Moment of inertia of a hollow cylinder derivation

[Carpetfizz]

Homework Statement

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.

Homework Equations

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

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.

 

[PeroK]

Homework Statement

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.

Homework Equations

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

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.

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

 

[Carpetfizz]

It's a thin ring, sorry.

 

[PeroK]

It's a thin ring, sorry.

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.