Angular velocity at a general point

Question

I have been reading circular motion for a while and the magnitude of angular velocity is defined as component of linear velocity (perpendicular to the position vector) divided by position vector.

But why only the perpendicular component of velocity is used and not the parallel component?

Answer 1

The first tool you should use when describing circular motion is the polar basis. It can be used for other motions but, its two vectors have the following properties for circular motion:

• $\hat{u}_r$ is alongside a radius of the circle
• $\hat{u}_\theta$ is tangent to the circle

Knowing this, you can decompose velocity on this basis. Let's interpret both component:

• The orthoradial (=perpendicular to radius) component $\vec{v}.\hat{u}_\theta$ describes a motion following the circle.
• The orthoradial component $\vec{v}.\hat{u}_r$ describes a motion alongside a radius.

It's obvious that, if the motion is circular, the latter cannot exist. Therefore the whole velocity vector is orthoradial. Let's confirm this by calculation. The position vector is : $$\vec{r}=r\,\hat{u}_r$$ So velocity is, since $r$ is constant for a circular motion: $$\vec{v} =\frac{d\vec{r}}{dt} =r\frac{d\hat{u}_r}{dt} =r\omega\,\hat{u}_\theta$$ This confirms that velocity and position vector are orthogonal. But there was a slight mistake in your question, velocity isn't equal to the angular velocity, it's proportional ($v=r\omega$).