elapsed time (Equation 8.2). Therefore, the angular displacement is equal to the product of the average angular velocity and the elapsed time The elapsed time is given, so we need to determine the average angular velocity. We can do this by using the graph of angular velocity versus time that accompanies the problem. SOLUTION The angular displacement is related to the average angular velocity and the 3.0 s 8.0 s +15 rad/ s –9.0 rad/ s 0 Time (s) Angular velocity +3.0 rad/s
elapsed time t by Equation 8.2, t . The elapsed time is given as 8.0 s. To obtain the average angular velocity, we need to extend the graph that accompanies this problem from a time of 5.0 s to 8.0 s. It can be seen from the graph that the angular velocity increases by +3.0 rad/s during each second. Therefore, when the time increases from 5.0 to 8.0 s, the angular velocity increases from +6.0 rad/s to 6 rad/s + 3 (3.0 rad/s) = +15 rad/s. A graph of the angular velocity from 0 to 8.0 s is shown at the right. The average angular velocity during this time is equal to one half the sum of the initial and final angular velocities: 1 1 0 2 2 9.0 rad/s +15 rad/s 3.0 rad/s The angular displacement of the wheel from 0 to 8.0 s is 3.0 rad/s 8.0 s 24 rad t 31. REASONING According to Equation 3.5b, the time required for the diver to reach the water, assuming free-fall conditions, is 2 / y t y a . If we assume that the "ball" formed by the diver is rotating at the instant that she begins falling vertically, we can use Equation 8.2 to calculate the number of revolutions made on the way down. SOLUTION Taking upward as the positive direction, the time required for the diver to reach the water is t 2(–8.3 m) –9.80 m/s 2 1.3 s Solving Equation 8.2 for , we find (1.6 rev/s)(1.3 s)= 2.1 rev t 32. REASONING The time required for the change in the angular velocity to occur can be found by solving Equation 8.4 for t . In order to use Equation 8.4, however, we must know the initial angular velocity 0 . Equation 8.6 can be used to find the initial angular velocity. SOLUTION From Equation 8.6 we have 1 2 ( 0 ) t Solving for 0 gives 0 2 t
Since the angular displacement is zero, 0 = – . Solving 0 t (Equation 8.4) for t and using the fact that 0 = – give t 2 2( 25.0 rad/s) 4.00 rad/s 2 12.5 s 33. REASONING The angular displacement of the child when he catches the horse is, from Equation 8.2, c c t . In the same time, the angular displacement of the horse is, from Equation 8.7 with 0 0 rad/s, h 1 2 t 2 . If the child is to catch the horse c h ( / 2) . SOLUTION Using the above conditions yields 2 1 1 c 2 2 0 t t or 2 2 1 1 2 2 (0.0100 rad/s ) 0.250 rad/s rad 0 t t The quadratic formula yields t = 7.37 s and 42.6 s; therefore, the shortest time needed to catch the horse is t = 7.37 s .
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