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Moment of Inertia

  1. Dec 30, 2009 #1
    1. Derive the moment of inertia of a sphere with thick walls.

    2. Relevant equations
    I=(.5)(M)(R^2+R^2) ?
    one is for the inner radius the other is for the outer radius.

    3. The attempt at a solution
    I believe this can be found by integrating a series of infinitely thin cylinders with different size holes, but I am only have a high school physics knowledge.

    Answer: I=(2/5)(R^5-R^5)(R^3-R^3)
    Is there an name for this equation?
    Is there any website detailing the steps to deriving the moment of inertia of this object without Calculus III? with Calculus III?

    Thanks for any help.
  2. jcsd
  3. Dec 30, 2009 #2


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    There is a very simple way to calculate this. Subtract the moment of inertia of a sphere made of the same material that has the radius of the empty inside from the moment of inertia of a solid sphere with larger radius.
  4. Dec 31, 2009 #3
    Thank You for the reply.

    I have tried this method before, but I am not sure how they end up with I=(2/5)(R^5-R^5)/(R^3-R^3).

    I am just using (2/5)(M)(R^2)-(2/5)(M)(R^2)
    (the two radii are different)
    Last edited: Jan 1, 2010
  5. Dec 31, 2009 #4


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    Well, for one thing, [itex]\tfrac{2}{5}(R^5 - R^5)(R^3 - R^3)[/itex] works out to zero. You really need to use different letters for the different radii.

    You seem to be on the right track with something like
    [tex]\frac{2}{5}(M)(R_1^2) - \frac{2}{5}(M)R_2^2[/tex]
    but one correction: the smaller sphere (the one that would fit in the hole in the middle) will not have the same mass as the larger sphere.
  6. Dec 31, 2009 #5
    solved. thank you guys so much.
    Last edited: Jan 1, 2010
  7. Dec 31, 2009 #6


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  8. Jan 1, 2010 #7
    solved. thank you guys so much.
    Last edited: Jan 1, 2010
  9. Jan 1, 2010 #8


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    The M's in the two equations don't represent the same quantity. The M in the first equation is the mass of a solid sphere of radius R whereas the M in the second equation is the mass of the hollow sphere. Fix the mass in your first equation, and you'll get the result you want.
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