# Metal disk problem!

by NasuSama
Tags: disk, metal
 P: 304 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
Mentor
P: 40,240
 Quote by NasuSama $I = mr^{2}/4$
How did you arrive at this result?
P: 304
 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.

Mentor
P: 40,240

## 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.
P: 304
 Quote by Doc Al 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...
Mentor
P: 40,240
 Quote by NasuSama 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!)
P: 304
 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 mrČ. 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:

$T = 2\pi √((3mr^{2}/2)/(mgr))$
$T = 2\pi √(3r/(g))$
 P: 304 Nvm. My answer is right. Thanks for your help by the way!
 Mentor P: 40,240 Good!

 Related Discussions General Physics 3 General Discussion 3 Introductory Physics Homework 2 Computing & Technology 2