Final answer:
To find the new angular velocity of the merry-go-round after the child gets on, we use conservation of angular momentum and calculate the final angular velocity with the formula ω_final = MR²ω_initial / (MR² + mr²).
Step-by-step explanation:
The problem describes a situation where a child climbs onto a rotating merry-go-round, and we need to find the new angular velocity. This is a classic physics problem involving the conservation of angular momentum because no external torques are acting on the system.
The initial angular momentum of the system must equal the final angular momentum of the system. We can use the formula L = I × ω to find the moment of inertia (I) and the angular velocity (ω).
Initially, the merry-go-round has a moment of inertia I equal to MR² (mass times radius squared), and an angular velocity ω given in the problem.
When the child, with mass m, gets onto the merry-go-round at the radius r, the new moment of inertia is the sum of the merry-go-round's moment of inertia and the child's moment of inertia (mr²). If 'L_initial' is the initial angular momentum and 'L_final' is the final angular momentum, then L_initial = L_final because angular momentum is conserved.
Therefore, MR²ω_initial = (MR² + mr²)ω_final. To find the new angular velocity, we solve the equation for ω_final, which gives us ω_final = MR²ω_initial / (MR² + mr²).
Plugging in the given values (M = 165 kg, R = 1.75 m, m = 356.0 kg, ω_initial = 0.600 rad/s), we can calculate ω_final. Finally, we get the new angular velocity of the merry-go-round after the child gets on.