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$ \require{enclose} $ The relation between angular velocity and linear velocity is given by this equation (from openstax: angular velocity):

$$ \mathbf{v} = \boldsymbol{\omega} \times \mathbf{r} $$

I can rearrange $\boldsymbol{\omega}$ to the left while keeping the right-hand-rule: $$ \enclose{horizontalstrike}{\boldsymbol{\omega} = \mathbf{r} \times \mathbf{v}} $$

But Wikipedia defines $\boldsymbol{\omega}$ in three-dimension to be this instead:

$$ \boldsymbol{\omega} = \frac{\mathbf{r} \times \mathbf{v}}{r^2} $$

Why is the $ \boldsymbol{\omega} $ scaled down by a factor of $ \frac{1}{r^2} $ in three-dimension?

There is a somewhat related question in Physics SE: Is Wikipedia's definition of angular velocity incorrect? but it doesn't address the additional $\frac{1}{r^2}$ factor.

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    $\begingroup$ How do you go from the first equation to the second? $\endgroup$
    – jacob1729
    Commented Nov 15, 2022 at 11:38
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    $\begingroup$ Re I can rearrange $\boldsymbol\omega$ to the left while keeping the right-hand-rule: No, you can't. What you've done is invalid. Look at the units. $\endgroup$ Commented Nov 15, 2022 at 13:50
  • $\begingroup$ You can write it as, $\omega\hat z=\frac{v}{r}\hat r\times \hat{\theta}$. $\endgroup$ Commented Nov 15, 2022 at 16:02

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As you likely know, in one dimensional circular motion, $\omega=\frac{v}{r}$, so the three dimensional equation you have written is a correct generalization of that.

We can do two fast sanity checks to verify that the equation on wikipedia is correct, and yours is not.

First, we can check if the wikipedia equation is dimensionally consistent:

\begin{equation} [\vec{\omega}] = {\rm s}^{-1} = \frac{{\rm m\ s^{-1}}}{{\rm m}} = \left[\frac{v}{r}\right] = \left[\frac{\vec{r} \times \vec{v}}{r^2}\right] \end{equation} where the notation $[x]$ means "the dimension of the quantity $x$". If you check your proposed equation, you will find it is not consistent.

Second, we can check the self consistency of the two equations $\vec{v}=\vec{\omega}\times \vec{r}$ and $\vec{\omega}=\frac{\vec{r}\times\vec{v}}{r^2}$. Actually, if you closely read your source, you will find that $\vec{v} = \vec{\omega} \times \vec{r}$ only holds for purely circular motion. In general, this equation only holds for the tangential part of velocity (the velocity with the radial component removed). With this caveat, we can check the self-consistency as follows: \begin{eqnarray} \vec{v}_\perp &=& \vec{\omega} \times \vec{r} \\ &=& \left(\frac{\vec{r}\times \vec{v}}{r^2}\right) \times \vec{r} \\ &=& -\frac{1}{r^2} \vec{r} \times \left(\vec{r} \times \vec{v}\right) \\ &=& - \frac{1}{r^2} \left[\left(\vec{r} \cdot \vec{v}\right) \vec{r} - \left(\vec{r} \cdot \vec{r}\right) \vec{v}\right] \\ &=& \vec{v} - v_r \hat{r} \\ &=& \vec{v}_\perp \end{eqnarray} where to go from line three to line four I used the vector triple product, and in the last two lines I defined the unit vector $\hat{r}=\vec{r}/r$, the radial component of velocity $v_r= \vec{v}\cdot \hat{r}$, and the tangential component of velocity (the part of the velocity that is not radial), $\vec{v}_\perp = \vec{v} - v_r \hat{r}$.

If you try a similar analysis with your proposed equation for $\vec{\omega}$, you will find it is not consistent.

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  • $\begingroup$ In the first equation, how do you go from $ v/r$ to $ (\mathbf{r} \times \mathbf{v}) / r^2$? That also implies that $ v = (\mathbf{r} \times \mathbf{v}) / r $ ... how could a cross product results in a scalar $v$? I don't think I understand what the first equation means... $\endgroup$
    – KMC
    Commented Nov 15, 2022 at 14:10
  • $\begingroup$ @KMC The brackets $[ \cdot ]$ refer to the dimensions of the quantity. Andrew isn't saying that $v/r = \vec r\times \vec v/r^2$, but rather that their dimensions are the same (both of them have dimensions of inverse time). $\endgroup$
    – J. Murray
    Commented Nov 15, 2022 at 14:23
  • $\begingroup$ @KMC J. Murray is correct. I added a note to clarify. All I'm saying with that equation is that $\vec{\omega}=\vec{r}\times \vec{v}/r^2$ could be correct based on dimensional analysis, while $\vec{\omega}=\vec{r}\times\vec{v}$ cannot. $\endgroup$
    – Andrew
    Commented Nov 15, 2022 at 14:56
  • $\begingroup$ @KMC, you got a very nice answer by J. Murray. If I could add a small contribution, we can write $- \mathbf{r} \times ( \mathbf{r} \times \mathbf{v}) = - \left[ \mathbf{r}_{\times} \mathbf{r}_{\times} \right] \cdot \mathbf{v} = \left[ \mathbb{I} - \mathbf{\hat{r}} \otimes \mathbf{\hat{r}} \right] \cdot \mathbf{v} = \mathbb{P}_{\perp \mathbf{r}} \cdot \mathbf{v}$, where $\mathbb{P}_{\perp \mathbf{r}}$ is nothing more than the orthogonal projector in the directions that are perpendicular to the vector $\mathbf{\hat{r}}$ $\endgroup$
    – basics
    Commented Nov 15, 2022 at 15:15

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