# Moment of Inertia (Disk)

Not a homework question per se, but I'm having some issues with moments of inertia. Say I wanted to calculate the I for a ring. What I would do is:

$$I = \int r^2dm$$

$$m = \lambda L$$

$$dm = \lambda dL$$

$$I_{ring} = \int_{0}^{L}\lambda r^2dL$$

And that would give the requiside mr2. My question is, why can't I just integrate THAT up to get the I of a disk. I mean something like this:

$$I_{disk} = \int_{0}^{A}\int_{0}^{L}\lambda r^2dLdA$$

Where A is the area of the disk. Doesn't figuring out the I for a ring essentially skip the first step for determining the I for a disk?

What's A, the area of what ?

No, I think you cant do that..since in your formula for Idisk still using $$\lambda$$ as mass distribution in length, while your formula is related to area. you should use $$\sigma$$ as mass distribution per area and start over.

What's A, the area of what ?
A is the area of the disk formed by integrating up all the rings

you should use $$\sigma$$ as mass distribution per area and start over.
But what if I want to write it all out in one double integral? That must be possible...

A is the area of the disk formed by integrating up all the rings

But what if I want to write it all out in one double integral? That must be possible...
If A is the area, the you have to integrate first along the angle to get a ring, then along the radius to get a disk.
You got to be careful with the mass definition, as Lepton pointed out.

you have to integrate first along the angle to get a ring, then along the radius to get a disk
$$dm = \lambda dL$$

$$L=r\theta$$

$$dL=rd\theta$$

$$dm = \lambda rd\theta$$

$$I_{ring}=\int_{0}^{2\pi}\lambda r^3d\theta$$

$$I_{disk}=\int_{0}^{R}\int_{0}^{2\pi}\lambda r^3d\theta dr$$

I think I'm missing something here.

The integral is correct.
But not dm
$$dm = \lambda \ r \ d\theta \ dr$$

Okay. So you basically just have to define dm in terms of some small change in theta and some small change in r preemptively so you can just integrate them both up right away.