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Proving the moment of inerta of a sphere: can someone help me please?

  1. Nov 10, 2005 #1
    I'm having a difficulty in proving the moment of inertia of a sphere.
    Using I=p(x^2)dv, we have to show that

    I = (3m/r^3) * the integral (from 0 to r) of (x^3)((r^2) - (x^2))^(1/2) dx

    where r=radius, x=the axis

    I've been trying to prove it and yet no success. if anyone can show me how to do it, i'd greatly appreciate that.
    thanks a lot
     
  2. jcsd
  3. Nov 12, 2005 #2

    HallsofIvy

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    It would seem more reasonable to me to use spherical coordinates:
    [tex]x= \rhocos(\theta)sin(\phi)[/tex] and [tex]dV= \rho^2 sin(\phi)d\rho\d\thetad\phi[/tex]
    [tex]\int\rho x^2dV= p\int_{\phi=0}^\pi\int_{\theta= 0}^{2\pi}\int_{\rho= 0}^R(\rho^3cos^2(\theta)sin^3(\phi)d\rhod\thetad\phi[/tex]

    (oops! Thanks, Fermat.)
     
    Last edited: Nov 12, 2005
  4. Nov 12, 2005 #3

    Fermat

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    Ah, now I can see what it says :smile:
     
    Last edited: Nov 12, 2005
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