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chaney (glc568) Rotations and Gravity murthy (21118) 1 This print-out should have 47 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 6 from the vertical when the car has a speed of 33 . 3 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: 1 . 03002 m / s 2 . Explanation: For the plumb bob, along the vertical direc- tion we have T cos = mg along the horizontal direction we have T sin = mv 2 r From these two equations the centripetal ac- celeration is a c = v 2 r = g tan , the radius of the curve is r = v 2 g tan , and the speed of the car v = radicalbig g r tan . The deflection angle is 1 = 6 , so a c = g tan 1 = ( 9 . 8 m / s 2 ) tan6 = 1 . 03002 m / s 2 . 002 (part 2 of 3) 10.0 points What is the radius of the curve? Correct answer: 1076 . 57 m. Explanation: The radius is r = v 2 g tan 1 = (33 . 3 m / s) 2 (9 . 8 m / s 2 ) tan6 = 1076 . 57 m . 003 (part 3 of 3) 10.0 points What is the speed of the car if the plumb bobs deflection is 12 . 2 while rounding the same curve? Correct answer: 47 . 7606 m / s. Explanation: Here the deflection angle is 2 = 12 . 2 so v 2 = radicalbig g r tan 2 = radicalBig (9 . 8 m / s 2 ) (1076 . 57 m) tan12 . 2 = 47 . 7606 m / s . 004 10.0 points A highway curves to the left with radius of curvature R = 44 m. The highways surface is banked at = 18 so that the cars can take this curve at higher speeds. Consider a car of mass 982 kg whose tires have static friction coefficient = 0 . 59 against the pavement. The acceleration of gravity is 9 . 8 m / s 2 . chaney (glc568) Rotations and Gravity murthy (21118) 2 top view R = 44 m 18 rear view = . 5 9 How fast can the car take this curve without skidding to the outside of the curve? Correct answer: 22 . 0925 m / s. Explanation: Let : R = 44 m , = 18 , m = 982 kg , and = 0 . 59 . Basic Concepts: (1) To keep an object moving in a circle requires a net force of magnitude F c = ma c = m v 2 r directed toward the center of the circle. (2) Static friction law: |F| N . Solution: Consider the free body diagram for the car. Looking from the rear of the car, we have: N F mg a c Let the x axis go parallel the highway sur- face , to the left and 18 below the horizontal, and let the y axis be perpendicular to the sur- face, 18 leftward from vertically up. The cars centripetal acceleration is directed hor- izontally to the left, so in our coordinates it has components... View Full Document