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Moment of inertia

  1. May 2, 2005 #1
    moment of inertia for a ball shell

    Is ther anyone that can explain how to calculate moment of inertia for a ball shell?? I know that the answer is 2/3mR^2, but how can it be showen with integral??
     
    Last edited: May 2, 2005
  2. jcsd
  3. May 2, 2005 #2

    Galileo

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    Draw a picture and use spherical coordinates. Set the z-axis as the axis of rotation.

    [tex]I=\int_S r^2 dm = \int_S (R\sin \theta)^2 \sigma da[/tex]

    where R is the radius of the sphere and [itex]\sigma[/itex] the surface mass density.

    Note that [itex]da = R^2\sin \theta d\theta d\phi[/itex]:

    [tex]I= \sigma R^4 \int_0^{2\pi}\int_0^\pi \sin^3 \theta d\theta d\phi=\frac{8}{3}\pi \sigma R^4=\frac{2}{3}MR^2[/tex]
     
    Last edited: May 2, 2005
  4. May 2, 2005 #3
    What do you mean by [itex]dV = R^2\sin \theta d\theta d\phi[/itex] ??
     
  5. May 2, 2005 #4
    It's an element of volume in spherical coordinates.
     
  6. May 2, 2005 #5

    Galileo

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    Yeah, although it should be AREA element: [itex]da[/itex], I'll change it now...
     
  7. May 2, 2005 #6
    I don't understand this calculation with [itex]da = R^2\sin \theta d\theta d\phi[/itex]
    ??
    Do you have a picture to show this??
     
  8. May 2, 2005 #7

    SpaceTiger

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    Here's one I found on a google search. The area element is just the top face of the shaded box.
     
  9. May 2, 2005 #8

    Galileo

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    http://mathworld.wolfram.com/SphericalCoordinates.html

    It is shown at eqn 14. Note that the notation for the angle of declination (zenith) and the azimuth are interchanged w.r.t. mine ([itex]\phi \leftrightarrow \theta[/itex])

    EDIT: Ah! Spacetiger has exactly the kind of picture I was looking for.
     
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