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

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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 = ± 1

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0 LL xy ∂∂ == , L mg z =− From equations 10.4.5 … 0 ii Ld L qd tq  −=   ± , 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 22 2 2 1111 2 7 2222 5 1 0 x T mx I mx ma mx a =+=+ = ± 2 ± For V at the initial position of the sphere, 0 = Vm sin g x θ 2 7 sin 10 LTV m x m g x =−= + ± 7 5 L mx x = ± ± , 7 5 dL mx dt x = ± sin L mg x = 7 sin 5 mx mg = 5 sin 7 xg = From equations 8.6.11 - 8.6.13 … 5 sin 7 cm = 10.4 (a) For x the distance of the hanging block below the edge of the table: 22 11 xm =+= 2 x ± x Tm and g x LT 2 V m g x + ± 2 L mx x = ± ± , 2 mx dt x = ± L mg x = 2 mx mg = 2 g x = 2
(b) 22 1 m x m x m x 2 Tm  =+   ±± ± and 2 x xm g x m g m g x x ll =− g Vm mm LTV m g x x l ′′ =−= + + + ± g () 2 L mmx x ± ± , 2 dL dt x ± Lm mg x g x l 2 mg mmxm g x l += + 2 gm lm x x m + = + 10.5 The four masses have positions: m 1 : 12 x x + m 2 : 11 lxx 2 + m 3 : 3 + m 4 : 223 lxlx 3 +− ( ) ( ) 21 1 2 Vg m l x x + 2 m x x + + ( ) ( ) 32 2 3 42 2 3 3 ml x x ml x l x + + − + ()( 2 2 1 2 1 2 x x x + + ) ) ( 323 423 mxx mxx +−+ + −− 2 2 1 2 3 2 3 1 2 mx x m x x =−= + + −+ + 2 ( ) 2 1 2 1234 3 34 1 . 2 mxx g x mmg x mmmmg x mmc o n s + − + +−−+ t ( 2 2 1 2 1 L m x x x + ± ) ) 2 ( 1 12 mmx mmx ( 1 dt x ± ) 2 1 L gm m x ( ) 2 mmx mmx gmm ++ − =− 3

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() ( ) ( 11 2 2 1 2 3 2 3 4 2 3 2 L mx x m x x m m x =+ + + + ±± ± ) 1234 2 L gm m m m x ( ) ( ) 12 1 1234 2 43 3 mmx mmmmx mmx gmmmm + +++ + − = +−− ( 323 423 3 L mxxmxx x =− ± ) 34 3 L x ( ) 2 3 mmx mmx gmm −+ +=− For , mm , , and 1 = 2 4 = 3 2 m = m 4 = : , 53 3 mx mx mg −= 3 5 x xg = 3 38 2 mx mx mx mg −+ −= , 23 3 mx mx mg −+ = 32 1 3 x = + Substituting into the second equation: 22 2 99 82 55 33 x g xxg −++−−= g 2 88 8 15 15 x g = , 2 11 g x = 1 31 0 6 51 1 1 1 x gg  =   3 2 4 1 1 1 x == Accelerations: : 1 m 5 11 x += : 2 m 7 11 x : 3 m 3 11 x : 4 m 5 11 x −− = 10.6 See figure 10.5.3, replacing the block with a ball. The square of the speed of the ball is calculated in the same way as for the block in Example 10.5.6.
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## This note was uploaded on 03/09/2008 for the course PHYS 301 taught by Professor Mokhtari during the Fall '04 term at UCLA.

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

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