# oldhw 12 - alexander(jra2623 oldhomework 12 Turner(92510...

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alexander (jra2623) – oldhomework 12 – Turner – (92510) 1 This print-out should have 11 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 curve oF radius r is banked at angle θ so that a car traveling with uniForm speed v can round the curve without relying on Friction to keep it From slipping to its leFt or right. The acceleration oF gravity is 9 . 8 m / s 2 . m μ 0 θ ±ind the component oF the net Force parallel to the incline s V F b . 1. F = m v 2 r tan θ 2. F = m g cos θ 3. F = m v 2 r tan θ 4. F = mg cot θ 5. F = m v 2 r cos θ correct 6. F = m r g 2 + v 4 r 2 7. F = m v 2 r sin θ 8. F = m v 2 r cos θ 9. F = m g tan θ 10. F = m v 2 r sin θ Explanation: Basic Concepts: To keep an object mov- ing in a circle requires 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: V F = s i V F i Solution: Solution in an Inertial Frame: Watching From the “Point oF View oF Some- one Standing on the Ground”. m g F net F b N r y The car is perForming circular motion with a constant speed, thus 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 s V F b = 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 b = F net cos θ = F c cos θ = m v 2 r cos θ

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alexander (jra2623) – oldhomework 12 – Turner – (92510) 2 002 (part 2 of 3) 10.0 points If r = 52 m and v = 62 km / hr, what is θ ? Correct answer: 30 . 2009 . Explanation: F b is the component of the weight of the car parallel to the incline. Thus m g sin θ = F b = m v 2 r cos θ tan θ = v 2 g r = (62 km / hr) 2 (9 . 8 m / s 2 )(52 m ) × p 1000 m km P 2 p hr 3600 s P 2 = 0 . 582035 θ = arctan(0 . 582035) = 30 . 2009 003 (part 3 of 3) 10.0 points With what frictional force must the road push on a 1200 kg car if the driver exceeds the speed for which the curve was designed by Δ v = 14 km / hr? Correct answer: 2973
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oldhw 12 - alexander(jra2623 oldhomework 12 Turner(92510...

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