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Moment of inertia of a uniform solid sphere

  1. Apr 7, 2006 #1
    This is not a homework question but I figured this was the most appropriate place to post it-

    Taking a uniform solid sphere of radius R and mass M, with the centre of mass at the origin, I divided it into infinitesimal disks of thickness dx, and radius y. I need to find the moment of inertia about the x-axis, so taking an arbitrary disk at some horizontal distance x from the centre of mass, I obtain ;

    y^2 + x^2 = R^2, (fairly obviously),

    density, rho = dm/dV,

    dV = (pi)(y^2)dx => dm = (rho)(pi)(y^2)dx

    So using the standard definition for moment of inertia :

    I = integral of (y^2)dm

    I = integral of (y^2)(rho)(pi)(y^2)dx -with x limits R and -R

    = (rho)(pi) integral of ((R^2 - x^2)^2) dx

    which simplifies down to I = (16/15)(pi)(rho)R^5,

    and using M = (4/3)(pi)R^3, I obtain I = (4/5)MR^2.

    Of course my textbook is telling me it should be (2/5)MR^2, and as far as my understanding goes, this is a consequence of each infinitesimal disk having a moment of inertia of (1/2)dm(r^2).

    Logically then, using dI = (1/2)dm(r^2), such that :

    I = integral of (pi)(rho)((R^2 - x^2)^2)dx with x limits R and 0, the answer comes out correctly as (2/5)MR^2.

    Unfortunately, I am not a particularly sophisticated mathematician and I am worried that my own method, using I = integral of (y^2)dm as described, is giving me an answer which is out by a factor of 2.

    I fear I may have made a trivial mistake, but if not, I'd greatly appreciate some insight as to the cause of the discrepancy.

    Many thanks!

  2. jcsd
  3. Apr 7, 2006 #2


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  4. Apr 7, 2006 #3

    Doc Al

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