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HW12_Solutions - hiarker(srh959 oldhomework 12 Turner(58220...

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hiarker (srh959) – oldhomework 12 – Turner – (58220) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find 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 Find the component of the net force parallel to the incline summationdisplay vector F bardbl . 1. F = m v 2 r sin θ 2. F = m v 2 r tan θ 3. F = m g cos θ 4. F = m g tan θ 5. F = m v 2 r cos θ 6. F = m v 2 r tan θ 7. F = m v 2 r cos θ correct 8. F = m v 2 r sin θ 9. F = mg cot θ 10. F = m radicalbigg g 2 + v 4 r 2 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: vector F = summationdisplay i vector 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 bardbl 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 summationdisplay vector F bardbl = 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 bardbl = F net cos θ = F c cos θ = m v 2 r cos θ
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hiarker (srh959) – oldhomework 12 – Turner – (58220) 2 002 (part 2 of 3) 10.0 points If r = 56 m and v = 62 km / hr, what is θ ? Correct answer: 28 . 3895 . Explanation: F bardbl is the component of the weight of the car parallel to the incline. Thus m g sin θ = F bardbl = m v 2 r cos θ tan θ = v 2 g r = (62 km / hr ) 2 (9 . 8 m / s 2 )(56 m ) × parenleftbigg 1000 m km parenrightbigg 2 parenleftbigg hr 3600 s parenrightbigg 2 = 0 . 540461 θ = arctan(0 . 540461) = 28 . 3895 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 = 17 km / hr? Correct answer: 3486 . 65 N. Explanation: The speed of the car is greater, so its cen- tripetal acceleration is greater. Thus, the net force parallel to the incline is F bardbl = m g sin θ + F fr where F fr is the friction force. On the other hand, the component of the acceleration par- allel to the incline is still a bardbl = ( v + Δ v ) 2 r cos θ Then, m g sin θ + F fr = m a bardbl = m ( v + Δ v ) 2 r cos θ or F fr = m bracketleftbigg ( v + Δ v ) 2 r cos θ - g sin θ bracketrightbigg = (1200 kg) bracketleftbigg (62 km / hr + 17 km / hr) 2 56 m × cos 28 . 3895 - (9 . 8 m / s 2 ) sin 28 . 3895 bracketrightbigg = 3486 . 65 N .
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