MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
8.01
W15D2-9 Physical Pendulum Solution
A physical pendulum consists of a ring of radius R and mass m . The ring is pivoted
(assume no energy is lost in the pivot). The ring is pulled out such that its center of mass
makes an angle  0 from the vertical and released from rest. The gravitational constant is
g.
a) First assume that  0  1 . Show that the angle the center of mass makes with the
vertical satisfies a simple harmonic oscillator differential equation. What is the
angular frequency and period of oscillation? What is the angular speed of the
object at the bottom of its swing?
b) Now assume that  0 is large enough that you cannot make the small angle
approximation. What is the angular speed of the object at the bottom of its swing?
Solution: (a) We shall calculate torque about the pivot point (the pivot forces therefore
do not contribute to the torque). A torque diagram is shown in the figure below.
Using the torque method we have that
d 2
MgRsin   I P 2 .
dt
(1)
We use the parallel axis theorem to calculate the moment of inertia about the pivot point,
I P  I cm  MR2  MR2  MR2  2MR2 .
(2)
Therefore the equation of motion (Eq. (1)) becomes
d 2
g

sin   0
2
dt
2R
(3)
Using the small angle approximation that sin ;  , we have that the pivoted ring
undergoes approximate simple harmonic motion
d 2
g

 ; 0,
2
dt
2R
(4)
where the angular frequency of small oscillations is given by
0 
g
2R
(5)
The general solution for the angle  (t) and the component of the initial angular velocity
are given by
&
(6)
 (t)  0 cos( 0t)  0 sin( 0t)
0
d
(t)   00 sin( 0t)  &0 cos( 0t)
dt
(7)
d
(t  0) .From
where we denote the component of the initial angular velocity by &0 
dt
d
(t  0)  0 . Therefore the general solutions simply
the description of the problem &0 
dt
to
 (t)   0 cos( 0t)
(8)
d
(t)   00 sin( 0t)
dt
(9)
The ring is at the bottom of its swing when  (t f )  0 cos( 0t f )  0 . This first occurs
when  0t f   / 2 . At that instant the component of the angular speed is
d
g
(t f )   0 0  
 .
dt
2R 0
(10)
b) Now assume that  0 is large enough that you cannot make the small angle
approximation. What is the angular speed of the object at the bottom of its swing?
When  0 is large enough that we cannot use the small angle approximation so we shall
use the energy method to find the component of the angular speed at the bottom of the
swing.
When the string is at an angle  0 with respect to the vertical, the initial energy is the
gravitational potential energy (relative to a choice of zero potential energy at the bottom
of the swing where   0 ) is given by
E  U 0  MgR 1  cos 0 
(11)
At the bottom of the swing the energy is rotational kinetic energy.
E  Kf 
1
1
I P  f 2  2MR 2  f 2 .
2
2
(12)
Because the energy is constant
MgR 1  cos 0  
1
2MR 2  f 2
2
(13)
So the component of the angular velocity at the bottom of the swing is given by
f 
d
g
(t f )  
1  cos0 
dt
R
(14)
where we have taken the negative square root because at the bottom of the swing
component of the angular velocity is negative. When  0 is small, we can approximate
1  cos 0 ; 1  (1 
0 2
2
)
0 2
2
Therefore the component of the angular velocity at the bottom of the swing is given by
f 
in agreement with our earlier result.
d
g
(t f )  
0
dt
2R