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Calculating rotational inertia of a sphere

  1. Dec 7, 2004 #1
    Just how do you calculate the rotational inertia of a sphere?
    Assuming the sphere lies at the center of the xyz coordinate system, I divided the sphere into a series of cross-sections of verticle width dz and area pi*y^2. I then multiplied these together and multiplied this by z^2, and multiplied this by density (M/V, or M/(4/3*pi*R^3)), and then tried to integrate with respect to z from -R to R. I wasn't sure whether or not to include z itself in the integration (z=(R^2-Y^2)^(1/2)). I have a feeling I completely messed the entire problem up; however, I'm not sure where. :yuck: Did I go about doing it in an entirely wrong way? Should I use double integrals (would that be easier)? Do I have to use spherical coordinates or something?
    Any help is appreciated. :smile:
  2. jcsd
  3. Dec 7, 2004 #2


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    Your answer lies here

    Hopefully Arildno and Krabs' approach are within your level of understanding...
  4. Dec 8, 2004 #3
    Thank you thank you thank you! That was REALLY helpful...I've been trying to understand the process of calculating inertia for days...my textbook was of no help. I think I finally understand it.
  5. Dec 8, 2004 #4

    Andrew Mason

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    Divide the solid sphere into thin disks of thickness dz and mass dm. For the thin disk is [itex]I = \frac{1}{2}MR^2[/itex]

    The moment of inertia of each disk is
    [tex]dI = \frac{1}{2}x^2dm[/tex] where [tex]dm = \rho \pi x^2 dz[/tex]

    So [tex]dI = \frac{1}{2}\rho \pi x^4 dz[/tex]

    Then integrate dI from z = -R to R (note: [itex]x^2 = R^2 - z^2[/itex])

    That will give you I in terms of [itex]\rho[/itex] which is M/V (where V is the volume of the sphere and M is its mass) so just replace [itex]\rho[/itex] with M/V.

    The integration looks a little tough because of the [itex](R^2 - z^2)^2[/itex] Good luck.

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