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# hw12 - Overton Mays Homework 12 Due 4:00 am Inst Turner...

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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 – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A curve of radius 59 . 3 m is banked so that a car traveling with uniform speed 63 km / hr can round the curve without relying on fric- tion to keep it from slipping to its left or right. The acceleration of 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 free body diagram for the car. The forces acting on the car are the normal force, the force due to gravity, and possibly friction. μ N N N cos θ m g N sin θ x y To keep an object moving in a circle re- quires a force directed toward the center of the circle; the magnitude of the force is F c = m a c = m v 2 r . Also remember, ~ F = X i ~ F i . Using the free-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 = m g (2) ( m g ) k = m g sin θ (3) m a k = m v 2 r cos θ (4) and , if 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 performing circular motion with a constant speed, so its acceleration is just the centripetal acceleration, a c = v 2 r . The net force on the car is F net = m a c = m v 2 r The component of this force parallel to the incline is X ~ F k = m g sin θ = F net cos θ = m v 2 r cos θ . In this reference frame, the car is at rest, which means that the net force on the car (taking in consideration the centrifugal force) is zero. Thus the component of the net “real” force parallel to the incline is equal to the component of the centrifugal force along that direction. Now, the magnitude of the cen- trifugal force 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 m g 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) × 1000 m km 2 hr 3600 s 2 = 0 . 526981 θ = arctan(0 . 526981) = (0 . 484999 rad) 180 deg π rad = 27 . 7884 .
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hw12 - Overton Mays Homework 12 Due 4:00 am Inst Turner...

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