A few kids decide to go to the park to go on the merry-go-round. They are the only ones there so the merry-go-round is not moving yet. They push it for 6.00 s and it moves over an angle of 129°. What is the final angular velocity of the merry-go-round?

Answer 1

assume constant torque = constant angular acceleration alpha

omega = alpha * t
theta = omega initial*t +(1/2)alpha t^2

129 * 2pi/360 = 2.25 radians
so
2.25 = 0 + .5 alpha (36)
alpha = .125 rad/s^2

omega = alpha t = .125*6 = .75 rad/s