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Moment of inertia of a hollow cylinder derivation

  1. Dec 9, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that a hollow cylinder of radius R_1, outer radius R_2, and mass M, is I=1/2M(R_1^2+R_2^2) if the rotation axis is through the center along the axis of symmetry.

    2. Relevant equations

    $$dm = \rho dV$$
    $$dV = (2 \pi R)(dR)(h)$$

    3. The attempt at a solution

    I was mainly confused about why dV is expressed as (2piR)(dR)(h) since the Volume of a cylinder is 2pir^2h. I know that the variable of integration is R so there has to be a dR in there somewhere, but I'm having trouble understanding the rationale.
     
  2. jcsd
  3. Dec 9, 2016 #2

    PeroK

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    ##dV## is not a cylinder. It is a small piece of a cylinder. Can you see what shape it is?
     
  4. Dec 9, 2016 #3
    It's a thin ring, sorry.
     
  5. Dec 9, 2016 #4

    PeroK

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    It has length ##h##, so it is effectively a thin hollow cylinder. Using this ##dV## would be one way to show that the volume of a cylinder is ##\pi R^2 h##. You could also use this to get the volume of your hollow cylinder.

    You could try that as a preliminary exercise before you do the MoI calculation. I find that can be a useful test that you've set up your integration properly, as you know the answer in advance.
     
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