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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 (7.13). 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 yoyo falls straight down because there are no horizontal forces. As noted in the problem, T = 1 2 mv 2 + 1 2 Iω 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 yoyo 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 Iω 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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 Fall '07
 EdgarKnobloch
 Energy, Kinetic Energy, Sin, Lagrangian mechanics, dy, Euler–Lagrange equation

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