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Question

A playground carousel is free to rotate about its center on frictionless bearings. The carousel has an angular speed of 3.14 rad/s, a moment of inertia of 124 kg m^2 and a radius of 1.80m. A 40.0kg person, standing still next to the carousel jumps on very close to the outer edge. Find the resulting angular speed of the carousel and person.

I've been at this question forever....anyone who can answer this shall have my unending love and gratitude :P

Answer 1

The basic equation you need here is the conservation of angular momentum - Iω = constant, where I is the moment of inertia and ω (omega) is the angular velocity. The final angular velocity is reduced in proportion to the change in moment of inertia: ω(2) = ω(1) ·I(1)/I(2). The moment of inertia of the person on the carousel is his mass times the square of his distance from the center. You just add that to the initial moment of inertia to get I(2). It's significant that the person jumps on standing still; a running person would add some angular momentum too, instead of just inertia. Hope that helps.

Answer 2

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