A uniform rod of mass 3.10×102 kg and length 0.390 m rotates in a horizontal pla

Question

A uniform rod of mass 3.10×102 kg and length 0.390 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.180 kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 5.10×102 m on each side from the center of the rod, and the system is rotating at an angular velocity 33.0 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

a)

M = mass of rod = 0.0310 kg

L = length of rod = 0.390 m

Ir = moment of inertia of rod = ML2 /12 = (0.0310) (0.390)2/12 = 0.000393 kgm2

ri = initial distance of rings = 0.051 m

Total initial moment of inertia = Ii = Ir + 2 m ri2 = 0.000393 + 2 (0.180) (0.051)2 = 0.00133 kgm2

rf = final distance of rings = 0.390/2 = 0.195 m

Total initial moment of inertia = If = Ir + 2 m rf2 = 0.000393 + 2 (0.180) (0.195)2 = 0.0141 kgm2

Wi = initial angular velcity = 33 rev/min

using conservation of angular momentum

Ii Wi = If Wf

(0.00133) (33) = (0.0141) Wf

Wf = 3.11 rev/min

B)

when the rings leave , If = final moment of inertia = Ir = 0.000393

using conservation of angular momentum

Ii Wi = If Wf

(0.00133) (33) = (0.000393) Wf

Wf = 111.7 rev/min

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.