Please show how to work! A uniform rod of mass 3.50 x10^-2 kg and length 0.350 m

Question

Please show how to work! A uniform rod of mass 3.50 x10^-2 kg and length 0.350 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.220 kg, are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 4.50 x 10^-2 m on each side from the center of the rod, and the system is rotating at an angular velocity 34.0 rev/mm . 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?

Explanation / Answer

initial angular speed , wi = 34 rev/min

Now, let the final angular velocity is wf

as there is no external torque acting on the system ,

the angular momentum of system is conserved ,

Ii * wi = If * wf

(3.5 *10^-2 * 0.35^2/12 + 2 * 0.22 * (4.5 *10^-2)^2 ) * 34 = (3.5 *10^-2 * 0.35^2/12 + 2 * 0.22 * ( 0.35/2)^2 ) *wf

solving for wf

wf = 3.07 rev/min

the final angular speed is 3.07 rev/min

part B)

when the disks leave it ,

Ii * wi = If * wf

(3.5 *10^-2 * 0.35^2/12 + 2 * 0.22 * (4.5 *10^-2)^2 ) * 34 = (3.5 *10^-2 * 0.35^2/12 ) *wf

solving for wf

wf = 118.8 rev/min

the final angular speed is 118.8 rev/min

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