Final answer:
Depending on how the child exits the merry-go-round, the angular velocity can increase, decrease, or remain the same due to conservation of angular momentum. Jumping off radially keeps it the same, jumping backward or straight up would increase it, and jumping off forward would decrease it.
Step-by-step explanation:
When considering the scenarios where a child gets off a rotating merry-go-round, we can deduce the following effects on the angular velocity of the merry-go-round:
(a) If the child jumps off radially, the angular velocity will remain the same because the child is moving in a direction perpendicular to the tangential velocity of the merry-go-round and isn't affecting the rotational inertia substantially.
(b) Jumping backward to land motionless implies that the child exerts a forward force on the merry-go-round. This would increase the angular velocity of the merry-go-round as a reaction to the child's action, due to conservation of angular momentum.
(c) If the child jumps straight up and holds onto an overhead tree branch, they remove their moment of inertia from the system. This increases the angular velocity as the total moment of inertia of the system decreases, conserving angular momentum.
(d) Jumping off forward, tangential to the edge, will decrease the angular velocity. The child imparts an opposite momentum to the system, which slows down the rotation due to conservation of angular momentum.