Moment of Inertia with Variable Density Function

  1. 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.
     
  2. jcsd
  3. Dick

    Dick 25,810
    Science Advisor
    Homework Helper

    It looks just fine to me.
     
  4. Thanks
     
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