Is angular velocity parallel to axis of rotation?

Question

I'm reading the Wikipedia page on angular velocity. It says here of the angular velocity vector in three dimensions that “[t]he magnitude is the angular speed, and the direction describes the axis of rotation”. There, the angular velocity vector $\vec \omega$ is defined as $\boldsymbol\omega = \frac{d\phi}{dt}\mathbf{u}$ which may be written as $$ \boldsymbol\omega=\frac{|\mathbf{v}|\sin \theta}{|\mathbf{r}|} \mathbf{u} $$ which, by the definition of the cross product, can be re-written as

$$ \boldsymbol{\omega} = \frac{\mathbf{r} \times \mathbf{v}}{|\mathbf{r}|^2} ~. \label{a}\tag{1} $$

Here $\mathbf{r}$ is vector from the origin to the particle position, $\mathbf{v}$ is the velocity, $\hat{\mathbf{u}}$ is the axis of rotation, and $\theta$ is the polar and $\phi$ is the azimuthal angle.

I imagine a particle spinning in a circle of unit radius about the $z$-axis (so rotation is parallel to the $xy$-plane). Suppose the particle has a non-zero $z$-coordinate, say $\mathbf{r}_z = 3 \, \hat{\mathbf{e}}_z$. Then it's clear that the formula ($\ref{a}$) will not give a vector parallel to the $z$-axis, despite the fact that this is the axis of rotation. So is the formula wrong?

Also, it occurred to me that it might be more natural to define the angular velocity vector to be parallel to the binormal of the path. I would think this would give the axis about which the particle is instantaneously rotating. Does any definition like this exist?

Finally, how would we define angular acceleration here? Is it really this? $$\alpha = \frac{d\omega}{dt}$$ With $\omega$ defined as above, that becomes quite a complicated expression which I'm not sure how to interpret.

Answer 1

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}

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^{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.}

Answer 2

Yes the formula is wrong.

I believe that it only holds if $r$ actually lies in the plane of rotation, such that $r$ and $\omega$ are perpendicular.

The definition of $\omega$ that I have come across is $$v = \omega \times r$$

This ensures that $v$ is perpendicular to both $r$ and $\omega$. The former has to happen since under rotations $r$ does not change it's length. That $v$ is perpendicular to $\omega$ naturally happens since $\omega$ defines the plane of rotation that $v$ lies in.

Note here that the cross product in non-invertible, that is all vectors perpendicular to $\omega$ will have $v=0$ (which is only natural). This means that in order to figure out $\omega$ you need to have access to (at least) two pairs of $(v,r)$-vectors.

Answer 3

Such a vectorial definition for $\omega$ was created to eliminate the following confusion. Let's suppose I am observing a circular motion in the X-Y plane, while standing in front of the plane. According to me, the body will either be having a clockwise or anticlockwise sense of rotation, but to a person standing behind the plane, his answer will always be the opposite of my answer. For example, if I claim that the motion is in the clockwise sense, his response would be that according to him the motion is in the anticlockwise sense. But if we both use the screw rule , we'll use the same unit vector to represent the sense of motion. So even though , upon first glance assigning vectorial notation to angular velocity seems quite counter intuitive, it still is quite reasonable to do so. What's also interesting to note, is that angular displacement is not a vector, yet angular velocity is. The reason is that vectorial addition is commutative , but in 3-D application angular displacement is not commutative. So then why do we claim that $\omega$, which is basically rate of change of angular displacement( ($\omega = \frac{d\theta}{dt}$), is a vector. We do so because a small angular displacement($d\theta$) does not have a sense of direction, and that the addition of such small angular displacements can be said to be commutative.