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Moment of Inertia (Disk)

  • #1
129
0
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:

[tex]I = \int r^2dm[/tex]

[tex]m = \lambda L[/tex]

[tex]dm = \lambda dL[/tex]

[tex]I_{ring} = \int_{0}^{L}\lambda r^2dL[/tex]

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:

[tex]I_{disk} = \int_{0}^{A}\int_{0}^{L}\lambda r^2dLdA[/tex]

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?
 

Answers and Replies

  • #2
557
1
What's A, the area of what ?
 
  • #3
26
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No, I think you cant do that..since in your formula for Idisk still using [tex]\lambda[/tex] as mass distribution in length, while your formula is related to area. you should use [tex]\sigma[/tex] as mass distribution per area and start over.
 
  • #4
129
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What's A, the area of what ?
A is the area of the disk formed by integrating up all the rings

you should use [tex]\sigma[/tex] 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...
 
  • #5
557
1
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.
 
  • #6
129
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you have to integrate first along the angle to get a ring, then along the radius to get a disk
[tex]dm = \lambda dL[/tex]

[tex]L=r\theta [/tex]

[tex]dL=rd\theta [/tex]

[tex]dm = \lambda rd\theta[/tex]

[tex]I_{ring}=\int_{0}^{2\pi}\lambda r^3d\theta[/tex]

[tex]I_{disk}=\int_{0}^{R}\int_{0}^{2\pi}\lambda r^3d\theta dr[/tex]

I think I'm missing something here.
 
  • #7
557
1
The integral is correct.
But not dm
[tex]dm = \lambda \ r \ d\theta \ dr[/tex]
 
  • #8
129
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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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