Moment of Inertia of Cavendish pendulum

[Pioneer1]

Hi,

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 a pendulum where the weights are suspended as in the case of the Cavendish pendulum? I couldn't find it online.

With this formula I got

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

Does this sound right for a pendulum of this dimensions?

Thanks for your help.

 

[jtbell]

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

For the benefit of those of us who haven't been following your previous discussions closely, can you provide a diagram of the specific geometry that you have in mind, so we can see how it differs from Example 1 in the link you provided (which is the case where $I = 2mr^2$)?

 

[Pioneer1]

For the benefit of those of us who haven't been following your previous discussions closely, can you provide a diagram of the specific geometry that you have in mind, so we can see how it differs from Example 1 in the link you provided (which is the case where $I = 2mr^2$)?

Thanks for your reply. Here's the drawing from Cavendish's original paper:

http://www.alphysics.com/cavendishexperiment/CavendishSchematic11.jpg

And here's a detail

http://www.alphysics.com/cavendishexperiment/CavendishSchematic111.jpg

Cavendish also gave the weight of the bar and the silver wire he used to strengthen it. I assume those will have to be taken into consideration as well.

 

[jtbell]

You've still got two masses revolving around a common central point, so their moment of inertia I is the same as when they're connected by a rod of negligible mass in a dumbbell configuration as in the page you first referenced. So you're OK so far.

What you need now is I for the rotating part of the suspension mechanism. You have the mass of the transverse rod, and I for a rod rotating transversely around its midpoint is a standard case. For some reason it's not on the page you first referenced, but you can find it on Hyperphysics:

http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#cmi

Just add the two I's together. In principle, you should also consider I for the rest of the rotating part of the apparatus, but the supporting wires are probably pretty light so you can safely neglect them, at least as a start in your analysis.

 

[Pioneer1]

Just add the two I's together. In principle, you should also consider I for the rest of the rotating part of the apparatus, but the supporting wires are probably pretty light so you can safely neglect them, at least as a start in your analysis.

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.