midterm 02_plus_solutions

midterm 02_plus_solutions - List midterm02 - Mar 06 2006 in...

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Unformatted text preview: List midterm02 - Mar 06 2006 in class PHY 303K-Florin-Spring 2008. 1 Question 1, chap 6, sect 3. part 1 of 1 0 points Barrel of Fun 05 (599) 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 and the radius of the cylinder is R . R What is the minimum tangential velocity needed to keep the person from slipping down- ward? 1. v = radicalBigg g R correct 2. v = radicalbig g R 3. v = 2 radicalbig g R 4. v = 1 2 radicalbig g R 5. v = radicalbig g R 6. v = 2 radicalbig g R 7. v = radicalbig 2 g R 8. v = radicalbig 2 g R 9. v = radicalbig 2 g R 10. v = 1 radicalbig g R Explanation: Basic Concepts: Centripetal force: F = mv 2 r Frictional force: f s N = f max s Solution: The maximum frictional force due to friction is f max = 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 mg so that the actual force, which is less than N , can take on the value mg in the positive vertical direction. Now, the normal force supplies the centripetal acceleration v 2 R on the person, so from Newtons second law, N = mv 2 R . Since f max s = N = mv 2 R mg , the minimum speed required to keep the per- son supported is at the limit of this inequality, which is mv 2 min R = mg, or v min = parenleftbigg g R parenrightbigg 1 2 . Question 2, chap 6, sect 99. part 1 of 1 0 points Ferris Wheel 01 (2443) The following figure shows a Ferris wheel that rotates 4 times each minute and has a diameter of 18 m. The acceleration of gravity is 9 . 8 m / s 2 . What force does the seat exert on a 40 kg rider at the lowest point of the ride? List midterm02 - Mar 06 2006 in class PHY 303K-Florin-Spring 2008. 2 Correct answer: 455 . 165 N (tolerance 1 %). Explanation: The period of the Ferris wheel is T = 60 s / 4 = 15 s . The speed of the wheel is v = 2 r T = 2 (9 m) 15 s = 3 . 76991 m / s , so the centripetal acceleration is a = v 2 r = (3 . 76991 m / s) 2 9 m = 1 . 57914 m / s 2 . The force exerted by the seat balances the gravity and provides the centripetal force, so F l = m [ g + a ] = (40 kg) (9 . 8 m / s 2 + 1 . 57914 m / s 2 ) = 455 . 165 N . Question 3, chap 6, sect 99. part 1 of 1 0 points Over the Crest (5229) A(n) 20 kg boy rides a roller coaster. The acceleration of gravity is 9 . 8 m / s 2 . With what force does he press against the seat when the car moving at 6 m / s goes over a crest whose radius of curvature is 10 m?...
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This note was uploaded on 02/03/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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midterm 02_plus_solutions - List midterm02 - Mar 06 2006 in...

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