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Unformatted text preview: khounvivongsy (sk27799) Practice Problems 1 Weathers (17104) 1 This printout should have 79 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. This set of problems is for your practice  it is strictly optional, and is not for credit. 001 10.0 points At t = 0, a wheel rotating about a fixed axis at a constant angular deceleration of 0 . 65 rad / s 2 has an angular velocity of 2 . 9 rad / s and an angular position of 7 . 8 rad. What is the angular position of the wheel after 3 s? Correct answer: 13 . 575 rad. Explanation: Let : = . 65 rad / s 2 , = 2 . 9 rad / s , = 7 . 8 rad , and t = 3 s . The angular position is f = + t + 1 2 t 2 = 7 . 8 rad + (2 . 9 rad / s) (3 s) + 1 2 ( . 65 rad / s 2 ) (3 s) 2 = 13 . 575 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 19 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 . 818531 s. Explanation: Let : n = 19 , 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 (19) (0 . 055 m) 9 . 8 m / s 2 = . 818531 s . 003 10.0 points A tire placed on a balancing machine in a ser vice station starts from rest and turns through 2 . 76 rev in 1 . 77 s before reaching its final an gular speed. Find its angular acceleration. Correct answer: 11 . 0706 rad / s 2 . Explanation: khounvivongsy (sk27799) Practice Problems 1 Weathers (17104) 2 Let : = 0 rev / s , = 2 . 76 rev , and t = 1 . 77 s . There is a constant angular acceleration, so = t + 1 2 t 2 = 1 2 t 2 = 2 t 2 = 2 (2 . 76 rev) (1 . 77 s) 2 2 rad rev = 11 . 0706 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 = 3 s 1 , = 0 . 9 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 9 . 1 cm from the axis? Correct answer: 99 . 1667 cm / s 2 . Explanation: Let : r = 9 . 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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 Spring '09
 weathers
 mechanics

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