Question

Three children are riding on the edge of a merry-go-round that is 122 kg, has a 1.60 m radius, and is spinning at 17.3 rpm. The children have masses of 19.9, 29.5, and 40.8 kg. If the child who has a mass of 29.5 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm?

Answer 1

Mass of the merry-go-round = M = 122 kg

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

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

I = MR²/2

I = (122)(1.6)²/2

I = 156.16 kg.m²

Mass of the first child = m₁ = 19.9 kg

Mass of the second child = m₂ = 29.5 kg

Mass of the third child = m₃ = 40.8 kg

Initial angular velocity of the merry-go-round = ₁ = 17.3 rpm

Converting the angular velocity to rad/s,

₁ = 1.812 rad/s

Angular velocity of the merry-go-round after the the 29.5 kg child moves to the center = ₂

Initially all three children area at the edge that is at a distance equal to the radius from the center and after that the 29.5 kg child moves to the center.

By conservation of angular momentum,

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

[156.16 + (19.9)(1.6)² + (29.5)(1.6)² + (40.8)(1.6)²](1.812) = (156.16 + (19.9)(1.6)² + (40.8)(1.6)²]₂

₂ = 2.251 rad/s

Converting to revolutions per minute,

₂ = 21.49 rpm

New angular velocity of the merry-go-round = 21.49 rpm