A 89 kg solid sphere with a 12 cm radius is suspended by a vertical wire. A torque of 0.37 N·m is required to rotate the sphere through an angle of 0.93 rad and then maintain that orientation. What is the period of the oscillations that result when the sphere is then released?

Question

A 89 kg solid sphere with a 12 cm radius is suspended by a vertical wire. A torque of 0.37 N·m is required to rotate the sphere through an angle of 0.93 rad and then maintain that orientation. What is the period of the oscillations that result when the sphere is then released?
Please help i have no idea how to do this problem

drwls · Answer

Think of this as a torsional spring, with a torsional stiffness coefficient
k = (0.37/0.93) = 0.398 N*m/radian

The equation of motion is
I*x" = -kx,

where x is the angular deflection in radians and I is the moment of inertia of the sphere. x" is the second derivative of x, which is the angular acceleration.

The solution to this differential equation leads to a period
P = 2 pi sqrt(I/k)

Look up and compute the moment of inertia of a sphere rotating about its axis. I think you will find that
I = (2/5)M R^2

Solve for period, P. The units will be Hz if I is in kg*m^2