Chapter 9
Rotation of Rigid Bodies
Goals for Chapter 9
• To study rotational kinematics
• To relate linear to angular kinematics
• To define moments of inertia and determine
rotational kinetic energy
• To calculate the moment of inertia.
Introduction
As you do anything that involves objects
that turn in circles, you’re already
involved in the process of rotational
dynamics and rotational kinematics.
Rotation occurs at all scales, from the motion of electrons in
atoms to the motions of entire galaxies. We need to develop some
general methods for analyzing the motion of a rotating body. In
this chapter and the next we consider bodies that have definite size
and definite shape, and that in general can have rotational as well
as translational motion.
Classification of rigid body motion
“Real-world” rotations can be very complicated . We can make a
substantial understanding by studying the ideal situation of rigid
bodies rotating.
Translational motion
Motion of rigid body
Fixed axis
Rotation
Unfixed axis
A speedometer as our starting model
A car’s analog speedometer gives
us a very good example to begin
defining rotational motion.
Consider the clockwise (or
counterclockwise) motion of a
rigid, fixed-length speedometer
needle about a fixed pivot point.
A speedometer as our starting model
The angu1ar coordinate θ of a rigid body rotating around a fixed
axis can be positive or negative. If we choose positive angles to be
measured counterclockwise from the positive x-axis, then the
angle θ is positive. If we instead choose the positive rotation
direction to be clockwise, then θ is negative.
When we discuss rotation around a fixed axis, it's essential to
specify the direction of positive rotation.
Angular motions in revolutions, degrees, and radians
One complete cycle of 360º is
one revolution or 2π radians.
Relating
the
two,
360º=2π
radians or 1 radian = 57.3º.
To describe rotational motion, the most
natural way to measure the angle θ is
not in degrees, but in radians.
s

r
or
1rad  57.3
s  r
Angular displacement is the angle being swept out
Like a second hand sweeping around a clock, a radius vector will
travel through a displacement of degrees, radians, or revolutions.
We denote angular displacement as θ. It is the angular equivalent
of x or y in earlier chapters.
Angular velocity
The angular velocity is the angle swept out divided by the time it
took to sweep out the angular displacement.
Angular velocity is denoted by the symbol ω (omega).
Angular velocity is measured in radians per second (SI standard) as
well as r.p.m. (revolutions per minute).
We define the average angular velocity ωav as
av  z 
 2 z  1z
t2  t1
 z

t
The instantaneous angular velocity ωz is the limit of ωav-z as Δt
approaches zero.
Example 9.1
The angular position θ of a 0.36m diameter flywheel is given
by    2.0rad / s 3  t 3 .
a.Find θ at t=2.0s and 5.0s in radians and degrees.
b. find the distance that a particle on the flywheel rim moves
from t=2.0s to 5.0s
c. find the average angular velocity in rad/s and rev/min.
d. find the instantaneous angular velocity at t=5.0s .
Example 9.1

  2.0s 
  2.0rad / s   5.0s 
Execute: a)   2.0rad / s 3
1
2
b)
c)
3
3
 16rad
3
 250rad
s  r 2  1    0.18m 234rad   42m
av 
 2 z  1z
t2  t1
250rad  16rad

 78rad / s
5.0s  2.0s
d d
z 
  2.0rad / s 2  t 3    6.0rad / s 2  t 2
dt dt
d) At t=5.0s
z  150rad / s
Angular velocity is a vector
You can visualize the position of the vector
with the fingers of your right hand. The
position of your thumb will be the position of
the angular velocity vector. -“right-hand
rule.”
At any instant, every part of a rotating
rigid body has the same angular velocity.
If the angle θ is in radians, the unit of
angular velocity is the radian per second
(rad/s). Revolution per minute (rpm) is
often used.
1rev / s  2 rad / s
2
1rev / min  1rpm 
rad / s
60
Angular acceleration
The angular acceleration is the change of angular velocity
divided by the time interval during which the change occurred.
