# Derivation of the moment of inertia of a cylinder

1. Nov 10, 2009

### 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 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!

2. Nov 11, 2009

### Staff: Mentor

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

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

3. Nov 11, 2009

### 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,
$$I=\int{d}^{2}dm=\rho\int_{0}^{R}\int_{0}^{\pi}\int_{0}^{2\pi}r^{4}\sin^{3}\phi{d\theta}{d\phi}dr=2\pi\rho\int_{0}^{R}\int_{0}^{\pi}r^{4}(\sin\theta-\sin\theta\cos^{2}\theta)d\phi{dr}=2\pi\rho\int_{0}^{R}r^{4}(2-\frac{2}{3})dr=\frac{8\pi\rho}{3}\int_{0}^{R}r^{4}dr=\frac{2}{5}Mr^{2}$$

Last edited: Nov 11, 2009