Final answer:
When a child jumps onto a stationary merry-go-round, they transfer angular momentum to the system, causing it to start rotating. The child's position relative to the center affects the rotational inertia and, therefore, the angular velocity of the merry-go-round. Overall, the system must conserve angular momentum.
Step-by-step explanation:
When a child runs and jumps onto a stationary merry-go-round, the conservation of angular momentum for the system must be considered. Suppose a child gets off a rotating merry-go-round. We would discuss what happens to the angular velocity of the merry-go-round in various scenarios, such as when jumping off radially or tangentially.
Initially, when the child is not on the merry-go-round, the system (child + merry-go-round) has no angular momentum because the merry-go-round is stationary. When the child jumps onto the merry-go-round, he or she brings angular momentum into the system. If the child jumps on tangentially and holds on, the total angular momentum of the system must be conserved, and since the child imparts a rotational impulse to the merry-go-round, the merry-go-round starts to rotate in the opposite direction to conserve angular momentum.
The child's movement towards or away from the center will affect the system’s rotational inertia. The closer the child is to the center, the lower the rotational inertia becomes, allowing the system to spin faster to conserve angular momentum (b). Conversely, as the child moves away from the center, the rotational inertia of the system increases, and the merry-go-round slows down to conserve angular momentum (d).