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Unformatted text preview: agnew (ja26385) – oldmidterm 02 – Turner – (58220) 1 This print-out should have 25 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 g R 3. v = radicalBigg g R μ correct 4. v = 2 radicalbig g R 5. v = μ radicalbig 2 g R 6. v = 1 2 radicalbig g R 7. v = μ radicalbig 2 π g R 8. v = 2 μ radicalbig g R 9. v = radicalbig g R 10. v = radicalbig 2 μg R Explanation: Basic Concepts: Centripetal force: F = mv 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 mg so that the actual force, which is less than μ 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 Newton’s second law, N = mv 2 R . Since f max s = μ N = μmv 2 R ≥ mg , the minimum speed required to keep the per- son supported is at the limit of this inequality, which is μmv 2 min R = mg, or v min = parenleftbigg g R μ parenrightbigg 1 2 . 002 10.0 points A(n) 24 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 7 . 9 m / s goes over a crest whose radius of curvature is 13 m? Correct answer: 119 . 982 N. Explanation: The boy is executing circular motion at the top of the crest, so the net force is centripetal. Two forces act on him: (1) his weight acting down and (2) the supporting normal force from the seat which acts up vector F net = M vectora agnew (ja26385) – oldmidterm 02 – Turner – (58220) 2 vector F g − N = M v 2 r ˆ k , so N = vector F g − M v 2 r ˆ k = M parenleftbigg g − v 2 r parenrightbigg ˆ k = (24 kg) × bracketleftbigg (9 . 8 m / s 2 ) − (7 . 9 m / s) 2 (13 m) bracketrightbigg ˆ k = 119 . 982 N . The force with which he presses against the seat is equal but opposite to the normal force. 003 10.0 points An object attached to the end of a string swings in a vertical circle of radius 1 . 7 m as shown. At an instant when θ = 34 ◦ , the speed of the object is 9 m / s and the tension in the string is 34 N....
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This note was uploaded on 05/11/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
- Spring '08