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Moment of Inertia of a Sphere!

  1. Nov 3, 2008 #1
    I know how to take out the moment of inertia of a sphere about an axis passing through a diameter. My method is the same old one of choosing an elemental sheet and integrating. Can someone please explain to me the things I saw on Mathworld please. The method looks fascinating to me and I wish to learn it and use it for other 3 bodies.
    The original link is:
  2. jcsd
  3. Nov 3, 2008 #2


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    Hi ritwik! :smile:

    Equations (15) and (16) are using your elemental sheet method.

    Equation (17) is using the same method, but with an elemental sphere instead of a flat sheet.

    Equation (18) is just a way of writing the moment of inertia for all possible diameters instead of just one … but the method of getting them is the same. :smile:

    (moment of inertia of a rigid body is a 3x3 tensor, so if you write it in coordinates, it looks like a 3x3 matrix)
  4. Nov 4, 2008 #3
    Could you please derive what those guys had written. I now possess an overwhelming desire to see that. Three integrals at a time seem so fascinating.


    Apart from this here are one of my wildest thoughts:

    What if we take the small element of sphere to be literally reduced to a point! The volume of this would be dy*dx*dz


    Coordinates of the sphere= R cos a cos b [tex]\hat{i}[/tex] + R cos a sin b [tex]\hat{j}[/tex] + R sin a [tex]\hat{k}[/tex]

    where a is the angle made by the position vector with its projection in xy plane. b is the angle made by this projection with the x axis.

    Is it possible to integrate thus!!! If yes, please do tell me how.

  5. Nov 4, 2008 #4


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    Hi Ritwik! :smile:

    If you use three integrals in the exam when you could simplify it to two or one, you will lose marks.

    The whole point of dividing a volume into slices (what you call "elemental sheets") is to make the integration easier … you choose a slice that you already know the area of, and that saves you two integrations, and leaves you with only one.
    Sorry, but that's hopeless … you have to slice the whole volume in a genuine way. :frown:
  6. Nov 5, 2008 #5
    Oh, thats not the way here. We are given ample time in an exam- specially the Sciences and Maths. And my teacher shall appreciate it if I do the question in numerous ways.

    By the way it doesnt matter. Could you please explain it to me? Please! I shall be very grateful.

    Oh! That was just a thought! I already had the inkling that it might be wrong!
  7. Nov 5, 2008 #6


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    ok … in (15), the sphere is divided into vertical slices of thickness dx, from x = -R to R,

    then each vertical slice is divided into vertical strips of thickness dy, from y = -√(R2 - x2) to √(R2 - x2),

    and each vertical strip is divided into horizontal blocks of thickness dz, from z = -√(R2 - x2 - y2) to √(R2 - x2 - y2).

    Then you add the volumes of all the blocks, which is the same as the integral in (15).

    (16) is very simi;ar and I expect you can work this out for yourself.

    (17) divides the sphere into spherical shells of thickness dr, from r = 0 to R,

    then divides each shell into vertical lunes of angular thickness dtheta, from theta = 0 to π,

    then divides each lune into blocks of thicknes r sinphi, from phi = 0 to 2π
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