Chapter 6 Rotation
In this chapter, we deal with the rotation of a rigid body
about a fixed axis. The first of these restrictions means that we
shall not examine the rotation of such objects as the Sun,
because the Sun-a ball of gas-is not a rigid body. Our second
restriction rules out objects like a bowling ball rolling down a
bowling lane. Such a ball is in rolling motion, rotating about a
moving axis.
6.1 The Rotational Variables
1. Translation and Rotation: The motion is the one of pure
translation, if the line connecting any two points in the object
is always parallel with each other during its motion.
Otherwise, the motion is that of rotation. Rotation is the
motion of wheels, gears, motors, the hand of clocks, the rotors
of jet engines, and the blades of helicopters.
2. The nature of pure rotation: The
right figure shows a rigid body of
arbitrary shape in pure rotation
around a fixed axis, called the axis
of rotation or the rotation axis.
(1). Every point of the body moves in a circle whose center
1
lies on the axis of the rotation.
(2). Every point moves through the same angle during a
particular time interval.
3. Angular position: The above figure shows a reference line,
fixed in the body, perpendicular to the axis, and rotating with
the body. We can describe the motion
of the rotating body by specifying the
angular position of this line, that is,
the angle of the line relative to a fixed
direction. In the right figure, the
angular position  is measured relative to the positive
direction of the x axis, and  is given by

s
r
(radian
measure).
Here s is the length of the arc (or the arc distance) along a
circle and between the x axis and the reference line, and r is a
radius of that circle.
An angle defined in this way is measured in radians (rad)
rather than in revolutions (rev) or degree. They have relations
1 rev  360 o 
2 r
 2 rad
r
4. If the body rotates about the rotation
axis as in the right figure, changing
the angular position of the reference
2
line from  1 to  2 , the body undergoes an angular
displacement

given by
   2  1
The definition of angular displacement holds not only for the
rigid body as a whole but also for every particle within the
body. The angular displacement

of a rotating body can
be either positive or negative, depending on whether the body
is rotating in the direction of increasing  (counterclockwise)
or decreasing  (clockwise).
5. Angular velocity
(1). Suppose that our rotating body is at angular position  1
at time
and at angular position  2 at time
t1
t2 .
We define
the average angular velocity of the body in the time interval
t
from
t1
to
to be
t2

In which

 2  1
t 2  t1


t
is the angular displacement that occurs during
t .
(2). The (instantaneous) angular velocity  , with which we
shall be most concerned, is the limit of the average angular
velocity as
t
is made to approach zero. Thus
 d

t 0 t
dt
  lim
If we know
 (t ) ,
we can find the angular velocity  by
3
differentiation.
(3). The unit of angular velocity is commonly the radian per
second (rad/s) or the revolution per second (rev/s).
(4). The magnitude of an angular velocity is called the
angular speed, which is also represented with  .
(5). We establish a
direction for the vector
of the angular velocity


by
using
a
rule,
as
right-hand
shown in the figure.
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.
6. Angular acceleration
(1). If the angular velocity of a rotating body is not constant,
then the body has an angular acceleration. Let  2 and 1 be
the angular velocity at times
t2
and
t1 ,
respectively. The
average angular acceleration of the rotating body in the
interval from
t1
to
t2

In which

is defined as
 2  1
t 2  t1


t
is the change in the angular velocity that occurs
4
during the time interval
t .
(2). The (instantaneous) angular acceleration  , with which
we shall be most concerned, is the limit of this quantity as
t
is made to approach zero. Thus
  lim
t  0
 d

t
dt
above equations hold not only for the rotating rigid body as a
whole but also for every particle of that body.
(3). The unit of angular acceleration is commonly the radian
per
second-squared
(rad/s2)
or
the
revolution
per
second-squared (rev/s2).
(4). The angular acceleration also is a vector. Its direction
depends on the change of the angular velocity.
7. Rotation with constant angular acceleration:
d
   d  dt     0   t
dt
d
1
  0   t  d  ( 0   t )dt     0 t   t 2
dt
2
Here we suppose that at time
t  0,  0  0 .
We also can get a
parallel set of equations to those for motion with a constant
linear acceleration.
8. Relating the linear and angular variables: They have relations
as follow:
Angular displacement:

