Question

Three children are riding on the edge of a merry‑go‑round that has a mass of 105 kg and a radius of 1.60 m . The merry‑go‑round is spinning at 22.0 rpm. The children have masses of 22.0, 28.0, and 33.0 kg. If the 28.0 kg child moves to the center of the merry‑go‑round, what is the new angular velocity in revolutions per minute? Ignore friction, and assume that the merry‑go‑round can be treated as a solid disk and the children as point masses.

final angular velocity: ________ rpm

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

All three children are initially at the edge of the merry-go-round that is they are at a distance equal to the radius of the merry-go-round from the center.

Initial angular speed of the merry-go-round = ₁ = 22 rpm

Converting the angular speed to rad/s

₁ = 2.304 rad/s

Now the 28 kg child moves to the center of the merry-go-round.

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

By conservation of angular momentum,

(I + m₁R² + m₂R² + m₃R²)₁ = (I + m₁R² + m₃R²)₂

[134.4 + (22)(1.6)² + (28)(1.6)² + (33)(1.6)²](2.304) = [134.4 + (22)(1.6)² + (33)(1.6)²]₂

₂ = 2.904 rad/s

Converting to revolutions per minute,

₂ = 27.73 rpm

New angular velocity of the merry-go-round = 27.73 rpm