# Moment of inertia of a hollow cylinder derivation

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1. Dec 9, 2016

### Carpetfizz

1. The problem statement, all variables and given/known data
Show that a hollow cylinder of radius R_1, outer radius R_2, and mass M, is I=1/2M(R_1^2+R_2^2) if the rotation axis is through the center along the axis of symmetry.

2. Relevant equations

$$dm = \rho dV$$
$$dV = (2 \pi R)(dR)(h)$$

3. The attempt at a solution

I was mainly confused about why dV is expressed as (2piR)(dR)(h) since the Volume of a cylinder is 2pir^2h. I know that the variable of integration is R so there has to be a dR in there somewhere, but I'm having trouble understanding the rationale.

2. Dec 9, 2016

### PeroK

$dV$ is not a cylinder. It is a small piece of a cylinder. Can you see what shape it is?

3. Dec 9, 2016

### Carpetfizz

It's a thin ring, sorry.

4. Dec 9, 2016

### PeroK

It has length $h$, so it is effectively a thin hollow cylinder. Using this $dV$ would be one way to show that the volume of a cylinder is $\pi R^2 h$. You could also use this to get the volume of your hollow cylinder.

You could try that as a preliminary exercise before you do the MoI calculation. I find that can be a useful test that you've set up your integration properly, as you know the answer in advance.