Moment of Inertia Tensor for a Flat Rigid Body

  • #1
1. Problem
I need to find the principal moments of inertia about the center of mass of a flat rigid 45 degree right triangle with uniform mass density.

2. Useful Formulae
[tex]I_{xx}=\int_{V} \rho (r^{2}-x^{2}) dV[/tex]
[tex]I_{jk}=\int_{V} \rho (r^{2} \delta_{jk} - x_{j} x_{k}) dV[/tex]

3. Attempt at a Solution
My strategy is to set my axes so that the hypotenuse of the triangle is centered on the x-axis, with the 'right-corner' on the positive y-axis. That way, I can find the elements of the moment of inertia tensor [tex]I_{jk}[/tex] about the origin, and then translate it to the CM (1/3 up the y-axis) using the parallel-axis theorem.

If the length of one side of the triangle is "a", then using the equation for an increasing/decreasing line for the integration boundaries:

[tex]y=\pm \frac{1}{2} x + \sqrt{\frac{a}{2}}[/tex]

So,

[tex]I_{xx} = \rho \int^{\sqrt{\frac{a}{2}}}_{0} \int^{- \frac{1}{2} x + \sqrt{\frac{a}{2}}}_{\frac{1}{2} x + \sqrt{\frac{a}{2}}} y^{2} dy dx = -\frac{25}{192} a^{2} [/tex]

That should be the "x,x" element in the Moment of Inertia tensor, right?

Basically, am I setting up my integrals correctly?

(If so, I can proceed with calculating all nine tensor elements, and then diaganolizing the tensor, and transforming it to the center of mass, right?)
 

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