Tuesday, October 18, 2016

homework and exercises - Moment of inertia of a hollow sphere wrt the centre?




I've been trying to compute the moment of inertia of a uniform hollow sphere (thin walled) wrt the centre, but I'm not quite sure what was wrong with my initial attempt (I've come to the correct answer now with a different method). Ok, here was my first method:


Consider a uniform hollow sphere of radius R and mass M. On the hollow sphere, consider a concentric ring of radius r and thickness dx. The mass of the ring is therefore dm=M4πR22πrdx. Now, use r2=R2x2: dm=M4πR22π(R2x2)1/2dx

and the moment of inertia of a ring wrt the centre is I=MR2, therefore: dI=dmr2=M4πR22π(R2x2)3/2dx
Integrating to get the total moment of inertia: I=RRM4πR22π(R2x2)3/2 dx=3MR2π16


which obviously isn't correct as the real moment of inertia wrt the centre is 2MR23.


What was wrong with this method? Was it how I constructed the element? Any help would be appreciated, thanks very much.



Answer



The mass of the ring is wrong. The ring ends up at an angle, so its total width is not dx but dxsinθ


You made what I believe was a typo when you wrote


dm=M4πR22π(R2x2)dx


because based on what you wrote further down, you intended to write



dm=M4πR22π(R2x2)dx


This problem is much better done in polar coordinates - instead of x, use θ. But the above is the basic reason why you went wrong.


In essence, sinθ=rR so you could write


dm=M4πR22πrsinθ dx=M4πR22πrrR dx=M4πR22πR dx=M2R dx


Now we can substitute this into the integral:


I=RRM2R(R2x2) dx=M2R[2R323R3]=23MR2


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