A uniform rod of mass 2.55×102 kg and length 0.360 m rotates in a horizontal pla

Question

A uniform rod of mass 2.55×102 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 0.240 kg , are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 4.50×102 m on each side from the center of the rod, and the system is rotating at an angular velocity 27.0 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.

What is the angular speed of the system at the instant when the rings reach the ends of the rod? system = rev/min

What is the angular speed of the rod after the rings leave it? rod = rev/min

Explanation / Answer

Given:-

m1 = 2.55 x 10^-2 kg

m2 = 0.240 kg

L = 0.360 m

r = 4.50 x 10^-2 m

I1 = (1/12)(m1)L^2 + 2(m2)r^2

I1 = 0.00027540 + 0.000972

I1 = 0.0012474

I2 = (1/12)(m1)L^2 + 2(m2)(L/2)²

I2 = 0.00027540 + 0.015552

I2 = 0.015827

2 = 1(I1/I2) = 27*(0.0012474 / 0.015552) = 2.1656 rpm

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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