Question

The angular velocity of a particle moving in a circle relative to the center of the circle is equal to $\alpha$. Find the angular velocity of the particle relative to a point on the circular path.

Answer 1

${\omega }_{BA}=\frac{v_{BA\perp }}{AB}$

Where $v_{BA\perp }=v\cos \theta$ and $AB=AC\cos \theta$ because $ABC$ is right-angled triangle.

Then,${\omega }_{BA}=\frac{v}{AC}=\frac{v}{2R}$

Substituting $\frac{v}{R}={\omega }_{BO},wehave$\omega_{BA}=\dfrac{1}{2}\omega_{BO}\left(-\dfrac{1}{2}\omega\right)$

Alternative procedure:

$\varphi =2\theta$

Then $\frac{d\varphi }{dt}(-{\omega }_{BO})=2\frac{d\theta }{dt}$

Where $\frac{d\theta }{dt}{\omega }_{BA}$

This given ${\omega }_{BO}=2{\omega }_{BA}.$