00:01
For this problem on the topic of rotation, we are given a disc which has a mass of 14 kilograms, a diameter of 30 centimeters, and a thickness of 8 centimeters mounted on a rough horizontal axis, as we can show, as you can see in the left of the figure.
00:15
Now, the disk is initially addressed, and a constant force of 70 neutrons is then applied to the edge of the disc at an angle of 37 degrees, as shown on the right in the figure.
00:24
And after two seconds the force is then reduced to 24 newton's and the disk spins with a constant angular velocity.
00:31
We want to find the magnitude of the torque due to the friction between the disc and the axle, the angular velocity of the disk after two seconds, as well as the kinetic energy of the disk after two seconds.
00:44
Now from newton's second law, we know that the sum of tors is equal to the applied torque tor a minus the frictional torque tor f and these are equal for constant angular velocity so this is equal to zero which means that the torque tor a is equal to the frictional torque tor f which is r f2 sine theta and so if we put our values into this this is 0 .15 meters times say force of 24 newton's times the sign of 37 degrees.
01:45
And so we get this torque to have a magnitude of 2 .17 newton meters.
01:59
For part b, the net torque initially from 0 to 2 seconds is not 0...