Example: Cube Rotating about a Corner

We have computed the inertia tensor for a cube with the origin at a corner and the axes along the edges.

$$I_{ij}'=Ms^2\begin{pmatrix}{2\over 3}&-{1\ove... ...\over 3}&-{1\over 4}\cr -{1\over 4}&-{1\over 4}&{2\over 3}\end{pmatrix}$$

Lets solve the equation to find the principal axes.

$$\left\vert\begin{matrix}{2\over 3}-x&-{1\over... ...ver 4}\cr -{1\over 4}&-{1\over 4}&{2\over 3}-x\end{matrix}\right\vert=0$$

(The moment of inertia will be $xMs^2$ with this definition of $x$.) To facilitate the solution of the cubic equation, lets get some zeros in this determinant by subtracting the second row from the first and the third.

$$\left\vert\begin{matrix}{11\over 12}-x&x-{11\... ...-{1\over 4}\cr 0&x-{11\over 12}&{11\over 12}-x\end{matrix}\right\vert=0$$

If we let $y={11\over 12}-x$, we can write the determinant.

$$y^2\left(y-{1\over 4}\right)-{1\over 4}y^2-{1\over 4}y^2=y^2\left(y-{3\over 4}\right)=0$$

So there are three roots, $y=0$, $y=0$, and $y={3\over 4}$; corresponding to three values for $x$ of $x={11\over 12}$, $x={11\over 12}$, and $x={1\over 6}$.

Knowing that two equal moments of inertia along principal axes indicate a symmetry, lets find the axis for the third value $I={1\over 6}Ms^2$.

$$I_{ij}\omega_j={1\over 6}Ms^2\omega_i$$

$$Ms^2\begin{pmatrix}{2\over 3}&-{1\over 4}&-{1\over 4}\cr -{1\over... ...trix}={1\over 6}Ms^2\begin{pmatrix}\omega_1\cr \omega_2\cr\omega_3\end{pmatrix}$$

$$\begin{pmatrix}{2\over 3}&-{1\over 4}&-{1\over 4}\cr -{1\over 4}&... ...{pmatrix}={1\over 6}\begin{pmatrix}\omega_1\cr \omega_2\cr\omega_3\end{pmatrix}$$

$$\begin{pmatrix}{1\over 2}&-{1\over 4}&-{1\over 4}\cr -{1\over 4}&... ...r 2}\end{pmatrix} \begin{pmatrix}\omega_1\cr \omega_2\cr\omega_3\end{pmatrix}=0$$

$$\begin{pmatrix}2&-1&-1\cr -1&2&-1\cr -1&-1&2\end{pmatrix}\begin{pmatrix}\omega_1\cr \omega_2\cr\omega_3\end{pmatrix}=0$$

One can see that all three of these equations are solved for the three components of $\vec{\omega}$ being equal. So this principal axis is the diagonal of the cube starting from the origin and going toward $(1,1,1)$.

The other roots just give the same equation three times $\omega_1+\omega_2+\omega_3=0$. There are many possible solutions. The equation defines a plane perpendicular to the cube diagonal. We can choose and orthogonal pair of axes in that plane.