Use the symbol α to denote radians per second2.
We can define the average angular
acceleration αav-z over the interval Δt=t2tl as the change in angnlar velocity
divided by Δt
 av  z 
2 z  1z
t2  t1
 z

t
The instantaneous angular acceleration αz, is
the limit of αav-z as Δt→0:
Angular acceleration is a vector
The angular acceleration
vector will be parallel or
antiparallel to the angular
velocity vector (as
determined by the Right-
hand rules.
In rotational motion, if the angular acceleration αz is positive, then
the angular velocity ωz is increasing; if αz is negative, then ωz is
decreasing. The rotation is speeding up if αz and ωz have the same
sign and slowing down if αz and ωz have opposite signs.
Fundamental equations for angular kinematics
In Chapter 2 we found that straight-line motion when the acceleration
is constant. When the angular acceleration is constant, we can derive
similar equations for angular velocity and angular position using
exactly the same procedure.
The circular motion of an audio CD—Example 9.3
You have just finished watching a movie on DVD and the disc is
slowing to stop. The angular velocity of the disc at t=0 is 27.5rad/s
and its angular acceleration is a constant -10.0rad/s2. A line PQ on the
surface of the disc lies along the +x-axis at t=0. (a) What is the disc's
angular velocity at t=0.300s?
(b) What angle does the line PQ make with the +x-axis at this time?
The circular motion of an audio CD—Example 9.3
Identify: The angular acceleration of the disc is constant, so we can
use any of the equations derived in this section.
Set up: We are given the initial angular velocity ω=27.5rad/s, the
initial angle θ0 = 0 between the line PQ and the +x-axis, the angular
acceleration αz= -10.0rad/s2, and the time t=0.300s.
The circular motion of an audio CD—Example 9.3
Execute: (a) From Eq. (9.7), at t = 0.300 s we have
(b) From Eq. (9.11),
Linear and angular quantities related
When a rigid body rotates about a fixed axis,
every particle in the body moves in a
circular path. The circle lies in a plane
perpendicular to the axis and is centered on
the axis. The speed of a particle is directly
proportional to the body's angular velocity
v=rω; In Figure, point P is a constant
distance r from the axis of rotation.
Linear and angular quantities related
At any time, the angle θ (in radians) and the arc length s are
related by s=rθ. We take the time derivative of this, noting that
r is constant for any specific particle, and take the absolute
value of both sides:
ds
d
r
 v  r
dt
dt
Relationship between linear and
angular speeds
More on Angular acceleration
A rigid body whose rotation is speeding up. The acceleration of point
P has a component arad toward the rotation axis (perpendicular to v)
and a component atan along the circle that point P follows.
The quantity α=dω/dt is the rate of change of
the angular speed. It is not quite the same as
az=dωz/dt. For example, consider a body
rotating so that its angular velocity vector
points in the -z-direction. If α=10rad/s2, then
αz=-10rad/s2.
More on Angular acceleration
arad
v2
2
  r
r
The vector sum of the centripetal and tangential components of
acceleration of a particle in a rotating body is the linear acceleration
a.
Remember that s=rθ, is valid only when θ is measured in radians.
Example 9.4 An athlete throwing the discus
A athlete moves the discus in a circle of radius 80.0cm. At a certain
instant, the thrower is spinning at an angular speed of 10.0rad/s and
the angular speed is increasing at 50.0rad/s2. At this instant, find the
tangential and centripetal components of the acceleration of the
discus and the magnitude of the acceleration.
Example 9.4 An athlete throwing the discus
Identify: We model the discus as a particle traveling on a circular
path (Fig.a), so we can use the ideas developed in this section.
Set up: We are given the radius r=0.800m, ω=10.0rad/s, and the
angular speed α=50.0rad/s2. Given these components of the
acceleration vector, we’ll find its magnitude a (the third target
variable) .
