# When does the 'standard' angular velocity formula not hold?

I have read that the formula for angular velocity: $$\dot {\vec r}=\vec \omega \times\vec r \tag{1}$$ does not hold in some situations, but the book does not specify what situation so please could you produce a list of when this formula does not hold.

If this formula does not hold is it also true that: $$\vec \omega= \frac{\vec r \times \vec v}{|\vec r|^2} \tag{2}$$ does not hold?

• Where did you read this? The context might be helpful. Also, I assume your $w$ is meant to be $\omega$ (written $\omega$ in MathJax)? – Floris Feb 3 '15 at 15:21
• @Floris It was in 'Introduction to classical mechanics' by David Morin, the context is in general circular motion, in 3d. – Quantum spaghettification Feb 3 '15 at 15:26
• I wonder whether this is talking about the situation where $\vec\omega$ is not constant ($\dot{\vec\omega}≠0$) - I think an additional term might appear in that case. – Floris Feb 3 '15 at 15:30
• @Floris I think I am, like when we have a torque acting – Quantum spaghettification Feb 3 '15 at 15:34
• I don't think there is a term for $\dot{\vec{\omega}}\ne0$, but there is a term if your rotating object has a non-zero translational velocity. There could also be a term if your object is doing some motion on top of a rotation. For example if you are driving a car on earth, your velocity relative to the stars is the sum of the earth's velocity, plus the $\vec{\omega} \times \vec{r}$ velocity that comes from earth's rotation, plus your velocity relative to the surface of the earth from you driving the car. – Brian Moths Feb 3 '15 at 15:42

Electron spin is not the result of a rotation of the electron around itself. In this case, of course (2) also doesn't hold.

In fact, one can show that there is a double implication as follows:

1) if $\vec v$ is defined as in (1) one gets

$$\frac {\vec r \times \vec v}{r^2} = \vec {\omega} - \vec r \frac {(\vec r \cdot \vec {\omega})}{r^2}. \tag{I}$$

So, as $\vec {\omega}$ is perpendicular to $\vec r$ the equality (2) is implied.

2) On the other hand if the equality (2) is true it implies

$$r^2 (\vec {\omega} \times \vec r) = (\vec r \times \vec v) \times \vec r = \vec v \ r^2 - \vec r (\vec v \cdot \vec r). \tag{II}$$

So, if your equality (2) is true, and $\vec v$ is defines as tangential velocity, then it implies (1). Therefore if $\vec v$ is defines as tangential velocity, and (1) isn't true, (2) cannot be true, otherwise it would imply that (1) is true.

• I go the second equation from wiki en.wikipedia.org/wiki/… – Quantum spaghettification Feb 3 '15 at 16:03
• it can't be true? why not? and you are saying it doesn't hold... – chuse Feb 3 '15 at 16:04
• @Joseph sorry, I made indeed a mistake in the calculus – Sofia Feb 3 '15 at 16:07
• @chuse , yes, thank you, it was a mistake in my calculus. I corrected it. – Sofia Feb 3 '15 at 16:09
• I got stuck (and stopped reading) at $\vec{\omega}$ is perpendicular to $\vec{r}$. That seems incorrect to me. $\vec{\omega}$ is perp to $\vec{v}$, and $\vec{r}$ is perp to $\vec{v}$, but in general $\vec{\omega}$ is not perp to $\vec{r}$. – garyp Feb 3 '15 at 18:14

I have read that the formula for angular velocity: $$\dot {\vec r}=\vec \omega \times\vec r \tag{1}$$ does not hold in some situations, but the book does not specify what situation so please could you produce a list of when this formula does not hold.

That expression is only true in the case of circular motion. It fails whenever the radial component of velocity is non-zero.

If this formula does not hold is it also true that: $$\vec \omega= \frac{\vec r \times \vec v}{|\vec r|^2} \tag{2}$$ does not hold?

That expression is tautologically true; that's one way the angular velocity of a point mass is defined.

It helps to define a set of unit vectors: \begin{aligned} \hat r &= \frac {\vec r}{||\vec r||} \\ \hat \omega &= \frac {\vec \omega}{||\vec \omega||} \\ \hat \theta &= \hat \omega \times \hat r \end{aligned}

The above unit vectors are well-defined and are mutually orthogonal so long as $\vec r \times \vec v$ is non-zero. Denoting $r = ||\vec r||$ and $\omega = ||\vec \omega||$, the above yields $$\vec v = \frac {d\vec r}{dt} = \frac {d}{dt}(r \hat r) = \frac {dr}{dt} \hat r + r \frac {d\hat r}{dt} = \dot r \hat r + r\omega \hat \theta = \frac{\dot r} r \vec r + \vec \omega\times\vec r$$ With this, your equation (1) becomes $$\vec v = \frac{\dot r} r \vec r + \vec \omega\times\vec r \tag{1'}$$ The above reduces to your equation (1) when $\dot r = 0$.