Two children (m = 40 kg each) stand opposite each other on the edge of a merry-go-round. The merry-go-round, which has a mass M of 200 kg and a radius of 2.0 m, is spinning at a constant rate of 12 rad/s. Treat the two children and the merry-go-round as a system. Both children walk half the distance toward the center of the merry-go-round. Calculate the final angular speed of the system.

Initially the two boys were sitting on the periphery, total moment of inertia

= 1/2 M r² + 2mr²; M is mass of the merry go round, m is mass of each boy and r is the radius

1/2 x 200 x 2² + 2 x 40 x 2²

= 400 + 320

I₁ = 720 kg m²

Finally the two boys were sitting at the middle, total moment of inertia

= 1/2 M r² + 2m (r/2) ²; M is mass of the merry go round, m is mass of each boy and r is the radius

1/2 x 200 x 2² + 2 x 40 x 1²

= 400 + 80

I₂ = 480

Now the system will obey law of coservation of angular momentum because no torque is acting on the system.

I₁ω₁ = I₂ω₂, I₁ and ω₁ are moment of inertia and angular velocity of first case and I₂ and ω₂ are of second case.

720 X 12 = 480 ω₂

ω₂ = 18 rad / s