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HW 9-solutions

# HW 9-solutions - kuruvila(lk5992 HW 9 opyrchal(11113 This...

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kuruvila (lk5992) – HW 9 – opyrchal – (11113) 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 θ ( t ) = θ 0 e βt , where β = 4 s - 1 , θ 0 = 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 0 e βt ω ( t ) = d θ ( t ) dt = β θ 0 e βt and α ( t ) = d ω ( t ) dt = β 2 θ 0 e βt . At t = 0, since e 0 = 1, ω = β θ 0 and α = β 2 θ 0 , so the tangential and radial accelerations are a t = α r = β 2 θ 0 r and a r = ω 2 r = β 2 θ 2 0 r , and the magnitude of the total linear acceler- ation is a = radicalBig a 2 t + a 2 r = radicalBig β 4 θ 2 0 r 2 + β 4 θ 4 0 r 2 = β 2 θ 0 r radicalBig 1 + θ 2 0 = (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 0 + α t = 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 = 0 . 307057 m / s 2 . 003 (part 2 of 3) 10.0 points When the record is at its final speed, what is the tangential velocity of the bug?

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