two 3kg weights at a distance.75m from the center of iis by In a classroom demon
ID: 2037899 • Letter: T
Question
two 3kg weights at a distance.75m from the center of iis by In a classroom demonstration Nkam is holding He is initially rotating at 3 rad/s -0.3 radis 3kg 0.75 m 0.25 m Initial Final 4. If he pulls the weights in so that they are only.25m from the center of his body, what is his new angular velocity (Ignore his moment of inertia)? (a) ? 1.0 rad/s (b) ?-1.4 rad/s (c) o 2.7 rad/s (d) ? = 3.2 rad/s 5. How much work did he do in pulling the weights inward? (a) 1.2J (b) 2.4J (e) 3.6J (d) 4.8J (e) 6.0J 6 He now lets the weights return to their initial position. What is the final angular momentum of the weights? (a) 51 kg-m'/s2 (b) 68 kg-mls (c) 82 kg-mls (d) 1.0 kg-m'ls (e) 1.2 kg-mls?Explanation / Answer
angular momentum=moment of inertia*angular speed
moment of inertia of a point mass m at a distance d from the axis =m*d^2
4:
as there are no external forces, total angular moment will be conserved.
if final angular speed is w rad/s,
then 2*mass of each weight*initial distance from body^2*initial angular speed
=2*mass of each weight *final distance from body^2*final angular speed
==>2*3*0.75^2*0.3=2*3*0.25^2*w
==>w=(2*3*0.75^2*0.3/(2*3*0.25^2))=2.7 rad/sec
5:
work done=final rotational kinetic energy-initial rotational kinetic energy
=2*(0.5*final moment of inertia of each weight*final angular speed^2-0.5*initial moment of inertia of each weight
*initial angular speed^2)
=2*(0.5*3*0.25^2*2.7^2-0.5*3*0.75^2*0.3^2)=1.215 J
6:
final angular momentum of the weights=2*3*0.75^2*0.3=1.0125 kg.m^2.rad/s
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