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If a disk-shaped merry-go-round of radius 2.93 m and mass 165 kg rotates freely with an angular speed of 0.641 rev/s, and a 64.7 kg person running tangential to the rim of the merry-go-round at 3.04 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim.

a. Angular speed increases
b. Angular speed decreases
c. Angular speed remains the same
d. Not enough information to determine

1 Answer

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Final answer:

Upon jumping onto a merry-go-round, if a person is moving tangentially in the same direction as the merry-go-round's rotation, the angular speed decreases due to the conservation of angular momentum and an increase in the system's moment of inertia.

Step-by-step explanation:

When a person jumps onto a merry-go-round that is already in motion, there is a momentum exchange which affects the angular speed of the merry-go-round depending on the direction and speed at which the person was moving prior to jumping on. In this specific scenario, the person is running tangential to the rim of the merry-go-round in the same direction as its rotation. Using conservation of angular momentum, we have to consider the rotational inertia of both the merry-go-round and the person as they come together as a single system post-jump. The total angular momentum before the person jumps on must equal the total angular momentum afterward, since no external torques are acting on the system.

If the person's tangential velocity is such that their angular momentum is in the same direction and sense as that of the merry-go-round, when they jump onto the merry-go-round, the system's total angular momentum is conserved but the moment of inertia increases. This would result in a decrease in the angular speed to conserve angular momentum (L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity). Therefore, the correct answer is b. Angular speed decreases.

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