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The moment of inertia of a group of seven pennies

  1. Aug 28, 2012 #1
    1. The problem statement, all variables and given/known data

    See attached photo

    2. Relevant equations



    3. The attempt at a solution

    I figured I would use the parallel axis theorem. I'm stuck between two different methods of doing the question, both of which are choices in the answers.

    My gut instinct says to take the moment of inertia of the middle penny, and then use the parallel axis theorem 6 times and add it to the MoI of the middle one.

    So if I take the moment of inertia of a thin disc, I have the MoI of the middle penny is 1/2 Mr^2.

    Then the MoI of each outer penny would be found by the parallel axis equation I=Icm + md^2.

    In this case, Icm would be 1/2 Mr^2, and Md^2 would be M(2r)^2 = 4Mr^2, making the total parallel axis moment 9/2 Mr^2

    Since there are six outer pennies, 54/2 Mr^2, then add back in the original MoI from the middle penny, 55/2 Mr^2

    Did I do this correctly?
     

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  2. jcsd
  3. Aug 28, 2012 #2

    Doc Al

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    Staff: Mentor

    Looks good to me.
     
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