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Unformatted text preview: Morby, Grant Homework 12 Due: Feb 20 2006, noon Inst: Drummond 1 This printout should have 11 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 2) 10 points The coefficient of static friction between the person and the wall is 0 . 75 . The radius of the cylinder is 5 . 36 m . The acceleration of gravity is 9 . 8 m / s 2 . 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. 5 . 36 m What is the minimum angular velocity min needed to keep the person from slip ping downward? Correct answer: 1 . 56135 rad / s. Explanation: Let : R = 5 . 36 m and = 0 . 75 . Basic Concepts: Centripetal force F = mv 2 r . Frictional force f N = f max . Solution: The maximum force due to static friction is f max = N , where N is the inward directed normal force exerted by the wall of the cylinder on the person. To support the person vertically, the maximal friction force must be larger than the force of gravity mg , so that the actual force, which is equal to or less than the maximum N , is allowed to take on the value mg in the pos itive vertical direction. In other words, the ceiling N on the frictional force has to be raised high enough to allow for the value mg . The normal force supplies the centripetal ac celeration v 2 R on the person, so from Newtons second law, N = mv 2 R . Since f max = 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 = s g R . From this we immediately find the angular speed min v min R = r g R = s 9 . 8 m / s 2 (0 . 75)(5 . 36 m) = 1 . 56135 rad / s . 002 (part 2 of 2) 10 points Suppose the person, whose mass is m , is be ing held up against the wall with an angular velocity of = 2 min . The magnitude of the frictional force be tween the person and the wall is 1. F = 2 mg . 2. F = 1 5 mg . Morby, Grant Homework 12 Due: Feb 20 2006, noon Inst: Drummond 2 3. F = mg . correct 4. F = 1 3 mg . 5. F = 4 mg . 6. F = 1 2 mg . 7. F = 5 mg . 8. F = 3 mg . 9. F = 1 4 mg . Explanation: As discussed above, f f max = N ....
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This note was uploaded on 03/02/2009 for the course PHY 58235 taught by Professor Kleinman during the Spring '09 term at University of Texas at Austin.
 Spring '09
 KLEINMAN
 Physics, Work

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