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Unformatted text preview: homework 09 – HETTIGER, CHRISTOF – Due: Feb 7 2008, 4:00 am 1 Question 1, chap 4, sect 5. part 1 of 1 10 points A ball on the end of a string is whirled around in a horizontal circle of radius 0 . 274 m. The plane of the circle is 1 . 49 m above the ground. The string breaks and the ball lands 2 . 79 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: 93 . 426 m / s 2 (tolerance ± 1 %). 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 = 93 . 426 m / s 2 . Question 2, chap 4, sect 5. part 1 of 2 10 points If the rotation of a planet of radius 3 . 49 × 10 6 m and freefall acceleration 9 . 7 m / s 2 increased to the point that the centripetal ac celeration was equal to the gravitational ac celeration at the equator, what would be the tangential speed of a person standing at the equator? Correct answer: 5818 . 33 m / s (tolerance ± 1 %). Explanation: From the formula for the centripetal accel eration a = v 2 R we obtain that the velocity v at the equator is v = radicalbig g R = radicalBig (9 . 7 m / s 2 ) (3 . 49 × 10 6 m) = 5818 . 33 m / s ....
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This note was uploaded on 05/09/2010 for the course PHYSICS PHY 301k taught by Professor Turner during the Spring '10 term at University of TexasTyler.
 Spring '10
 Turner
 Physics, Work

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