105Sol8 - Physics 105 Solutions to Problem Set 8 By Stephen...

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Unformatted text preview: Physics 105 Solutions to Problem Set 8 By Stephen Glassman, Fall 2007 7.8 (a) The Lagrangian is defined in equations (7.1-3). L = T- U = 1 2 m x 2 1 + 1 2 m x 2 2- 1 2 k ( x 1- x 2- l ) 2 where T is the usual kinetic energy and U is the spring potential as specified in the problem. (b) Our new variables, as noted in the problem statement, are: X = 1 2 ( x 1 + x 2 ) x = x 1- x 2- l Solving for x 1 and x 2 , we find: x 1 = X + 1 2 x + 1 2 l x 2 = X- 1 2 x- 1 2 l Substituting these into our Lagrangian, we find: L = 1 2 m X + 1 2 x 2 + 1 2 m X + 1 2 x 2- 1 2 kx 2 = m X 2 + 1 4 m x 2- 1 2 kx 2 Our Lagrangian equations of motion for these coordinates are given by equation (7.52): d dt L X = L X = d dt 2 m X = 0 = X = 0 1 d dt L x = L x = d dt 1 2 m x =- kx = x =- 2 k m x (c) The solutions for X are just the solutions to a free particle in one dimension, while x satisfies the equations of motion for an harmonic oscillator. X ( t ) = X + V t x ( t ) = A sin r 2 k m t + B cos r 2 k m t Basically, the center of mass just travels along at constant velocity while the two masses oscillate around the center of mass. 7.14 The yo-yo falls straight down because there are no horizontal forces. As noted in the problem, T = 1 2 mv 2 + 1 2 I 2 with I = 1 2 mR 2 . Of course, U = mgh where h =- x . We now just need to express in terms of x . In the lab frame, the CM has speed x while the point of contact between the string and the yo-yo is motionless. In the CM frame, then, that point of contact has speed x . From our study of circular motion, we conclude that = R- 1 x . L = T- U = 1 2 mv 2 + 1 2 I 2 + mgx = 1 2 m x 2 + 1 4 m x 2 + mgx = 3 4 m x 2 + mgx Our Lagrangian equations of motions are: d dt L x = L x = d dt 3 2 m x = mg 2 = 3 2 m x = mg = x...
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105Sol8 - Physics 105 Solutions to Problem Set 8 By Stephen...

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