A uniform rod of mass 2.5010^?2 kg and length 0.400 m rotates in a horizontal pl

Question

A uniform rod of mass 2.5010^?2 kg and length 0.400 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.160 kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 4.5010?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? w_system= Part B What is the angular speed of the rod after the rings leave it? w_rod=

Explanation / Answer

a)

MOI of rod alone = mL^2/12 = 2.5*10^-2 *(0.4)^2/12 = 3.33*10^-4 kg-m^2

Initial MOI of rod + rings = 3.33*10^-4 + 2*m*r1^2 = 3.33*10^-4 + 2*0.16*0.045^2 = 9.813*10^-4 kg-m^2

Final MOI of rod + rings = 3.33*10^-4 + 2*m*r2^2 = 3.33*10^-4 + 2*0.16*(L/2)^2 = 3.33*10^-4 + 2*0.16*(0.4/2)^2 = 0.01313 kg-m^2

Angular momentum conservation: I1*1 = I2*2

9.813*10^-4 *26 = 0.01313*2

2 = 1.943 rev/min (or 2*3.14*1.943/60 rad/s = 0.2033 rad/s)

b)

Angular momentum conservation:

I2*2 = I3*3

0.01313 *1.943 = 3.33*10^-4 *3

3 = 76.63 rev/min (or 2*3.14*76.63/60 rad/s = 8.021 rad/s)

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