Example: The Inertia Tensor for a Cube

We wish to compute the inertia tensor for a uniform density cube of mass $M$ and side $s$. The density is simply $\rho={M\over s^3}$.

$$I_{11}={M\over s^3}\int\limits_{-{s\over 2}}^{s\over 2}\int\limits_{-{s\over 2}}^{s\over 2} \int\limits_{-{s\over 2}}^{s\over 2}(r^2-x^2)dxdydz$$

$$I_{11}={M\over s^3}\int\limits_{-{s\over 2}}^{s\over 2}\int\limits_{-{s\over 2}}^{s\over 2}(y^2+z^2)dydz\int\limits_{-{s\over 2}}^{s\over 2}dx$$

$$I_{11}={M\over s^2}\int\limits_{-{s\over 2}}^{s\over 2}\int\limits_{-{s\over 2}}^{s\over 2}(y^2+z^2)dydz$$

$$I_{11}={M\over 3s}(\left[y^3\right]_{-{s\over 2}}^{s\over 2}+\left[z^3\right]_{-{s\over 2}}^{s\over 2})$$

$$I_{11}={M\over 3s}{s^3\over 2}= {Ms^2\over 6}$$

$$I_{12}={M\over s^3}\int\limits_{-{s\over 2}}^{s\over 2}\int\limits_{-{s\over 2}}^{s\over 2} \int\limits_{-{s\over 2}}^{s\over 2}(-xy)dxdydz=0$$

$$\mathbb{I}=M{s^2\over 6}\begin{pmatrix}1&0&0\cr 0&1&0\cr 0&0&1\end{pmatrix}$$

There are only two kinds of integrals. The rest are symmetric. The inertia tensor is diagonal so rotation about these axes will have the angular momentum parallel to the axis. The angular momentum then does not change with time and no torque is needed to rotate the cube.

Now lets consider rotating a cube about one of its corners. We can compute the new inertia tensor by using the parallel axis theorem with a translation of $\vec{a}=(-{s\over 2},-{s\over 2},-{s\over 2})$.

$$I_{ij}'=I_{ij}+M\left[\delta_{ij}a^2-a_ia_j\r... ...ver 4}\right] =I_{ij}+Ms^2\left[\delta_{ij}{3\over 4}-{1\over 4}\right]$$

Note that the final ${1\over 4}$ has no indices so it is subtracted from all 9 terms in the tensor.

$$I_{ij}'=Ms^2\left[\delta_{ij}\left({1\over 6}... ...\over 3}&-{1\over 4}\cr -{1\over 4}&-{1\over 4}&{2\over 3}\end{pmatrix}$$

This inertia tensor is not diagonal so torque will be required to keep a cube rotating about a corner if the axis of rotation in one of the edges of the cube. (Note that angular momentum and torque are measured relative to some origin so the location of the origin is important in the calculation of the inertia tensor.)

For any rotating mechanical part, it is important that it rotate about a principal axis, otherwise it will exert torque and cause vibrations.