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Unformatted text preview: billing (cab4763) HW 9 opyrchal (11107) 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points The angular position of an object that rotates about a fixed axis is given by ( t ) = e t , where = 4 s 1 , = 0 . 6 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 . 7 cm from the axis? Correct answer: 130 . 987 cm / s 2 . Explanation: Let : r = 11 . 7 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 . At t = 0, since e = 1, = and = 2 , so the tangential and radial accelerations are a t = r = 2 r and a r = 2 r = 2 2 r , and the magnitude of the total linear acceler ation is a = radicalBig a 2 t + a 2 r = radicalBig 4 2 r 2 + 4 4 r 2 = 2 r radicalBig 1 + 2 = (4 s 1 ) 2 (0 . 6 rad) (11 . 7 cm) radicalBig 1 + (0 . 6 rad) 2 = 130 . 987 cm / s 2 . 002 (part 1 of 3) 10.0 points What is the tangential acceleration of a bug on the rim of a 78 rpm record of diameter 5 . 35 in . if the record moves from rest to its final angular speed in 4 s? The conversion between inches and meters is 0 . 0254 m / in. Correct answer: 0 . 138746 m / s 2 . Explanation: Let : = 78 rpm , r = (2 . 675 in) (0 . 0254 m / in) = 0 . 067945 m t = 4 s . = + t = t = t , so a t = r = r t = (0 . 067945 m) (78 rpm) 4 s 2 1 rev 1 min 60 s = . 138746 m / s 2 ....
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This note was uploaded on 02/04/2011 for the course PHYSICS 121 taught by Professor Fayngold during the Spring '10 term at NJIT.
 Spring '10
 Fayngold
 Physics, Acceleration

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