A horizontal 810-N merry-go-round of radius 1.70 m is started from rest by a constant horizontal force of 55 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go-round after 2.0 s. (Assume it is a solid cylinder.)

Answer 1

The kinetic energy is (1/2) I w^2 , where I is the moment of inertia, which is (1/2) M R^2 for a solid cylinder. Therefore KE = (1/2) M R^2 w^2

w is the angular velocity in radians per second. To determine its value after 2.0 seconds, you will need to use the equation of motion

T = I*dw/dt
where T is the torque, which in this case is 55 * 1.70 = 93.5 N m/s. You will need to compute dw/dt, the rate of change of angular velocity.

The value of w at 2.0 s is
w (@t=2) = dw/dt * 2.0 radians per second.

Now put all that information together and do the calculation.