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Unformatted text preview: Pickering, Tracey Homework 8 Due: Feb 8 2006, noon Inst: Drummond 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A train is moving along a circular track with r = 100 m . At A , v = k ~v k = 10 m/s . It is slowing down with a tangential decelera tion of magnitude a tangential = k ~a tangential k = 1 m/s 2 . Sketch ~a total at A . O r A v Quadrants II I III IV Which quadrant should it be in? 1. III correct 2. I 3. IV 4. II Explanation: a radial a tangential a total ~a total = ~a radial + ~a tangential 002 (part 1 of 5) 10 points A flea is at point A on a horizontal turntable 12 . 8 cm from the center. The turntable is ro tating at 30 . 7 rev / min in the counterclockwise direction. The flea jumps vertically upward to a height of 7 . 39 cm and lands on the turntable at point B . Place the coordinate origin at the center of the turntable with the positive x axis fixed in space and initially passing through A . The acceleration of gravity is 9 . 8 m / s 2 . Find the linear displacement of the flea. Correct answer: 0 . 101072 m. Explanation: The angular frequency of the turntable is given by = (30 . 7 rev / min) 2 rad rev 1 min 60 s = 3 . 2149 rad / s . Therefore, the speed at point A is v A = r A = (3 . 2149 rad / s) (0 . 128 m) = 0 . 411507 m / s . We note that, since point A coincides initially with the x axis, and the turntable is rotating counterclockwise, the velocity at point A is ~v A = (0 . 411507 m / s) ~. This is the y component of the velocity of the flea as it jumps vertically up. Now we need to get the time to jump to a height of 0 . 0739 m and to land. Note: According to our axes, the vertical direction is along the z axis, and the acceleration due to gravity is pointing in the negative z axis. Along the z axis we have z ( t ) = v z t 1 2 g t 2 , and v z ( t ) = v z g t. At the maximum height, v z = 0, and we obtain v z = g t. Substituting this expression in z ( t ) we obtain for the maximum height h = g t 2 1 2 g t 2 = 1 2 g t 2 ....
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