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oldmidterm2 #1

# oldmidterm2 #1 - oldmidterm2 01 Turner(58185 This print-out...

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This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that any person inside is held up against the wall when the floor drops away (see figure). The coefficient of static friction between the person and the wall is μ and the radius of the cylinder is R . ω R What is the minimum tangential velocity needed to keep the person from slipping down- ward? 1. v = 1 μ radicalbig g R 2. v = μ radicalbig 2 g R 3. v = 2 radicalbig g R 4. v = radicalbig 2 μ g R 5. v = 1 2 radicalbig g R 6. v = μ radicalbig g R 7. v = radicalBigg g R μ correct 8. v = 2 μ radicalbig g R 9. v = μ radicalbig 2 π g R 10. v = radicalbig g R Explanation: Basic Concepts: Centripetal force: F = m v 2 r Frictional force: f s μ N = f max s Solution: The maximum frictional force due to friction is f max = μ N , where N is the inward directed normal force of the wall of the cylinder on the person. To support the person vertically, this maximal friction force f max s must be larger than the force of gravity m g so that the actual force, which is less than μ N , can take on the value m g in the positive vertical direction. Now, the normal force supplies the centripetal acceleration v 2 R on the person, so from Newton’s second law, N = m v 2 R . Since f max s = μ N = μ m v 2 R m g , the minimum speed required to keep the per- son supported is at the limit of this inequality, which is μ m v 2 min R = m g, or v min = parenleftbigg g R μ parenrightbigg 1 2 . 002 (part 1 of 4) 10.0 points The following figure shows a Ferris wheel that rotates 2 times each minute and has a diame- ter of 19 m. The acceleration of gravity is 9 . 8 m / s 2 . R ω oldmidterm2 01 – Turner – (58185) 1

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hinojosa (jlh3938) – oldmidterm2 01 – Turner – (58185) 2 What is the centripetal acceleration of a rider? Correct answer: 0 . 416717 m / s 2 . Explanation: Let : r = D 2 = 9 . 5 m . The period of the Ferris wheel is T = one minute n = 60 s 2 = 30 s and the speed of the wheel is v = 2 π r T = 2 π (9 . 5 m) 30 s = 1 . 98968 m / s , so the centripetal acceleration is a = v 2 r = (1 . 98968 m / s) 2 9 . 5 m = 0 . 416717 m / s 2 . 003 (part 2 of 4) 10.0 points What force does the seat exert on a 33 kg rider at the lowest point of the ride? Correct answer: 337 . 152 N. Explanation: The force exerted by the seat balances the gravity and provides the centripetal force, so F l = m ( g + a ) = (33 kg) (9 . 8 m / s 2 + 0 . 416717 m / s 2 ) = 337 . 152 N . 004 (part 3 of 4) 10.0 points What force does the seat exert on a 33 kg rider at the highest point of the ride? Correct answer: 309 . 648 N. Explanation: The gravity is partly balanced by the force exerted by the seat and this resultant provides the centripetal force, so F l = m ( g a ) = (33 kg) [(9 . 8 m / s 2 ) (0 . 416717 m / s 2 )] = 309 . 648 N . 005 (part 4 of 4) 10.0 points What force (magnitude) does the seat exert on a rider when the rider is halfway between top and bottom?
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