A uniform rod of mass 2.55×10 −2 kg and length 0.450 m rotates in a horizonta

Question

A uniform rod of mass 2.55×10−2 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 0.180kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 5.50×10−2m on each side from the center of the rod, and the system is rotating at an angular velocity 33.0rev/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? wsystem=_____rev/min Part B What is the angular speed of the rod after the rings leave it? Wrod=____rev/min A uniform rod of mass 2.55×10−2 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 0.180kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 5.50×10−2m on each side from the center of the rod, and the system is rotating at an angular velocity 33.0rev/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? wsystem=_____rev/min Part B What is the angular speed of the rod after the rings leave it? Wrod=____rev/min What is the angular speed of the system at the instant when the rings reach the ends of the rod? wsystem=_____rev/min Part B What is the angular speed of the rod after the rings leave it? Wrod=____rev/min wsystem=_____rev/min Part B What is the angular speed of the rod after the rings leave it? Wrod=____rev/min wsystem=_____rev/min Part B What is the angular speed of the rod after the rings leave it? Part B What is the angular speed of the rod after the rings leave it? Wrod=____rev/min

Explanation / Answer

a)

Ii = 1/12 M L^2 + 2 (m d^2) = (1/12)*2.55e-2*0.45*0.45+2*0.180*5.5e-2*5.5e-2 = 0.0015193

If = 1/12 M L^2 + 2 (m D^2) = (1/12)*2.55e-2*0.45*0.45+2*0.180*0.45/2*0.45/2 = 0.018655

Ii wi = If wf ==> 0.0015193 * 33 = 0.018655*wf ==> wf = 2.69 rev/min

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