Angular velocity basics

Question

Is angular velocity only defined for circular motion or can it be described for lets say projectile motion as well?

Answer 1

For any two-dimensional motion we can switch to polar coordinates: $$ x,y\rightarrow r,\phi,\\ x=r\cos\phi, y=r\sin\phi. $$ Then instead of linear velocities $v_x=\dot{x},v_y=\dot{y}$ we will have radial and angular velocities: $$ v_r=\dot{r}, \omega=\dot{\phi}. $$

Similar approach is possible in in spherical coordinates, but, since in that case we have two angles characterizing position, term angular velocity becomes ambiguous. It is thus customary to speak about the velocity of rotation about a selected axis, whose direction is specified in three dimensions.

Answer 2

Absolutely, the angular momentum of a particle or object is defined regardless of the shape of its trajectory. The angular momentum of a particle is simply $\vec{r} \times \vec{p}$, where $\vec{r}$ is its position vector and $\vec{p}$ is its momentum vector.

If you are reasoning about the angular momentum of a particle which is moving in a circular orbit, then it would be natural to define the origin of your coordinate system at the center of that orbit. But all the laws of mechanics which concern angular momentum remain true no matter where the origin is.

Answer 3

This is a good question because angular velocity for a particle in circular motion only makes sense if you consider a virtual rigid body with zero mass, except for this one particle and you track this particle's velocity as it moves around in space but he body is pinned to the origin.

In the context of a particle being part of a rigid body you can write the famous equation for the velocity vector $\vec{v}$ in terms of the position $\vec{r}$ and a constant vector $\vec{\omega}$ which we call angular velocity:

$$ \vec{v} = \vec{\omega} \times \vec{r} \tag{1}$$

This angular velocity vector designates this axis of rotation, at least in terms of its magnitude and direction.

Many textbooks call this virtual rigid body a rotating frame. It is the same thing as the parts of the rotating frame move together, just as the parts of a rigid body move together is it moves trough space.

Now for your question, as I understand it. What if the path of the particle isn't a circle, and its distance from the origin isn't fixed, and hence cannot be considered part of a rigid body that is pinned to the origin?

There is the school of thought that at any instant you can recover some angular velocity $\vec{\omega}$ from the velocity vector $\vec{\omega}$ and the position $\vec{r}$ with

$$ \vec{\omega}_{\perp} = \frac{\vec{r}\times\vec{v}}{\|\vec{r}\|^{2}} \tag{2}$$

but this only recovers the components of $\vec{\omega}$ that are perpendicular to $\vec{r}$.

Some say that you can always attach a rigid body to a moving particle, but this body cannot be pinned to the ground. The most general case happens when at any instant this rigid body is rotating about some arbitrary point and having a parallel velocity $\vec{v}_\parallel$ along the axis of rotation such that the velocity of the particle is decomposed as:

$$ \vec{v} = \vec{v}_\parallel + \vec{\omega}\times \vec{r} \tag{3} $$

with the location of the axis of rotation found by

$$ \vec{r} = \frac{ \vec{v} \times \vec{\omega} }{\| \vec{\omega} \|^2} \tag{4}$$ or at least the point on this axis of rotation closest to the origin.

With this view, the rotational velocity $\vec{\omega}$ is a well-defined quantity that arises from the requirement that all particles within a rigid body must maintain fixed distances with each other. This leads to the transformation law for velocities between two points A and B as

$$ \vec{v}_A = \vec{v}_B + \vec{\omega} \times (\vec{r}_A - \vec{r}_B) \tag{5}$$