angular momentumangular velocitymoment of inertiarotational-kinematics
I have a quiz with the following question:
How can the angular momentum vector not be parallel to the angular velocity vector? That would mean they don't have the same direction, right?
We have this relation between angular momentum and angular speed:
Since they only differ by a scalar, how can they not have the same direction?
The essence of the question is the definition of the angular velocity and its relation to the axis of rotation.
First of all, to describe the rotation of a single particle, one requires an axis of rotation $\hat{\mathbf{n}}$, and a rate of change of ‘orientation’ (or ‘angular position’) $\frac{d \theta( \hat{\mathbf{n}}) }{d t}$ around that axis. Without these 2 pieces of information, “rotation” is meaningless. Having the axis of rotation and a rate of change of orientation, one can define the angular velocity $\vec{\omega}$ as $$ \boldsymbol{\omega} = \frac{d \theta}{d t} \, \hat{\mathbf{n}} $$ which includes both pieces of information. Notice that the angular velocity is parallel to the axis of rotation by definition, and has the same amount of ‘information’ as the particle velocity $\mathbf{v}$.
To clarify this further, consider the following figure which depicts the rotation in 3d space, and conforms to the example given in the question,
where the spherical-polar coordinates $\theta$ and $\phi$ are used as the azimuthal and polar angles, respectively and the zenith ($z$-axis) is taken to be parallel to $\hat{\mathbf{n}}$.
In the limit of infinitesimal change in time, $\Delta t \rightarrow 0$, $$ \Delta \mathbf{r} \approx r \, \sin \phi \, \Delta \theta ~, $$ and thus, $$ \left| \frac{d \mathbf{r}}{d t} \right| = r \sin \phi \frac{d \theta}{d t} ~. $$
One clearly observes that both the magnitude and the direction of $\frac{d \mathbf{r}}{d t}$ (which is perpendicular to the plane defined by the particle position $\mathbf{r}$ and the rotation axis $\hat{\mathbf{n}}$) are given correctly by the cross product
$$ \frac{d \mathbf{r}}{d t} = \hat{\mathbf{n}} \times \mathbf{r} \frac{d \theta}{d t} ~. $$
Since $\frac{d \mathbf{r}}{d t} \equiv \mathbf{v}$ and $\hat{\mathbf{n}} \frac{d \theta}{d t} \equiv \vec{\omega}$, one obtains
$$ \mathbf{v} = \frac{d \mathbf{r}}{d t} = \boldsymbol{\omega} \times \mathbf{r} ~. $$
Therefore, the instantaneous axis of rotation $\hat{\mathbf{n}}(t)$ and the instantaneous angular velocity $\vec{\omega}(t)$ are indeed parallel.
Finally, as shown by Gary Godfrey in a comment, one can obtain the correct relation for the angular velocity $\boldsymbol{\omega}$ in terms of the velocity $\mathbf{v}$ and position $\mathbf{r}$ as
\begin{align} \mathbf{r} \times \mathbf{v} &= \mathbf{r} \times ( \boldsymbol{\omega} \times \mathbf{r} ) \stackrel{\text{BAC-CAB}}{=} \boldsymbol{\omega} \, (\mathbf{r} \cdot \mathbf{r} ) - \mathbf{r} \, (\mathbf{r} \cdot \boldsymbol{\omega}) \\ \Rightarrow \boldsymbol{\omega} &= \frac {(\mathbf{r} \times \mathbf{v}) + \mathbf{r} \, (\mathbf{r} \cdot \boldsymbol{\omega})}{r^2} ~. \end{align}
This answer is based on Kleppner, D., and R. Kolenkow. “An introduction to mechanics”, (2ed, 2014), pp. 294–295. The figure is taken from the same source.
Consider a thin rectangular block with width $w$, height $h$ resting along the xy plane as shown below.
The mass of the block is $m$. The mass moment of inertia (tensor) of the block about point A is
$$ {\bf I}_A = m \begin{vmatrix} \frac{h^2}{3} & -\frac{w h}{4} & 0 \\ -\frac{w h}{4} & \frac{w^2}{3} & 0 \\ 0 & 0 & \frac{w^2+h^2}{3} \end{vmatrix} $$
This was derived from the definition (as seen on https://physics.stackexchange.com/a/244969/392)
If this block is rotating along the x axis with a rotational velocity $$ \boldsymbol{\omega} = \begin{pmatrix} \Omega \\ 0 \\ 0 \end{pmatrix} $$ then the angular momentum about point A is
$${\bf L}_A = m \Omega\,\begin{pmatrix} \frac{h^2}{3} \\ -\frac{w h}{4} \\ 0 \end{pmatrix} $$
As you can see, there is a component of angular momentum in the y direction. The angular momentum vector forms an angle $\psi = -\tan^{-1} \left( \frac{3 w}{4 h} \right)$
In the figure below you see the direction of angular momentum, and the circle about which the center of mass is going to orbit due to precession.
Best Answer
The comment by QuantumBrick really tells you all you need to know: "$I$ is not a scalar, but a tensor". However, sometimes it's hard to get an intuition for this. Let me try the following:
Imagine a rod - some small diameter, long object. For the sake or argument let's say that the moment of inertia about the axis perpendicular to the rod is 10x greater than the moment of inertia about the parallel axis. Rotating that rod about an axis that is 45° to its length, I have equal angular velocity about the axis perpendicular to the rod, and the axis parallel to the rod. But the angular momentum parallel to the axis will be 1/10th of the angular momentum perpendicular to the axis.
This means that while the angular velocity vector points in the 45° direction, the angular momentum vector will be almost perpendicular to the rod. See this diagram: