# exam01_sol_f06 - 1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

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Unformatted text preview: 1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Physics 8.01 Fall Term 2006 Exam 1 Solutions Part I Concept Questions: (20 points). Circle your answer. There are five questions and each question is worth 4 points. 1) (4 points) A pendulum bob swings down and is moving fast at the lowest point in its swing. T is the tension in the string, T ! W = mr " 2 W is the gravitational force exerted on the pendulum bob. Which free-body diagram below best represents the forces exerted on the pendulum bob at the lowest point? The lengths of the arrows represent the relative magnitudes of the forces. Solution: d). The bob is undergoing circular motion. It is accelerating towards the center. Newton’s Second Law gives T ! W = mr " 2 hence T = W + mr ! 2 so T > W . 2 2) (4 points) Which of the following expressions gives a reasonable formula for the period of the moon orbiting around the earth? Let G be the universal gravitational constant, r e , m be the average distance between the earth and the moon, m m be the mass of the moon, and m e be the mass of the earth. a) T ! G m e m m / r e , m 2 . b) T ! G m e m m / r e , m . c) T ! r e , m 3 / Gm m . d) T ! r e , m 3 / Gm e . e) T ! r e , m 2 / Gm e . f) T ! r e , m 2 / Gm m . Solution: (d). The answer cannot depend on mass of the moon by the Equivalence Principle (the moon’s mass cancels from each side of Newton’s second Law). The period must be some product of the given T ! r e , m a G b m e c dim[ G ] = dim[force ! m 2 ! kg-2 ] = dim[kg ! m ! s-2 ! m 2 ! kg-2 ] = dim[s-2 ! m 3 ! kg-1 ] dim[ m e ] = kg dim[( r e , m 3 / Gm e...
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## This note was uploaded on 10/15/2011 for the course PHYSICS 0207239 taught by Professor Mamar during the Spring '09 term at Jordan University of Science & Tech.

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exam01_sol_f06 - 1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

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