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Unformatted text preview: madrid (tmm2353) – HW 5-3 – Antoniewicz – (56445) 1 This print-out should have 13 questions. Multiple-choice 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 stiffness k s and relaxed length L . The other end of the spring is fastened to a fixed point on a low-friction 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 = radicalBigg 4 π 2 mR 2 k s ( R − L ) 2. T = radicalBigg 2 π mR k s ( R − L ) 3. T = radicalBigg 4 π 2 mR k s ( R − L ) correct 4. T = radicalBigg 4 π mR k s ( R − L ) 5. T = radicalBigg 3 π 2 mR 2 k s ( R − L ) Explanation: The net force on the ball is equal to the force by the spring on the ball, which is vextendsingle vextendsingle vextendsingle vector F spring vextendsingle vextendsingle vextendsingle = 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, vextendsingle vextendsingle vextendsingle vector F net vextendsingle vextendsingle vextendsingle = m | vectorv | 2 R . So we have vextendsingle vextendsingle vextendsingle vector F spring vextendsingle vextendsingle vextendsingle = k s s and vextendsingle vextendsingle vextendsingle vector F net vextendsingle vextendsingle vextendsingle = | F spring | ⇒ m | vectorv | 2 R = k s ( R − L ) ⇒ | vectorv | = radicalbigg k s R ( R − L ) m . For circular motion, | vectorv | = 2 π R/T . We substitute: 2 π R T = radicalbigg k s R ( R − L ) m ⇒ T = radicalBigg 4 π 2 mR k s ( R − L ) . 002 20.0 points You put a 32 kg object on a bathroom scale at the North Pole, and the scale reads exactly 32 kg. At the North Pole, you are 6357 m from the center of the Earth. At the equator, the scale reads a different value due to two effects: 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 effects, 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....
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This note was uploaded on 11/15/2011 for the course PHY 303K taught by Professor Turner during the Fall '08 term at University of Texas.
- Fall '08