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Homework Help: Deriving the moment of inertia for a sphere

  1. Dec 2, 2007 #1
    1. The problem statement, all variables and given/known data

    Derive the moment of inertia for a solid sphere with a uniform mass

    2. Relevant equations

    [Tex]I= \sum mr^{2}[/tex]

    3. The attempt at a solution

    I decided to change everything to polar coordinates. Since the polar coordinate substitution is

    [tex]\int\int\int_{v} Fr^{2}sin(\phi)drd\phi d\theta[/tex]

    I figured that all you should do is plug in the moment of inertia equation into the integral giving you:

    [tex]m\int^{2\pi}_{0}\int^{\pi}_{0}\int^{R}_{0} r^{4}sin(\phi)drd\phid\theta =\frac{4\pi R^{5}}{5}[/tex]

    however this does not seem to be the correct answer. Can anyone tell me what I'm doing wrong? Thanks in advance for your time and any help.
    Last edited: Dec 2, 2007
  2. jcsd
  3. Dec 2, 2007 #2
    For some reason the images are coming up in the wrong places, and one image is actually missing. I don't know how to fix that. Sorry.
  4. Dec 2, 2007 #3


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    The last image is probably not showing because it has a syntax error: there is something like
    in it, and LaTeX doesn't know the symbol \phid. Insert a space and it will work.
    In the first formula you quoted, what is this F and where does the sine come from? Also, did you put in the mass density
    [tex]\rho = \frac{m}{V} = \frac{m}{\tfrac43 \pi R^3} [/tex]
    correctly (I think not, which would account for the R^5 instead of R^2 as you should have found).

    [edit]I think I see where the [itex]r^2 \sin\theta[/itex] comes from, is that the square of the shortest distance from a point on the sphere to some axis? I calculate that distance to be r sin(theta), so in the formula there is also a square missing probably, and the integration there is over the volume dV, the dr dtheta dphi only come in when you switch to polar coordinates[/edit]
    Last edited: Dec 2, 2007
  5. Dec 2, 2007 #4


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    I've corrected your tex to make it easy to read. Your problem was that that the first tags were [ Tex] [ /tex] (plus the problems that compuchip mentioned). See the quote.
    Last edited: Dec 2, 2007
  6. Dec 2, 2007 #5
    I too am not so sure about what you are doing. If you use an inertia tensor the formula goes as

    [tex]I_{ij} = \int_V \rho(\mathbf{r}) (\delta_{ij} \sum_k x_k^2 -x_i x_j) dv[/tex]

    so, for example, the first element would be

    [tex]I_{11} = \int_V \frac{M}{4/3 \pi R^3} (x^2 + y^2 + z^2 - x^2) dv[/tex]

    where you can use coordinate and jacobian (for the volume differential) conversions to get it into spherical form (which I think is what you meant because you certainly didn't show a polar jacobian). From symmetry all the off diagonals will be zero, and all the diagonals will be the same.
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