MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
W14D3-2 Table Problem Suspended Gyroscope Solution
A gyroscope wheel is at one end of an
axle of length d . The other end of the
axle is suspended from a string of length
s . The wheel is set into motion so that it
executes uniform precession in the
horizontal plane. The string makes a
fixed angle β with the vertical. The
wheel has mass M and moment of
inertia about its center of mass I cm . Its
spin angular speed is ω . Neglect the
mass of the shaft and the mass of the

string. Assume ω >> Ω . What is the
M , ω , β , s , and g as needed. Hint:
Choose the center of mass as the point
for calculating torque and angular
momentum.
direction and magnitude of the
precession angular velocity? Express
your answer in terms of in terms of d ,
Solution: We shall choose as our system the wheel and axle but not the string. The force
diagram and coordinate system are shown in figure below.
The torque on the wheel with respect to the center of mass of the wheel is given by



τ cm = rcm ,T × T ,
In the coordinate system shown in figure below,

T = −T sin β î + T cos β ĵ . So the torque in Eq. (1) becomes
(1)

rcm,T = −d î
and

τ cm = −d î × (−T sin β î + T cos β ĵ) = −dT cos β ( î × ĵ) = −d T cos β k̂ .
(2)
The direction of the torque is into the plane of figure (negative z-direction). This means
that the direction of the change of the angular momentum of the spinning wheel is also
into the plane the figure. Since the angular momentum is pointing outward (in the
positive x-direction at the instant shown in the figure below on the lerft), the change in
the angular momentum points in the negative z-direction as seen in the overhead view in
the figure below on the right.
The wheel will precess counterclockwise about the vertical axis. Denote the wheel’s
precessional angular frequency by Ω = dθ / dt . Note we choose the angle θ to be
increasing in the counterclockwise direction and so Ω > 0 .
The spin angular momentum is directed along the axis of rotation and is equal in
magnitude to the product of the moment of inertia about the center of mass and the spin
angular velocity. At the instant depicted in the figure above, the angular momentum
about the center of mass is

Lcm = I cmω s î + Lcm, z k̂ .
(3)

dL cm
= I cmω s Ω (−k̂) .
dt
(4)


dL cm
τ cm =
,
dt
(5)
The time derivative is then
Because
substitute Eq. (2) into the left hand side of Eq. (5), and substitute Eq. (4) into the right
hand side of Eq. (5) yielding,
(6)
−dT cos β k̂ = I cmω s Ω (− k̂) .
In Eq. (6), we have still not determined the tension of the string. We can use Newton’s
Second Law applied to the center of mass


F = ma cm .
(7)
The weight of the wheel must be balanced by the vertical component of the force
supplied by the string,
T cos β − mg = 0 .
(8)
Therefore the tension is
T=
mg
.
cos β
(9)
We can substitute Eq. (9) into Eq. (6) and solve for the precessional frequency,
Ω=
d mg
.
I cmω s
(10)
Note that we are assuming the precession is uniform. This is our gyroscopic
approximation and requires us to ignore any “nutational” effects as well as requiring that
ω s >> Ω .