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

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Homework Help Overview

The discussion revolves around calculating the moment of inertia for a disk, with initial considerations stemming from the moment of inertia for a ring. Participants explore the integration methods and mass distribution definitions relevant to the problem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the integration process for calculating the moment of inertia, questioning the use of linear versus area mass distributions. There is an exploration of whether a double integral can be effectively used to derive the moment of inertia for a disk from that of a ring.

Discussion Status

The conversation is ongoing, with participants providing insights into the correct definitions of mass elements and integration strategies. Some guidance has been offered regarding the need to redefine mass distribution when transitioning from a ring to a disk, but no consensus has been reached.

Contextual Notes

There are questions about the definitions and assumptions related to mass distribution (linear vs. area) and the integration limits necessary for the calculations. Participants are also considering the implications of integrating over angles and radii in their approach.

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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