Question

A uniform rod of mass 3.40×10⁻² kg and length 0.410 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.190 kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 5.50×10⁻² m on each side from the center of the rod, and the system is rotating at an angular velocity 28.0 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.

(A) What is the angular speed of the system at the instant when the rings reach the ends of the rod?

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

Answer 1

initial angular momentum L1 = ((1/12)*Mrod*L^2 + 2*m*r^2) *w1

after the rings move to the end of the rod

L2 = ( (1/12)*Mrod*L^2 + 2*m*(L/2)^2 ) *w2

total angular momentum is conserved

L2 = L1

( (1/12)*Mrod*L^2 + 2*m*(L/2)^2 ) *w2 = ((1/12)*Mrod*L^2)+ 2*m*r^2) *w1

( (1/12)*3.4*10^-2*0.41^2 + 2*0.19*(0.41/2)^2 )*w2 = ( (1/12)*3.4*10^-2*0.41^2 + 2*0.19*(5.5*10^-2)^2 )*28

angular speed w2 = 2.77 rev/min

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B)

after the rings leave the the rod

the angualr velocity remins same as w2 = 2.77 rev/min