Analytical Mech Homework Solutions 162

Analytical Mech Homework Solutions 162 - 10.30 L = 1 212 mx...

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10.30 22 11 Lm x k =− ± x 2 1 0 tt t t Ldt Ldt mx kx dt δδ δ  ===   ∫∫ ∫ ± () 2 1 0 t t mx x kx x dt ±± d x x dt = ± () () 2 1 t t d mx xdt mx x dt mxd x dt == ∫∫ ± ± Integrating by parts: 2 1 t t mx xdt mx x x d mx ± ± 0 x = at and 1 t 2 t m x t m xd t d dt dm x d ± ± mx x dx x mx dt ± ± 2 1 0 t t mx x kx x dt 0 mx kx += 10.31 (a) 2 2 0 2 1 v c c V Let 2 2 1 1 v c γ = 0 x L mx p x =≡ ± ± This is the generalized momentum for part (b). Thus, Lagrange’s equations for the x-component are … 0 x dL L d V p dt x x dt x ∂∂ −= − = ± … and so on for the y and z components.
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