A uniform rod of mass 3.45×10^2 kg and length 0.360 m rotates in a horizontal pl

Question

A uniform rod of mass 3.45×10^2 kg and length 0.360 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 4.70×10^2 m on each side from the center of the rod, and the system is rotating at an angular velocity 31.0 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?

Explanation / Answer

M =3.45x10^-2 , R = 0.36 m, m=0.19 kg, d = 4.7x10*-2m,

w1 =31 rev/min = 31*2pi/60 = (31*2*3.14)/60 =3.245 rad/s

(a) From conservation of angular momentum

I1w1 =I2w2

[(ML2/12)+(md2)]w1 =[(ML2/12)+(m(L/2)2)]w2

[(3.45x10-2x0.36x0.36/12)+(0.19x4.7x10-2x4.7x10-2)]3.245 =[(3.45x10-2x0.36x0.36/12)+(0.19x0.36x0.36/4)]w2

[3.726x10-4+4.1971x10-4]3.245 =[3.726x10-4+61.56x10-4]w2

25.7x10-4 =(65.286x10-4)w2

w2 =25.7/65.286 = 0.394 ras/s

w2 = (0.394*60)/(2*3.14) =3.764 rev/min

(b) From conservation of angular momentum

I1w1 =I2w2

[(ML2/12)+(md2)]w1 =[ML2/12]w2

[(3.45x10-2x0.36x0.36/12)+(0.19x4.7x10-2x4.7x10-2)]3.245 =[(3.45x10-2x0.36x0.36)/12]w2

[3.726x10-4+4.1971x10-4]3.245 =[3.726x10-4]w2

25.7x10-4 =(3.726x10-4)w2

w2 =25.7/3.726 = 6.898 ras/s

w2 = (6.898*60)/(2*3.14) =65.9 rev/min

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