solution_6 - ouseph(jao888 HW06 tsoi(92940 This print-out...

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ouseph (jao888) – HW06 – tsoi – (92940) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 of 2) 10.0 points A record has an angular speed oF 19 rev / min. What is its angular speed? Correct answer: 1 . 98968 rad / s. Explanation: Let : ω = 19 rev / min . ω = (19 rev / min) p 2 π rad rev Pp 1 min 60 s P = 1 . 98968 rad / s . 002 (part 2 of 2) 10.0 points Through what angle does it rotate in 0 . 94 s? Correct answer: 1 . 8703 rad. Explanation: Let : t = 0 . 94 s . θ = ω t = (1 . 98968 rad / s) (0 . 94 s) = 1 . 8703 rad . 003 (part 1 of 2) 10.0 points A racing car travels on a circular track oF radius 310 m, moving with a constant linear speed oF 70 . 9 m / s. ±ind its angular speed. Correct answer: 0 . 22871 rad / s. Explanation: Let : v = 70 . 9 m / s and R = 310 m . The linear speed is v = R ω ω = v R = 70 . 9 m / s 310 m = 0 . 22871 rad / s . 004 (part 2 of 2) 10.0 points ±ind the magnitude oF its acceleration. Correct answer: 16 . 2155 m / s 2 . Explanation: With the car moving at a constant speed, there is no tangential acceleration, so the ac- celeration is purely radial: a r = v 2 R = (70 . 9 m / s) 2 310 m = 16 . 2155 m / s 2 . 005 10.0 points Initially a wheel rotating about a fxed axis at a constant angular deceleration oF 0 . 6 rad / s 2 has an angular velocity oF 0 rad / s and an angular position oF 9 . 5 rad. What is the angular position oF the wheel aFter 2 . 9 s? Correct answer: 6 . 977 rad. Explanation: Let : α = 0 . 6 rad / s 2 , ω 0 = 0 rad / s , θ 0 = 9 . 5 rad , and t = 2 . 9 s . ±rom kinematics, θ = ω 0 t + 1 2 αt 2 + θ 0 = (0 rad / s) (2 . 9 s) + 1 2 ( 0 . 6 rad / s 2 ) (2 . 9 s) 2 + 9 . 5 rad = 6 . 977 rad . 006 (part 1 of 2) 10.0 points A dentist’s drill starts From rest. AFter 1 . 28 s
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ouseph (jao888) – HW06 – tsoi – (92940) 2 of constant angular acceleration it turns at a rate of 20240 rev / min. Find the drill’s angular acceleration. Correct answer: 1655 . 88 rad / s 2 . Explanation: Let : t = 1 . 28 s and ω f = 20240 rev / min . Since ω 0 = 0, α = ω f ω 0 t = ω f t = 20240 rev / min 1 . 28 s · 2 π rev · 1 min 60 s = 1655 . 88 rad / s 2 . 007 (part 2 of 2) 10.0 points Throughout what angle does the drill rotate during this period? Correct answer: 1356 . 5 rad. Explanation: θ = ω 0 t + 1 2 α t 2 = 0 + 1 2 (1655 . 88 rad / s 2 ) (1 . 28 s) 2 = 1356 . 5 rad . 008 10.0 points An automobile moves at constant speed down one hill and up another hill along the smoothly curved surface as shown below. car Which of the following diagrams best rep- resents the directions of the velocity and the acceleration of the automobile at the instant that it is at the lowest position as shown? 1. v a correct 2. v a 3. v a 4. v a 5. v a Explanation: At the lowest position, the instantaneous velocity of the automobile is directed horizon- tally to the right. Since the speed of the automobile is un- changed, there is no tangential accelera- tion. Also since the path of the automobile is curved, there is centripetal acceleration, pointing upward. 009 (part 1 of 5) 10.0 points A car travels at a speed of 22 m / s around a curve of radius 42 m.
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solution_6 - ouseph(jao888 HW06 tsoi(92940 This print-out...

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