Help Please! A uniform rod of mass 3.45 times 10-2kg and length 0.360m rotates i
ID: 1266413 • Letter: H
Question
Help Please!
A uniform rod of mass 3.45 times 10-2kg and length 0.360m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.190kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 5.20 times 10-2m on each side from the center of the rod, and the system is rotating at an angular velocity 30.0rev/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
a) moment of inertia of the rod = ml^2/12
=3.45*10^-2*0.36^2/12
=3.726*10^-4
so, conserving momentum,
(3.726*10^-4+2*0.19*(5.2*10^-2)^2)*30=(3.726*10^-4+2*0.19*0.36^2)*w
or w=0.8464 rev/min ( answer 1)
B) again conserving angular momentum,
(3.726*10^-4+2*0.19*(5.2*10^-2)^2)*30=(3.726*10^-4)*w
or w=112.73 rev/min( answer 2)
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