OldMidterm2 - oldmidterm 02 PAPAGEORGE MATT Due Mar 3 2008...

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oldmidterm 02 – PAPAGEORGE, MATT – Due: Mar 3 2008, 6:00 pm 1 Question 1, chap 6, sect 3. part 1 of 1 10 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 = radicalbig g R 2. v = μ radicalbig 2 π g R 3. v = 1 2 radicalbig g R 4. v = μ radicalbig 2 g R 5. v = 2 radicalbig g R 6. v = radicalBigg g R μ correct 7. v = 1 μ radicalbig g R 8. v = radicalbig 2 μ g R 9. v = μ radicalbig g R 10. v = 2 μ 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 . Question 2, chap 6, sect 99. part 1 of 1 10 points The following figure shows a Ferris wheel that rotates 3 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 37 kg rider at the lowest point of the ride? Correct answer: 395 . 466 N (tolerance ± 1 %). Explanation:
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oldmidterm 02 – PAPAGEORGE, MATT – Due: Mar 3 2008, 6:00 pm 2 The period of the Ferris wheel is T = 60 s / 3 = 20 s . The speed of the wheel is v = 2 π r T = 2 π (9 m) 20 s = 2 . 82743 m / s , so the centripetal acceleration is a = v 2 r = (2 . 82743 m / s) 2 9 m = 0 . 888264 m / s 2 . The force exerted by the seat balances the gravity and provides the centripetal force, so F l = m [ g + a ] = (37 kg) (9 . 8 m / s 2 + 0 . 888264 m / s 2 ) = 395 . 466 N . Question 3, chap 6, sect 99. part 1 of 1 10 points A(n) 21 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 8 m / s goes over a crest whose radius of curvature is 11 m? Correct answer: 83 . 6182 N (tolerance ± 1 %). 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 = Mvectora vector F g − N = Mv 2 r ˆ k ⇒ N = vector F g M v 2 r ˆ k = M parenleftbigg g v 2 r parenrightbigg ˆ k = (21 kg) × bracketleftbigg 9 . 8 m / s 2 (8 m / s) 2 11 m bracketrightbigg ˆ k = 83 . 6182 N The force with which he presses against the seat is equal but opposite to the normal force.
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