d
 

ds  d  r
Angular velocity：


  
v r
5

Angular acceleration：
 

at    r

 
an    v

6.2 Kinetic Energy of Rotation
1. To discuss kinetic energy of a rigid body, we cannot use the
familiar formula
K  mv 2 / 2
directly because it applies only to
particles. Instead, we shall treat the object as a collection of
particles-all with different speeds. We can then add up the
kinetic energies of these particles to find kinetic energy of the
body as a whole. In this way we obtain, for the kinetic energy
of a rotating body,
K
In which
1
1
1
1
m1v12  m2 v 22  m3 v32     mi vi2
2
2
2
2
mi
is the mass of the ith particle and
vi
is its speed.
The sum is taken over all the particles in the body.
2. The problem with above equation is that
vi
is not the same
for all particles. We solve this problem by substituting for
in the equation with
 r,
v
so that we have
1
1
K   mi ( ri ) 2  ( mi ri 2 )
2
2
2
In which  is the same for all particles.
3. The quantity in parentheses on the right side of above
equation tells us how the mass of the rotating body is
distributed about its axis of rotation.
(1). We call that quantity the rotational inertia (or moment of
6
inertia) I of the body with respect to the axis of rotation. It’s a
constant for a particular rigid body and for a particular
rotation axis. We may now write
I   mi ri2
(2). The SI unit for I is the kilogram-square meter ( kg  m 2 ).
(3). The rotational inertia of a rotating body depends not only
on its mass but also on how that mass is distributed with
respect to the rotation axis.
4. We can rewrite the kinetic energy for the rotating object as
K
1
I
2
2
Which gives the kinetic energy of a rigid body in pure
rotation. It’s the angular equivalent of the formula
2
K  Mv cm
/2,
which gives the kinetic energy of a rigid body in pure
translation.
7
6.3 Calculating the Rotational Inertia
1. If a rigid body is made up of discrete particles, we can
calculate its rotational inertia from
I   mi ri2 .
2. If the body is continuous, we can replace the sum in the
equation with an integral, and the definition of rotational
inertia becomes
I   r 2 dm .
In general, the rotational inertia of
any rigid body with respect to a rotation axis depends on (1).
The shape of the body, (2). The perpendicular distance from
the axis to the body’s center of mass, and (3). The orientation
of the body with respect to the axis.
The table gives the rotational inertias of several common
bodies, about various axes. Note how the distribution of mass
relative to the rotational axis affects the value of the
rotational inertia I. We would like to give the example of
rotational inertia for
a thin circular plate
I   r 2 rd dr  

R
0
r 3 dr 
2
0
1
1
1
d   R 4 2  (  R 2 ) R 2  mR 2
4
2
2
a thin rod
(1)
(2)
I1  
l
2
l

2
1
r  dr   r 3
3
2
l
2

l
2
1
l
1
1
  ( ) 3  2  ( l )l 2  ml 2
3 2
12
12
1 3 1
l  ( l )l 2
0
3
3
l
1
3
4
1
I 2  I 1  m( ) 2  ml 2  ml 2  ml 2  ml 2
2
12
12
12
3
l
I 2   r 2  dr  
8
3. The parallel-axis theorem: If you know the rotational inertia
of a body about any axis that passes through its center of
mass, you can find its rotational inertia about any other axis
parallel to that axis with the parallel-axis theorem:
I  I cm  Mh 2
Here M is the mass of the body and h is the perpendicular
distance between the two parallel axes.
4. Proof of the parallel-axis theorem: Let O be the center of
mass of the arbitrarily shaped
body shown in cross section
in the figure. Place the origin
of coordinates at O. Consider
an
axis
through
O
perpendicular to the plane of
the figure, and another axis of P parallel to the first axis, Let
the coordinates of P be a and b.
Let dm be a mass element with coordinates x and y. The
rotational inertia of the body about the axis through P is then
I   r 2 dm   [( x  a) 2  ( y  b) 2 ]dm
Which we can rearrange as
I   ( x 2  y 2 )dm  2a  xdm  2b  ydm   (a 2  b 2 )dm
 I cm  0  0  Mh 2  I cm  Mh 2
9
6.4 Newton’s Second Law for Rotation
1. Torque: The following figure 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 F is applied at
point P, whose position relative to O is defined by a position
vector r. Vector F and r make an angle

with each other.
(For simplicity, we consider only forces that have no
component parallel to the rotation axis: thus, F is in the plane
of the page). We define the torque  as a vector cross
product of the position vector and the force



  r F
Discuss the direction and the magnitude of the torque.
2. Newton’s second law for rotation
(1). The figure shows a simple case of
rotation about a fixed axis. The
rotating rigid body consists of a single
particle of mass m fastened to the end
10
of a massless rod of length r. A force F acts as shown,
causing the particle to move in a circle about the axis. The
particle has a tangential component of acceleration
governed by Newton’s second law:
acting on the particle is

Ft  mat
at
. The torque

  r  F  Ft r  mat r  m( r )r  (mr 2 )
.
The quantity in parentheses on the right side of above
equation is the rotation inertia of the particle about the
rotation axis. So the equation can be reduced to
  I .
(2) For the situation in which more than one force is applied to
the particle, we can extend the equation as   I . Where
 is the net torque (the sum of all external torques) acting
on the particle. The above equation is the angular form of
Newton’s second law.
(3) Although we derive the angular form of Newton’s second law
for the special case of a single particle rotating about a fixed
axis, it holds for any rigid body rotating about a fixed axis,
because any such body can be analyzed as an assembly of
single particles.
11
6.5 Work and Rotational Kinetic Energy
1. Work-kinetic energy theorem: Let’s again consider the
situation of the figure, in which
force F rotates a rigid body
consisting of a single particle of
mass m fastened to the end of a
massless rod. During the rotation,
Force F does work on the body. Let us assume that the only
energy of the body that changed by F is the kinetic energy.
Then we can apply the work-kinetic energy theorem to get
K  K f  K i  W
 K 
1 2 1 2
I f  I i  W
2
2
Above equation is the angular equivalent of the work-kinetic
energy theorem for translational motion. We derive it for a
rigid body with one particle, but it holds for any rigid body
rotated about a fixed axis.
2. We next relate the work W done on the body in the figure to
the torque  on the body due to force F. If the particle in Fig.
11-17 were move a differential distance ds along its circular
d
,
 
dW  F  ds  Ft ds  Ft rd   d
.
path, the body would rotate through differential angle
with
ds  rd
. We would get
Thus the work done during a finite angular displacement
from
i
to
f
is then
f
W    d .
i
12
Above equation holds for
any rigid body rotating about a fixed axis.
3. We can find the power P for rotational motion
P
dW
d

 
dt
dt
13