For a particle executing uniform circular motion:

The position vector of a particle is $\vec{r} = a \cos \omega t \hat{i} + a \sin \omega t \hat{j}$. Calculate the angle between its position vector and the velocity vector. $(\cos 0 = 1, \cos 90 = 0, \cos 60 = 0.5, \sin 30 = 0.5, \sin 0 = 0)$ (in $^\circ$)

$A$ particle is moving on a circular path of radius $r$ with uniform speed $v$. What is the displacement of the particle after it has described an angle of $60^{\circ}$?

$A$ particle $P$ is moving in a circle of radius $a$ with a uniform speed $v$. $C$ is the centre of the circle and $AB$ is a diameter. When passing through $B$,the angular velocity of $P$ about $A$ and $C$ are in the ratio:

$A$ body moves along a circular path of diameter $30 \text{ cm}$. It starts from one end of a diameter, moves along the circular path, and reaches the other end of the diameter in $3 \text{ seconds}$. The angular speed of the body in radians per second is:

If the linear speed of a particle moving in a circular path is constant,then the angular speed is proportional to (where $r$ is the radius of the circle):