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Old Jun 23rd 2012, 07:21 AM   #1
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Inertial moment of pendulum watch

Hello guys, how are you?

Well, here is my doubt:

"The pendulum of a watch is formed by a thin rod of 207g and 20cm lenght and by a disc of 398g and 11cm of diameter. When the pendulum rotates around the extremity of the rod, with the plane of the disc perpendicular to the rotate axis, it's inertial moment is 0.016kg.m^2. How far from the rotate axis, is the center of the disc attached to the rod?"

First, I have to say that I found the question a little confusing... but if I understand well, I think it is this what it wants we to find:


As I said, this is only my interpretation. This figure wasn't on the question.

Well, my attempt:

m: mass of the rod
M: mass of the disc

Using the Steiner's Theorem:


Solving this equation I found x=7.13cm.

I'd like to know if I did the correct resolution, because I'm not very confident with the way I did the inertial moment and all the stuff.

Thanks in advance,
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Old Jun 25th 2012, 10:18 AM   #2
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You're on the right track, but I would offer a couple of pointers:

1. The formula that you used for the rod, I = mL^2/12, is valid if the pivot point is at the center if the rod. But for a rod whose pivot point is at one end (like on a pendulum), the moment of inertia is much larger: I = mL^2/12.

2. You need to apply the Parallel Axis theorem to the disc only, as the dimension 'x' is the distance from the pivot point to the center of the disc (the rod has a given length of 20 cm and teg pivot is at one end, so its moment of inertia is not affected by 'x').

FYI, I get an answer this is a over twice the value of yours.
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inertial, moment, pendulum, watch

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