If a particle is moving on a circular path with constant speed, then the angle between the direction of acceleration and its position vector w.r.t. centre of circle will be ............ 

A cycle wheel of radius 0.4 m completes one revolution in one second then the acceleration of a point on the cycle wheel will be

A man is running with constant speed along a circular path of radius $2 \sqrt 2\, m$. He completes $1$ round in $10\, second$. Find instantaneous speed at $2.5 \,sec.$

A particle moving in a circle of radius $R$ with uniform speed takes time $\mathrm{T}$ to complete one revolution. If this particle is projected with the same speed at an angle $\theta$ to the horizontal, the maximum height attained by it is equal to $4 R$. The angle of projection $\theta$ is then given by :

A small block slides down from rest at point $A$ on the surface of a smooth cylinder, as shown. At point $B$, the block falls off (leaves) the cylinder. The equation relating the angles $\theta_1$ and $\theta_2$ is given by

A cyclist starts from centre 0 of a circular park of radius $1\, km$ and, moves along the path $OPRQO$ as shown in figure.

If he maintains constant speed of $10\, ms^{-1}$, what is his acceleration at point $R$ in magnitude and direction ?