A uniform rod of mass 2_70 times 10^-2 kg and length 0.440 m rotates in a horizo

Question

A uniform rod of mass 2_70 times 10^-2 kg and length 0.440 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 4.70 times 10^-2 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

part A)

m = mass of each ring = 0.210 kg

ri = initial distance of each from center = 0.047 m

rf = final distance of each from center = 0.44/2 = 0.22 m

M = mass of rod = 0.027 kg

L = length of rod = 0.44 m

Ii = initial total moment of inertia = 2 mri2 + ML2/12

If = final total moment of inertia = 2 mrf2 + ML2/12

Wi = initial angular velocity = 26 rev/min

Wf = final angular velocity

using conservation of momentum

Ii Wi = If Wf

(2 mri2 + ML2/12) Wi = (2 mrf2 + ML2/12) Wf

(2 (0.210) (0.047)2 + (0.027)(0.44)2/12) (26) = (2 (0.210) (0.22)2 + (0.027)(0.44)2/12) Wf

Wf = 1.71 rev/min

b)

using conservation of momentum

(2 (0.210) (0.047)2 + (0.027)(0.44)2/12) (26) = (ML2/12) Wf

(2 (0.210) (0.047)2 + (0.027)(0.44)2/12) (26) = ((0.027)(0.44)2/12) Wf

Wf = 81.4 rev/min

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