A particle moves along a circle of radius `R` with a constant angular speed `omega` . Its displacement (only magnitude) in time `t` will be

A particle moving along a circule path of radius 'r' with uniform angular velocity omega . Its angular acceleration is

A particle moves along a semicircular path of radius R in time t with constant speed. For the particle calculate

A particle of mass m is moving along a circle of radius r with a time period T . Its angular momentum is

A particle moves over three quarters of a circle of radius r . What is the magnitude of its displacement ?

A particle moves over three quarters of a circle of radius r . What is the magnitude of its displacement ?

An alpha is moving along a circle of radius R with a constant angular velocity omega . Point A lies in the same plane at a distance 2R from the centre. Point A records magnetic field produced by the alpha -particle. If the minimum time interval between two successive time at which A records zero magnetic field is t, the angular speed omega , in terms of t, is

A particle moves along an arc of a circle of radius R according to the law l=a sin omegat , where l is the displacement from the initial position measured along the arc, and a and omega are constants. Assuming R=1.00m , a=0.80m , and omega=2.00rad//s , find: (a) the magnitude of the total acceleration of the particle at the points l=0 and l=+-a , (b) the minimum value of the total acceleration w_(min) and the corresponding displacement l_m .

If a particle is moving along a circle of radius 3 m with a constant speed 9 m//s , then it covers a quarter of the circle in time of

A particle moves in circle of radius R with a constant speed v. Then, find the magnitude of average acceleration during a time interval (pi R)/(2v) .

If the body is moving in a circle of radius r with a constant speed v , its angular velocity is