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Ch5-HW2_corrected-solutions

Ch5-HW2_corrected-solutions - louey(cal2859 Ch5-HW2...

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louey (cal2859) – Ch5-HW2 corrected – florin – (56930) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of2)10.0points By “weight” we usually mean the gravita- tional force exerted on an object by the Earth. However, when you sit in a chair your own perception of you own “weight” is based on the contact force the chair exerts upward on your rear end rather than on the gravitational force. The smaller this contact force is, the less “weight” you perceive, and if the contact force is zero, you feel peculiar and “weight- less” (an odd word to describe a situation when the only force acting on you is the gravi- tational force exerted by the Earth!). Also, in this condition you internal organs no longer press on each other, which presumably con- tributes to the odd sensation in your stomach. How fast must a roller coaster car go over the top of a circular arc for you to feel “weight- less”? The center of the car moves along a circular arc of radius R as in the following figure). vectorv R 1. | vectorv | = radicalbig 2 g R 2. | vectorv | = 2 g R 3. | vectorv | = radicalbig g R correct 4. | vectorv | = radicalbigg g R 2 5. | vectorv | = g R Explanation: The force by the seat on the rider can be up- ward (+ y direction) or downward ( y direc- tion) depending on how fast the roller coaster goes over the hill. The gravitational force by the Earth on the rider is downward ( y direc- tion). At the top of the hill, the net force on the rider is in the downward direction, toward the center of the circle. To feel weightless, the force by the seat on the rider must be zero at the top of the hill: vector F net = vector F seat + vector F grav ( 0 , m | vectorv | 2 R , 0 ) = ( 0 , m g, 0 ) m | vectorv | 2 R = m g ⇒ | vectorv | = radicalbig g R . 002(part2of2)10.0points How fast must a roller coaster car go through a circular dip for you to feel three times as “heavy” as usual, due to the upward force of the seat on your bottom being three times as large as usual? The center of the car moves along a circular arc of radius R as in the figure below. vectorv R 1. | vectorv | = radicalbig g R 2. | vectorv | = g R 3. | vectorv | = 2 g R 4. | vectorv | = radicalbig 2 g R correct 5. | vectorv | = radicalbigg g R 2 Explanation: At the bottom of a hill (or a “dip”), the net force on the rider is upward. The force by the seat on the rider is also upward, and in this case has a magnitude of 3 m g . The net force on the rider is
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