Twisting the object in a horizontal plane produces a torsional restoring force (moment) in the rod, M, which is proportional to the stiffness, k, and the angular displacement, theta:
M=k x theta
This moment acts in the direction opposite to the angular displacement, theta.
From Netwon's Second Law of Motion, we know that the moment is proportional to the objects MOI and its angular acceleration, alpha:
-M = I x alpha or -k x theta = I x alpha
Re-arranging the equation, we find:
alpha + (k/I) theta = 0
Note that this equation is in the standard form for simple harmonic motion having a frequency, f:
alpha + f^2 theta = 0, where f = ( k / I )^0.5
The period, t, is a function of the frequency:
t = 2Pi / f = 2Pi / ( k / I )^0.5 = 2Pi ( I / k )^0.5
By placing a calibration bar having a known MOI, I1, into the pendulum and determining the period of the system, t1, we find:
t1 = 2Pi [(I1 + Ip)/k]^0.5
The empty pendulum creates the second set of conditions, and has a period, t2:
t2 = 2Pi ( Ip / k )^0.5
We now have a system of two equations in two unknowns. If we re-arrange the equations to solve for Ip, we find:
Ip = k( t1 / 2Pi )^2 - I1 & Ip = k( t2 / 2Pi )^2
Therefore:
k( t1 / 2Pi )^2 - I1 = k( t2 / 2Pi )^2
We can now solve for the torsional stiffness, k:
k = I1 / [( t1 / 2Pi )^2 - ( t2 / 2Pi )^2]
And, once k is known, we can solve for Ip:
Ip = k( t1 / 2Pi )^2 - I1
If we replace a calibration bar with a gun, determine the average period of the gun and pendulum, t, we can re-arrange the equation, and solve for the gun's MOI, Ig:
Ig = k( t / 2Pi )^2 - Ip