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Linear velocity vector is $\vec{v}$ and the distance vector is $\vec{r} $.

How is it that angular velocity vector is $\vec{ω}$ = $\frac {\vec{r} × \vec{v}}{|\vec{r}|^2}$

The equation I am aware of is $\vec{v}$ $=$ $\vec{r}×\vec{ω}$. I tried taking a dot product on both sides, cross product etc but I wasn't able to derive the above equation. How is it possible ?

Shaurya Goyal
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2 Answers2

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The triple vector-product identity gives $$ {\bf r}\times {\bf v}= {\bf r}\times({\boldsymbol \omega}\times {\bf r} )=-({\boldsymbol \omega}\cdot {\bf r}){\bf r}+ |{\bf r}|^2 {\boldsymbol \omega}. $$ But $({\boldsymbol \omega}\cdot {\bf r})$ is not necessarily zero, and so in general ${\boldsymbol \omega}\ne ({\bf r}\times {\bf v})/|r|^2$. It's only true if ${\bf r}$ is perpendicular to ${\boldsymbol \omega}$

In my original answer I miscopied fom a piece of paper and had $-({\boldsymbol \omega}\cdot {\bf v}){\bf r}$ in the first term, and this is zero, but wrong!

mike stone
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The correct definition of linear velocity is $\vec{v} = \vec{\omega}\times\vec{r}$ and your relation for $\vec{\omega}$ is then a simple application of the vector triple product, see https://en.wikipedia.org/wiki/Triple_product.

(Or the other way around if you prefer that.)

Edit: As @mikestone correctly notes in his answer this is only valid for circular motion where the position vector is orthogonal to the angular velocity vector.

NDewolf
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  • I made an error in my answer! It is not generally true that $\omega= r\times v/|r|^2$. You should probably undo the green tick of approval you gave me! – mike stone Sep 04 '20 at 17:20
  • @mikestone I think you meant this to be a comment to the question, right? Furthermore, i gave a similar answer to you. So this would also invalidate mine. Could you elaborate? – NDewolf Sep 04 '20 at 18:21
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    I edited my anwer, forgot to save it, and then wrote my comment. I have now, at last, put what I believe to be correct answer! – mike stone Sep 04 '20 at 20:28