Analytical Mech Homework Solutions 154

Analytical Mech Homework Solutions 154 - 2g d ( sin ) and...

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Hence: () ( 2 sin g x d a ) dy θ = and ( ) sin 0 0 2 sin yx yx g x d a θθ = = = ∫∫ D but at so 2 0 x == ± 0 = D 2 sin g yx a = 1 2 2 2s i n 1s i n m g x a f   ± Now we can solve for ±± and plug 2 ± , into (3) to obtain ( ) λθ 1 2 2 i 21 s m dg d da d f n i n     ± After some algebra – yields 1 1 2 2 2 i n cos i n m m f g a f + = The solution for is thus determined from equation (3) 1 22 1 sin cos cos sin m g f am a λ −+ + ± ± = collecting terms: 11 1 1 cos sin cos sin mm g ff a −− + = ± Plug , ± into the above --- plus --- a lot of algebra yields … 1 1 2 2 1 i n s i n c o s 1 s i n sin i n m m f mg f −= + where 2 2 12 m m f = + As a check, let so it is immoveable … then and 2 m →∞ 2 1 m f 1 0 m f and we have 1 3sin −→ … which checks 10.21 (a) 2 2 Tm v m RR z φ ++ ± ± ± 2 2 m LTV R R z V =−= + + −
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