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Superposition of moments of Inertia

  1. Jul 11, 2007 #1
    1. The problem statement, all variables and given/known data

    Consider a thin rod of length L which is pivoted at one end. A uniform density spherical object (whose mass is m and radius is r = 1/6L) is attached to the free end of the rod. The moment of inertia of the rod about an end if I = 1/3 mL^2. The moment of inertia of the sphere about its center of mass is I = 2/5 mr^2. Determine the moment of inertia, I, of the rod plus mass sytem with respect to the pivot point.

    2. Relevant equations

    I system = I rod + I sphere + parallel axis contribution


    3. The attempt at a solution

    1/3ML^2 + 2/5 M(1/6L)^2 + M(L+1/6L)^2

    This is what I got, but is not quite right. Can someone please tell me what terms I left out? Thank you!
     
  2. jcsd
  3. Jul 11, 2007 #2

    Dick

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    The center of mass of the sphere is at the end of the rod at distance L. So doesn't that make the parallel axis contribution of the sphere ML^2? Why the L/6 part?
     
  4. Jul 11, 2007 #3
    because I thought that the radius 1/6L has to be added onto the length of the rod...
     
  5. Jul 11, 2007 #4

    Dick

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    I guess it depends on whether you glue the sphere to the end of the rod or drill a hole in the sphere to attach the rod at the center of the sphere. Are there any pictures that might suggest which?
     
  6. Jul 11, 2007 #5
    I think it's just connected end to end.
     
  7. Jul 11, 2007 #6

    Dick

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    End to outside of sphere? Not end to center? Then I think you are right. Either answer could be correct depending on how you connect them.
     
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