Metal disk problem!


by NasuSama
Tags: disk, metal
NasuSama
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#1
Nov29-12, 12:32 PM
P: 326
1. The problem statement, all variables and given/known data

A uniform metal disk (M = 8.21 kg, R = 1.88 m) is free to oscillate as a physical pendulum about an axis through the edge. Find T, the period for small oscillations.

2. Relevant equations

[itex]I = mr^{2}/4[/itex]
[itex]T = 2\pi √(I/mgd)[/itex]

3. The attempt at a solution

I combined the formula together to get:

[itex]T = 2\pi √((mr^{2}/4)/(mgr))[/itex]
[itex]T = 2\pi √(r/(4g))[/itex]

But the answer is incorrect
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Doc Al
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#2
Nov29-12, 01:24 PM
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Quote Quote by NasuSama View Post
[itex]I = mr^{2}/4[/itex]
How did you arrive at this result?
NasuSama
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Nov29-12, 01:56 PM
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Quote Quote by Doc Al View Post
How did you arrive at this result?
I am thinking that I need to use the moment of inertia of the disk.

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Nov29-12, 02:15 PM
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Metal disk problem!


Quote Quote by NasuSama View Post
I am thinking that I need to use the moment of inertia of the disk.
Of course you do, but that's not the correct formula.
NasuSama
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#5
Nov29-12, 06:29 PM
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Quote Quote by Doc Al View Post
Of course you do, but that's not the correct formula.
Then, it's something like I = mrČ/2, rotating to its center. However, the disk oscillates through its edge.

I am not sure which path to go for...
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Nov29-12, 06:31 PM
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Quote Quote by NasuSama View Post
Then, it's something like I = mrČ/2, rotating to its center.
Right.
However, the disk oscillates through its edge.
Use the parallel axis theorem. (Look it up!)
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#7
Nov29-12, 07:01 PM
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Quote Quote by Doc Al View Post
Right.

Use the parallel axis theorem. (Look it up!)
Hm.. By the Parallel Axis Theorem, I would assume that:

[itex]I = I_{center} + md^{2}[/itex]
[itex]I = mr^{2}/2 + mr^{2}[/itex] [Since the disk rotates about an axis through the edge, we must add the inertia by mrČ. r is the distance between the center and the edge of the disk.]
[itex]I = 3mr^{2}/2[/itex]

Is that how I approach this? Let me know where I go wrong. Otherwise, I can just plug and chug this expression:

[itex]T = 2\pi √((3mr^{2}/2)/(mgr))[/itex]
[itex]T = 2\pi √(3r/(g))[/itex]
NasuSama
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#8
Nov29-12, 07:18 PM
P: 326
Nvm. My answer is right. Thanks for your help by the way!
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#9
Nov29-12, 07:33 PM
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Good!


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