Simple Example: Inertia Tensor for Dumbbell

As a simple example of this phenomenon, consider two equal (point) masses $m$, connected by a massless rod of length $\ell$. Assume it rotates about a fixed axis at an angle $\theta$ to the rod. First, note that since

$$L=\sum\limits_\alpha m_\alpha\vec{r}_\alpha\times\vec{v}_\alpha$$

it must be perpendicular to the $\vec{r}_\alpha$ and therefore perpendicular to the rod. $\vec{L}$ is not parallel to $\vec{\omega}$. $\vec{L}$ changes direction as the dumbbell rotates so a torque is required to keep it rotating about the fixed axis $\vec{\omega}$.

We might also note that if the dumbbell rotates about any axis perpendicular to the rod, $\vec{L}$ is parallel to $\vec{\omega}$.

Now lets compute the inertia tensor in two coordinate systems. Both should have the origin at the center of mass, in the middle of the rod. First, with the rod in the $xy$ plane at an angle $\phi$ from the $x$ axis. Note that we are working in the rotating (body) frame in which the masses are at rest.

$$I_{ij} \equiv \sum\limits_\alpha m_\alpha\left[\delta_{ij}r_\alpha^2-r_{\alpha i} r_{\alpha j}\right]$$

$$\vec{r}_2=-\vec{r}_1$$

$$I_{ij} = 2m{\ell^2\over 4}\left[\delta_{ij}-\hat{r}_{i} \hat{r}_{j}\right]$$

$$\hat{r}=(\cos\phi,\sin\phi,0)$$

$$I_{11} = {m\ell^2\over 2}\left[1-\cos^2\phi\right]$$

$$I_{12} = I_{21} = {m\ell^2\over 2}\left[-\cos\phi\sin\phi\right]$$

$$I_{22} = {m\ell^2\over 2}\left[1-\sin^2\phi\right]$$

$$I_{33} = {m\ell^2\over 2}\left[1\right]$$

$$I_{31} = I_{13} = {m\ell^2\over 2}\left[0\right]$$

$$I_{23} = I_{32} = {m\ell^2\over 2}\left[0\right]$$

$$\mathbb{I}={m\ell^2\over 2}\begin{pmatrix}1-\cos^2\phi & -\cos\phi\sin\phi & 0 \cr -\cos\phi\sin\phi & 1-\sin^2\phi & 0 \cr 0&0&1\end{pmatrix}$$

The inertia tensor will be diagonal for $\phi=0$ or $\phi={\pi\over 2}$ when the rod is along the $x$ or $y$ axis. The diagonal element for the axis along the rod is zero because we have assumed point masses and all the mass is on the axis.