Question

A merry-go-round rotates at the rate of 0.20 rev/s with an 75 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 1.0 m from the center? Assume that the merry-go-round is a solid 25 kg cylinder of radius 2.0 m.

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

Answer 1

Given,

w1 = 0.2 rav/s = 1.26 rad/s ; M = 75 kg ; r = 2 m ; m = 25 kg ; r = 2 m

a) r' = 1 m

I1 = 1/2 m r^2 + m r^2

I1 = 0.5 x 25 x 2^2 + 75 x 2^2 = 350 kg-m^2

I2 = 0.5 x 25 x 2^2 + 75 x 1^2 = 125 kg-m^2

from conservation of angular momentum

Li = Lf

I1 w1 = I2 w2

w2 = (I1/I2) w1

w2 = (350/125) x 1.26 = 3.528 rad/s

Hence, w2 = 3.528 rad/s = 0.56 rev/min

b)KEi = 1/2 I1 w1^2

KEi = 0.5 x 350 x 1.26^2 = 277.83 J

KEf = 0.5 x 125 x 3.53^2 = 778.81 J

change in KE

delta-KE = 778.81 - 277.83 = 501 J

Hence, delta-KE = 501 J