Chapter 8 – Rotational Equilibrium and Rotational Dynamics
This chapter deals with the forces that produce rotational motion. It builds onto the
material developed in Ch 6 on rotational kinematics.
Torque
F
Torque causes an object to rotate. It
depends on the force and the point at which
the force is applied relative to the axis of
rotation. Specifically, torque is the product
of the force and the nearest distance between
the line of action of the force and the axis or
rotation,
  Fd
pivot

r
d
(units are N-m)
F
F = F sin
If r is the distance from the point of
application of the force to the pivot, then d =
r sin (called the ‘lever arm’). We can also
write
pivot
r

F = F cos
  F r
where F = F sin is the component of F that is perpendicular to r.
The torque due to the force shown in the diagram would produce a counter-clockwise
rotation about the pivot and is assumed to be positive. A torque that would produce a
clockwise rotation is negative.
Example:
A disk pivoted about an axis through its center is
subjected to two forces, as shown to the right. Given:
F1 = 10 N, F2 = 5 N, d1 = 8 cm, d2 = 5 cm. What is the
net torque?
 = F1d1 – F2d2 = (10 N)(0.08 m) – (5 N)(0.05 m)
= 0.8 N-m – 0.25 N-m = 0.55 N-m
1
d1 d2
F1
F2
Center of Gravity
Gravity acts on all the particles that make up an extended object. The net torque due to
gravity and the resulting rotation of the object about a pivot depend on how the mass is
distributed. Effectively, we can think of all these individual weights as acting at a single
point located so as to give the same net torque. In a uniform gravitational field this
‘center of gravity’ is the same as the ‘center of mass’. In 2-D the location of the center of
mass is given by
y (m)
m2
 mi xi
 mi yi
3
xcm 
, ycm 
m3
 mi
 mi
2
m1
Example:
1
What is the location of the center of mass of the
three particles in the figure to the right? Given:
1
2
3
4 x (m)
m1 = 10 g, m2 = 20 g, m3 = 30 g.
m x  m2 x 2  m3 x3
xcm  1 1
m1  m2  m3
(0.01kg)(1m)  (0.02kg)( 2m)  (0.03kg)( 4m)
 2.83 m
0.01kg  0.02kg  0.03kg
m y  m2 y 2  m3 y3 (0.01kg)(1m)  (0.02kg)(3m)  (0.03kg)( 2m)
y cm  1 1

m1  m2  m3
0.01kg  0.02kg  0.03kg

 2.17 m
If a regularly shaped object has a uniform mass density, then its center of mass is the
same as its geometrical center.
Equilibrium
An object is in equilibrium if both the net force and net torque on it are zero. Thus, for
equilibrium
F  0
1st condition for equilibrium
  0
2nd condition for equilibrium
(Note: The location of the axis of rotation is arbitrary when the object is in equilibrium.)
2
Example:
Masses m1 = 1 kg and m2 = 2 kg hang from opposite ends of a 1-m rod. The mass of the
rod is 3 kg. Where would you lift the rod to keep it in balance?
x
Pick the axis of rotation to be the left end (any position will work). The resulting forces
are shown below.
F
x
L/2
m1 g
mrod g
m2 g
From the 1st condition for equilibrium,
 F  0  F  m1 g  m2 g  mrod g
F  (m1  m2  mrod ) g  (6kg)(9.8m / s 2 )  58.8 N
From the 2nd condition, using the left end as the pivot,
  0  Fx  m1 g (0)  m2 gL  mrod L / 2
m2 gL  mrod gL / 2 (2kg)(9.8m / s 2 )(1m)  (3kg)(9.8m / s 2 )(0.5m)
x

 0.58 m
F
58.8 N
3
Example:
A sign hangs from a horizontal beam as shown to the
right. The left end of the beam is attached to a
building and a cable supports the right end of the
beam at an angle of 30o with respect to the beam. The
beam is 2 m long and its mass is 5 kg. The sign is
attached 1.5 m from the building and its mass if 10
kg. What is the tension in the cable?

The forces acting on the beam are shown below.
T
R

mbg
msg
Picking the left end as the pivot,
   0  T sin  (2m)  mb g (1m)  ms g (1.5m)
mb g (1m)  ms g (1.5m) (5kg)(9.8m / s 2 )(1m)  (10kg)(9.8m / s 2 )(1.5m)
T

sin  (2m)
(sin 30 o )( 2m)
 196 N
Note: By picking the left end as the pivot, we don’t need to worry about solving for R,
the force exerted on the left end by the building.
Torque and Angular Acceleration
For a single point mass going in a circle subject to a tangential acceleration,
Ft  mat  mr (it also has a centripetal component of force and acceleration)
The torque is then
Ft r  mr 2  ,
or
  mr 2
4
For a rigid body consisting of many masses, each mass goes in a circle and has the same
angular acceleration, . Thus, the net torque is


