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Center of mass of physical pendulum |
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| Nov13-08, 09:22 PM | #1 |
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Center of mass of physical pendulum
1. The problem statement, all variables and given/known data
A pendulum consists of a uniform disk with radius r=0.100m and mass 0.500 kg attached to the end of a uniform rod with length L=0.500 m and m 0.250 kg. It pivots at the other end of the rod. a) Calculate the rotational inertia of the pendulum about the pivot point. b) What is the distance between the pivot point and the center of mass of the pendulum? c)Calculate the period of oscillation 2. Relevant equations I=Icom+mh^2 T=2pi(sqrt(I/(mgh))) 3. The attempt at a solution I got the first part using the parallel axis theorum for both the disk and the rod and adding them together. I=0.205 kg*m^2 (checked, correct). I can't figure out how to get the distance between the pivot point without having the period. I tried setting I=mL^2 and solving for L, but this was incorrect. The correct answer is 0.477 m. I can get part c) once I have some clues to b. I'm completely lost. Please help! |
| Nov14-08, 09:42 AM | #2 |
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So I take it no one else can figure this one out either?
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| Nov14-08, 10:10 AM | #3 |
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| center of mass, physical pendulum |
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