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 mxm x ω ′′ = ±± () tt x tA eB e =+ x e =− ± Boundary Conditions: 2 0 2 l x AB ==+ () ( ) 00 x == ± 4 l A = 4 l B = (a) cosh 2 l x = sinh 2 l x = ± (b) cosh 222 lll x TT =+= when the bead reaches the end of the rod cosh 2 T = or 1 11 cosh 2 T . 3 1 7 (c) sinh 2 l x = ± 1 sinh cosh 2 1.732 0.866 22 ll l ωω  =  o r 1 2 2 cosh 1 3 0.866 Tl −= = 5.11 mph ˆ 400 vj = G ˆ 586.67 j = ft s 1 G 5 ˆ ˆ 7.27 10 cos41 sin 41 j k + DD s 1 GG ( ) 5 ˆ 7.27 10 586.67 sin 41 vi ×= × D ft s 2 2 cor grav mv F Fm g −× = ( ) 5 2 7.27 10 586.67 sin 41 32 × = D 0.0017 cor grav F F = The Coriolis force is in the
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This note was uploaded on 01/06/2012 for the course PHYSICS 4360C taught by Professor Johnanderson during the Fall '11 term at UNF.

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