Phys 325 Spring 2011 Lecture 19 - Physics 325 Lecture 19...

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Physics 325 Lecture 19 More Examples using Lagrangians Example : Moveable plane pendulum m 1 moves along x without friction. 2 m swings from a massless rod of length that is suspended from 1 m and restricted to move in the x - y plane. For generalized coordinates we choose 1 x and . With ( 0) 0 Uy  , we have 22 2 cos U m gy mg  where cos and sin yx  The kinetic energy of m 1 is 2 1 1 1 1 2 T m x For m 2 we have a contribution from , and a contribution from the motion of m 1 ( 1 x ). This gives   21 1 ˆ ˆ ˆˆ cos sin v x x x x y  so that   2 2 2 2 2 2 2 2 1 1 1 2 1 2 cos 2 T m v m x x x y m 1 m 2
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The total kinetic energy is T 1 +T 2 , and the Lagrangian is   2 2 2 1 2 1 2 2 1 2 11 cos cos 22 L T U m m x m m x m g  The partial derivatives are as follows       2 1 2 1 1 2 1 2 1 1 2 1 1 2 21 0 sin sin cos cos cos LL m x m g x L m m x m m x m x x L mx      The Lagrange equation for x 1 again gives conservation of linear momentum. We have       1 1 2 1 1 2 cos 0 xx dd m x m x p p dt dt We can work out the time derivative to give       2 1 2 1 2 2 cos sin 0 m m x m m   (19.1) The Lagrange equation for provides     2 2 1 1 2 1 cos sin sin 0 m x dx m g x or 1 cos sin 0 xg (19.2) Equations (19.1) and (19.2) are the differential equations of motion for the system. They are second order (as always – think F mx ), and they are non-linear.
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Phys 325 Spring 2011 Lecture 19 - Physics 325 Lecture 19...

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