A uniform rod of mass 3.00×102 kg and length 0.350 m rotates in a horizontal pla
ID: 1471428 • Letter: A
Question
A uniform rod of mass 3.00×102 kg and length 0.350 m rotates in a horizontal plane about a fixed axis through its center and perpendicular to the rod. Two small rings, each with mass 0.230 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 35.0 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?
w_system = ?
b ) rev/min What is the angular speed of the rod after the rings leave it?
w_rod = ? rev/min
Explanation / Answer
given
Mass of rod, M = 0.03 kg
L = 0.35 m
mass of one ring, m = 0.23 kg
r = 0.055 m
w1 = 35 rev/min
Initial moment of Inertia of the system, I1 = I_rod + I_rings
= M*L^2/12 + m*r^2
= 0.03*0.35^2/12 + 0.23*0.055^2
= 0.00102 kg.m^2
A) at the instant when the rings reach the ends of the rod,
moment of Inertia of the system, I2 = I_rod + I_rings
= M*L^2/12 + m*r^2
= 0.03*0.35^2/12 + 0.23*(0.35/2)^2
= 0.00735 kg.m^2
Apply conservation of angular momentum
I2*w2 = I1*w1
w2 = I1*w1/I2
= 0.00102*35/0.00735
= 4.86 rev/min
b) when the rings leave the rod,
moment of Inertia of the system, I2 = I_rod + I_rings
= M*L^2/12 + 0
= 0.03*0.35^2/12 + 0
= 0.00030625 kg.m^2
Apply conservation of angular momentum
I2*w2 = I1*w1
w2 = I1*w1/I2
= 0.00102*35/0.00030625
= 116.6 rev/min
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