# Moment of Inertia with Variable Density Function

by Piamedes
Tags: density, function, inertia, moment, variable
 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 $$I = \int_{object} \rho (r,\theta) r^3 dr d\theta$$ 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: $$I = \int_{0}^{2 \pi} \int_{0}^{R} k r^4 dr d\theta$$ $$I = k \int_{0}^{2 \pi} d\theta \int_{0}^{R} r^4 dr$$ $$I = 2 \pi \frac{r^{5}}{5} ]_{0}^{R}$$ $$I = \frac{2k\pi R^{5}}{5}$$ 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: $$M = \int dm$$ $$M = \int \rho dA$$ $$M = \int_{0}^{2 \pi} \int_{0}^{R} kr * rdrd\theta$$ $$M = k \int_{0}^{2 \pi} d\theta \int_{0}^{R} r^2 dr$$ $$M = 2\pi k \frac{r^3}{3} ]_{0}^{R}$$ $$M = \frac{2k\pi R^3}{3}$$ Solving for K: $$k = \frac{3M}{2\pi R^{3}}$$ Now plugging that back into the equation for I, $$I = \frac{2\pi R^{5}}{5} k$$ $$I = \frac{2\pi R^{5}}{5} \frac{3M}{2\pi R^{3}}$$ $$I = \frac{3MR^{2}}{5}$$ Is this the proper way to solve a moment of inertia problem of variable density? Thanks for any and all help.
 Sci Advisor HW Helper Thanks P: 25,175 It looks just fine to me.
 P: 41 Thanks

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