A uniform rod of mass 3.35×10 2 kg and length 0.430 m rotates in a horizontal pl

Question

A uniform rod of mass 3.35×102 kg and length 0.430 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.170 kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 5.00×102 m on each side from the center of the rod, and the system is rotating at an angular velocity 35.0 rev/min . Without otherwise changing the system, the catches are released, and the rings slide outward along the rod and fly off at the ends

Part A

What is the angular speed of the system at the instant when the rings reach the ends of the rod?

Part B

What is the angular speed of the rod after the rings leave it?

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Explanation / Answer

moment of inertia of rod = (1/12) * mass * length^2

moment of inertia of rod = (1/12) * 3.35 * 10^-2 * 0.43^2

moment of inertia of rod = 0.00051617916 kg.m^2

moment of inertia of both the ring = 2 * 0.170 * (5 * 10^-2)^2

moment of inertia when ring will reach at end = 2 * 0.170 * 0.43^2

initial velocity = 35 rev/min or 3.665 rad/sec

by conservation of momentum

initial momentum = final momentum

(0.00051617916 + 2 * 0.170 * (5 * 10^-2)^2) * 3.665 = (0.00051617916 + 2 * 0.170 * 0.43^2) * w

w = 0.07899 rad/sec

angular speed of the system at the instant when the rings reach the ends of the rod = 0.07899 rad/sec

(0.00051617916 + 2 * 0.170 * 0.43^2) * 0.07899 = 0.00051617916 * w

w = 9.6992 rad/sec

angular speed of the rod after the rings leave it = 9.6992 rad/sec

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