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Homework Help: Rotational Inertia of a hollow sphere(having trouble!)

  1. Mar 12, 2005 #1
    Hello,mates.i`,ve been struggling to demonstrate the procedure to
    calculate the rotational inertia of a hollow/empty sphere around an axe that passes through its diameter.Please,could
    you shed me some light or show me the procedure(Even show me any online papers on it)I`d be Glad.Please help me! :bugeye:

    The result is I=2/3(M*R^2).

    THANK YOU!
    Jason,physicist21@yahoo.com
     
  2. jcsd
  3. Mar 12, 2005 #2

    dextercioby

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    [tex] I=\int r^{2}dm [/tex] In the case of a sphere of radius R,the moment of inertia becomes

    [tex] I=\int R^{2}\sin^{2}\vartheta \left( \frac{M}{4\pi R^{2}}dS\right) =\frac{1}{4\pi}MR^{2}\int_{0}^{2\pi}d\varphi\int_{0}^{\pi}d\vartheta \ \sin^{3}\vartheta [/tex]

    Compute the integrals and get the result.

    "r" is in the general case the distance between the mass element "dm" & the rotation axis.In this case,i've chosen spherical coordinates & the axis of rotation as Oz...

    Daniel.
     
  4. Mar 12, 2005 #3
    hey Daniel,thanks!!!But,i`m still confused about the choice of dm and the use of coordinates to use on the integrals.could ya elaborate some more?(my professor wants things with so much detail! :cry: ).Once more,i thank you and wait for yer answer. Jason.
     
  5. Mar 12, 2005 #4

    dextercioby

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    I've chosen a completely arbitrary mass element of the sphere.I assumed that the density of the sphere is constant & equal to the ratio between the sphere's mass & its surface.Then the mass "dm" of the surface element "dS" and the element of the surface "dS" are linked through

    [tex] \mbox{mass}=\mbox{density}\cdot \mbox{surface} [/tex]

    [tex] dm=\frac{M}{4\pi R^{2}} dS [/tex]

    That's all there is to it.

    The parametrization of the 2 sphere is of course:

    [tex] \left\{ \begin{array}{c}x=R\sin\vartheta\cos\varphi\\y=R\sin\vartheta\sin\varphi\\z=R\cos\vartheta \end{array} \right [/tex]

    which induces the surface element

    [tex] dS=R^{2}\sin\vartheta \ d\vartheta \ d\varphi [/tex]

    Daniel.
     
  6. Mar 12, 2005 #5
    Daniel,i`m almost there.i just have one more doubt.Where the result [tex] dS=R^{2}\sin\vartheta \ d\vartheta \ d\varphi [/tex] comes from?I wont bother ya anymore;) thanks!
     
  7. Mar 12, 2005 #6

    dextercioby

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    It's the surface element in spherical coordinates...It's famous and it is deduced in a course on calculus/analysis.It can be even justified by geometrical arguments.

    Daniel.
     
  8. Mar 12, 2005 #7
    Sorry daniel:) i had forgoten for a moment this deduction) which is so easy.Now,i`ve made it.I`m just tired of demonstrating every inertia possible (big homework!).Ya helped a lot! thank ya! :cool:
     
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