Chapter 10

Rotation
In this chapter we will study the rotational motion of rigid bodies
about a fixed axis. To describe this type of motion we will introduce
the following new concepts:
-Angular displacement
-Average and instantaneous angular velocity (symbol: ω )
-Average and instantaneous angular acceleration (symbol: α )
-Rotational inertia, also known as moment of inertia (symbol I )
-Torque (symbol τ )
We will also calculate the kinetic energy associated with rotation,
write Newton’s second law for rotational motion, and introduce the
work-kinetic energy theorem for rotational motion.

(10-1)
 The Rotational Variables
In this chapter we will study the rotational motion of rigid
bodies about fixed axes. A rigid body is defined as one
that can rotate with all its parts locked together and without
any change of its shape. A fixed axis means that the object
rotates about an axis that does not move. We can describe
the motion of a rigid body rotating about a fixed axis by
specifying just one parameter. Consider the rigid body of
the figure.
We take the z-axis to be the fixed axis of rotation. We define a reference line
that is fixed in the rigid body and is perpendicular to the rotational axis. A top
view is shown in the lower picture. The angular position of the reference line at
any time t is defined by the angle θ(t) that the reference lines makes with the
position at t = 0. The angle θ(t) also defines the position of all the points on the
rigid body because all the points are locked as they rotate. The angle θ is
related to the arc length s traveled by a point at a distance r from the axis via
s
the equation   . Note: The angle θ is measured in radians.
r (10-2)
 t2 Angular Displacement :
In the picture we show the reference line at a time t1 and
at a later time t2 . Between t1 and t2 the body undergoes
t1 an angular displacement    2  1 . All the points of the
rigid body have the same angular displacement because they
rotate locked together.
Angular Velocity :
We define as average angular velocity for the time interval  t1 , t2  the ratio
 2  1 
avg   . The SI unit for angular velocity is radians/second.
t2  t1 t

We define as the instantaneous angular velocity the limit of as t  0,
t
d
 
  lim . This is the definition of the first derivative with t : dt
t  0 t

Algebraic sign of angular frequency: If a rigid body rotates counterclockwise

(CCW),  has a positive sign. If on the other hand the rotation is clockwise (CW),
 has a negative sign. (10-3)
 t2
ω2 Angular Acceleration :
ω1 If the angular velocity of a rotating rigid object changes
t1 with time we can describe the time rate of change of 
by defining the angular aceleration.

In the figure we show the reference line at a time t1 and at a later time t2 .
The angular velocity of the rotating body is equal to 1 at t1 and 2 at t2 .
We define as average angular acceleration for the time interval  t1 , t2  the ratio
2  1 
 avg   . The SI unit for angular velocity is radians/second 2 .
t2  t1 t

We define as the instantaneous angular acceleration the limit of as t  0,
t

  lim . This is the definition of the first derivative with t : d
t 0 t 
dt
(10-4)
 Angular Velocity Vector :
For rotations of rigid bodies about a fixed axis
we can describe accurately the angular velocity
by assigning an algebraic sign: positive for
counterclockwise rotation and negative for
clockwise rotation.

We can actually use the vector notation to describe rotational motion, which
is more complicated. The angular velocity vector is defined as follows:
The direction of  is along the rotation axis.
The sense of  is defined by the right-hand rule (RHL).
Right-hand rule: Curl the right hand so that the fingers point in the direction
of the rotation. The thumb of the right hand gives the sense of .

(10-5)
 Rotation with Constant Angular Acceleration :
When the angular acceleration  is constant we can derive simple expressions
that give us the angular velocity  and the angular position  as a function of
time. We could derive these equations in the same way we did in Chapter 2.
Instead we will simply write the solutions by exploiting the analogy between
translational and rotational motion using the following correspondence
between the two motions.

