Question

Three children are riding on the edge of a merry-go-round that is 105 kg, has a 1.60-m radius, and is spinning at 16.0 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

Mass of the merry-go-round = M = 105 kg

Radius of the merry-go-round = R = 1.6 m

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

I = MR²/2

I = (105)(1.6)²/2

I = 134.4 kg.m²

Mass of the first child = m₁ = 22 kg

Mass of the second child = m₂ = 28 kg

Mass of the third child = m₃ = 33 kg

Initially all three children are at the edge of the merry-go-round.

Initial moment of inertia of the system = I₁

I₁ = I + m₁R² + m₂R² + m₃R²

I₁ = 134.4 + (22)(1.6)² + (28)(1.6)² + (33)(1.6)²

I₁ = 346.88 kg.m²

Initial angular speed of the merry-go-round = ₁ = 16 rpm = 16 x (2/60) rad/s = 1.6755 rad/s

Now the 28 kg child moves to the center of the merry-go-round while the other two children are still at the edge.

New moment of inertia of the system = I₂

I₂ = I + m₁R² + m₃R²

I₂ = 134.4 + (22)(1.6)² + (33)(1.6)²

I₂ = 275.2 kg.m²

New angular velocity of the merry-go-round = ₂

By conservation of angular momentum,

I₁₁ = I₂₂

(346.88)(1.6755) = (275.2)₂

₂ = 2.112 rad/s

Converting from rad/s to rpm,

₂ = 20.17 rpm

New angular velocity of the merry-go-round = 20.17 rpm