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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 x x ω = ±± t t x Ae Be ω ω = + Assuming a particular solution to have the form sin p x C t ω = , 2 2 sin sin sin C t C t g t ω ω ω ω ω = − 2 2 g C ω = 2 sin 2 t t g x Ae Be t ω ω ω ω = + + At time t , 0 = x x = D and 0 x = ± x A B = + D 0 2 g A B ω ω ω = + 2 1 2 2 g A x ω = D 2 1 2 2 g B x ω = + D ( ) ( ) 2 2 1 1 2 2 2 2 t t t t g g x x e e e e t ω ω ω ω sin ω ω ω = + + D From Appendix B, we use the identities for hyperbolic sine and cosine to obtain 2 2 cosh sinh sin 2 2 g g x x t t t ω ω ω ω ω = + 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 ω ω = G . ( ) ( ) ˆ ˆ ˆ ˆ ˆ v v r ix jy k ix jy ω ω = + × = + + × + G G G G ± ± G
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