Analytical Mech Homework Solutions 148

# Analytical Mech Homework Solutions 148 - 2 1 2 2 ml ml 2...

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() () 2 22 2 1 cos 2 ml ml ω θω ωθ θ  =+ + + +   ±± 0 LdL dt θθ ∂∂ −= ± () ( ) sin sin 0 ml ml θωθ −++ ± ± = (a) 2 sin 0 += (b) The bead executes simple harmonic motion ( ) 0 > ² about a point diametrically opposite the point of attachment. (c) The effective length is 2 g l "" = 10.14 vj ˆˆˆ cos sin a t li j + G ± ˆˆ cos sin il j at l + ± 2 2 1 cos 2 sin sin m Tm v v l a ta t l l =⋅ = ++ + GG ± 2 1 cos 2 Vm g a t l  =−   2 2s i n c o s ma LTV l a t a t l m g l t =−= + + 2 sin L ml matl ± ± , 2 sin cos dL ml mal matl dt + ± ± cos sin L matl mgl ± 2 sin cos cos sin ml mal matl matl mgl = ± ± sin 0 ag l + For small oscillations, sin 0 l + 2 2 l T π == + D 10.15 (a) 2 11 x I 2 ± ± and 2 2 5 I
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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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