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Analytical Mech Homework Solutions 56

Analytical Mech Homework Solutions 56 - 5.10(See Example...

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5.10 (See Example 5.3.3) 2 m x mx ω = ±± ( ) t t x t Ae Be ω ω = + ( ) t t x t Ae Be ω ω ω ω = ± Boundary Conditions: ( ) 2 0 2 l x A B = = + ( ) ( ) 0 0 x A B ω = = ± 4 l A = 4 l B = (a) ( ) cosh 2 l x t t ω = ( ) sinh 2 l x t t ω ω = ± (b) ( ) cosh 2 2 2 l l l x T T ω = + = when the bead reaches the end of the rod cosh 2 T ω = or 1 1 1 cosh 2 T .317 ω ω = = (c) ( ) sinh 2 l x T T ω ω = ± ( ) 1 sinh cosh 2 1.732 0.866 2 2 l l l ω ω ω = = = or 1 2 2 cosh 1 3 0.866 2 2 T l ω l l ω ω ω = = 5.11 mph ˆ 400 v j = G ˆ 586.67 j = ft s 1 G ( ) 5 ˆ ˆ 7.27 10 cos41 sin 41 j k ω = × + D D s 1 G G ( ) ( ) ( ) 5 ˆ 7.27 10 586.67 sin 41 v i ω × = − × D ft s 2
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