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Moment of Inertia of a Torus

  • Thread starter naptor
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  • #1
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Homework Statement


To calculate the moment of inertia of a solid torus through the z axis(the torus is on the xy plane), using the parallel and perpendicular axis theorem.


Homework Equations





The Attempt at a Solution



Well, first I divided the torus into tiny little disks and calculated the moment of inertia of one of these disks about an axis parallel to the z axis, using the perpendicular theorem. Then by the parallel axis theorem I found that the moment of inertia of the disk about the axis passing through the center of the torus is:
Idisk=dItorus=[itex]dma^{2}/4[/itex]+[itex]R^{2}dm[/itex]

Then in order to find the moment of inertia of the torus I should integrate that expression, but it doesn't work. I used the relation : [itex]dm/Rd\theta = M/2\pi R[/itex]
to substitute for dm. Whats wrong ?
[
 
Last edited:

Answers and Replies

  • #2
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A torus is given by two radii. You only mention one, R. Which of the two is it? And what's the other one?
 
  • #3
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A torus is given by two radii. You only mention one, R. Which of the two is it? And what's the other one?
Sorry I didn't write the equations correctly. a is the radius of the little disk and R is the distance from the center of the torus to the center of the little disk, that is : Inner Radius = R-a.
 
  • #4
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So what does the integral look like?
 
  • #5
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Itorus=[itex]\int \frac{(a^{2}M)}{8\pi} d\theta[/itex] + [itex]\int \frac{R^{2}M}{2\pi} d\theta[/itex], from 0 to 2[itex]\pi[/itex] and got :

[itex] \frac{(a^{2}M)}{4} [/itex] + [itex]R^{2}M[/itex]

Well the substitution was silly because all I had was a bunch of constants and no functions of theta .
 
  • #6
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Is that the correct answer?
 
  • #7
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Is that the correct answer?
Nope. But thinking about that integral was helpful I think I got it, at least according to wikipedia. Thanks.
 

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