Question

Three children, each of mass 15 kg, ride at the edge of a frictionless merry-go-round of mass 163 kg and radius 1.2 meters.  The children jump off outward, directly away from the center, taking zero angular momentum with them.   If the angular velocity of the merry-go-round before they jumped was  = 0.3 rad/s, what is its angular velocity after they jump?

Treat the merry-go-round as a solid cylinder.

2 sig figs, units (rad/s)

Answer 1

Data provided in the question;

Mass of each child = m = 15 kg

Mass of merry-go-round = M = 163 kg

Radius of merry-go-round = R = 1.2 m

Initial angular velocity (Before jumping) = ₁ = 0.3 rad/s

​​​​​​In the initial case (before jumping off the merry-go-round), children and merry-go-round both are moving together and children are sitting on the edge of the merry-go-round.

Thus, the total moment of inertia initially (I_{​​​​​​1}​​​​​) will be the summation of moments of inertia of the merry-go-round (solid cylinder) and all the children.

Substituting, the values in the above equation;

And the initial angular velocity

​​​​​​After the children jumping off, only merry-go-round will be in motion and hence, the total moment of inertia (I_{​​​​​​2}) in the later case will be equal to the moment of inertia of the merry-go-round.

Substituting the values;

And we need to calculate, the final angular velocity ().

​​​​​​This, using the law of conservation of angular momentum i.e.

​​​​​​​​​​​​Substituting all the values, we get

​​​​​​Hence, the angular velocity of the merry-go-round after the children jumping off it is equal to 0.47 rad/s.