HW 9-solutions

HW 9-solutions - billing (cab4763) HW 9 opyrchal (11107) 1...

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Unformatted text preview: billing (cab4763) HW 9 opyrchal (11107) 1 This print-out should have 11 questions. Multiple-choice 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.

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HW 9-solutions - billing (cab4763) HW 9 opyrchal (11107) 1...

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