(6 marks) A 80 kg man is sitting on the edge of a flat disks that weighs 220 kg
ID: 1403747 • Letter: #
Question
(6 marks) A 80 kg man is sitting on the edge of a flat disks that weighs 220 kg with radius 8 m. The disc and the man are spinning at 0.5 revolutions per second about an axis through the centre of the disc. The moment of inertia for the disc is M R^2/2 and the man can be treated as a point mass. (a) Calculate the total angular momentum and kinetic energy of the system. (b) A dog that weighs 8 kg touches down on the disc (straight from above) and lands on the edge. What is the final kinetic energy of the dog, man, and disc system? What was the change in total kinetic energy? You can treat the dog as a point mass.Explanation / Answer
8)
mass of man , m = 80 kg
mass of disk , M = 220 Kg
radius , r = 8 m
w1 = 0.5 rev/s
a)
total angular momentum = I*w
total angular momentum = (0.5 * 220 * 8^2 + 80 * 8^2) * (0.5 * 6.282)
total angular momentum = 38194.6 Kg.m^2/s
the total angular momentum is 38194.6 Kg.m^2/s
total kinetic energy = 0.5 * I*w^2
total kinetic energy = (0.5 * 220 * 8^2 + 80 * 8^2) * (0.5 * 6.282)^2
total kinetic energy = 59984.6 J
the total kinetic energy is 59984.6 J
b)
let the final angular speed is wf
wf * (0.5 * 220 * 8^2 + 80 * 8^2 + 8 * 8^2) = 0.5 * 6.282 * (0.5 * 220 * 8^2 + 80 * 8^2)
wf = 3.014 rad/s
final kinetic energy = 0.5 * (0.5 * 220 * 8^2 + 80 * 8^2 + 8 * 8^2) * 3.014^2
final kinetic energy = 57557 J
the final kinetic energy is 57557 J
--------------------------------
change in kinetic energy = 57557 - 59984.6
change in kinetic energy = -2427 J
the change in kinetic energy is -2427 J
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