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Unformatted text preview: banuelos (ojb93) – Homework 08 – florin – (58140) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Consider a conical pendulum, where a string with length ℓ is attached to a mass m . The angle between the string and the ver- tical is θ . The orbit is in the horizontal plane with radius r and tangential velocity vectorv . v r g ℓ m θ What is the free body diagram for mass m ? The acceleration of gravity is 9 . 8 m / s 2 . 1. θ correct 2. θ 3. θ 4. θ 5. θ Explanation: There are only two forces exerted on the ball, the gravity and the tension on the string. T mg θ 002 10.0 points A highway curves to the left with radius of curvature R = 42 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 1055 kg whose tires have static friction coefficient μ = 0 . 69 against the pavement. The acceleration of gravity is 9 . 8 m / s 2 . banuelos (ojb93) – Homework 08 – florin – (58140) 2 top view R = 42 m 2 7 ◦ rear view μ = . 6 9 How fast can the car take this curve without skidding to the outside of the curve? Correct answer: 27 . 5938 m / s. Explanation: Let : R = 42 m , θ = 27 ◦ , m = 1055 kg , and μ = 0 . 69 . 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 27 ◦ below the horizontal, and let the y axis be perpendicular to the sur- face, 27 ◦ 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 θ, a y = + v 2 R sin θ. (1) At the same time, totalling all the forces act- ing on the car and applying the Second Law, we have ma x = F net x = + F + mg sin θ, (2) ma y = F net y = + N - mg cos θ. (3) Combining eqs. (1–3) and solving for the nor- mal force N and the friction force F we find N = m parenleftbigg + g cos θ + v 2 R sin θ parenrightbigg , (4) F = m parenleftbigg- g sin θ + v 2 R cos θ parenrightbigg . (5) To be precise, these are the normal force and the static friction force on the car which does not skid. To make sure this is possible we must also verify the static friction law, namely |F| ≤ μ N = ⇒ F ≤ μ N and-F ≤ μ N . (6) Substituting eqs. (4–5) into this formula yields m parenleftbigg v 2 R cos θ- g sin θ parenrightbigg (7) ≤ μm parenleftbigg g cos θ + v 2 R sin θ parenrightbigg , and m parenleftbigg g sin θ- v 2 R cos θ parenrightbigg (8) ≤ μm parenleftbigg g cos θ + v 2 R sin θ parenrightbigg , and we may further simplify these conditions by dividing both sides by mg...
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This note was uploaded on 04/11/2010 for the course PHYSICS 58100 taught by Professor Florin during the Spring '10 term at University of Texas.
- Spring '10