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MoI of a Sphere using Spherical Coordinates

  1. Jul 31, 2013 #1

    Ackbach

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    Gold Member

    1. The problem statement, all variables and given/known data

    Calculate the moment of inertia of a uniformly distributed sphere about an axis through its center.

    2. Relevant equations

    I know that
    $$I= \frac{2}{5} M R^{2},$$
    where ##M## is the mass and ##R## is the radius of the sphere. However, for some reason,
    when I do this integration in spherical coordinates, I do not get this result.

    3. The attempt at a solution

    The density is
    $$ \rho= \frac{M}{V}= \frac{3M}{4 \pi R^{3}}.$$
    Then the moment of inertia is
    $$I= \int r^{2} \,dm= \rho \iiint_{V}r^{2} \,dV
    = \rho \int_{0}^{2 \pi} \int_{0}^{ \pi} \int_{0}^{R} r^{4} \sin(\theta) \, dr \, d\theta \, d\phi
    =4 \pi \rho \frac{R^{5}}{5}
    $$
    $$=4 \pi \frac{3M}{4 \pi R^{3}} \frac{R^{5}}{5}= \frac{3}{5} MR^{2}.$$
    Where is my error?

    I know, I know: all the derivations of this result I can find split the sphere up into
    disks. I understand those derivations: I want to know where this one is wrong.
     
  2. jcsd
  3. Jul 31, 2013 #2

    WannabeNewton

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    Science Advisor

    The ##r## in the moment of inertia integral is the perpendicular distance from the axis of rotation to a given mass element ##dm##. The ##r## in spherical coordinates is the distance to ##dm## from the center of the 2-sphere. The two only agree for mass elements that lie on the equatorial plane.
     
  4. Jul 31, 2013 #3

    WannabeNewton

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    Science Advisor

    If you really want to use spherical coordinates, then note that the perpendicular distance from the axis of rotation to a given mass element will be given by ##r_{\perp } = r\sin\theta## hence the moment of inertial integral becomes ##I = \rho \int _{S^{2}}r^{2}\sin^{2}\theta dV = \frac{3}{10}MR^{2} \int_{0}^{\pi}\sin^{3}\theta d\theta = \frac{3}{10}\frac{4}{3}MR^{2} = \frac{2}{5}MR^{2}##.
     
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