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Analytical Mech Homework Solutions 139

# Analytical Mech Homework Solutions 139 - CHAPTER 10...

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CHAPTER 10 LAGRANGIAN MECHANICS 10.1 Solution … () ( ) ( ) 0, x txt t αη =+ 0, x t ±± ± where () 0, sin x tt ω = and 0, cos x = ± 2 1 2 Tm = ± x 22 11 Vk x m == 2 x so: 2 1 2 2 t t m Jx αω =− ± x d t ( 2 1 2 cos sin 2 t t m  +  ± ) d t 2 2 2 cos sin cos sin ttt mm J t t dt m t t dt dt α η +− + ∫∫∫ 2 1 t Examine the term linear in : ( ) 2 2 1 1 cos sin cos sin sin 0 t t t t dt t t tdt tdt ηω −=+− ∫∫ ± t t (1 st term vanishes at both endpoints: ( ) ( ) 21 0 ηη = = ) so 2 cos 2 Jm t d t m ± 2 2 d t [] 2 1 222 2 sin 2 sin 2 42 t t mt t m ωω + ± d t which is a minimum at 0 = 10.2 Vm g z = 1 2 x y z + ± 1 2 LTV mx y z m g z =−= + + − ± L mx x = ± ± , L my y = ± ± , L mz z = ± ± dL mx dt x  =   ± , my dt y = ± , mz dt z
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