Execute: From Eqs. (9.14) and (9.15),
Using the Pythagorean theorem the magnitude of the acceleration
vector is:
An airplane propeller
You are asked to design an airplane propeller to turn at 2400rpm.
The forward airspeed of the plane is to be 75.0m/s and the speed of
the tips of the propeller blades through the air must not exceed
270m/s. (This is about 0.80 times the speed of sound in air)(a) What
is the maximum radius the propeller can have? (b) With this radius,
what is the acceleration of the propeller tip?
An airplane propeller
Identify: The object of interest in this example is a particle at the
tip of the propeller; our target variables are the particle's distance
from the axis and its acceleration. Note that the speed of this
particle through the air (which cannot exceed 270m/s) is due to both
the propeller's rotation and the forward motion of the airplane.
An airplane propeller
Set up: As Fig.b shows, the velocity vtip of a particle at the propeller
tip is the vector sum of its tangential velocity due to the propeller's
rotation and the forward velocity of the airplane. The rotation plane
of the propeller is perpendicular to the direction of flight, so these
two vectors are perpendicular and we can use the Pythagorean
theorem to relate vtan and vplane to vtip. We will then set vtip=270m/s
and solve for the radios r.
An airplane propeller
Execute: We first convert ω to rad/s:
The tangential acceleration is zero because the angular speed is
constant
Bicycle pedals and gears
How are the angular
speeds of the two
bicycle sprockets
related to the number
of teeth on each
sprocket?
The chain does not slip or stretch, so it moves at the same tangential
speed v on both sprockets.
For chain sprockets the teeth must be equally spaced on the
circumferences of both sprockets, for the mesh of chain be properly
the same. Let Nfront and Nrear be the numbers of teeth; the condition
that the tooth spacing is the same on both sprockets is
9.4 Energy in rotational motion
A rotating rigid body consists of mass in motion, so it has kinetic
energy. we can express this kinetic energy in terms of the body's
angular speed and a new quantity, called moment of inertia, that
depends on the body's mass and how the mass is distributed.
1
1
2
2 2
mi vi  mi ri i
2
2
1
1 2 2 1
1 2
2
2
2
K   mi vi   mi ri i    mi ri    I 
2
2
2
2
I   mi ri 2
Moment of inertia
Rotational energy
Just like linear kinetic energy is ½ mv2, the
angular energy will be determined by ½ Iω2.
Equation (9.17) gives a simple physical
interpretation of moment of inertia:
So the greater a body's moment of inertia,
the harder it is to start the body rotating if
it's at rest and the harder it is to stop. For
this reason, I is also called the rotational
inertia.
Rotational energy changes if parts shift and I changes
An engineer is designing a machine part consisting of three heavy
disks linked by lightweight struts. (a) What is the moment of inertia
of this body about an axis through the center of disk A,
perpendicular to the plane of the diagram? (b) What is the moment
of inertia about an axis through the centers of disks B and C? (c) If
the body rotates about an axis through A perpendicular to the plane
of the diagram, with angular speed ω =4.0rad/s, what is its kinetic
en energy?
Identify: We consider the disks
as massive particles and the
lightweight struts as massless
rods. So we use the ideas of this
section to calculate the moment
of this collection of three
particles.
Rotational energy changes if parts shift and I changes
Set up: In parts (a) and (b), we'll use Eq.(9.16) to find the moments
of inertia for each of the two axes. Given the moment of inertia for
axis A, we'll use Eq.(9.17) in part (c) to find the rotational kinetic
energy.
Execute:
Finding the moment of inertia for common shapes
Calculating rotational energy
A light, flexible, nonstretching cable is wrapped several times around
a winch drum, a solid cylinder of mass 50kg and diameter 0.120m,
which rotates about a stationary horizontal axis held by frictionless
bearings. The free end of the cable is pulled with a constant 9.0N
force for a distance of 2.0m. It unwinds without slipping and turns
the cylinder. If the cylinder is initially at rest, find its final angular
speed and the final speed of the cable.
