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Moment of Inertia of a Torus

  1. Sep 29, 2012 #1
    1. The problem statement, all variables and given/known data
    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.


    2. Relevant equations



    3. 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: Sep 29, 2012
  2. jcsd
  3. Sep 29, 2012 #2
    A torus is given by two radii. You only mention one, R. Which of the two is it? And what's the other one?
     
  4. Sep 29, 2012 #3
    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.
     
  5. Sep 29, 2012 #4
    So what does the integral look like?
     
  6. Sep 29, 2012 #5
    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 .
     
  7. Sep 29, 2012 #6
    Is that the correct answer?
     
  8. Sep 29, 2012 #7
    Nope. But thinking about that integral was helpful I think I got it, at least according to wikipedia. Thanks.
     
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