A uniform rod of mass 3.40×10 2 kg and length 0.450 m rotates in a horizontal pl

Question

A uniform rod of mass 3.40×102 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 0.250 kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 5.40×102 m on each side from the center of the rod, and the system is rotating at an angular velocity 27.0 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?

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

Explanation / Answer

M = mass of rod = 0.034 kg

L = length of rod = 0.45 m

m = mass of ring = 0.250

ri = initial distance of rings from axis of rotation = 0.054 m

Wi = initial angular velocity = 27 rev/min = 27 (2 x 3.14/60) = 2.83 rad/s

Wf = final angular velocity = ?

when the rings leave :

using conservation of angular momentum

Ii Wi = If Wf

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

(0.034 (0.45)2/12 + 2 (0.25) (0.054)2) (2.83) = (0.034 (0.45)2/12) Wf

Wf = 10.02 rad/s

when the rings at at the end :

rf = final distance of rings from axis of rotation = 0.225 m

using conservation of angular momentum

Ii Wi = If Wf

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

(0.034 (0.45)2/12 + 2 (0.25) (0.054)2) (2.83) = (0.034 (0.45)2/12 + 2 (0.25) (0.225)2) Wf

Wf = 0.22 rad/s

Wf = 10.02 rad/s

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