A uniform rod of mass 3.30 times 10^-2 kg and length 0.350 m rotates in a horizo

Question

A uniform rod of mass 3.30 times 10^-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 0.210 kg, are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 4.60 times 10^-2 in on each side from the center of the rod, and the system is rotating at an angular velocity 32.0 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? Part B What is the angular speed of the rod after the rings leave It?

Explanation / Answer

Net moment of inertia= ml^2/12 + 2× m'r^2

=0.033×0.350×0.350/12. +2×. 0.210×0.046×0.046

=0.00033+0.00088=0.00121

Moment of inertia when rings reach end=0.00033+2×0.210×0.350×0.350/4=0.0131

Angular velocity at the point mentioned in 1sr subpart of question= MI1/MI2. × angular speed. Initial=

= 0.00121×32/0.0131=2.95 rev per minute

This obtained by conserving angular momentum in initial and final position

In second case, initial case is when the rings are just attend and final point is when both rings leave

So required angular speed in second part of question=

MI2×angukar speed/MI rod>

=117.1 rev per minute

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