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Proof of disk moment of inertia using area density

  1. Nov 7, 2014 #1
    1. The problem statement, all variables and given/known data

    Disk with radius R

    σ = M/A

    I = ∫ mr2

    2. Relevant 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.

    3. 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
     
  2. jcsd
  3. Nov 7, 2014 #2
    Never mind I think I have it now
     
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