Question

. In a park, four children are riding on a merry-go-round with a radius of 2.2 m and a mass of 300 kg. The children have an average mass of 45 kg each, and they are initially each standing 2.0 m from the center of the merry-go-round. The merry-go-round is rotating at a speed of 0.25 revolutions per second. The children then move inwards towards the center of the merry-go-round, so they are standing 0.5 m from the center. What is the final rotational velocity of the merry-go-round?

Answer 1

Mass of the merry-go-round = M = 300 kg

Radius of the merry-go-round = R = 2.2 m

Mass of one child = m = 45 kg

Initial distance of the children from the center of the merry-go-round = r₁ = 2 m

Final distance of the children from the center of the merry-go-round = r₂ = 0.5 m

Initial angular speed of the merry-go-round = ₁ = 0.25 rev/s = 1.57 rad/s

Final angular speed of the merry-go-round = ₂

Initial speed of the children = V₁ = ₁r₁

Final speed of the children = V₂ = ₂r₂

Moment of inertia of the merry-go-round = I

I = MR²/2

I = (300)(2.2)²/2

I = 726 kg.m²

By angular momentum conservation,

I₁ + 4mV₁r₁ = I₂ + 4mV₂r₂

I₁ + 4m₁r₁² = I₂ + 4m₂r₂²

(726)(1.57) + 4(45)(1.57)(2)² = ₂[726 + 4(45)(0.5)²]

₂ = 2.944 rad/s

₂ = 0.469 rev/s

Final rotational velocity of the merry-go-round = 2.944 rad/s = 0.469 rev/s