Final answer:
The angular velocity of the merry-go-round will increase when the 28.0-kg child moves to the center due to the conservation of angular momentum, making the correct answer 25.0 rpm (option c).
Step-by-step explanation:
When the 28.0-kg child moves to the center of the merry-go-round, the overall moment of inertia of the system decreases. Since no external torques are acting on the system, the angular momentum (L) is conserved. Angular momentum is the product of the moment of inertia (I) and the angular velocity (ω), so if I decreases and L is conserved, ω must increase.
The original angular velocity of the merry-go-round is 20.0 rpm. To find the new angular velocity, we can set up the conservation of angular momentum equation L_initial = L_final, which in this case simplifies to I_initial × ω_initial = I_final × ω_final. Since I_final is less than I_initial due to the child moving to the center, and ω_initial is known, we can solve for ω_final. The child's mass moving closer to the axis does not change the angular momentum but it does reduce the moment of inertia, which means the merry-go-round will spin faster, thus increasing its angular velocity.
Without the detailed moment of inertia calculations for the initial and final states, we cannot compute the exact new angular velocity. However, logic dictates that since the moment of inertia decreases, the angular velocity must increase to conserve angular momentum. Therefore, the only logical choice among the given options that represents an increase in angular velocity is 25.0 rpm (option c).