Proof of disk moment of inertia using area density

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Homework Statement

Disk with radius R

σ = M/A

I = ∫ mr²

Homework Equations

Today we learned how to derive various moments of inertia via density equations (M/L, M/A, M/V). I understand all of them except on how to get MR²/2 for a disk.

The Attempt at a Solution

I = ∫mr²

σ = M/A

dM = σdA

A = πr(dr) <--- I know my problem is here and that it should be 2πr(dr). My question is why is this true? The area of a circle is πr^2 so why would an individual section have an are of 2πr(dr)?

I = σdAr²

I = ∫ σ(2πr)(r²)(dr)

I = σ(2π) ∫ r³ (dr)

∫ r³ (dr) = r⁴/4

Add back constants and substitute R in for r because integration is from 0 to R

I = R⁴/4 (σ) (2π) = R⁴/4 (M/A) (2π)

I = R⁴/4 (M/(πR²)) (2π)

I = MR²/2

 

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Never mind I think I have it now