Translational Motion Rotational Motion

x  
v  
a  
v  v0  at    0   t (eq. 1)
at 2 t2
x  xo  v0t       0 t  (eq. 2)
2 2
v 2  v02  2a  x  xo    2  02  2    0  (eq. 3) (10-6)
 Relating the Linear and Angular Variables :
Consider a point P on a rigid body rotating about
a fixed axis. At t  0 the reference line that connects
s the origin O with point P is on the x-axis (point A).
O θ
A
During the time interval t , point P moves along arc AP
and covers a distance s. At the same time, the reference
line OP rotates by an angle  .
Relation between Angular Velocity and Speed :
The arc length s and the angle  are connected by the equation
ds d  r  d
s  r , where r is the distance OP. The speed of point P is v   r .
dt dt dt
v  r
circumference 2 r 2 r 2
The period T of revolution is given by T     .
speed v r 
2 1
T T   2 f
 f (10-7)
 The Acceleration :
The acceleration of point P is a vector that has two
components. The first is a "radial" component along
r
O the radius and pointing toward point O. We have
enountered this component in Chapter 4 where we
called it "centripetal" acceleration. Its magnitude is

v2
ar    2 r
r
The second component is along the tangent to the circular path of P and is thus
known as the "tangential" component. Its magnitude is
dv d  r  d
at   r  r at  r
dt dt dt

The magnitude of the acceleration vector is a  at2  ar2 .

(10-8)
 Kinetic Energy of Rotation :
vi
Consider the rotating rigid body shown in the figure.
mi
We divide the body into parts of masses m1 , m2 , m3 ,..., mi ,....
ri
O The part (or "element") at P has an index i and mass mi .
The kinetic energy of rotation is the sum of the kinetic
1 1 1
energies of the parts: K  m1v12  m2v22  m3v32  ....
2 2 2
1 1
K   mi vi2 The speed of the ith element vi   ri  K   mi  ri  .
2

i 2 i 2

1 2 2 1 2
K    mi ri    I  The term I   mi ri 2 is known as
2 i  2 i

rotational inertia or moment of inertia about the axis of rotation. The axis of
rotation must be specified because the value of I for a rigid body depends on
its mass, its shape, as well as on the position of the rotation axis. The rotational
inertia of an object describes how the mass is distributed about the rotation axis.
1 2
I   mi ri 2 I   r dm
2 K I
i 2 (10-9)
In the table below we list the rotational inertias for some rigid bodies.
I  r
2
dm

(10-10)
 Calculating the Rotational Inertia :
The rotational inertia I   mi ri 2 . This expression is useful for a rigid body that
i

has a discreet distribution of mass. For a continuous distribution of mass the sum
becomes an integral, I   r 2 dm.
Parallel - Axis Theorem :
We saw earlier that I depends on the position of the rotation axis.
For a new axis we must recalculate the integral for I . A simpler
method takes advantage of the parallel-axis theorem.
Consider the rigid body of mass M shown in the figure.
We assume that we know the rotational inertia I com
about a rotation axis that passes through the center
of mass O and is perpendicular to the page.
The rotational inertia I about an axis parallel to the axis through O that passes
through point P, a distance h from O, is given by the equation
I  I com  Mh 2 (10-11)
 A Proof of the Parallel - Axis Theorem : We take the origin O
to coincide with the center of mass of the rigid body shown
in the figure. We assume that we know the rotational inertia
I com for an axis that is perpendicular to the page and passes
through O.

We wish to calculate the rotational inertia I about a new axis perpendicular

to the page and passing through point P with coordinates  a, b  . Consider
an element of mass dm at point A with coordinates  x, y  . The distance r

between points A and P is r   x  a   y  b

2 2
.

Rotational Inertia about P : I   r 2 dm    x  a    y  b   dm.

