Inertia – Calculating the Tensor of Inertia of a Hollow Cube

homework-and-exercisesmoment of inertia

I have found the tensor of inertia of a rectangle of sides $a$ and $b$ and mass $m$, around its center, to be $$I_{11}=ma^2/12,$$ $$I_{22}=mb^2/12,$$ $$I_{33}=(ma^2 + mb^2)/12.$$ All other elements of that tensor are equal to zero. I would now like to use this result to determine the tensor of inertia of a hollow cube of side a around its center of mass.

I realise I have to use the parallel axis theorem. I also knoww that the correct equation is I$_{11}$=I$_{22}$=I$_{33}$=ma$^2$/12+ma$^2$/12+ma$^2$/6+4(ma$^2$/12 + m(a/2)$^2$)=5/3*ma$^2$
I simply do not understand why this is correct. Could anyone please explain why this is the correct way to calculate the desired tensor of inertia? Also, why would I be summing all the diagonal elements in my tensor for the rectangle?

Best Answer

We will work in units where the mass of each face is $1$ and where the length of the side of the cube is $1$.

The contribution to the moment of inertia of each of the top and bottom faces is, using your result for the moment of inertia of a rectangle, $\frac{1}{6}$.

By symmetry, each of the four other faces has the same contribution to the moment of inertia. Let's calculate the contribution of one of them. Let's project the system along the axis about which we are calcuating the moment of inertia. This operation has no effect on the moment of inertia. Our face now becomes a rod. We know the moment of inertia through the center of the rod is $\frac{1}{12}$, but we are calculating the moment of inertia at a distance $\frac{1}{2}$ away from the center of the rod. Using the parallel axis theorem, we get that the moment of inertia is $\frac{1}{12} + (\frac{1}{2})^2 = \frac{4}{12} = \frac{1}{3}$. Then our total moment of inertia is $\frac{1}{6} + \frac{1}{6} + 4*\frac{1}{3}= \frac{5}{3}$. (The first two terms come from the top and bottom face, the last term comes from the four side faces.)

Putting units back in we get the moment of inertia is $I=\frac{5}{3} M a^2$, where $M$ is the mass of a face and $a$ is the side length of the cube (also the side length of a face).