I am trying to compute the moment of inertia of the Cavendish pendulum. I used

I = 2(m r^2)

r = gyration arm
m = the weight attached to the pendulum

But this formula is for a dumbell type of torsion pendulum where the weights are attached to the bar.

Does anyone know the formula for the case where the weights are suspended as in the case of the Cavendish pendulum?

With this formula I got

I = 2 ( 729.8 * 93.1^2) = 12,651,243.56 g cm^2

Does this sound right?

Posted this to Physics Forum:

Ok. Thanks. From hyperphysics I got 1/12 M L^2 and added it to the previous result:

I = 2 m r^2 + 1/12 M L^2

L = length of the rod
r = half length of the rod

So,

Total moment of inertia for the Cavendish experiment = r^2 (2m + 1/3M) = 13,137,851.84 cm2 g

m = weight of each ball = 729.8 g

M = Rod weight = 155.5 + 11 + 2.9 = 169.40 g (including deal rod, silver supporting wire, vernier attached to the arm)

L = Rod length = 73.3 inches = 186.18 cm

r = Half rod length = 93.09 cm

I also added another picture with a little more detail about the arm. (Figure 3 here.)

Does this look good?

Thanks again for your help.