Analytical Mech Homework Solutions 144

Analytical Mech Homework Solutions 144 - The solution to...

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The solution to the homogeneous equation , is 2 0 xx ω −= ±± tt x Ae Be =+ Assuming a particular solution to have the form sin p x Ct = , 22 sin sin sin CtCtgt ωω −− = 2 2 g C = 2 sin 2 g x Ae Be t =+ + At time t , 0 = x x = D and 0 x = ± x AB D 0 2 g =−+ 2 1 g Ax  =−   D 2 1 g Bx D () 11 gg e e t s i n −−  +   D From Appendix B, we use the identities for hyperbolic sine and cosine to obtain cosh sinh sin x xt t t + D 10.8 In order that a particle continues to move in a plane in a rotating coordinate system, it is necessary that the axis of rotation be perpendicular to the plane of motion. For motion in the xy plane, ˆ k = G . ( ) ˆ ˆˆ vv r i xj y k i y ′′ =+×= + + × + GG GG G ( vix y jy x ) =− + + 2 2 1 m Tm v v xx y y yy =⋅ = − ++ + + GG ± 2 2 LTV L mx y x ± ± ,
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This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.

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