MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
W14D2-1 Gyroscope on Rotating Platform Solution
A gyroscope consists of an axle of
negligible mass and a disk of mass M and
radius R mounted on a platform that rotates
with angular speed ! as shown in the figure
below. The gyroscope is spinning with
angular speed ! . Forces Fa and Fb act on
the gyroscopic mounts. The goal of this
problem is to find the magnitudes of the
forces Fa and Fb . You may assume that the
moment of inertia of the gyroscope about an
axis passing through the center of mass
normal to the plane of the disk is given by
In .
a) Calculate the torque about the center of mass of the gyroscope.
b) Calculate the angular momentum about the center of mass of the gyroscope.
c) Use Newton’s Second Law find a relationship between Fa and Fb , the mass M of
the gyroscope, and the gravitational constant g .
d) Use the torque equation and Newton’s Second Law to find expressions for Fa and
Fb .
Solution Figure 1 shows a choice of coordinate system and force diagram on the gyroscope.
Figure 1
The vertical forces sum to zero since there is no vertical motion
Fa + Fb ! Mg = 0
(1.1)
Using the coordinate system depicted in the Figure 1 above, torque about the center of mass is
!
! cm = d(Fa " Fb )#̂
(1.2)
The only part of the angular momentum about the center of mass that is changing in time is the
horizontal component. Looking down on the gyroscope from above (Figure 2), the horizontal
component of the angular momentum about the center of mass is rotating counterclockwise
(positive z-direction).
Figure 2
The magnitude of the horizontal component of the angular momentum about the center of mass
!
is L cm, H = I cm! . Denote the angular velocity of the rotating platform by
! d"
!=
k̂ # !k̂
dt
Therefore the derivative of the angular momentum about the center of mass is
!
dL cm, H
dt
We can now apply the torque law
!
d! ˆ
= L cm, H
! = I cm" # !ˆ
dt
(1.3)
!
dL cm, H
!
! cm =
.
dt
(1.4)
Substitute Eqs. (1.2) and (1.3) into Eq. (1.4) and just taking the component of the resulting vector
equation yields
d(Fa ! Fb ) = I cm" #
(1.5)
We can divide Eq. (1.5) by the quantity d yielding
Fa ! Fb =
I cm" #
d
(1.6)
We can now use Eqs. (1.6) and (1.1) to solve for the forces Fa and Fb ,
Fa =
I !" &
1#
Mg + cm
%
2$
d ('
(1.7)
Fb =
I cm" # '
1$
Mg
!
2 &%
d )(
(1.8)
Note that if ! = Mgd / I cm" then Fb = 0 and one could remove the right hand support in Figure
1. This condition was satisfied by the simple pivoted gyroscope that we already analyzed in
W14D1. The forces we just found are the forces that the mounts must exert on the gyroscope in
order to cause it to move in the desired direction. It is important to understand that the gyroscope
is exerting equal and opposite forces on the mounts , i.e. the structure that is holding it. This is a
manifestation of Newton’s Third Law