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Moment of Inertia of a Disk

  1. Dec 25, 2005 #1
    I am having a bit of trouble deriving the moment of inertia for a disk with uniform density:

    [tex]I=\int r^{2}\,dm=\int \rho r^{2}\,dV[/tex]

    For a disk, I just used dA instead of dV. Now, to calculate the density:

    [tex]\rho = \frac{\text{mass}}{\text{volume}}=\frac{m}{\pi r^{2}}[/tex]

    So now we have:

    [tex]I=\int \rho r^{2}\,dA=\frac{m}{\pi}\int \,dA=\boxed{mr^{2}}[/tex]

    However, I know that the moment of inertia for a disk is [itex]\frac{1}{2}mr^{2}[/itex]. Where did I go wrong?

    Thank you.
  2. jcsd
  3. Dec 25, 2005 #2
    Nevermind I see where I went wrong (I used r twice, whereas I should have used R and r, where r ranges from 0 to R).
  4. Dec 25, 2005 #3


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    In the integral, one would have [itex]r^2\,\frac{m}{\pi\,r^2} r dr d\theta[/itex] the r2 terms cancel, and that leaves dA = r dr d[itex]d\theta[/itex].
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