Three children are riding on the edge of a merry–go–round that is

Question

Three children are riding on the edge of a merry–go–round that is 90 kg, has a 1.80–m radius, and is spinning at 22.0 rpm. The children have masses of 17.0, 21.0, and 25.0 kg. If the child who has a mass of 21.0 kg moves to the center of the merry–go–round, what is the new angular velocity in rpm? 

Answer 1

Let

M = 90 kg be the mass of the merry-go-round,

r = 1.8m be its radius,

m₁ = 17kg,

m₂ = 21kg,

m₃ = 25kg be the masses of the three children,

ω = 22 rpm be the angular velocity with all children at the edge,

ω' (to be calculated) be the angular velocity after m₂ is moved to the center.

This can be solved by conservation of angular momentum --

angular momentum before the move must be the same as after the move.

But we need to make two assumptions.

1) The merry-go-round is a uniform disk.

This is not a realistic assumption, but if the merry-go-round has a more common

complicated shape, the problem becomes complicated.

2) The children are riding at the edge of the merry-go-round.

The problem can be solved easily regardless where the children are,

but in the absence of any information I will assume it is at the edge.

In general, a mass m at the edge contributes mr²ω to the angular momentum.

(If you do not know why that is, let me know and I can explain it.)

The rotating uniform disk contributes Mr²ω/2 to the angular momentum.

(If you do not know why that is, let me know and I can explain it.)

Now we can equate the total angular momentum after the move to the total

angular momentum before the move.

Mr²ω'/2 + m₁r²ω' + m₃r²ω' = Mr²ω/2 + m₁r²ω + m₂r²ω + m₃r²ω

From that express

ω' = ω(M/2 + m₁ + m₂ + m₃)/(M/2 + m₁ + m₃)

Substituting actual numbers

ω' = 22×(90/2 + 17 + 21 + 25)/(90/2 + 17 + 25) = 27.31 rpm

Please note that the result does not depend on the radius r.