Assignment 10 Problem 10.86 Problem 10.86 A uniform rod of mass 2.70x10-2 kg and
ID: 1794754 • Letter: A
Question
Assignment 10 Problem 10.86 Problem 10.86 A uniform rod of mass 2.70x10-2 kg and length 0.370 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.210 kg, are mounted so that they can slide along the rod. They are initially held by catches at positions a distance 5.50 10-2 m on each side from the center of the rod, and the system is rotating at an angular velocity 30.0 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 PaExplanation / Answer
Ii = (2.70 x 10^-2)(0.370^2) / 3 + 2 (0.210 x 0.055^2)
{ Irod = m L^2 / 3 }
Ii = 2.5026 x 10^-3 kg m^2
wi = 30 rev/min
IF = (2.70 x 10^-2)(0.370^2) / 3 = 1.2321 x 10^-3 kg m^2
Applying angular momentum conservation,
Ii wi = If wf
wf = 60.8 rev/min