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Moment of inertia of a sphere

  1. Jan 12, 2006 #1
    I am having trouble with the following question.

    Q. Derive the expression for the moment of inertia of a spherical ball of mass m and radius r about the axis z through its centre of gravity.

    Ans: [itex]I_G = \frac{2}{5}mr^2 [/itex].

    I did the following.

    [tex]I_G = \int\limits_m^{} {r^2 } dm[/tex]

    [tex]dm = \rho dV[/tex] where rho is the constant density.

    V = \frac{4}{3}\pi r^3 \Rightarrow dV = 4\pi r^2 dr

    I_G = 4\pi \rho \int\limits_0^r {r^4 dr} = \frac{{4\pi r^5 \rho }}{5} = \frac{3}{5}\left( {\frac{4}{3}\pi r^3 \rho } \right)r^2 = \frac{3}{5}mr^2 \ne \frac{2}{5}mr^2

    I can get the answer using a triple integral but the question seems to only require a single integral. At the moment I can't see where I've gone wrong. Can someone help me out? Thanks.
  2. jcsd
  3. Jan 12, 2006 #2


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    To calculate the moment of inertia you need to integrate over the square of the distance from the axis of rotation which is [itex]r^2 \sin^2 \theta[/itex] and not just [itex]r^2[/itex]. This also requires you to integrate over the polar angle.
  4. Jan 12, 2006 #3
    Oh ok. I just looked up the definition on mathworld and it appears to say that a perpendicular distance needs to be considered. Hopefully these will become more easy to do as I get more practice. Thanks for the help.
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