Derivation of the moment of inertia of a cylinder

[SHawking]

First, this is not a homework assignment even though it may seem like a homework type question.

I have no problem with the derivation of the moment of inertia of a cylinder but am having more trouble with a sphere. I completely understand the often referenced disk method, but, I would like to take a shell method approach. THe following work is wrong conceptually, I think mathematically it is fine, and I hope someone can point me in the right direction. I don't see what one would need to do it using discs, and am assuming that some slight variation on the following should work:
I=$$\int$$r²dm
p=dm/dv
dvp=dm
I=$$\int$$r²pdv
v=4/3$$\pi$$r³
dv=4$$\pi$$r²dr
I=$$\int$$r²p4$$\pi$$r^2dr
I=4$$\pi$$p$$\int$$r^4dr (Evluated between 0 and R)
I=4$$\pi$$pR⁵/5
p=m/v=m/(4/3 $$\pi$$ r³)
Plugging that in and simplifying I get
I=3/5mr², though I need 2/5mr².
I think the problem lies in my set up, conceptually, and that I need to set up an equation somewhere subtracting two values (R from r?) though I am not sure.

If you could help me out I would really appreciate it.


And again, I understand and have no problem with the disk method for this, I am just trying to figure out what this method is not working.


Thanks!

 

[Doc Al]

The problem is that you are mixing up two meanings of 'r'.

I=$$\int$$r²dm

Here 'r' refers to the distance from the axis of rotation.

v=4/3$$\pi$$r³

Here 'r' refers to the radius of the sphere. Big difference!

That's why using a 'shell method' is not such a good idea. What's the moment of inertia of a shell? The various parts are all at different distances from the axis, so it's not a trivial integration.

 

[arildno]

For the record, here is one way how you should proceed to determine the moment of inertia of a ballof radius R:

Volume element:
$$dv=r^{2}\sin\phi{d\phi}{d\theta}dr$$
Distance d to axis:
$$d=r\sin\phi$$
Thus,
$$I=\int{d}^{2}dm=\rho\int_{0}^{R}\int_{0}^{\pi}\int_{0}^{2\pi}r^{4}\sin^{3}\phi{d\theta}{d\phi}dr=2\pi\rho\int_{0}^{R}\int_{0}^{\pi}r^{4}(\sin\theta-\sin\theta\cos^{2}\theta)d\phi{dr}=2\pi\rho\int_{0}^{R}r^{4}(2-\frac{2}{3})dr=\frac{8\pi\rho}{3}\int_{0}^{R}r^{4}dr=\frac{2}{5}Mr^{2}$$