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Unformatted text preview: pasha (sep635) – HW 53 – Antoniewicz – (56445) 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 20.0 points A small block of mass m is attached to a spring with sti ff ness k s and relaxed length L . The other end of the spring is fastened to a fixed point on a lowfriction table. The block slides on the table in a circular path of radius R > L . How long does it take for the block to go around once? 1. T = s 4 π 2 mR 2 k s ( R L ) 2. T = s 3 π 2 mR 2 k s ( R L ) 3. T = s 4 π mR k s ( R L ) 4. T = s 2 π mR k s ( R L ) 5. T = s 4 π 2 mR k s ( R L ) correct Explanation: The net force on the ball is equal to the force by the spring on the ball, which is ~ F spring = k s s. The distance the spring stretches is s = R L . Because the speed of the ball is constant and is much less than the speed of light, ~ F net = m  ~ v  2 R . So we have ~ F spring = k s s and ~ F net =  F spring  ⇒ m  ~ v  2 R = k s ( R L ) ⇒  ~ v  = r k s R ( R L ) m . For circular motion,  ~ v  = 2 π R/T . We substitute: 2 π R T = r k s R ( R L ) m ⇒ T = s 4 π 2 mR k s ( R L ) . 002 20.0 points You put a 20 kg object on a bathroom scale at the North Pole, and the scale reads exactly 20 kg. At the North Pole, you are 6357 m from the center of the Earth. At the equator, the scale reads a di ff erent value due to two e ff ects: 1) the Earth bulges at the equator due to its rotation, and you are 6378 m from the center of the Earth, and 2) you are mov ing in a circular path due to the rotation of the Earth. Taking into account both of these e ff ects, what does the scale read at the equa tor? Use G = 6 . 67 × 10 11 N · m 2 / kg 2 and M e = 6 × 10 24 kg. Correct answer: 19 . 8685 kg. Explanation: Let R 1 be the Earth’s radius at the North Pole and let R 2 be the radius at the equator. Then we can write that the force at the North Pole is F 1 = G M m R 2 1 = mg 1 , where g 1 = GM/R 2 1 . At the Equator the force on the object is F 2 = G M m R 2 2 m v 2 R 2 = mg 1 R 2 1 R 2 2 m ω 2 R 2 Then we have, for the change in apparent mass, F 2 F 1 g 1 = m " R 2 1 R 2 2 1 2 π T 2 R 2 g 1 # = . 131486 kg...
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This note was uploaded on 11/17/2011 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
 Fall '08
 Turner
 Mass

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