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Moment of Inertia with Variable Density Function

by Piamedes
Tags: density, function, inertia, moment, variable
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Sep24-09, 11:41 AM
P: 41
1. The problem statement, all variables and given/known data
There are two parts to question, the first asks for you to find the moment of inertia I for a thin disk of uniform density, a relatively trivial problem.

My problem centers around that second part, "Repeat the case where the density increases linearly with r, starting at 0 at the center, but the object has the same mass as the original disk."

2. Relevant equations
[tex] I = \int_{object} \rho (r,\theta) r^3 dr d\theta [/tex]

3. The attempt at a solution

Assuming that the density function is p=kr, where k is some constant I'll work out later, then the moment of inertia would be:

[tex] I = \int_{0}^{2 \pi} \int_{0}^{R} k r^4 dr d\theta [/tex]

[tex] I = k \int_{0}^{2 \pi} d\theta \int_{0}^{R} r^4 dr [/tex]

[tex] I = 2 \pi \frac{r^{5}}{5} ]_{0}^{R} [/tex]

[tex] I = \frac{2k\pi R^{5}}{5}[/tex]

With this in mind I now would need to find k. I know that it must have units of kg/m^3 in order to make the moment of inertia have the proper units. My guess on how to do this is to integrate to find the total mass, which I know to be M, solve for k in terms of M and than back substitute:

[tex] M = \int dm [/tex]

[tex] M = \int \rho dA [/tex]

[tex] M = \int_{0}^{2 \pi} \int_{0}^{R} kr * rdrd\theta [/tex]

[tex] M = k \int_{0}^{2 \pi} d\theta \int_{0}^{R} r^2 dr [/tex]

[tex] M = 2\pi k \frac{r^3}{3} ]_{0}^{R} [/tex]

[tex] M = \frac{2k\pi R^3}{3} [/tex]

Solving for K:

[tex] k = \frac{3M}{2\pi R^{3}} [/tex]

Now plugging that back into the equation for I,

[tex] I = \frac{2\pi R^{5}}{5} k [/tex]

[tex] I = \frac{2\pi R^{5}}{5} \frac{3M}{2\pi R^{3}} [/tex]

[tex] I = \frac{3MR^{2}}{5} [/tex]

Is this the proper way to solve a moment of inertia problem of variable density?

Thanks for any and all help.
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Sep24-09, 02:58 PM
Sci Advisor
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P: 25,228
It looks just fine to me.
Sep24-09, 03:13 PM
P: 41

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