# hw14 - hinojosa(jlh3938 – homework 14 – Turner...

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Unformatted text preview: hinojosa (jlh3938) – homework 14 – Turner – (58185) 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 8 ◦ from the vertical when the car has a speed of 24 . 6 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 . 3773 m / s 2 . Explanation: For the plumb bob, along the vertical direc- tion we have T cos θ = mg along the horizontal direction we have T sin θ = mv 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 = 8 ◦ , so a c = g tan θ 1 = ( 9 . 8 m / s 2 ) tan8 ◦ = 1 . 3773 m / s 2 . 002 (part 2 of 3) 10.0 points What is the radius of the curve? Correct answer: 439 . 381 m. Explanation: The radius is r = v 2 g tan θ 1 = (24 . 6 m / s) 2 (9 . 8 m / s 2 ) tan8 ◦ = 439 . 381 m . 003 (part 3 of 3) 10.0 points What is the speed of the car if the plumb bob’s deflection is 10 . 1 ◦ while rounding the same curve? Correct answer: 27 . 6948 m / s. Explanation: Here the deflection angle is θ 2 = 10 . 1 ◦ so v 2 = radicalbig g r tan θ 2 = radicalBig (9 . 8 m / s 2 ) (439 . 381 m) tan10 . 1 ◦ = 27 . 6948 m / s . 004 10.0 points A highway curves to the left with radius of curvature R = 41 m. The highway’s surface is banked at θ = 27 ◦ so that the cars can take this curve at higher speeds. Consider a car of mass 1054 kg whose tires have static friction coefficient μ = 0 . 73 against the pavement. The acceleration of gravity is 9 . 8 m / s 2 . hinojosa (jlh3938) – homework 14 – Turner – (58185) 2 top view R = 41 m 2 7 ◦ rear view μ = . 7 3 How fast can the car take this curve without skidding to the outside of the curve? Correct answer: 28 . 1603 m / s. Explanation: Let : R = 41 m , θ = 27 ◦ , m = 1054 kg , and μ = 0 . 73 . 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| ≤ μ N ....
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hw14 - hinojosa(jlh3938 – homework 14 – Turner...

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