hw12 - Overton, Mays Homework 12 Due: Feb 21 2005, 4:00 am...

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Overton, Mays – Homework 12 – Due: Feb 21 2005, 4:00 am – Inst: Turner 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – fnd all choices be±ore answering. The due time is Central time. 001 (part 1 o± 1) 10 points A curve o± radius 59 . 3 m is banked so that a car traveling with uni±orm speed 63 km / hr can round the curve without relying on ±ric- tion to keep it ±rom slipping to its le±t or right. The acceleration o± gravity is 9 . 8 m / s 2 . 2 . 9 Mg μ 0 θ What is θ ? Correct answer: 27 . 7884 . Explanation: Let : m = 2900 kg , v = 63 km / hr , r = 59 . 3 m , and μ 0 . Basic Concepts: Consider the ±ree body diagram ±or the car. The ±orces acting on the car are the normal ±orce, the ±orce due to gravity, and possibly ±riction. N N cos θ mg N sin θ x y To keep an object moving in a circle re- quires a ±orce directed toward the center o± the circle; the magnitude o± the ±orce is F c = ma c = m v 2 r . Also remember, ~ F = X i ~ F i . Using the ±ree-body diagram, we have X i F x N sin θ - μ N cos θ = m v 2 r (1) X i F y N cos θ + μ N sin θ = mg (2) ( mg ) k = mg sin θ (3) ma k = m v 2 r cos θ (4) and , μ = 0 , we have tan θ = v 2 g r (5) Solution: Solution in an Inertial Frame: Watching from the Point of View of Some- one Standing on the Ground. The car is per±orming circular motion with a constant speed, so its acceleration is just the centripetal acceleration, a c = v 2 r . The net ±orce on the car is F net = ma c = m v 2 r The component o± this ±orce parallel to the incline is X ~ F k = mg sin θ = F net cos θ = m v 2 r cos θ . In this re±erence ±rame, the car is at rest, which means that the net ±orce on the car (taking in consideration the centri±ugal ±orce) is zero. Thus the component o± the net “real” ±orce parallel to the incline is equal to the component o± the centri±ugal ±orce along that direction. Now, the magnitude o± the cen- tri±ugal ±orce is equal to F c = m v 2 r , so F k = F net cos θ = F c cos θ = m v 2 r cos θ
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Overton, Mays – Homework 12 – Due: Feb 21 2005, 4:00 am – Inst: Turner 2 F k is the component of the weight of the car parallel to the incline. Thus mg sin θ = F k = m v 2 r cos θ tan θ = v 2 g r = (63 km / hr) 2 (9 . 8 m / s 2 )(59 . 3 m) × µ 1000m km 2 µ hr 3600s 2 = 0 . 526981 θ = arctan(0 . 526981) = (0 . 484999 rad) 180 deg π rad = 27 . 7884 . 002
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This note was uploaded on 03/22/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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hw12 - Overton, Mays Homework 12 Due: Feb 21 2005, 4:00 am...

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