- #1
Sentin3l
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Homework Statement
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.
Homework Equations
[tex] I = \iiint r^{2}_{\bot} \rho dV [/tex]
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.
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