Analytical Mech Homework Solutions 9

Analytical Mech Homework Solutions 9 - 1.24 d dr d r ( v a...

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1.24 () dd r d rva va r va dt dt dt ⋅× = ⋅×+ ×   K KKK KK K KK dv da vva r a v dt dt  =⋅ × +⋅ × + ×   K K KKK K K K ( ) 00 rv a =+⋅ + × K ± () () d dt ⋅× = ⋅× ± 1.25 ˆ vv τ = K and ˆˆ n aa a n =+ K va v a ⋅= , so a v = 22 n aaa 2 , so 1 2 n =− 1.26 For 1.14, 23 2 1 2 2 2 2 2 cos sin sin cos 4 cos sin 4 bt t t a c t ωω ω −⋅ +⋅ + = ++ c t 2 1 2 2 2 4 4 ct a bc t = + 1 42 2 2 2 2 16 4 4 n ab c t + For 1.15, 222 2 2 2 2 1 2 1 2 2 kt kt kt kt bk k c e bcke be k c −+ == + k e k c + 1 1 2 2 2 2 2 22 2 2 2 2 2 kt kt kt n ekc b k ekc b c e kc − + = +  1.27 ˆ = K , 2 v n av ρ K ± n va va v ×= ⋅ =
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