This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: hinojosa (jlh3938) homework 14 Turner (58185) 1 This printout should have 12 questions. Multiplechoice 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 = 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 = 8 , so a c = g tan 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 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 bobs deflection is 10 . 1 while rounding the same curve? Correct answer: 27 . 6948 m / s. Explanation: Here the deflection angle is 2 = 10 . 1 so v 2 = radicalbig g r tan 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 highways surface is banked at = 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 = 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 = . 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 , = 27 , m = 1054 kg , and = 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 N ....
View Full
Document
 Spring '08
 Turner
 Physics, Work

Click to edit the document details