Fowles10 - CHAPTER 10 LAGRANGIAN MECHANICS 10.1 Solution x...

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CHAPTER 10 LAGRANGIAN MECHANICS 10.1 Solution … ( ) ( ) ( ) 0, x t x t t αη = + ( ) ( ) ( ) 0, x t x t t αη = + ± ± ± where ( ) 0, sin x t t ω = and ( ) 0, cos x t t ω ω = ± 2 1 2 T m = ± x 2 2 1 1 2 2 V kx m ω = = 2 x so: ( ) ( ) 2 1 2 2 2 2 t t m J x α ω = ± x dt ( ) ( 2 1 2 2 2 cos sin 2 t t m t t ω ω αη ω ω αη = + + ± ) dt ( ) ( ) ( ) ( ) 2 2 1 1 2 2 2 2 2 2 2 cos sin cos sin 2 2 t t t t t m m J t t dt m t t dt dt α α ω ω ω αω η ω ωη ω η ω η = + + ± ± 2 1 t Examine the term linear in α : ( ) ( ) 2 2 2 2 1 1 1 1 cos sin cos sin sin 0 t t t t t t t t dt t t tdt tdt η ω ωη ω η ω ωη ω ωη ω = + ± t t (1 st term vanishes at both endpoints: ( ) ( ) 2 1 0 t t η η = = ) so ( ) ( ) 2 2 1 1 2 2 2 1 1 cos2 2 2 t t t t J m tdt m α ω ω α η ω η = + ± 2 2 dt [ ] ( ) 2 1 2 2 2 2 2 1 1 1 sin 2 sin 2 4 2 t t m t t m ω ω ω α η ω η = + ± dt which is a minimum at 0 α = 10.2 V mgz = ( ) 2 2 2 1 2 T m x y z = + + ± ± ± ( ) 2 2 2 1 2 L T V m x y z mgz = = + + ± ± ± L mx x = ± ± , L my y = ± ± , L mz z = ± ± d L mx dt x = ±± ± , d L my dt y = ±± ± , d L mz dt z = ±± ± 1
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0 L L x y = = , L mg z = − From equations 10.4.5 … 0 i i L d L q dt q = ± , 0 mx = ±± mx const = ± , 0 my = ±± my const = ± mz mg = − ±± 10.3 Choosing generalized coordinate x as linear displacement down the inclined plane (See Figure 8.6.1), for rolling without slipping … x a ω = ± 2 2 2 2 2 1 1 1 1 2 7 2 2 2 2 5 10 x T mx I mx ma mx a ω  = + = + =   ± ± ± 2 ± For V at the initial position of the sphere, 0 = V m sin gx θ = − 2 7 sin 10 L T V mx mgx θ = = + ± 7 5 L mx x = ± ± , 7 5 d L mx dt x = ±± ± sin L mg x θ = 7 sin 5 mx mg θ = ±± 5 sin 7 x g θ = ±± From equations 8.6.11 - 8.6.13 … 5 sin 7 cm x g θ = ±± 10.4 (a) For x the distance of the hanging block below the edge of the table: 2 2 1 1 2 2 x mx m = + = ± ± 2 x ± x T m and V m g = − x L T 2 V mx mgx = = + ± 2 L mx x = ± ± , 2 d L mx dt x = ±± ± L mg x = 2 mx mg = ±± 2 g x = ±± 2
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(b) 2 2 1 2 2 m x m x m x 2 T m = + = + ± ± ± and 2 2 2 x x m gx m g mgx x l l = − = − g V m 2 2 2 2 m m L T V m x mgx x l = = + + + ± g ( ) 2 L m m x x = + ± ± , ( ) 2 d L m m x dt x = + ±± ± L m mg x g x l = + ( ) 2 m g m m x mg x l + = + ±± 2 g ml m x x l m m + = + ±± 10.5 The four masses have positions: m 1 : 1 2 x x + m 2 : 1 1 l x x 2 + m 3 : 2 2 l x x 3 + m 4 : 2 2 3 l x l x 3 + ( ) ( ) 2 1 1 2 V g m l x x = − + 1 1 2 m x x + + ( ) ( ) 3 2 2 3 4 2 2 3 3 m l x x m l x l x + + + + ( ) ( 2 2 1 1 2 2 1 2 1 2 T m x x m x x = + + + ± ± ± ± ) ) ( ) ( 2 2 3 2 3 4 2 3 m x x m x x + + + ± ± ± ± ( ) ( ) ( ) 2 2 1 1 2 2 1 2 3 2 3 1 1 1 2 2 2 L T V m x x m x
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