A uniform rod of mass 2.55×102 kg and length 0.370 m rotates in a horizontal pla

Question

A uniform rod of mass 2.55×102 kg and length 0.370 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.150 kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 4.80×102 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.

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

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

Explanation / Answer

moment of inertia of the rod is I1 = (1/12)*M*L^2 = (1/12)*2.55*10^-2*0.37^2 = 2.90*10^-4 kg-m^2

moment of inertia of thwo rings are I2 = 2*m*r^2 = 2*0.15*(4.8*10^-2)^2 = 6.912*10^-4 kg-m^2

angualar velocity is wi = 35 rev/min = 35*2*3.142/60 = 3.665 rad/s

initial angular momentum is Li = (I1+I2)*wi = (2.9+6.912)*10^-4*3.665 = 3.6*10^-3 kg m^2/sec

final angular momentum is Lf = (I1+I2)*w2

Now I2 = 2*m*r^2 = 2*0.15*(0.37/2)^2 = 102.67*10^-4 kg -m^2

Lf = (2.9+102.67)*10^-4*w2

Li = Lf

3.6*10^-3 = (2.9+102.67)*10^-4*w2

w2 = 0.341 rad/s^2 = 0.341*(60/(2*3.142)) = 3.25 rev/min

after leaving the rings

now moment of inertia is I1 = 2.9*10^-4 kg-m^2

then Li= Lf

3.6*10^-3 = 2.9*10^-4*w2

w2= 12.4 rad/sec = 12.4*60/(2*3.142) = 118.4 rev/min

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