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Moment of inertia of a cylinder about perpendicular axes

  1. Dec 4, 2013 #1
    1. The problem statement, all variables and given/known data

    Calculate the moment of inertia of a uniform, solid cylinder about it's perpendicular axes. The cylinder has length L, radius R, and total mass M. It is centered on the origin with the z-axis running through the center of it's circular faces.

    2. Relevant equations

    [tex] I = \iiint r^{2}_{\bot} \rho dV [/tex]

    3. The attempt at a solution

    I can immediately see that due to symmetry [itex] I_{x} = I_{y}[/itex].

    I'll choose the x-axis to work with (so our cylinder is spinning about the x-axis). I can see that [itex] r_{\bot} = \sqrt{z^{2}+y^{2}}[/itex] so [itex] r^{2}_{\bot} = z^{2} +y^{2}[/itex].

    Now I need to work out my bounds for the integration. I chose a cartesian coordinate system (only as I couldn't see if cylindrical would give any simplification, but I'm sure I could've missed something) and I determined the bounds by drawing several diagrams and after some scratch work I found:

    [tex] -R \leq x\leq R [/tex]
    [tex] -L/2 \leq y\leq L/2 [/tex]
    [tex] -\sqrt{R^{2}-x^{2}} \leq z \leq \sqrt{R^{2}-x^{2}}[/tex]

    Which gives an integral:

    [tex] I = \rho \int_{-R}^{R} \int_{-L/2}^{L/2} \int_{-\sqrt{R^{2}-x^{2}}}^{\sqrt{R^{2}-x^{2}}} (z^{2}+y^{2})\:dzdydx[/tex]

    [tex] I = \rho\int^{R}_{-R} \: [\:\frac{8L}{3}(\sqrt{R^{2}-x^{2}})^{3}\:+\:\frac{L^{3}}{6}\sqrt{R^{2}-x^{2}}\:] \:dx [/tex]

    And this is where I'm not so sure about it. The integral looks particulary nasty (wolfram gave me a solution with several arctan's). I don't see where I went wrong here.
     
    Last edited: Dec 4, 2013
  2. jcsd
  3. Dec 4, 2013 #2

    TSny

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    This looks ok. But it appears that you have let the y-axis "run through the center of its circular faces" rather than the z axis. It won't make any difference.

    I don't quite get the numerical factor of 8/3 for the first term.

    Yes, the integral can be expressed in terms of arctan's. It's not too hard to evaluate the expression at the limits.

    Alternate approach (if you can use the parallel axis theorem): slice the cylinder into thin circular disks.
     
  4. Dec 5, 2013 #3
    The 8/3 comes from evaluating the z and y integrals. I could have got the constant wrong but it's still a constant.

    How would you use the parallel axis theorem?
     
  5. Dec 5, 2013 #4

    TSny

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    You can use the parallel axis theorem to express the moment of inertia of a disk about the green axis in terms of the moment of inertia about a diameter of the disk (blue axis) and the distance Z.
     

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  6. Dec 5, 2013 #5
    That seems significantly easier. Thanks.
     
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