Moment of inertia of the Cavendish pendulum
Published by admin August 18th, 2007 in Cavendish ExperimentI 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?
With this formula I got
I = 2 ( 729.8 * 93.09^2) = 12,648,797.7 g cm^2
Does this sound right?
Physics Forum discussion on this topic.
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I don’t know anything about the “Cavendish pendulum”, but when I looked it up on wikipedia it sure seemed that it was the same sort of torsion pendulum where the weights are attached to the bar.
The result of the calculation are correct, but a lot of physics or engineering professors will mark you wrong for giving it to too many decimal places of accuracy. It doesn’t bother the mathematicians because to them, “729.8″ means the same thing as “729.8000000000″.
Thanks. Looking at it again, I see that the result should be a little less. Cavendish gave r as 36.65 inches. That’s 93.09 cm. I rounded it to 93.1 but since it is squared it makes a difference.
According to a comment at Physics Forum, the suspending lead weights are equivalent to the dumbell configuration. So the calculation may be in the right track.