A rotating uniform-density disk of radius 0.8 m is mounted in the vertical plane
ID: 2263284 • Letter: A
Question
A rotating uniform-density disk of radius 0.8 m is mounted in the vertical plane. The axle is held up by supports that are not shown, and the disk is free to rotate on the nearly frictionless axle. The disk has mass 4.8 kg. A lump of clay with mass 0.6 kg falls and sticks to the outer edge of the wheel at location A, < -0.6, 0.529, 0 > m. (Let the origin of the coordinate system be the center of the disk.) Just before the impact the clay has a speed 8 m/s, and the disk is rotating clockwise with angular speed 0.75 radians/s.
Explanation / Answer
a) Just before the impact the angular momentum of the combined system = mvr + Iw
= 0.6*sqrt(0.6^2 + 0.529^2)*8 + 4.8*0.8^2/2*0.75
= 4.9915 Kg.m^2/s
b) Since the axle is frictionless no external moment acts on the system. And so from Newton's second law,
Final angular momentum = 4.9915 Kg.m^2/s
c) From conservation of angular momentum
New moment of Inertia = m*r^2/2 + m*sqrt(0.6^2 + 0.529^2)
= 4.8*0.8^2/2 + 0.6*sqrt(0.6^2 + 0.529^2)
= 2.01594
2.01594*w = 4.9915
w = 2.476 radians/s
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