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Moment of Inertia for a Thick Spherical Shell

  1. Mar 13, 2016 #1
    1. The problem statement, all variables and given/known data
    A)
    Consider a hollow sphere of uniform density with an outer radius [itex]R[/itex] and inner radius [itex]\alpha R[/itex], where [itex]0\leq\alpha\leq1[/itex]. Calculate its moment of inertia.
    B) Take the limit as [itex]\lim_{\alpha\to1}[/itex] to determine the moment of inertia of a thin spherical shell.
    2. Relevant equations
    Moment of Inertia: [itex]I = \int r^2 dm[/itex]
    3. The attempt at a solution
    [itex] dm = \rho dV[/itex]. Where rho is density. The volume element for a sphere is [tex] dV=r^2sin\theta d\theta d\varphi dr[/tex]
    So I think I would integrate over a sphere but instead from inner radius to the outer radius? [tex] I = \rho \int_{\alpha R}^{R} r^4 dr \int_{0}^{\pi} sin\theta d\theta \int_{0}^{2\pi} d\varphi [/tex]
    Which yields [tex]\frac{4\pi}{5} \rho (R^5 - \alpha R^5) [/tex]
    If [tex] \rho = \frac{m}{V} = \frac{m}{\frac{4\pi (R^3 - \alpha R^3)}{3}} [/tex]
    Then the equation for moment of inertia becomes
    [tex] I = \frac{3}{5}mR^2 \frac{1-\alpha^5}{1-\alpha^3} [/tex]

    The problems is now when I take the limit as alpha approaches 1, and apply l'Hopital's rule, I get that moment of inertia is [itex] mR^2[/itex], when there should be a factor of [itex]\frac{2}{3}[/itex]?
     
  2. jcsd
  3. Mar 13, 2016 #2

    BvU

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    In your I calculation, the r in your relevant equation is not the distance from the origin, but the distance from the axis of rotation ...
     
  4. Mar 13, 2016 #3
    Thanks BvU, that was a stupid mistake on my part. I replaced r with rsin(theta) in spherical coordinates and I got the correct answer.
     
  5. Mar 13, 2016 #4

    BvU

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    well done!
     
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