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

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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

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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 0 + V 0 t x ( t ) = A sin 2 k m t + B cos 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 2 with I = 1 2 mR 2 . Of course, U = mgh where h = - x . We now just need to express ω 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 ω = R - 1 ˙ x . L = T - U = 1 2 mv 2 + 1 2 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 = 2 3 g as required.

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