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Inertia with the parallel axis theorem

  1. Aug 9, 2012 #1
    1. The problem statement, all variables and given/known data
    A uniform solid sphere has a moment of inertia I about an axis tangent to its surface. What is the moment of inertia of this sphere about an axis through its center?


    2. Relevant equations

    Ip = Icm + Md2

    Isolid sphere = 2/5MR^2

    3. The attempt at a solution

    This is what I tried:

    Ip = 2/5MR^2 + MR^2
    =7/5MR^2

    I'm not sure what I'm supposed to do with this. I know that the correct answer is 2/7 but I don't know how you get that answer.
     
  2. jcsd
  3. Aug 9, 2012 #2

    cepheid

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    Gold Member

    In this case you're told that Ip = "I", where "I" is just meant to be some given value. You've already worked out from the parallel axis theorem that if the moment of inertia around the parallel axis is I, then the moment of inertia around the centre is I - Md^2, where d = r here.

    You ended up with I = 7/5MR^2. If 7/5MR^2 = I, then what is Icm = 2/5MR^2 equal to in terms of the given value, I?
     
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