Question

a particle is moving with constant speed v in xy plane as shown in figure. the magnitude of its angular velocity about point O is

Answer 1

Final Answer :
$\frac{vb}{a^2+b^2}$

Steps:
1) Let break velocity 'v' into two components.
$v \cos(\theta)$ will contribute in angular motion.
But, $v\sin(\theta)$ is parallel to $\vec{r}$ ,and hence not contribute in angular velocity.

2) We have,
$\cos(\theta) = \frac{b}{\sqrt{a^2+b^2}} \\ r = \sqrt{a^2 +b^2 }\\ And\: \\ \omega = \frac{v \cos(\theta)}{r} \\ => \omega = \frac{vb}{a^2+b^2}$

Hence,
The magnitude of Angular Velocity is
$\frac{vb}{a^2+b^2}$