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Moment of Inertia!

  1. Jan 4, 2005 #1
    I've been working on deriving the moment of inertia of a hollow sphere (ie basketball) with uniform density for a while now with no success... Can anyone show me this derivation?

    The moment of inertia is I = 2/3 MR^2 . If anyone can help me get to this step, it would be greatly appreciated.
  2. jcsd
  3. Jan 4, 2005 #2


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  4. Jan 4, 2005 #3


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    Okay,that's a link to a good calculation for the moment of inertia of a full sphere.
    You need for a hollow sphere.
    [tex] I=\int r^{2} dm [/tex] (1)
    ,where 'r' is the distance between the mass element 'dm' and the axis of rotation.
    [tex] dm=\rho_{surf.sphere}dS [/tex] (2)
    [tex] \rho_{surf.sphere}=\frac{M_{sphere}}{S_{sphere}}=\frac{M}{4\pi R^{2}} [/tex] (3)

    The distance 'r' to the axis of rotation chosen as Oz is
    [tex] r=R\sin \theta [/tex](4)
    The sphere surface element is
    [tex] dS=R^{2}d\Omega=R^{2}\sin\theta d\theta d\phi [/tex](5)

    Then I becomes
    [tex] I=\int_{0}^{2\pi} d\phi\int_{0}^{\pi} d\theta (R^{2}\sin^{2}\theta)(\frac{M}{4\pi R^{2}}) R^{2}\sin\theta [/tex] (6)

    Make simplifications and integrate after [itex] \phi [/itex],simplify again and u'll get
    [tex] I=\frac{MR^{2}}{2}\int_{0}^{\pi} d\theta \sin^{3}\theta =\frac{MR^{2}}{2}(-)\int_{1}^{-1} (1-u^{2}) du =\frac{MR^{2}}{2}(u-\frac{u^{3}}{3})|_{-1}^{+1}=\frac{MR^{2}}{2}\frac{4}{3}=\frac{2MR^{2}}{3}[/tex](7)
    ,where i made use of
    [tex] \sin^{3}\theta=\sin\theta(1-\cos^{2}\theta) [/tex](8)
    and the substitution
    [tex] \cos\theta\rightarrow u [/tex] (9)
    ,under which the limits of integration transform in the prescribed way.


    [tex] I=
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