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Determining Moment of Inertia of a sphere.

  1. Feb 1, 2007 #1
    Determining Moment of Inertia of a Sphere

    I'm having some troubles determining the moment of inertia of a sphere about it's central axis. My original question was to calculate it for a cylinder, which I've done, but I'd like to know how to find it for a sphere.
    Here is the problem solved for a cylinder:

    Problem:
    A uniform solid cylinder has radius R, mass M, and length L. Calculate its moment of inertia about its central axis (the z axis).

    Solution:
    I divided the cylinder into infinitesimally small layers because I knew that [tex]dV = (2\pi*r*dr)*L[/tex]. From here I calculated the integral [tex]I = \int \rho*r^2 dV = \int_{0}^R \rho*r^2*(2\pi*r*L)dr = 2*\pi*\rho*L*R^4 [/tex]
    I substituted [tex]\frac {M}{\pi*R^2*L}[/tex] (or [tex]\frac{M}{V}[/tex]) for [tex]\rho[/tex] into the equation to get [tex] I = \frac{1}{2}*\pi*(\frac {M}{\pi*R^2*L})*L*R^4 = \frac{1}{2}*M*R^2[/tex]

    I understand this, but when I tried to get it as a sphere I ended up getting the wrong answer. Could anyone please show me how to start the problem with a sphere?
     
    Last edited: Feb 1, 2007
  2. jcsd
  3. Feb 1, 2007 #2

    berkeman

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    Staff: Mentor

    Do it in spherical coordinates, being careful to take the sin(theta) for each little volume piece (since the spherical r value is from the origin, not from the axis of rotation). And be careful to use the correct value for dV in spherical coordinates.
     
  4. Feb 1, 2007 #3
    For the solid spere, I like to add up the moment of inertia of a bunch of disks form -R to R.
     
  5. Feb 1, 2007 #4

    cristo

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    Staff Emeritus
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    How did you try to perform your integral for the sphere? Think of your choice of coordinate system--hint: spherical polar coordinates.

    damn: really should have refreshed quicker!
     
  6. Feb 2, 2007 #5
    Couldn't I divide the sphere into a hemisphere and multiply it by 2? E.G. [tex]2*\pi*\int_{0}^R (\sqrt{R^2-r^2}*r^2)*r^2 dr[/tex] Where R is the radius of the sphere. So essentially I'm adding a bunch of [tex]\pi*r^2[/tex] (circles) to get a half of a sphere. Then multiplying the whole thing times two.
     
    Last edited: Feb 2, 2007
  7. Feb 2, 2007 #6
    hmm, why dont you show a little more work so that it is easier to follow.
    one thing though, you might want to reconsider muliplying it by two--think about it.
     
  8. Feb 6, 2007 #7
    I have no idea how to integrate that, so I don't have much work to show. The idea makes sence in my head, but I dont know how to follow through with it.
     
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