help help help?

Question

physics homework, i need the steps to get through this?
a merry go round rotates at a .20 rev/s with a 80kg man standing at a point 2.0 m from the center. if he moves to a point 1.0 m from the center, what will be the new angular speed of the ride?
how much does the rotational kinetic energy change during the motion? assume the ride is a uniform solid 25 kg cylinder and its radius is 2.0 m and that the man can be modeled as a uniform solid sphere for moment of inertia measurements?

answer is 3.6 rad/s and 540 J , how do i get here?

Answer 1

What a badly worded question! Assuming the man isn't sliding on the merry go round, then his angular speed does not change with distance from the center. At 2 m or 1 m or 0.01 m from the center, he is still going around at the same RPM. The mass and shape of the man or the merry go round makes no difference. I am not sure what rotational kinetic energy is, but since the instantaneous liner speed is slower nearer the center, the kinetic energy is less too. Kinetic energy is related to speed squared and speed is distance over time. The time is not changing but the distance decreases as you get closer to the center. It is just the circumference at the specified radius. So speed decreases linearly and kinetic energy decreases as the square of the speed change.