How Is the Moment of Inertia Calculated for a Solid Torus on the XY Plane?

[naptor]

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:
I_disk=dI_torus=dma^{2}/4+R^{2}dm

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 : dm/Rd\theta = M/2\pi R
to substitute for dm. Whats wrong ?
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[voko]

A torus is given by two radii. You only mention one, R. Which of the two is it? And what's the other one?

 

[naptor]

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.

 

[voko]

So what does the integral look like?

 

[naptor]

I_torus=\int \frac{(a^{2}M)}{8\pi} d\theta + \int \frac{R^{2}M}{2\pi} d\theta, from 0 to 2\pi and got :

\frac{(a^{2}M)}{4} + R^{2}M

Well the substitution was silly because all I had was a bunch of constants and no functions of theta .

 

[voko]

Is that the correct answer?

 

[naptor]

Is that the correct answer?

Nope. But thinking about that integral was helpful I think I got it, at least according to wikipedia. Thanks.