A particle starts moving at t = 0 in a circle of radius `R = 2m` with constant angular acceleration of `a - 3 "rad/sec"^(2)`. Initial angular speed of the particle is 1 rad/sec. At time `t_(0)` the angle between the acceleration vector and the velocity vector of the particle is `37^(@)`.
What is the value of `t_(0)` ?

A particle starts moving at t = 0 in a circle of radius R = 2m with constant angular acceleration of a - 3 "rad/sec"^(2) . Initial angular speed of the particle is 1 rad/sec. At time t_(0) the angle between the acceleration vector and the velocity vector of the particle is 37^(@) . What is the magnitude of the acceleration of the particle at t_(0) ?

A particle moves along a circular path a radius 15 cm with a constant angular acceleration of 4 "rad/s"^(2). If the initial angular speed of the particle was 5 rad/s. Find its angular displacement in 5 seconds.

A particle is revolving in a circular path of radius 2 m with constant angular speed 4 rad/s. The angular acceleration of particle is

A particle is revolving in a circular path of radius 2 m with constant angular speed 4 rad/s. The angular acceleration of particle is

A particle starts moving on circular path of radius (9)/(4)m with tangential acceleration 3ms^(-2) . Find angle between acceleration of the particle and radius vector at time t=1 sec.

Starting from rest, a particle rotates in a circle of radius R = sqrt 2 m with an angular acceleration = pi/4 rad/s2.The magnitude of average velocity of the particle over the time it rotates quarter circle is:

Starting from rest, a particle rotates in a circle of radius R  2 m with an angular acceleration  = /4 rad/s2.The magnitude of average velocity of the particle over the time it rotates quarter circle is: