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Unformatted text preview: s =0 = F Î¾ 2 Â· âˆ‚ Î¾ 1 f ( Î¾ 1 ( C )) Â· d ds Î¾ 1 ( C ( s )) Â¯ Â¯ Â¯ Â¯ s =0 and thus F Î¾ 1 = F Î¾ 2 Â· âˆ‚ Î¾ 1 f ( Î¾ 1 ( C )) . 6 Proposition 20 Consider a coordinate chart ( C , Î¾ ) , where Î¾ = ( Î¾ 1 , Î¾ 2 ) T and F Î¾ = ( F 1 , F 2 ) . Suppose that Î¾ 1 = g ( Î¾ 2 ) on a restricted collection Ëœ C âŠ‚ C of allowable con f gurations, such that Â³ Ëœ C , Î¾ 2 Â´ is a chart. Let Ëœ C ( s ) be a curve in Ëœ C that runs through C at s = 0 . It follows that F Î¾ Â· d ds Î¾ Â³ Ëœ C ( s ) Â´ Â¯ Â¯ Â¯ Â¯ s =0 = ( F 1 , F 2 ) Â· Î¼ âˆ‚ Î¾ 2 g ( Î¾ 2 ( C )) Â· d ds Î¾ 2 ( C ( s )) Â¯ Â¯ s =0 d ds Î¾ 2 ( C ( s )) Â¯ Â¯ s =0 Â¶ = Â¡ F 1 Â· âˆ‚ Î¾ 2 g ( Î¾ 2 ( C )) + F 2 Â¢ Â· d ds Î¾ 2 ( C ( s )) Â¯ Â¯ Â¯ Â¯ s =0 and thus F Î¾ 2 = F 1 Â· âˆ‚ Î¾ 2 g ( Î¾ 2 ( C )) + F 2 . 7...
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 Spring '08
 Weaver
 Force, Manifold, Lagrangian mechanics, Virtual displacement, Generalized coordinates, FÎ¾

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