solution_pdf3 - chaney (glc568) Rotations and Gravity...

Download Document
Showing pages : 1 - 2 of 21
This preview has blurred sections. Sign up to view the full version! View Full Document
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 – fnd 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: ±or 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 ±rom 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 = r g r tan θ . The de²ection angle is θ 1 = 6 , so a c = g tan θ 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 θ 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 bob’s de²ection is 12 . 2 while rounding the same curve? Correct answer: 47 . 7606 m / s. Explanation: Here the de²ection angle is θ 2 = 12 . 2 so v 2 = r g r tan θ 2 = R (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 highway’s surFace is banked at θ = 18 so that the cars can take this curve at higher speeds. Consider a car oF mass 982 kg whose tires have static Friction coe³cient μ = 0 . 59 against the pavement. The acceleration oF gravity is 9 . 8 m / s 2 .
Background image of page 1
chaney (glc568) – Rotations and Gravity – murthy – (21118) 2 top view R = 44 m 18 rear view μ = 0 . 59 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 , θ = 18 , m = 982 kg , and μ = 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| ≤ μ 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 car’s centripetal acceleration is directed hor- izontally to the left, so in our coordinates it has components a x = + v 2 R cos
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.