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Determine period for off center disk

  1. Feb 17, 2010 #1
    1. The problem statement, all variables and given/known data

    A disk of mass m and radius a is suspended horizontally by a rod bored through the disk at a location d distance from the center. When in equilibrium the center of the disk is directly below the rod. Obtain an approximate expression for the period, ignore disapative effects

    2. Relevant equations

    per newtonian

    I = 1/2 mr^2 + md^2
    I*theta: = -mg(rsin(theta))


    3. The attempt at a solution
    1/2 ma^2 +md^2

    Im missing something simple here....If someone could point me in the right direction it would be appreciated
     
  2. jcsd
  3. Feb 17, 2010 #2
    Wait, can I use alpha= to T/I , then relate that to 2pi to find period?
     
  4. Feb 18, 2010 #3
    If I did this right I get ;

    pi((a^2)+d^2)/g*d

    If someone would confirm that would be great
     
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