A uniform rod of mass 2.50×102 kg and length 0.450 m rotates in a horizontal pla

Question

A uniform rod of mass 2.50×102 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 0.160 kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 5.50×102 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?

Part B

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

Explanation / Answer

mass of the rod m1=2.5*10^-2 kg

length of the rod l=0.45 m

mass of the each ring m2=0.16 kg

positon of the ring r1=5.5*10^-2

initial angular speed w1=26 rev/min

final angular speed is w2

A)

by using law of conservation of momentum,

I1*w1=I2*w2

(Irod+Iring)*w1=(Irod+I'ring)*w2

(1/12*m1*l^2+2*m2*r1^2)*w1=(1/12*m1*l^2+2*m2*(l/2)^2)*w2

(1/12*0.025*0.45^2+2*0.16*0.055^2)*26=(1/12*0.025*0.45^2+2*0.16*(0.45/2)^2)*w2

===> w2=2.174 rev/min

final anular speed w2=2.174 rev/min

or

w2=0.23 rad/sec

B)

same as part A ,

w2=2.174 rev/min or w2=0.23 rad/sec

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