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darkar
Jan18-06, 03:54 PM
Here's the question:
Show that the moment of inertia of a spherical shell of radius R and mass m is 2mRČ/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 = 4πRČρdR.

What I did is using I = ∫rČdm=∫4πr⁴ρdr=4πρR⁵/5=3MRČ/5.

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

Thanks ~

DaleSpam
Jan18-06, 05:27 PM
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

darkar
Jan18-06, 06:10 PM
Well, the moment of inertia about the diameter is not 2MRČ/3 but is 2MRČ/5 instead. So i would say it is not about the diameter.

DaleSpam
Jan18-06, 10:24 PM
Read carefully. 2/5 is for a solid sphere and 2/3 is for a spherical shell. Both rotated about the diameter.

-Dale

darkar
Jan19-06, 04:29 AM
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 ~