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hw12 - lenoir(wml297 homework 12 Turner(58220 This...

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lenoir (wml297) – homework 12 – Turner – (58220) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A student builds and calibrates an accelerom- eter, which she uses to determine the speed of her car around a certain highway curve. The accelerometer is a plumb bob with a protrac- tor that she attaches to the roof of her car. A friend riding in the car with her observes that the plumb bob hangs at an angle of 19 from the vertical when the car has a speed of 30 . 4 m / s. The acceleration of gravity is 9 . 8 m / s 2 . At this instant, what is the centripetal ac- celeration of the car rounding the curve? Correct answer: 3 . 37441 m / s 2 . Explanation: For the plumb bob, along the vertical direc- tion we have T cos θ = m g along the horizontal direction we have T sin θ = m v 2 r From these two equations the centripetal ac- celeration is a c = v 2 r = g tan θ , the radius of the curve is r = v 2 g tan θ , and the speed of the car v = radicalbig g r tan θ . The deflection angle is θ 1 = 19 , so a c = g tan θ 1 = ( 9 . 8 m / s 2 ) tan 19 = 3 . 37441 m / s 2 . 002 (part 2 of 3) 10.0 points What is the radius of the curve? Correct answer: 273 . 873 m. Explanation: The radius is r = v 2 g tan θ 1 = (30 . 4 m / s) 2 (9 . 8 m / s 2 ) tan 19 = 273 . 873 m . 003 (part 3 of 3) 10.0 points What is the speed of the car if the plumb bob’s deflection is 6 . 7 while rounding the same curve? Correct answer: 17 . 7565 m / s. Explanation: Here the deflection angle is θ 2 = 6 . 7 so v 2 = radicalbig g r tan θ 2 = radicalBig (9 . 8 m / s 2 ) (273 . 873 m) tan 6 . 7 = 17 . 7565 m / s . 004 10.0 points A highway curves to the left with radius of curvature R = 48 m. The highway’s surface is banked at θ = 25 so that the cars can take this curve at higher speeds. Consider a car of mass 971 kg whose tires have static friction coefficient μ = 0 . 83 against the pavement. The acceleration of gravity is 9 . 8 m / s 2 .
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lenoir (wml297) – homework 12 – Turner – (58220) 2 top view R = 48 m 25 rear view μ = 0 . 83 How fast can the car take this curve without skidding to the outside of the curve? Correct answer: 31 . 5406 m / s. Explanation: Let : R = 48 m , θ = 25 , m = 971 kg , and μ = 0 . 83 . Basic Concepts: (1) To keep an object moving in a circle requires a net force of magnitude F c = m a c = m v 2 r directed toward the center of the circle. (2) Static friction law: |F| ≤ μ N . Solution: Consider the free body diagram for the car. Looking from the rear of the car, we have: N F m g a c Let the x axis go parallel the highway sur- face , to the left and 25 below the horizontal, and let the y axis be perpendicular to the sur- face, 25 leftward from vertically up. The car’s centripetal acceleration is directed hor-
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