MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
W14D3-3 Rotating Skew Rod Solution
Consider a simple rigid body consisting of two particles of mass m separated by a rod of
length 2l and negligible mass. The midpoint of the rod is attached to a vertical axis that
rotates with angular velocity    k̂ about the z -axis. The rod is skewed from the
vertical at an angle  . Set time t  0 when the rod is in the position shown in figure
below left. At t   /  the rod has rotated to the position shown in the figure below right.
a) Find the direction and magnitude of the angular momentum about the center of
mass at t  0 .
b) Find the direction and magnitude of the torque about the center of mass at time
t  0.
Solution:
a)
We use Lcm  r  p  r  mv for the angular momentum about the center of mass
for each particle:
Lcm  r1  p1  r2  p2  r1  mv1  r2  mv2
Since each particle travels at an angular speed  in a circular orbit of radius cos  , the
speed of each particle is given by v   l cos  . We choose a coordinate system shown in
the figure below
r
r
For particle 1: r1  l cos φ
i  l sin  kφ and v1  l cos φ
j . Thus
r
r
r
φ  (ml cos  φ
L cm,1  r1  mv1  (l cos  φ
i  l sin k)
j) .
After calculating the cross products, we have that the angular momentum about the center
of mass for particle 1 is
r
L cm,1  ml 2 cos  (cos  kφ  sin  φ
i) .
Note that for particle 2, r2  r1 and v 2   v1 , so
Lcm,2  r2  mv 2  r1  mv1  Lcm,1 .
Thus the angular momentum about the center of mass at time t  0 is given by
r
Lcm (0)  2ml 2 cos(cos kφ  sin  φ
i) .
The magnitude of the angular momentum about the center of mass is given by
Lcm  2ml 2 cos(cos2   sin2  )  2ml 2 cos .
b) At t  0 , the rod is rotating in the x  y plane. The figure below shows the orientation
of the rod as seen from above.
The z -component of the angular momentum about the center of mass is constant and the
x -component of the angular momentum about the center of mass is changing in time as
the rod rotates and is given by
r
L cm, x (0)  2ml 2 cos  sin  φ
i
The time derivative of the angular momentum about the center of mass is perpendicular
the angular momentum, points in the positive y -direction, and has a magnitude that is
r
equal to L cm,x (0)  . Therefore
r
r
dL cm
(0)  L cm,x (0)  φ
j  2ml 2 2 cos  sin  φ
j.
dt
The torque about the center of mass is given by
Therefore at t  0 , we that
r
dL cm
r
 cm 
dt
r
 cm  2ml 2 2 cos sin  φ
j.