I know that when the motion of a particle is circular about the origin then: $$\vec v=\vec \omega \times \vec r$$ But that this does not hold for any motion with a radial as well as tangential component. So how can we define angular velocity [mathematically] for general motion?
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asked Mar 31, 2015 at 5:13
Quantum spaghettificationQuantum spaghettification
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When you fix a reference point (take it to be the origin of your reference frame) you can write the position as
$$\vec{r} = r \hat{r}$$
where $\hat{r}$ is the unit vector pointing toward the particle. Deriving you obtain
$$\vec{v} = \frac{dr}{dt} \hat{r}+ r \frac{d \hat{r}}{dt}$$
The first term is the radial component of the velocity, the second one is proportional to the derivative of a vector with fixed modulus, so it can be represented as
$$\frac{d \hat{r}}{dt} = \vec{\omega} \times \hat{r}$$
and inserting this in the previous equation you get
$$\vec{v} = \frac{dr}{dt} \hat{r}+ \vec{\omega} \times \vec{r}$$
You can extract the second term from the velocity by projecting in the plane orthogonal to the radial vector,
$$\vec{v}-(\vec{v}\cdot \hat{r})\hat{r} = \vec{\omega} \times \vec{r}$$ which generalize your equation. The angular velocity is perpendicular to the plane defined by $\vec{r}$ and $\vec{v}$, and depends on the reference point. A preferred reference point is the center of the osculator circle of the trajectory, at a given time. In this case the radial component of the velocity is zero, but unless the trajectory is circular this preferred point is not fixed.
