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Unformatted text preview: 1 Physics 221 Fall 2008 Homework #9 Solutions Ch. 10:5,6; Ch. 11:1-3; Ch. 12:1-3 Due Tues, Oct 28, 2008 9.1 (a) Under what circumstance is the angular momentum of a system constant in time? The angular momentum of a system is conserved if there is no net torque on the system. (b) Consider a system initially rotating at constant angular velocity about a symmetry axis through the center of mass of the system. Under what condition does a net external torque on the system increase the magnitude of the angular momentum of the system versus time? We have = d L dt . Therefore, if is in the direction of L , then d L is also in the direction of L , which means that after a time dt , the angular momentum is now L + d L , and thus the magnitude of L increases with time. 9.2 The bottom disk #1 in the figure below with moment of inertia I 1 = 10.0 kg m 2 is freely spinning about its symmetry axis at initial angular speed 10 = 20.0 rad/s. It is supported from below on an axle at its center like a merry-go-round. Now a stationary disk #2 with moment of inertia I 2 = 8.0 kg m 2 is dropped onto disk #1, and friction between the two disks causes the two disks to attain the same final angular velocity. The system consists of both disks. (a) What is the initial angular momentum of the system? Let the z-axis point upwards. Then the initial angular momentum of the system is L 1z = I 1 10 = (10.0 kg m 2 )(20.0 rad/s) = 200 kg m 2 /s . I 1 I 2 2 (b) After disk #2 is dropped onto disk #1 and they attain the same angular velocity, what is this final angular velocity of the system? Using conservation of angular momentum, the final angular momentum of the system is the same as the initial angular momentum: L 2 z = ( I 1 + I 2 ) 2 z = L 1z . Thus, 2 z = L 1z /( I 1 + I 2 ) = (200 kg m 2 /s)/(10.0 kg m 2 + 8.0 kg m 2 ) = 11.1 rad/s . (c) What are the initial and final kinetic energies of the system? How much kinetic energy was lost after disk #2 was dropped onto disk #1? Where did the missing kinetic energy go? The initial (rotational) kinetic energy of the system is K 1rot = (1/2) I 1 10 2 = (1/2)(10.0 kg m 2 )(20.0 rad/s) 2 = 2000 J . The final (rotational) kinetic energy of the system is K 2rot = (1/2)( I 1 + I 2 ) 2 2 = (1/2)(10.0 kg m 2 + 8.0 kg m 2 )(11.1 rad/s) 2 = 1110 J . Thus, 890 J of kinetic energy was lost. This was converted to internal energy of the disks via the friction force between them when the top disk was dropped onto the bottom disk, and resulted in the temperatures of the disks increasing....
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