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Moment of Inertia of a cylinder

  1. May 1, 2009 #1
    When I am reading a book, I find it is listed that the moment of Inertia of a cylinder is
    MR^2/4 + Ml^2/12

    It is a cylinder with rotation axis passing through the curve surface and its centre of mass. And its density is constant. With the circile surface raius = R and height = l

    Can anybody show me the procedure of the integration? I have tried several times but fail. I just cannot get that answer. It is not homework but I am interested in the process. You know, usually, moment of Inertia is provided in the book, integration is not required. But I am really curious about it. Could anybody please help me?
     
  2. jcsd
  3. May 1, 2009 #2

    tiny-tim

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    Hi loup! :smile:

    Slice the cylinder into discs.

    Use the moment of inertia formula for a disc about its diameter, combined with the parallel axis theorem, and integrate. :wink:
     
  4. May 2, 2009 #3
    The problem is I don't know about how to integrate a disc.
     
  5. May 2, 2009 #4
    And I think the integration of disc actually comes from cylinder. I expected once I finished this cylinder I could do the disc.
     
  6. May 2, 2009 #5

    tiny-tim

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    Slice it into strips parallel to a diameter, and integrate …

    what do you get? :smile:
     
  7. May 2, 2009 #6
    The r requires a cosine and there are more than one variable, what I should do?
     
  8. May 2, 2009 #7
    I cannot use parallel axis theorem. I think it is too tricky.
     
  9. May 2, 2009 #8

    tiny-tim

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    uhh? what equation are you using? :confused:
     
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