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Unformatted text preview: khounvivongsy (sk27799) Practice Problems 1 Solutions Weathers (17104) 1 This printout should have 79 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points At t = 0, a wheel rotating about a fixed axis at a constant angular deceleration of 0 . 54 rad / s 2 has an angular velocity of 4 rad / s and an angular position of 6 . 6 rad. What is the angular position of the wheel after 4 s? Correct answer: 18 . 28 rad. Explanation: Let : = . 54 rad / s 2 , = 4 rad / s , = 6 . 6 rad , and t = 4 s . The angular position is f = + t + 1 2 t 2 = 6 . 6 rad + (4 rad / s) (4 s) + 1 2 ( . 54 rad / s 2 ) (4 s) 2 = 18 . 28 rad . 002 10.0 points A student throws a long stick of length . 11 m and mass 0 . 5 kg into the air in such a way that the center of mass rises vertically. At the moment it leaves his hand, the stick is horizontal and the speed of the end of the stick nearest to him is zero. When the center of mass of the stick reaches its highest point, the stick is horizontal, and it has made 28 complete revolutions. . 055 m . 11 m v cm . 5 kg v end How long did it take for the center of mass to reach its highest point? Assume the sticks cross sectional area and mass is uniform. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 0 . 993659 s. Explanation: Let : n = 28 , r = 0 . 055 m , m = 0 . 5 kg , and g = 9 . 8 m / s 2 . v = g t and = t = n (2 ) , so from the definition of angular velocity v = r g t = r n (2 ) t . t = radicalbigg 2 n r g = radicalBigg 2 (28) (0 . 055 m) 9 . 8 m / s 2 = . 993659 s . 003 10.0 points A tire placed on a balancing machine in a ser vice station starts from rest and turns through 4 . 04 rev in 0 . 86 s before reaching its final an gular speed. Find its angular acceleration. Correct answer: 68 . 6427 rad / s 2 . Explanation: khounvivongsy (sk27799) Practice Problems 1 Solutions Weathers (17104) 2 Let : = 0 rev / s , = 4 . 04 rev , and t = 0 . 86 s . There is a constant angular acceleration, so = t + 1 2 t 2 = 1 2 t 2 = 2 t 2 = 2 (4 . 04 rev) (0 . 86 s) 2 2 rad rev = 68 . 6427 rad / s 2 . 004 10.0 points The angular position of an object that rotates about a fixed axis is given by ( t ) = e t , where = 2 s 1 , = 1 . 1 rad, and t is in seconds. What is the magnitude of the total linear acceleration at t = 0 of a point on the object that is 11 . 1 cm from the axis? Correct answer: 72 . 6059 cm / s 2 . Explanation: Let : r = 11 . 1 cm . We can find the angular velocity and accel eration: ( t ) = e t ( t ) = d ( t ) dt = e t and ( t ) = d ( t ) dt = 2 e t ....
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This note was uploaded on 05/09/2010 for the course PHYS 1710 taught by Professor Weathers during the Spring '09 term at University of North Carolina Wilmington.
 Spring '09
 weathers
 mechanics

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