A uniform rod of mass 3.00x 102 kg and length 0.390 m rotates in a horizontal pl

Question

A uniform rod of mass 3.00x 102 kg and length 0.390 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.210 kg, are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 5.00x 102 m on each side from the center of the rod, and the system is rotating at an angular velocity 27.0 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? ANSWER: Wsystem rev /min Part B What is the angular speed of the rod after the rings leave it! ANSWER: Wrod rev/min

Explanation / Answer

solving 1st question of the page

Irod=Moment of inertia od rod about center=(1/12)M*L^2=3.8*10^(-4)

moment of inertia of ring when

they are at r=0.05 ,Iring=2*mr^2=10.5*10^(-4)

they are at r=0.195 ,I'ring=0.0159

Part (A)

let angular speed is w

conserving angulr momentum

(3.8+10.5)*10^(-4)*27=(3.8*10^(-4) +0.0159)*w

w=2.36 rev/min

Part B

when they are away from rod

(3.8+10.5)*10^(-4)*27=(3.8*10^(-4)*w

w=101.6 rev/min

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