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Unformatted text preview: lim (kl9356) – Homework 5 – Weathers – (17104) 1 This printout should have 9 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A ball on the end of a string is whirled around in a horizontal circle of radius 0 . 254 m. The plane of the circle is 1 . 24 m above the ground. The string breaks and the ball lands 2 . 77 m away from the point on the ground directly beneath the ball’s location when the string breaks. The acceleration of gravity is 9 . 8 m / s 2 . Find the centripetal acceleration of the ball during its circular motion. Correct answer: 119 . 371 m / s 2 . Explanation: In order to find the centripetal acceleration of the ball, we need to find the initial velocity of the ball. Let y be the distance above the ground. After the string breaks, the ball has no initial velocity in the vertical direction, so the time spent in the air may be deduced from the kinematic equation, y = 1 2 g t 2 . Solving for t , ⇒ t = radicalbigg 2 y g . Let d be the distance traveled by the ball. Then v x = d t = d radicalbigg 2 y g . Hence, the centripetal acceleration of the ball during its circular motion is a c = v 2 x r = d 2 g 2 y r = 119 . 371 m / s 2 . 002 10.0 points Before throwing a 0 . 808 kg discus, an ath lete rotates it along a circular path of radius 1 . 47 m. The maximum speed of the discus is 11 . 5 m / s. Determine the magnitude of its maximum radial acceleration. Correct answer: 89 . 966 m / s 2 . Explanation: The maximum radial acceleration is a r = v 2 r = (11 . 5 m / s) 2 1 . 47 m = 89 . 966 m / s 2 . 003 (part 1 of 3) 10.0 points A train slows down at a constant rate as it rounds a sharp circular horizontal turn. Its initial speed is not known. It takes 18 . 4 s to slow down from 81 km /...
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This note was uploaded on 09/07/2010 for the course PHYS 1710 taught by Professor Weathers during the Fall '08 term at North Texas.
 Fall '08
 Weathers
 Magnetism, Work

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