## Metal disk problem!

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

$I = mr^{2}/4$
$T = 2\pi √(I/mgd)$

3. The attempt at a solution

I combined the formula together to get:

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

But the answer is incorrect
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 Quote by NasuSama $I = mr^{2}/4$
How did you arrive at this result?

 Quote by Doc Al 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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## Metal disk problem!

 Quote by NasuSama 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.

 Quote by Doc Al Of course you do, but that's not the correct formula.
Then, it's something like I = mr²/2, rotating to its center. However, the disk oscillates through its edge.

I am not sure which path to go for...

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 Quote by NasuSama Then, it's something like I = mr²/2, rotating to its center.
Right.
 However, the disk oscillates through its edge.
Use the parallel axis theorem. (Look it up!)

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

$I = I_{center} + md^{2}$
$I = mr^{2}/2 + mr^{2}$ [Since the disk rotates about an axis through the edge, we must add the inertia by mr². r is the distance between the center and the edge of the disk.]
$I = 3mr^{2}/2$

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

$T = 2\pi √((3mr^{2}/2)/(mgr))$
$T = 2\pi √(3r/(g))$
 Nvm. My answer is right. Thanks for your help by the way!
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