Question

A merry-go-round rotates at the rate of 0.20 rev/s with an 80-kg man standing at a point 2.0 m from the axis of rotation.

(a) What is the new angular speed when the man walks to a point 0.8 m from the center? Assume that the merry-go-round is a solid 25-kg cylinder of radius 2.0 m.
rad/sec

(b) Calculate the change in kinetic energy due to this movement.

Answer 1

part A)

let the new angular speed of the ride is w

Using conservation of angular momentum

I1 * w1 = I2 * w2

(0.50 * 25 * 2^2 + 80 * 2^2) * 0.20 = w * (0.50 * 25 * 2^2 + 80 * 0.80^2)

solving for w

w = 0.73 rev/s

the new angular speed is 0.73 rev/s

part B)

change in kinetic energy = final KE - initial KE

change in kinetic energy = 0.50 * (0.50 * 25 * 2^2 + 80 * 0.80^2) * (0.73 * 2pi)^2 - 0.50 * (0.50 * 25 * 2^2 + 80 * 2^2) * (0.20 * 2pi)^2

change in kinetic energy = 772.4 J

the change in kinetic energy is 772.4 J