:
v!p!p
:,r:
! r!p ! r!mv,
! rp! ! rmv!
! ! rmv sin "
!
:
!
:
! r: ! p:! m( r: ! v:).
v:
p:,!
:
#:
F
:
r!r
:,
F
:
F!r
:,F
:
# ! rF sin " ! rF! ! r!F,
#:
r:F
:
#:! r: ! F
:
,
#:
acom, x ! $
g sin %
1 & Icom /MR2
.
a:com
law for a particle can be written in angular form as
(11-23)
where is the net torque acting on the particle and is the an-
gular momentum of the particle.
Angular Momentum of a System of Particles The angu-
lar momentum of a system of particles is the vector sum of the
angular momenta of the individual particles:
(11-26)
The time rate of change of this angular momentum is equal to the
net external torque on the system (the vector sum of the torques
due to interactions of the particles of the system with particles ex-
ternal to the system):
(system of particles). (11-29)
Angular Momentum of a Rigid Body For a rigid body
rotating about a fixed axis, the component of its angular
momentum parallel to the rotation axis is
L ! Iv (rigid body, fixed axis). (11-31)
Conservation of Angular Momentum The angular mo-
mentum of a system remains constant if the net external torque
acting on the system is zero:
(isolated system) (11-32)
or (isolated system). (11-33)
This is the law of conservation of angular momentum.
Precession of a Gyroscope A spinning gyroscope can pre-
cess about a vertical axis through its support at the rate
(11-46)
where M is the gyroscope’s mass, r is the moment arm, I is the rota-
tional inertia, and v is the spin rate.
' !
Mgr
I(
,
L
:
i ! L
:
f
L
:
! a constant
L
:
#:net !
dL
:
dt
L
:
! !
:
1 & !
:
2 & ) ) ) & !
:
n ! !
n
i!1
 !
:
i.
L
:
!
:
#:net
#:net !
d!
:
dt
,
halliday_c11_275-304hr2.qxd 29-09-2009 13:17 Page 296
377R EVI EW & S U M MARY
PART 2
(Here the top of the tank is represented by the left side of
the equation and the hole by the right side. The zero on the
right indicates that the hole is at our reference level.)
Before we solve Eq. 14-39 for v, we can use our result that
v0 v to simplify it: We assume that v20, and thus the term
in Eq. 14-39, is negligible relative to the other terms,
and we drop it. Solving the remaining equation for v then
yields
(Answer)
This is the same speed that an object would have when
falling a height h from rest.
v ! 12gh.
1
2"v
2
0
#
Additional examples, video, and practice available at WileyPLUS
Calculations: From Eq. 14-24,
RV ! av ! Av0
and thus
Because a A, we see that v0 v.To apply Bernoulli’s equa-
tion, we take the level of the hole as our reference level for
measuring elevations (and thus gravitational potential en-
ergy). Noting that the pressure at the top of the tank and at
the bullet hole is the atmospheric pressure p0 (because both
places are exposed to the atmosphere), we write Eq. 14-28 as
(14-39)p0 $
1
2"v
2
0 $ "gh ! p0 $
1
2"v
2 $ "g(0).
##
v0 !
a
A
 v.
Density The density r of any material is defined as the material’s
mass per unit volume:
(14-1)
Usually, where a material sample is much larger than atomic 
dimensions, we can write Eq. 14-1 as
(14-2)
Fluid Pressure A fluid is a substance that can flow; it conforms
to the boundaries of its container because it cannot withstand shear-
ing stress. It can, however, exert a force perpendicular to its surface.
That force is described in terms of pressure p:
(14-3)
in which %F is the force acting on a surface element of area %A. If the
force is uniform over a flat area,Eq.14-3 can be written as
(14-4)
The force resulting from fluid pressure at a particular point in a
fluid has the same magnitude in all directions. Gauge pressure is the
difference between the actual pressure (or absolute pressure) at a
point and the atmospheric pressure.
Pressure Variation with Height and Depth Pressure in a
fluid at rest varies with vertical position y. For y measured positive
upward,
p2 ! p1 $ rg(y1 & y2). (14-7)
The pressure in a fluid is the same for all points at the same level. If
h is the depth of a fluid sample below some reference level at which
the pressure is p0, Eq. 14-7 becomes
p! p0 $ rgh, (14-8)
where p is the pressure in the sample.
p !
F
A
.
p !
%F
%A
,
" !
m
V
.
" !
%m
%V
.
Pascal’s Principle A change in the pressure applied to an en-
closed fluid is transmitted undiminished to every portion of the
fluid and to the walls of the containing vessel.
Archimedes’ Principle When a body is fully or partially sub-
merged in a fluid,a buoyant force from the surrounding fluid acts on
the body.The force is directed upward and has a magnitude given by
Fb !mf g, (14-16)
where mf is the mass of the fluid that has been displaced by the
body (that is, the fluid that has been pushed out of the way by the
body).
When a body floats in a fluid, the magnitude Fb of the
(upward) buoyant force on the body is equal to the magnitude Fg
of the (downward) gravitational force on the body. The apparent
weight of a body on which a buoyant force acts is related to its ac-
tual weight by
weightapp ! weight& Fb. (14-19)
Flow of Ideal Fluids An ideal fluid is incompressible and
lacks viscosity, and its flow is steady and irrotational. A streamline
is the path followed by an individual fluid particle. A tube of flow is
a bundle of streamlines. The flow within any tube of flow obeys the
equation of continuity:
RV !Av! a constant, (14-24)
in which RV is the volume flow rate, A is the cross-sectional area of
the tube of flow at any point, and v is the speed of the fluid at that
point.The mass flow rate Rm is
Rm ! rRV ! rAv! a constant. (14-25)
Bernoulli’s Equation Applying the principle of conservation
of mechanical energy to the flow of an ideal fluid leads to
Bernoulli’s equation:
p$ rv2 $ rgy! a constant (14-29)
along any tube of flow.
1
2
F
:
b
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delta P = dgy + (1/2) d(v1ˆ2 - v2ˆ2) 
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