Calculating rotational energy
Identify: We will solve this problem using energy methods. There
is friction between the cable and the cylinder, but because the cable
doesn't slip, there is no mechanical energy is lost in friction.
Set up: The cylinder starts at rest, so the initial kinetic energy is
K1=0. Between points 1 and 2 the force F does work on the cylinder
over a distance s=2.0m. As a result, the kinetic energy at point 2 is
K2= Iω2/2.
Calculating rotational energy
Execute: The work done on the cylinder is Wother=Fs=(9.0N)(2.0
m) = 18 J. The moment of inertia is
The relationship K1+ U1+ Wother= K2+U2, then gives
The final tangential speed of the cylinder, and hence the final speed
of the cable, is
Example 9.9 an unwinding cable
We wrap a light, flexible cable around a solid cylinder with mass M
and radius R. The cylinder rotates with negligible friction about a
stationary horizontal axis. We tie the free end of the cable to a block
of mass m and release the object with no initial velocity at a distance
h above the floor. As the block falls, the cable unwinds without
stretching or slipping, turning the cylinder. Find the speed of the
falling block and the angular speed of the cylinder just as the
block strikes the floor.
Example 9.9 an unwinding cable
Identify: The cable doesn't slip and friction does no work. The
cable does no net work. Hence only gravity does work, and so
mechanical energy is conserved.
Set up: Figure a shows the situation just before the block begins to
fall. At this point the system has no kinetic energy, so K1=0, we take
the potential energy to be zero when the block is at floor level; then
U1=mgh and U2=0. the total kinetic energy K2 at that instant is
Execute: We use our expressions for K1, U1, K2 and U2 and the
relationship ω= v/R in the energy-conservation equation
K1+U1=K2+U2. We then solve for v:
The final angular speed of the cylinder is ω=v/R
Gravitational potential energy for an extended body
In Example 9.9 the cable was of negligible mass, so we could
ignore the gravitational potential energy.
If the mass is not negligible, the
gravitational potential energy is the
same as though all the mass were
concentrated at the center of mass
of the body.
Suppose we take the y-axis
vertically upward. Then for a body
with total mass M, the gravitational
potential energy U is simply
Parallel Axis Theorem
There is a simple relationship between the
moment of inertia Icm of a body of mass M
about an axis through its center of mass and
the moment of inertia Ip about any other
axis parallel to the original one but
displaceed from it by a distance d.
The relationship, called parallel-axis
theorem, stated that .
I cm   mi ri 2   mi  xi2  yi2 
I p   mi
 xi  a    yi  b 
2
2

  mi  xi2  yi2    2a  mi xi  2b mi yi    a 2  b 2   mi
 I cm  Md 2
2a mi xi  2b mi yi  2aMxcm  2bMycm  0
Example 9.10 using the parallel-axis theorem
A part of a mechanical linkage has a mass of 3.6kg. We measure its
moment of inertia about an axis 0.15m from its center of mass to be
Ip=0.132kg·m2. What is the moment of inertia Icm about a parallel
axis through the center of mass?
Moment-of-inertia calculations
If a rigid body is a continuous distribution of mass, like a solid
cylinder or a solid sphere, it cannot be represented by a few point
masses. In this case the sum of masses and distances that defines the
moment of inertia becomes an integral.
We call this distance r, as before. Then the moment of inertia is
I   r dm
2
When the object is effectively one-dimensional, such as the slender
rods, we can use a coordinate x along the length and relate dm to an
increment dx. For a three-dimensional object it is usually easiest to
express dm in terms of an element of volume dV so we may also
write
I   r  dV
2
I    r dV
2
Q9.1
The graph shows the
angular velocity and
angular acceleration
versus time for a
rotating body. At
which of the following
times is the rotation
speeding up at the
greatest rate?
