1.

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»

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}


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