A circular disk of mass 10.0 kg and radius 0.500 m rotates at 30.0 revolutions p

ID: 3894461 • Letter: A

Question

A circular disk of mass 10.0 kg and radius 0.500 m rotates at 30.0 revolutions per second on a vertical

axle, in the direction clockwise when viewed from above. Below it is a 30.0 kg disk of the same radius

that is rotating 30.0 revolutions per second, but in the opposite (counterclockwise) direction. The upper

disk is then allowed to drop onto the lower and, after friction forces have acted the two rotate together as

a single disk of mass 40.0 kg.

(a) What is the final angular velocity of the combination of disks?

(b) What fraction of the initial kinetic energy of the system of two disks is lost in this interaction?

(a) initial angular momentum about axis of rotation= [0.5(10*0.5^2)*(-30)*pi] + [0.5(30*0.5^2)*30*pi] ------ { taking counterclockwise to be +ve}

final angular momentum about same axis of rotation= 0.5(40*0.5^2)*w*pi

conservation of angular momentum for system of two disk ( friction is now internal force) =

[0.5(10*0.5^2)*(-30)*pi] + [0.5(30*0.5^2)*30*pi] = 0.5(40*0.5^2)*w*pi

=> 10*(-30) + 30*30 = 40*w

=> 600 = 40w=> w=15 revolution per second counterclock wise

(b) fraction of lost kinetic energy= 1- [0.5(40*0.5^2)*(15*pi)^2] / [[0.5(10*0.5^2)*((-30)*pi)^2] + [0.5(30*0.5^2)*(30*pi)^2] ]

=1- [ 40*15*15 ] / [ 10*900+ 30*900 ]

= 1- 1/4= 3/4

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