A particle is moving with uniform speed along the circumference of a circle of radius `R` under the action of a central fictitious force `F` which is inversely proportional to `R^(3)`. Its time period of revolution will be given by :

A particle moves along the circumference of a circle of radius r when its completes one rotation.

A particle of mass M moves with constant speed along a circular path of radius r under the action of a force F. Its speed is

A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely proportional to the n^(th) power of R. If the period of rotation of the particle is T, then :

A particle is revolving in a circle of radius R. If the force acting on it is inversely proportional to R, then the time period is proportional to

A particle is moving with speed v in a circle of radius R. Find the magnitude for average velocity of point for the half revolution

A particle is moving in a circular path of radius a under the action of an attractive potential U=-(k)/(2r^(2)) . Its total energy is :

A particle is moving with constant speed V m//s along the circumference of a circle of radius R meters as shown. A, B and C are three points on periphery of the circle and DeltaABC is equilateral. The magnitude of average velocity of particle, as it moves from A to C in clockwise sence, will be :

A particle moves with constant speed v along a circular path of radius r and completes the circle in time T. The acceleration of the particle is

A particle with the constant speed in a circle of radius r and time period T.The centripetal acceleration of a particle is

Two particles of equal mass go around a circle of radius R under the action of their mutual gravitational attraction. The speed v of each particle is