A swing moving in a children's garden is observed to move with an angular velocity given by, `omega-a(t^2)hat(i)+b(e^-1)hat(j)`. What will be the angle between angular acceleration and angular velocity at t=1 s given that a=b=1 unit?

What is the angular acceleration of a particle moving with constant angular velocity ?

What is the angular acceleration of a particle moving with constant angular velocity ?

What is the angular acceleration of a particle moving with constant angular velocity ?

The angular velocity of a particle is given by omega=1.5t-3t^(2)+2 , Find the time when its angular acceleration becomes zero.

The angular velocity of a particle is given by omega=1.5t-3t^(@)+2 , Find the time when its angular acceleration becomes zero.

The angular velocity of a particle is given by omega=1.5t-3t^(@)+2 , Find the time when its angular acceleration becomes zero.

The angular velocity of a particle is given by the equation omega =2t^(2)+5rads^(-1) . The instantaneous angular acceleration at t=4 s is

The angular displacement of a body is given by theta = 2t^(2) + 5t -3 . Find the value of the angular velocity and angular acceleration when t = 2s.

If the velocity is vec(v) = 2 hat(i) + t^(2) hat(j) - 9 vec(k) , then the magnitude of acceleration at t = 0 . 5 s is :

A particle is moving in a circular orbit of radius r_(1) with an angular velocity omega_(1) . It jumps to another circular orbit of radius r_(2) and attains an angular velocity omega_(2) . IF r_(2) = 0.5 r_(1) and assuming that no external torque is applied to that system, then, the angular velocity omega_(2) is given by