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previous | 11 of 25 next » Problem 10.88 Part A What is the angular speed of the system at the instant when the rings reach the ends of the rod? A uniform rod of mass 2.80x10-2 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 0.250 kg, are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 4.50x10 m on each side from the center of the rod, and the system is rotating at an angular velocity 25.0 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 Wsystem rev/min Submit My Answers Give Up Incorrect, Iry Again, 9 attempts remaining Part B What is the angular speed of the rod after the rings leave it? Wrod rev/min Submit My Answers Give Up

Explanation / Answer

here angular momentum is conserve

I1w1 = I2w2

I1 = Irod + Iring

I1 = ML^2/12 + 2*mr1^2

M = 2.8 x 10^-2 kg

L = 0.360 m

r1 = 4.5 x 10^-2

m = 0.250 kg

I1 = 3.024 x 10^-4 + 1.0125 x 10^-3 = 1.3149 x 10^-3 kg.m^2

I2 = Irod + Iring

Iring = mr2^2

r = 0.180 m

I2 = 0.0165025

w2 = I1w1/I2

w1 = 25 rev/min

w2 = 1.992 rev/min = 2 rev/min

part b )

The forces and torques that the rings and the rod exert on each other will vanish, but the common angular velocity will be the same,

= 2 rev/min

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