Question

If a point moves along a circle with constant speed, prove that its angular speed about any point on the circle is half of that about the centre.

Answer 1

Let, O be a point on a circle and P be the position of the particle at any time t, such that
$\angle POA=\theta .Then,\angle PCA=2\theta$
Here, C is the centre of the circle.
Angular velocity of P about O is
${\omega }_0=\frac{d\theta }{dt}$
and angular velocity of P about C is,
${\omega }_c=\frac{d}{dt}\left(2\theta \right)=2\frac{d\theta }{dt}$
or ${\omega }_c=2{\omega }_0$ Proved.