PROBLEM 10.88 A uniform rod of mass 3.15×10 2 kg and length 0.430 m rotates in a

Question

PROBLEM 10.88

A uniform rod of mass 3.15×102 kg and length 0.430 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 5.10×102 m on each side from the center of the rod, and the system is rotating at an angular velocity 26.0 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

Moment of inertia of Rod = mL^2/12 = (3.15×102 kg * 0.430^2)/12

Moment of Inertia of Ring = m*r^2 = 0.150 kg * (5.10×102 m)^2

Using conservation of angular momentum

I1 * W1 = I2 * W2

((3.15×10^2 kg * 0.430^2)/12 + 2 *0.150 kg * (5.10×10^2)^2 ) * 26.0 = ((3.15×10^2 kg * 0.430^2)/12 + 2 *0.150 kg * (0.43/2)^2 )* W2

solving for w2

w2 = 2.29 rev/min

angular speed of the system at the instant when the rings reach the ends of the rod = 2.29 rev/min

part B)

as the rings fly off the rod

again using conservation of angular momentum

((3.15×10^2 kg * 0.430^2)/12 + 2 *0.150 kg * (5.10×10^2)^2 ) * 26.0 = ((3.15×10^2 kg * 0.430^2)/12)* W3

solving for w3

w3 = 67.7 rev/min

angular speed of the rod after the rings leave it = 67.7 rev/min

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