Question

A horizontal circular platform (m = 112.1 kg, r = 3.01m) rotates about a frictionless vertical axle. a student (m = 66.3kg) walks slowly from the rim of the platform toward the center. the angular velocity, omega, of the system is 2.5 rad/s when the student is at the rim. find omega when the student is 1.73m from the center.

Answer 1

To find the angular velocity when the student is 1.73m from the center, use the principle of conservation of angular momentum. Substitute the given values into the formula ω2 = (Iω1) / (I - msrs2) and calculate the result.

Step-by-step explanation:

To find the angular velocity of the system when the student is 1.73m from the center, we can use the principle of conservation of angular momentum. The initial angular momentum of the system is equal to the final angular momentum. Let's assume the moment of inertia of the platform is I and the angular velocity is ω1 when the student is at the rim. When the student is 1.73m from the center, the moment of inertia of the system becomes I - msrs2, where ms is the mass of the student and rs is their distance from the center. The initial angular momentum is Iω1 and the final angular momentum is (I - msrs2)ω2, where ω2 is the angular velocity when the student is 1.73m from the center. Setting the initial and final angular momenta equal and solving for ω2, we get:

ω2 = (Iω1) / (I - msrs2)

Substituting the given values, we have:

ω2 = (112.1 kg * (2.5 rad/s) * (3.01m) / (112.1 kg * (3.01m)2 - 66.3 kg * (1.73m)2)

Calculating this expression gives us the angular velocity when the student is 1.73m from the center.