Mathematica: Inertia Tensor w/ 3-d Rectangle

Paul E.

Hey All,

I'm trying to create a 3-D rectangle in Mathematica with the following measurements: Mass M=1.5 kg, and sides of length a=10 cm (parallel to the x-axis), 2a (parallel to the y-axis), and 3a (parallel to the z-axis). Let one corner be at the origin, and let the three adjacent edges lie along the coordinate axes, so the corners of the cube lie at (0,0,0), (a,0,0), (0,2a,0), (0,0,3a), (0,2a,3a), (a,0,3a), (a,2a,0), and (a,2a,3a).

I'm trying to set it up in matrix form, as I have to find the principal moments of inertia and principal axes as well. What's the best way to solve for this with the eigenvalues and eigenvectors of the tensor?

Also have to answer which of the principal axes would free rotation of the block be stable?

There's a long list of things I have to do with this problem, so I will update as necessary.

Thanks!

tiny-tim

Hey Paul!
If your three coordinate axes are parallel to the axes of symmetry (in this case, the sides of the box), then the moment of inertia tensor is diagonal

Paul E.

Thanks Tim,
Since I'm fairly new to Mathematica... all I've seen is ParametricPlot3d for other plots, but never seen anything for a block or the inertia tensor. Anyone familiar with the programming?