A disk-shaped merry-go-round with a radius of 2.63 m and a mass of 152 kg rotates freely with an angular speed of 0.526 rev/s. A 51.7 kg person, running tangentially to the rim at 2.76 m/s, jumps onto the rim and holds on. Before jumping on, the person was moving in the same direction as the rim's rotation. What is the final angular speed of the merry-go-round?

To find the new angular speed of a merry-go-round after a person jumps on, apply the conservation of angular momentum to calculate the combined rotational effect of the person and merry-go-round system.

Explanation:

The student's question revolves around the concept of conservation of angular momentum in a physics context, specifically relating to a rotating disk-shaped merry-go-round. The scenario involves a person jumping onto a spinning merry-go-round and determining the new angular speed after this action. Assuming that no external torques act on the system, the initial angular momentum of the system (merry-go-round plus person running tangential to it) must equal the final angular momentum after the person jumps on.

To find the final angular speed of the merry-go-round, we will:
1. Calculate the initial angular momentum of the merry-go-round.
2. Convert the linear velocity of the person to angular momentum.
3. Combine these to find the total initial angular momentum.
4. Use the final moment of inertia (including the person) to calculate the final angular speed.

It is crucial to remember that this application of conservation of angular momentum indicates that the final angular speed will be less than the initial speed due to the increased moment of inertia when the person jumps on.