Calculating Moment of Inertia of a magnet

[Jdo300]

Hello all,

I have a ceramic doughnut magnet (from a speaker) and I'm wondering how to calculate its moment of inertia. I'm using a simulation program to do some mechanical analysis and I need to spin the doughnut magnet. It weighs roughly 185 grams, the inner diameter is 1.25in, the outer diameter is 2.75in, and it's 0.5in thick. Any help will be apprectiated.

Thanks,
Jason O

 

[kanato]

If I'm understanding the geometry correctly, you should be able to just take the moment of inertia of a cylinder, and take a large radius cylinder and subtract a small radius cylinder to find the moment of inertia for a donut.

 

[Nenad]

If I'm understanding the geometry correctly, you should be able to just take the moment of inertia of a cylinder, and take a large radius cylinder and subtract a small radius cylinder to find the moment of inertia for a donut.

Im not sure Moment of inertia works that way. A lot of people come up with solutions like these ones when the real solution is a little more complicated. The best way for you to find the moment of inertia is to use some simple mechanics formulas.

\Sigma\tau = I \omega^2
or
FR = I \omega^2 = I \frac{\alpha}{R}

It all depends if you know your torque on the system. Find out what it is and plug and play with the formulas. If you just want to know what the moment of inertia for a doughnut is, look it up on the net. There are tables of derived inertias for all kinds of objects.

 

[arildno]

I assume you've got a solid doughnut.
Let the mid-line circle have radius R, whereas the radius of the doughnut tube is r_{t}.
Let the (uniform) density be \rho
The mass M, is therefore M=2\rho\pi^{2}Rr_{t}^{2}
The moment of inertia is given by:
I=\int_{0}^{2\pi}\int_{0}^{2\pi}\int_{0}^{r_{t}}\rho{r}(R+r\cos\phi)^{3}drd\phi{d\theta}=M(R^{2}+\frac{3}{4}r_{t}^{2})
unless I made a mistake somewhere.

 

[Astronuc]

What is the cross section. Arildno gave the correct formula for a toroidal donut (torus) with a circular cross-section. However, it would appear that you are describing a square annulus rather than a round one.

 

[Jdo300]

What is the cross section. Arildno gave the correct formula for a toroidal donut (torus) with a circular cross-section. However, it would appear that you are describing a square annulus rather than a round one.

Hi, yes, the problem is for a square doughnut. I have a couple of questions too. As far as actually calculating the integrals, I am in second year Calculus now so I can do integration problems, but how do I do this since the variables are basically an integral inside of another one? are the limits dependent on each other or independent?

Also, I have one more question which I am sure will complicate the problem more. Inside of the doughnut, I have a plastic disk with three metal cylinders in it that are equally spaced around the disk. I would like to calculate the moment of inertia of this disk but since it doesn't have a uniform density, how would I go about it? I've made a simple diagram showing the rotor with only one of the three meatal cylinders in it for simplicity. I can say that the metal cylinder is the same thickness as the disk.

Thank you for any and all help in advance :-).