A. t = 1 s
B. t = 2 s
C. t = 3 s
D. t = 4 s
E. t = 5 s
A9.1
The graph shows the
angular velocity and
angular acceleration
versus time for a
rotating body. At
which of the following
times is the rotation
speeding up at the
greatest rate?
A. t = 1 s
B. t = 2 s
C. t = 3 s
D. t = 4 s
E. t = 5 s
Q9.2
A DVD is initially at rest so that the
line PQ on the disc’s surface is along
the +x-axis. The disc begins to turn
with a constant az = 5.0 rad/s2.
At t = 0.40 s, what is the angle
between the line PQ and the +x-axis?
A. 0.40 rad
B. 0.80 rad
C. 1.0 rad
D. 2.0 rad
A9.2
A DVD is initially at rest so that the
line PQ on the disc’s surface is along
the +x-axis. The disc begins to turn
with a constant az = 5.0 rad/s2.
At t = 0.40 s, what is the angle
between the line PQ and the +x-axis?
A. 0.40 rad
B. 0.80 rad
C. 1.0 rad
D. 2.0 rad
Q9.3
A DVD is rotating with an everincreasing speed. How do the
centripetal acceleration arad and
tangential acceleration atan
compare at points P and Q?
A. P and Q have the same arad
and atan.
B. Q has a greater arad and a
greater atan than P.
C. Q has a smaller arad and a greater atan than P.
D. P and Q have the same arad, but Q has a greater atan than P.
A9.3
A DVD is rotating with an everincreasing speed. How do the
centripetal acceleration arad and
tangential acceleration atan
compare at points P and Q?
A. P and Q have the same arad
and atan.
B. Q has a greater arad and a
greater atan than P.
C. Q has a smaller arad and a greater atan than P.
D. P and Q have the same arad, but Q has a greater atan than P.
Q9.5
You want to double the radius of a rotating solid sphere
while keeping its kinetic energy constant. (The mass
does not change.) To do this, the final angular velocity
of the sphere must be
A. 4 times its initial value.
B. twice its initial value.
C. the same as its initial value.
D. 1/2 of its initial value.
E. 1/4 of its initial value.
A9.5
You want to double the radius of a rotating solid sphere
while keeping its kinetic energy constant. (The mass
does not change.) To do this, the final angular velocity
of the sphere must be
A. 4 times its initial value.
B. twice its initial value.
C. the same as its initial value.
D. 1/2 of its initial value.
E. 1/4 of its initial value.
Q9.6
The three objects
shown here all have
the same mass M
and radius R. Each
object is rotating
about its axis of
symmetry (shown
in blue). All three
objects have the
same rotational
kinetic energy.
Which one is
rotating fastest?
A. thin-walled hollow cylinder
B. solid cylinder
C. thin-walled hollow cylinder
D. two or more of these are tied
for fastest
A9.6
The three objects
shown here all have
the same mass M
and radius R. Each
object is rotating
about its axis of
symmetry (shown
in blue). All three
objects have the
same rotational
kinetic energy.
Which one is
rotating fastest?
A. thin-walled hollow cylinder
B. solid cylinder
C. thin-walled hollow cylinder
D. two or more of these are tied
for fastest
Q9.7
A thin, very light wire is wrapped
around a drum that is free to rotate.
The free end of the wire is attached
to a ball of mass m. The drum has
the same mass m. Its radius is R and
its moment of inertia is I = (1/2)mR2.
As the ball falls, the drum spins.
At an instant that the ball has
translational kinetic energy K, the
drum has rotational kinetic energy
A. K.
B. 2K.
C. K/2.
D. none of these
A9.7
A thin, very light wire is wrapped
around a drum that is free to rotate.
The free end of the wire is attached
to a ball of mass m. The drum has
the same mass m. Its radius is R and
its moment of inertia is I = (1/2)mR2.
As the ball falls, the drum spins.
At an instant that the ball has
translational kinetic energy K, the
drum has rotational kinetic energy
A. K.
B. 2K.
C. K/2.
D. none of these
Homework:
64, 68, 82, 83