Analytical Mech Homework Solutions 60

# Analytical Mech Homework Solutions 60 - ( j Hence: a = a 2...

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Hence: ( ) 2 ˆ ˆˆ 23 2 a x i k i x ωω ω =∆ − × =− × + ˆ j y av G GG G ±± G ai 2 ˆ ˆ 322 xj y x i y i x ˆ j =+ = + ± ± So xy 2 x −− = ± 0 20 yx += ± 5.19 () mr qE q v B =+ × Equation 5.2.14 ( ) 2 r v r rr ′′ =+×+ ×+× × G G GG GG G GG ± GG GG Equation 5.2.13 vv r =+× 2 q B m =− G G so 0 = G ± () ( ) 2 q mr q B v B r qE q v r B −× × =+ + × × G G G G G G G () () 2 q mr q v B r B qE q v B q r B + × × × G G G G G G G G 2 q mr qE r B =+ ×× G ( ) 2 sin 22 2 qq q B rB r BB m ωθ  ××=   G G G Neglecting terms in 2 B , mr qE ′ = 5.20 For cos sin x xt yt =+ t sin cos yx t y + cos sin sin cos x x t y t ++ ± t sin cos cos sin x t y t y + ± cos sin x y ′′′ =++ ± x sin cos + ± cos sin sin cos x x t y t y + ± ± ± x sin cos cos sin x t y t + ± ± ± 2 cos sin 2 x ytyx + ± 2 y sin cos 2 y t x + + ± Substituting into Eqns 5.6.3: 2 cos sin 2 x ty t y x + ± cos sin 2 gg x ty t ll y + ± 2 sin cos 2 x t x y +− + ± sin cos 2 x x ± 10
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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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