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Moment of inertia about center of solid sphere

  1. Jan 18, 2006 #1
    Here's the question:
    Show that the moment of inertia of a spherical shell of radius R and mass m is 2mR²/3. This can be done by direct integration or, more easily, by finding the increase in the moment of inertia of a solid sphere when its radius changes. To do this, first show that the moment of inertia of a solid sphere of density ρ is I=(8/15)πρR⁵. Then compute the change dI in I for a change dR, and use the fact that the mass of this shell is dm = 4πR²ρdR.

    What I did is using I = ∫r²dm=∫4πr⁴ρdr=4πρR⁵/5=3MR²/5.

    what is wrong with my equation? And please note that I am not asked to ask moment of inertia about the diameter.

    Thanks ~
     
  2. jcsd
  3. Jan 18, 2006 #2

    Dale

    Staff: Mentor

    Just glancing at your equation it looks like you are essentially integrating in spherical coordinates. Usually cylindrical coordinates is easier and more direct.

    Although you are not asked about the moment of inertia about the diameter the formula given in the problem is for the moment of inertia about the diameter. It looks to me like you are trying to take the moment of inertia around the center, which is not possible since you can't rotate about the center.

    -Dale
     
  4. Jan 18, 2006 #3
    Well, the moment of inertia about the diameter is not 2MR²/3 but is 2MR²/5 instead. So i would say it is not about the diameter.
     
    Last edited: Jan 18, 2006
  5. Jan 18, 2006 #4

    Dale

    Staff: Mentor

    Read carefully. 2/5 is for a solid sphere and 2/3 is for a spherical shell. Both rotated about the diameter.

    -Dale
     
  6. Jan 19, 2006 #5
    Oh rite! Thanks for pointing the wrong. I think the question confused because of the second part which asked me to find the solid sphere first.

    Thanks Dale ~
     
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