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Moment of Inertia of a cylinder with varying density

  1. Apr 4, 2012 #1
    A cylinder with radius R and mass M has density that increases linearly with distance r from the cylinder axis, ρ = αr, where α is a positive constant.
    Calculate the moment of inertia of the cylinder about a longitudinal axis through its center in terms of M and R.

    I have I = ∫r^2dm
    and ρ = αr = m/V, I solved for m and took the derivative to get dm = α(dV = dr).
    Putting this into the Inertia equation, I get I = α∫r^2 (dV + dr). Using the equation for volume, i took the derivative and got dV = L * 2∏r dr. Plugging this into the original equation, and integrating, I get I = α(Lr^2)/2 + α(r^3)/2 with bounds of R2 and R1. Not sure where to go from here...
     
    Last edited: Apr 4, 2012
  2. jcsd
  3. Apr 4, 2012 #2

    K^2

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    dm=2πr L ρ dr - Because you can break the disk into thin rings distance r from center, each of which has volume 2πr L dr

    M=∫dm

    Try using these.
     
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