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Moment of inertia.

  1. Apr 24, 2005 #1
    This is not homework, but since it is related I thought it should go here. I have not yet had calculus III class, but I would like for someone to show me how the moment of inertia is derived for a solid sphere. From what I have heard from my physics professor, it is easiest to use sphereical coordinates to find it. Am I right to assume sphereical coordinates are something similar to polar coordinates in the third dimension?
  2. jcsd
  3. Apr 24, 2005 #2


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    Spherical coordinates involve one linear coordinate, the distance from the origin, and two angular coordinates. The angle theta is the same angle as used in cylindrical coordinates. The angle phi is the angle between the positve z-axis and the position in space.

    It may be easiest in spherical coordinates, but I'm not so sure. Because of the symmetry and because the distance involved in the integral is the distance from an axis, I think you will find polar coordinates to work just fine. In fact, since all mass at a given distance from the axis forms a cylindrical shell, all you need to do is figure out the moment of inertia for such a shell and treat the problem as concentric shells. Of course the height of each shell depends on the radius, so you still need to do an integral over r. You could instead set it up to integrate over z by treating the sphere as a many disks of varying radii. You then need to know the moment of inertia of a disk. Any way you look at it, you are doing a double integral, but the concentric shell approach involves one integral that is essentially done for you by virtue of the constant radius.

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