Question

Three children are riding on the edge of a merry-go-round that is 105 kg, has a 1.70-m radius, and is spinning at 24.0 rpm. The children have masses of 22.0, 28.0, and 33.0 kg. If the child who has a mass of 28.0 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm? Ignore friction, and assume that the merry-go-round can be treated as a solid disk and the children as points.

Answer 1

Moment of inerta of the merry-go-round = 1/2 Mr², M = 105 kg, r = 1.7 m
Moment of inertia of all the three children = m1r² + m2r² + m3r²
m1 = 22 kg, m2 = 28 kg, m3 = 33 kg
Total initial moment of inetia, Ii = [M/2 + m1 + m2 + m3] r²
= [105/2 + 22 + 28 + 33] x (1.7)²
= 391.595 kg m².
Initial angular velocity, i = 24 rpm

Final moment of inertia
When the child with mass 28 kg is movend to the center, he is not contributing to the net moment of inertia since r = 0 at the center.
If = = [105/2 + 22 + 33] x (1.7)²
= 310.675 kg m².
Take f is the final angular momentum.

Since there is no external torque, angular momentum is conserved.
Ii x i = If x f
f = [Ii x i] / If
= [391.595 x 24] / 310.675
30.25 rpm