A uniform rod of mass 2.85×10 2 kg and length 0.360 m rotates in a horizontal pl

Question

A uniform rod of mass 2.85×102 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 0.200 kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 4.60×102 m 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 B

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

120

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120

Explanation / Answer

as no external torque is applied about axis

so angular momentum will be conserved

I1 w1 = I2 w2

(2.85×10^(2) * (0.360^2)/12 + 2* 0.200 * (4.60×10^(2))^2 ) * 32 =

((2.85×10^(2) * (0.360^2)/12 + 2* 0.200 * (0.18)^2 ) ) w

w = 2.78376219117 rev/m

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