What is the Proper Method for Calculating Moment of Inertia for a Disk?

AI Thread Summary
To calculate the moment of inertia for a disk, one must first understand the distinction between mass distribution in length and area. The proper approach involves using area density (σ) for a disk rather than linear density (λ) used for a ring. The integration should be performed first along the angle to define a ring, then along the radius to construct the disk. The correct expression for the mass element (dm) incorporates both radial and angular components. Ultimately, defining dm accurately allows for a proper double integral to compute the moment of inertia for the disk.
Screwdriver
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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?
 
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What's A, the area of what ?
 
No, I think you can't 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...
 
Screwdriver said:
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.
 
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