Resource Quiz 6 - schroeder(cs39963 Resource 6 Circular...

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schroeder (cs39963) – Resource 6: Circular Motion and Gravitation – balasubramanya – (1401310) 1 This print-out should have 45 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points An ant of mass m clings to the rim of a flywheel of radius r , as shown. The flywheel rotates clockwise on a horizontal shaft S with constant angular velocity ω . As the wheel rotates, the ant revolves past the stationary points I , II , III , and IV . The ant can adhere to the wheel with a force much greater than its own weight. r S I II III IV Ant ω It will be most difficult for the ant to adhere to the wheel as it revolves past which of the four points? 1. II 2. It will be equally difficult for the ant to adhere to the wheel at all points. 3. I 4. III correct 5. IV Explanation: The sum of the ant’s gravity and its ad- hesion force is the centripetal force vector F c with magnitude m ω 2 r . Thus vector F c = mvectorg + vector F ad vector F ad = vector F c mvectorg . The maximum vector F ad is when vector F c and mvectorg are in opposite directions; i.e. , at III . 002 (part 2 of 2) 10.0 points What is the magnitude of the minimum adhe- sion force necessary for the ant to stay on the flywheel at point III ? 1. F = m ω 2 r + m g correct 2. F = m ω 2 r m g 3. F = m ω 2 r 2 4. F = m ω 2 r 2 + m g 5. F = m g Explanation: According to the explanation in the previ- ous part, the minimum adhesion force is bardbl vector G bardbl + bardbl vector F c bardbl = m ω 2 r + m g . 003 (part 1 of 2) 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 g R
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schroeder (cs39963) – Resource 6: Circular Motion and Gravitation – balasubramanya – (1401310) 2 3. v = radicalbig g R 4. v = μ radicalbig 2 π g R 5. v = radicalBigg g R μ correct 6. v = 2 radicalbig g R 7. v = 2 μ radicalbig g R 8. v = μ radicalbig 2 g R 9. v = radicalbig 2 μ g R 10. v = 1 2 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 .
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