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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.

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    $\begingroup$ Angular velocity is defined relative to a point that you choose, so if you choose a point that is different, it can result in a different angular velocity. In fact, your path need not be an orbit or curved in any way, as you can define an angular velocity for a straight line path, and it will even be non-zero if the point you are measuring it relative to is not in the path. $\endgroup$ – tmwilson26 Dec 10 '15 at 1:05
  • $\begingroup$ The critically important addendum to @tmwilson's point is that for extended bodies you have to use the same origin to evaluate every $\mathrm{r}$. $\endgroup$ – dmckee Dec 10 '15 at 3:28
  • $\begingroup$ Does this mean that the direction of $\omega$ does not necessarily point along the axis of rotation then? If so, where does the formula come from and why is it useful? $\endgroup$ – user100898 Dec 10 '15 at 3:32
  • $\begingroup$ I think a lot of these ambiguities go away if we define $\vec{r}$ to be the shortest vector pointing from the axis of rotation to the object in question. This is also the only way to ensure $v=r\omega$ is a valid formula. $\endgroup$ – Jahan Claes Dec 10 '15 at 5:43
  • $\begingroup$ @JahanClaes Again, that works for point masses, but not for extended bodies. For extended bodies there is simply no guarantee that $\omega$ is co-linear with the axis of rotation. That's why the tires on your car have to be balanced. $\endgroup$ – dmckee Dec 16 '15 at 4:10
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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,

Rotation in 3d space in spherical coordinates

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.

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  • $\begingroup$ Once I also deduced it the same way in one of my answers; it was based on A.P.French's Mechanics. I think that would answer the question.+1. $\endgroup$ – user36790 Dec 16 '15 at 17:52
  • $\begingroup$ This is what we call pseudo-vector. $\endgroup$ – user36790 Dec 16 '15 at 17:55
  • $\begingroup$ I'm not convinced that $v = \omega \times r$ is always true. Consider again a body rotating (in a circle with some angular speed) parallel to the xy-plane, but this time with radius $1$, and with center point at $(x,y,z)=(0,1,1)$ ($z$ is always $1$ on the circle). So one of the points on the circle is $(0,0,1)$. Clearly $\omega$ by your definition is pointing in the $z$-direction, but then at this point $\omega$ and $r$ are parallel, so $\omega \times r$ would be $0$ which is not equal to $v$. $\endgroup$ – user100898 Dec 19 '15 at 15:19
  • $\begingroup$ Also that formula for $\omega$ seems to have an $\omega$ on both sides :) $\endgroup$ – user100898 Dec 19 '15 at 15:24
  • $\begingroup$ I suggest adding this example of “body rotating (in a circle...) parallel to the xy-plane, ... with center point at $(x,y,z)=(0,1,1)$ ... ”, to the original post. @NotNotLogical $\endgroup$ – AlQuemist Dec 19 '15 at 22:57
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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.

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  • $\begingroup$ I'm not sure about that formula either. See my comment to another answer: Consider again a body rotating (in a circle with some angular speed) parallel to the xy-plane, but this time with radius $1$, and with center point at $(x,y,z)=(0,1,1)$ ($z$ is always $1$ on the circle). So one of the points on the circle is $(0,0,1)$. If $\omega$ is pointing in the $z$-direction, then at this point $\omega$ and $r$ are parallel, so $\omega \times r$ would be $0$ which is not equal to $v$. $\endgroup$ – user100898 Dec 19 '15 at 15:22
  • $\begingroup$ @NotNotLogical I don't agree with you description of the motion. If the motion is as you describe, then the orbit does not pass through $(0,0,1)$. Instead the velocity would be $v=(-\omega,0,0)$, which is what the above formula produces $\endgroup$ – Mikael Fremling Jan 1 '16 at 15:51
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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.

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  • $\begingroup$ I guess I'm not sure how this answers the question. What I'm asking is if $\omega$ should point along the axis of rotation, and if not, what it should be interpreted to mean. $\endgroup$ – user100898 Dec 10 '15 at 19:13
  • $\begingroup$ $\omega$ according to you will either be clockwise or anticlockwise right? But to avoid any confusion( as mentioned in answer), we chose $\omega$ to have a direction perpendicular to the plane of rotation or into the plane of rotation. Does angular velocity actually lie along the axis of rotation ( parallel to it) , no, but we claim so only to eliminate any possible physical conflict in solutions to problems. It's like a convention of sorts. As you can see in the formula you mentioned above, we basically use the screw rule to figure out the direction of angular velocity. $\endgroup$ – Garvit Sharma Dec 11 '15 at 12:10
  • $\begingroup$ I'm still not understanding. You say $\omega$ has "a direction perpendicular to the plane of rotation" and yet is not along the axis of rotation. I would think these are the same directions. And you're saying that the angular velocity vector does not lie along the axis of rotation, and yet we say it does? $\endgroup$ – user100898 Dec 19 '15 at 15:27

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