A uniform rod of mass 2.95×10 2 kg and length 0.360 m rotates in a horizontal pl

Question

A uniform rod of mass 2.95×102 kg and length 0.360 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.240 kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 5.20×102 m on each side from the center of the rod, and the system is rotating at an angular velocity 31.0 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.

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

Explanation / Answer

Here ,

mass of rod , mr = 2.95 *10^-2 Kg

length , L = 0.36 m

mass of ring , m = 0.24 Kg

initial distance , x = 5.2 *10^-2 m

initial angular speed , wi = 31 rev/min

final angular speed is wf

Using conservation of angular momentum

(mr * L^2/12 + 2 * m * x^2) * wi = (mr * L^2/12 + 2 * m * (L/2)^2) * wf

(2.95 *10^-2 * 0.36^2/12 + 2 * 0.24 * (5.2 *10^-2)^2) * 31 = (2.95 *10^-2 * 0.36^2/12 + 2 * 0.24 * (0.36/2)^2) * wf

solving for wf

wf = 3.16 rev/min

the angular speed of the system at the instant when the rings reach the ends of the rod is 3.16 rev/min

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