Final answer:
The new angular velocity of the merry-go-round when the children move to 0.75 m from the axis of rotation can be found using the conservation of angular momentum. After calculations, the new angular velocity is determined to be 15.0 rev/min.
Step-by-step explanation:
The question concerns the conservation of angular momentum in the context of a small merry-go-round. As there is no external torque acting on the system, the angular momentum of the system remains constant. Initially, eight children are positioned on the outer edge of the merry-go-round, which has a given moment of inertia, and the system is rotated with an initial angular velocity of 6.0 rev/min. When the children move toward the center, the total moment of inertia of the system decreases.
To find the new angular velocity when the children move inward, we use the conservation of angular momentum:
L_initial = L_final
I_initial * ω_initial = I_final * ω_final
Where ω represents the angular velocity.
Given that the initial moment of inertia (I_initial) of the system includes both the merry-go-round and the children at the edge, and the final moment of inertia (I_final) includes the merry-go-round and the children at a radius of 0.75 m, we can set the initial angular momentum equal to the final angular momentum and solve for the new angular velocity (ω_final).
By calculating this, we find the new angular velocity of the merry-go-round when the children are 0.75 m from the axis of rotation. The moment of inertia of the children when positioned at the edge is m*r² for each child, where m is the mass, and r is the radius from the center (4.0 m). When they move inwards, the new moment of inertia is m*r'², with r' being the new radius (0.75 m).
Using these values, we can calculate the new angular velocity, which will be option (c) 15.0 rev/min.