When is $v=r\omega $? I have seen a few posts about this relationship, but none seemed to get at the answer of when the above holds. I am thinking of orbits rather than rotational motion.
The expression is easily derived from assuming uniform circular motion. If a particle does not have uniform circular motion, what does this relationship become? I would think (but am not at all sure) that this would still apply to relating a particle's instantaneous angular velocity with the component of its linear velocity that is perpendicular to the position vector of the particle from some given origin point (about which it's angular momentum is calculated) at any given moment.
How does this apply to elliptic orbits, for example?
EDIT: perhaps it is related, but my lecturer also said that we only get stable orbits when the kinetic energy is half of the potential energy (and or non-circular orbits this was over an average of one orbit). I was also wondering why this was the case and found that it holds if I assume $v=r\omega $. I would also be grateful if, if it is a simple extension of the above question, someone could explain this as well.
 A: A simple diagram will help. Your expression had only magnitudes - but in reality, all three quantities are vectors (even rotation!). This means that you need to take into account that if the radius is changing, then so will the velocity.
The instantaneous position is given by $\vec{r}$; the velocity is the time derivative, 
$$\vec{v} = \frac{d\vec{r}}{dt}$$
You can think of this as being made up of two parts - the radial velocity, and the tangential velocity. The radial velocity is independent of rotation; the tangential velocity depends on the instantaneous rate of rotation.
The real equation, then, is
$$\vec{v} = \frac{d|r|}{dt}\hat{r} + \vec\omega\times\vec{r}$$
where $\hat{r}$ is the unit vector pointing along the position vector.
Now it is easy to see that when you have circular motion, the first term is zero (no change in length of the radial vector) - then the expression simplifies to the one you gave (although mine is still in vector form - showing that the velocity is at right angles to both the angular velocity vector (which points along the axis of rotation) and the radial vector).
So the short answer to your question: the expression holds when the distance of the particle to the axis of rotation doesn't change.
A: You're right, when you have a uniform circular motion you can use that equation, but the most general expresion is another one, and it's not so hard to understand.
In general cases you always can define your system (in 2 dimensions) with two vectors. We will name them $n$ as the normal vector and $l$ as the azimutal vector.
The expresion of these vector are:
$\vec{n}=(cos\theta,sin\theta)$
$\vec{l}=(-sin\theta,cos\theta)$
You can see that $\vec{\dot n}=\dot \theta·\vec{l}$.
Where $\dot \theta$ is $w$
Now you can define the position vector with the normal vector as a constant $r$ multiplied by $n$
$\vec{r}=r·\vec{n}$
If you derive
$\vec{\dot r}=\dot r·\vec{n}+r·\dot \theta·\vec{l}$
So if your particle doesn't move on the radial direction you can make $\dot r$=$0$ and you got the initial equation.
But in eliptical moves as you ask, you have moves on the radial direction, so the most general expresion for the lineal velocity $\vec{\dot r}$ is the one demonstrated.
Also you can obtain the most general expresion for the lineal acceleration if you derive again the velocity and considering that:
$\vec{\dot l}=-\dot \theta·\vec{n}$
I hope I've helped you.
