Solving Metal Disk Problem: Find T

[NasuSama]

Homework Statement

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.

Homework Equations

$I = mr^{2}/4$
$T = 2\pi √(I/mgd)$

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

 

[Doc Al]

$I = mr^{2}/4$

How did you arrive at this result?

 

[NasuSama]

How did you arrive at this result?

I am thinking that I need to use the moment of inertia of the disk.

 

[Doc Al]

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]

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...

 

[Doc Al]

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!)

 

[NasuSama]

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:

$T = 2\pi √((3mr^{2}/2)/(mgr))$
$T = 2\pi √(3r/(g))$

 

[NasuSama]

Nvm. My answer is right. Thanks for your help by the way!

 

[Doc Al]

Good!