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Unformatted text preview: kuruvila (lk5992) – HW 9 – opyrchal – (11113) 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 θ ( t ) = θ e βt , where β = 4 s 1 , θ = 1 . 3 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 2 . 5 cm from the axis? Correct answer: 85 . 2863 cm / s 2 . Explanation: Let : r = 2 . 5 cm . We can find the angular velocity and accel eration: θ ( t ) = θ e βt ω ( t ) = d θ ( t ) dt = β θ e βt and α ( t ) = d ω ( t ) dt = β 2 θ e βt . At t = 0, since e = 1, ω = β θ and α = β 2 θ , so the tangential and radial accelerations are a t = αr = β 2 θ r and a r = ω 2 r = β 2 θ 2 r , and the magnitude of the total linear acceler ation is a = radicalBig a 2 t + a 2 r = radicalBig β 4 θ 2 r 2 + β 4 θ 4 r 2 = β 2 θ r radicalBig 1 + θ 2 = (4 s 1 ) 2 (1 . 3 rad) (2 . 5 cm) radicalBig 1 + (1 . 3 rad) 2 = 85 . 2863 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 11 . 84 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 . 307057 m / s 2 . Explanation: Let : ω = 78 rpm , r = (5 . 92 in) (0 . 0254 m / in) = 0 . 150368 m t = 4 s . ω = ω + αt = α t α = Δ ω Δ t , so a t = r α = r Δ ω Δ t = (0 . 150368 m) (78 rpm) 4 s 2 π 1 rev 1 min 60 s = . 307057 m / s 2 ....
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This note was uploaded on 11/10/2010 for the course PHYS 111 taught by Professor Moro during the Fall '08 term at NJIT.
 Fall '08
 moro

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