Question

This is the question: After a merry-go-round is set in motion, its rotational speed decreases noticeably if another
person jumps on it. However, if a person riding the merry-go-round steps off, the rotational speed
seems not to change at all. Explain.

This is the explanation given in the textbook:

A person jumping on the merry-go-round right before landing on the carousel had zero rotational velocity. Her
landing increases the rotational inertia of the system, thus the rotational speed decreases. A person jumping off was
originally moving with the carousel and had rotational momentum, when she leaves, she still has this rotational
momentum, therefore the rotational momentum of the carousel decreases, to keep the total momentum of the system
person-carousel constant.

I am kind of confused as to why the rotational velocity of the person matters- all I understand is that shouldn't the velocity of the carousel increase because momentum has to stay the same- so when the rotational inertia goes down (when the person leaves), then shouldn't the velocity increase? A

Answer 1

When the person drops down, the person doesn't just drop with zero velocity on the ground, rather he goes down with the tangential velocity. The person gets stopped consequently by the friction of the ground. So the energy is lost there.That's why when a person drops off a carousel he has zero relative velocity relative to carousel but not with the ground. So here momentum of both bodies is remainig same after the drop but the rotational momentum of the person is converted into translatory momentum(tangential velocity). In case of a person jumping on the carousel the observation is quite obvious that initial momentum of carousel is distibuted to both carousel and the person.