A uniform rod of mass 3.10?10 ?2 kg and length 0.370m rotates in a horizontal pl

Question

A uniform rod of mass 3.10?10?2kg and length 0.370m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.170kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 5.20?10?2m on each side from the center of the rod, and the system is rotating at an angular velocity 30.0rev/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

M = mass of rod = 3.10 x 10-2 kg = 0.031 kg

L = length of rod = 0.37 m

m = mass of rings = 0.170 kg

moment of inertia of rod is given as ::

Irod = ML2 /12

Irod = (0.031) (0.37)2 /12

Irod = 0.000354 kgm2

moment of inertia of rings initially :

Iring = 2 m r2 = 2 (0.17) (0.052)2

Iring = 0.00092 kg m2

total moment of inertia initially

Iinitial = 0.000354 + 0.00092 = 1.274 x 10-3

moment of inertia of rings finally :

Iring = 2 m r2 = 2 (0.17) (0.185)2

Iring = 0.0116 kg m2

total moment of inertia finally

Ifinal = 0.000354 + 0.0116 = 1.195 x 10-2 kgm2

Winitial = 30 rev/min

Using conservation of momentum ::

Iinitial Winitial = Ifinal Wfinal

(1.274 x 10-3) (30) = (1.195 x 10-2) Wfinal

Wfinal = 3.2 rev/min

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