A uniform rod of mass 3.45×102 kg and length 0.450 m rotates in a horizontal pla

Question

A uniform rod of mass 3.45×102 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.160 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 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.What is the angular speed of the system at the instant when the rings reach the ends of the rod?

Explanation / Answer

conserving angular momentum as there are no non-conservative forces

I1w1 = I2w2

where i1 is the Moment of ienrtia of and W 1 is the angular v]2locity

I= 1/12 mr^2 +2 * (1/2) * mrw^2

0.0345 *0.45* 0.45/12 + 2*0.160*0.45 * 0.45 *2* 3.14 *20/60 =

1 *0.0345 *0.480 * 0.480/12 + 2*0.160*0.215 * 0.215*w

0.136 = 0.0154 W

W = 8.83 rad/s or 55.4 rev

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