# Derivation of the moment of inertia of a cylinder

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 dont 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!

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Doc Al
Mentor
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
Homework Helper
Gold Member
Dearly Missed
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,
$$I=\int{d}^{2}dm=\rho\int_{0}^{R}\int_{0}^{\pi}\int_{0}^{2\pi}r^{4}\sin^{3}\phi{d\theta}{d\phi}dr=2\pi\rho\int_{0}^{R}\int_{0}^{\pi}r^{4}(\sin\theta-\sin\theta\cos^{2}\theta)d\phi{dr}=2\pi\rho\int_{0}^{R}r^{4}(2-\frac{2}{3})dr=\frac{8\pi\rho}{3}\int_{0}^{R}r^{4}dr=\frac{2}{5}Mr^{2}$$

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