   mr 2 
The moment of inertia (or rotational inertia) is defined as
I   mr 2
Thus,
  I
This is in essence Newton’s 2nd law of motion for rotation of a rigid body. The torque
drives the angular acceleration and the rotational inertia resists the angular acceleration.
The rotational inertia depends not only on the mass but on where the mass is located with
respect to the axis or rotation.
y (m)
(0, 3m)
Example:
m1
Three masses, each 2 kg, are located at the corners
of a triangle consisting of light rigid rods, as
shown to the right. The location (x, y) of each
mass is shown. What is the rotational inertia
about the x-axis?
x (m)
m2
(-3m, -2m)
m3
(3m, -2m)
What matters here is the nearest distance of each
mass from the axis of rotation. So,
I x   mr 2  m1 y12 m2 y 2 2  m3 y3 2  (2)(3) 2  (2)( 2) 2  (2)( 2) 2  34 kg  m 2
Suggested exercise: Find the rotational inertia about the y-axis and about the z-axis
(which is perpendicular to the page).
Example:
What is the rotational inertia of a thin cylindrical shell or
radius R about its symmetry axis?
Since all the mass is the same distance from the axis,
then
I   mr 2  ( m) R 2  mR 2
5
Example:
m
What is the rotational inertia of a uniform rod
about an axis perpendicular to the rod at an end
of the rod?
x
L
I   mx 2
The mass of the rod is distributed uniformly from x = 0 to x = L, so one must use the
average value of x2. It can be shown that <x2> = L2/3. Thus,
I  1 M L2
3
Rotational Inertia of Various Rolling Objects:
The rotational inertia of various uniform ‘round’ objects can be obtained using calculus.
As they roll, they rotate about a symmetry axis through the center of the object. Some
results for rolling objects:
Object
cylindrical shell
Rotational Inertia
solid cylinder
I  1 M R2
I  M R2
spherical shell
I
solid sphere
I
2
2M
3
2M
5
R2
R2
Example:
A flywheel of radius R = 0.2 m and mass M = 25 kg is
mounted to rotate about its symmetry axis. A rope is
wrapped around the wheel and pulled with a force of 10 N.
What is the angular acceleration of the wheel? (Assume the
wheel is a solid cylinder.)
I  1 MR 2  1 (25kg)(0.2m) 2  0.5 kg  m 2
2


I
2

FR (10 N )(0.2m)

 0.4 rad / s 2
2
I
0.5kg  m
6
Rotational Kinetic Energy
For a rigid object rotating about an axis, all the pieces that make up the object rotate in a
circle with the same angular velocity, . Thus, the kinetic energy is


KErot   1 mv 2   1 m 2 r 2  1  mr 2  2
2
2
2
or,
KE rot  1 I 2
2
A rolling object will have kinetic energy related to the translational speed of its center of
mass, ½ mv2, and kinetic energy related to its rotation about an axis through its center of
mass, ½ I 2. It can also have gravitational potential energy. Thus, the total energy is
E  KEt  KEr  PE g  1 Mv 2  1 I 2  Mgh
2
2
Example:
A ball rolls from rest down an incline of height 3 m without slipping. What is its speed
when it reaches the base of the incline?
Use conservation of energy –
E f  Ei
1 Mv 2  1 I 2  Mgh
2
2
v
I  2 MR 2 ,  
5
R


1 Mv 2  1 2 MR 2  v 
 
2
2 5
R
2
 1 Mv 2  1 Mv 2  7 Mv 2  Mgh
2
5
10
v  10 gh  10 (9.8)(3)  6.5 m / s
7
7
Note that the final speed does not depend on either the mass of the ball or its radius. This
will be true for all rolling objects.
Question: What fraction of the final KE is due to rotation?
1 Mv 2
KErot
2
 5

7
KEtotal
Mv 2 7
10
7
Angular Momentum
From Newton’s 2nd law of motion applied to rotation,
  I  I
  0 I  I0

I

t
t
t
The angular momentum is defined as
L  I
Thus, we can write

L
t
This leads to the principle of conservation of angular momentum:
If net = 0, then L = constant.
Example:
A figure skater goes into a spin with her arms outstretched. She then pulls her arms in
against her body. If she is initially rotating at 2 rev/s, what is her final rotational speed?
Assume that by pulling in her arms she reduces her rotational inertia from 2 kg-m2 to 1.5
kg-m2.
L f  Li
I f  f  I i i
f 
I i i (2kg  m 2 )( 2rev / s)

 2.67 rev / s
If
1.5kg  m 2
How much kinetic energy was gained or lost when she pulled in her arms?


KE f  1 I f  f 2  1 1.5kg  m 2 (2 )( 2.67) rad / s 2  210.5 J
2
2


KEi  1 I f  f 2  1 2kg  m 2 (2 )( 2) rad / s 2  157.9 J
2
2
KE  KE f  KEi  210.5 J  157.9 J  52.6 J
Thus, there was a gain in KE. Note that we had to convert angular velocity to rad/s when
calculating KE.
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