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Moment of Inertia for a Solid Sphere

  1. Nov 12, 2003 #1
    So I am having an issue getting the correct moment here. Although this isn't homework, I thought this would be the best forum for the issue.
    So... I=r^2dm Assuming a const mass density: I=p r^2 dV, (p=mass/volume). For a sphere the natural choice of
    dV = r^2 dr d(omega). So this leads to:
    I=4(pi) p r^4 dr . Lets say that the sphere has a radius b.
    Therefore: I=4(pi) p b^5/5, with p=m/(4/3 pi b^3)
    This gives me a final answer of I=3/5 m b^2, which is according to ever text I have looked at (Serway, Halliday/Reznik) incorrect.
    The supposed correct answer would be: I=2/5 m b^2
    I should be able to do this but I seem to be unable to find the assumption that is leading to my error. Something is tickling the back of my mind from my undergrad years telling me there is a subtle point I am missing.
    Any help would be greatly appreciated.
  2. jcsd
  3. Nov 12, 2003 #2

    Doc Al

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    Staff: Mentor

    The r means different things in each of your formula; don't mix them up. In I=r^2dm, r means distance from the axis of rotation; in dV = r^2 dr d(omega), r means distance from the center.
  4. Nov 12, 2003 #3

    Yes of course. So now this is one ugly looking integral. Thanks a lot for the help.
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