2 2
 
I    x 2  y 2  dm  2a  xdm 2b  ydm    a 2  b 2  dm. The second
and third integrals are zero. The first integral is I com . The term  a 2  b 2   h 2 .
Thus the fourth integral is equal to h 2  dm  Mh 2  I  I com  Mh
2
(10-12)
 Torque:
In fig. a we show a body that can rotate about an axis through
point O under the action of a force F applied at point P a distance
r from O. In fig. b we resolve F into two components, radial and
tangential. The radial component Fr cannot cause any rotation
because it acts along a line that passes through O. The tangential
component Ft  F sin  on the other hand causes the rotation of the
object about O. The ability of F to rotate the body depends on the
magnitude Ft and also on the distance r between points P and A.
Thus we define as torque   rFt  rF sin   r F .
The distance r is known as the moment arm and it is the
perpendicular distance between point O and the vector F .
The algebraic sign of the torque is assigned as follows:
If a force F tends to rotate an object in the counterclockwise
direction the sign is positive. If a force F tends to rotate an
  r F object in the clockwise direction the sign is negative. (10-13)
 Newton's Second Law for Rotation :
For translational motion, Newton's second law connects
the force acting on a particle with the resulting acceleration.
There is a similar relationship between the torque of a force
applied on a rigid object and the resulting angular acceleration.

This equation is known as Newton's second law for rotation. We will explore
this law by studying a simple body that consists of a point mass m at the end
of a massless rod of length r. A force F is applied on the particle and rotates
the system about an axis at the origin. As we did earlier, we resolve F into a
tangential and a radial component. The tangential component is responsible
for the rotation. We first apply Newton's second law for Ft . Ft  mat (eq. 1)
The torque  acting on the particle is   Ft r (eq. 2). We eliminate Ft
between equations 1 and 2:   mat r  m  r  r   mr 2    I .
  I (compare with: F  ma)
(10-14)
 1 Newton's Second Law for Rotation :
2
3 We have derived Newton's second law for rotation
i
for a special case: a rigid body that consists of a point
ri mass m at the end of a massless rod of length r. We will
now derive the same equation for a general case.
O
Consider the rod-like object shown in the figure, which can rotate about an axis
through point O under the action of a net torque  net . We divide the body into
parts or "elements" and label them. The elements have masses m1 ,m2 , m3 ,..., mn
and they are located at distances r1 , r2 , r3 ,..., rn from O. We apply Newton's second
law for rotation to each element:  1  I1 (eq. 1),  2  I 2 (eq. 2),
 3  I 3 (eq. 3), etc. If we add all these equations we get
 1   2   3  ...   n   I1  I 2  I 3  ...  I n   . Here I i  mi ri 2 is the rotational inertia
of the ith element. The sum  1   2   3  ...   n is the net torque  net applied.
The sum I1  I 2  I 3  ...  I n is the rotational inertia I of the body.
Thus we end up with the equation  net  I
(10-15)
 Work and Rotational Kinetic Energy W :  K
In Chapter 7 we saw that if a force does work W
on an object, this results in a change of its kinetic energy
K  W . In a similar way, when a torque does work W
on a rotating rigid body, it changes its rotational kinetic
energy by the same amount.

Consider the simple rigid body shown in the figure, which consists of a mass m
at the end of a massless rod of length r. The force F does work dW  Ft rd   d .
The radial component Fr does zero work because it is at right angles to the motion.
f

The work W   Ft rd    d . By virtue of the work-kinetic energy theorem, we

i

1 2 1 2 1 2 2 1 2 2
have a change in kinetic energy K  W  mv f  mvi  mr  f  mr i 
2 2 2 2
W  K . f
1 2 1 2 W    d
K  I  f  I i
2 2 i (10-16)
 Power :
Power has been defined as the rate at which work is done
by a force and, in the case of rotational motion, by a torque.
We saw that a torque  produces work dW   d as it
rotates an object by an angle d .
dW d d
P   d      (Compare with P  Fv)
dt dt dt
Below we summarize the results of the work-rotational kinetic energy theorem:
f

W    d W    f  i  For constant torque

i

1 2 1 2
W  K  I  f  I i Work-Rotational Kinetic Energy Theorem
2 2

P  
(10-17)
 Analogies Between Translational and Rotational Motion

Translational Motion Rotational Motion

x  
v  
a  
v  v0  at    0   t
at 2 t2
x  xo  v0t      0  0 t 
2 2
v 2  v02  2a  x  xo    2  02  2    0 
mv 2 I 2
K  K
2 2
m  I
F  ma    I
F  
P  Fv  P  
(10-18)