Analytical Mech Homework Solutions 32

Analytical Mech Homework Solutions 32 - 3.25 + sin = 0 d 2...

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3.25 sin 0 θ += ±± 2 cos 0 2 d dt  =   Integrating: 2 0 cos 2 = D ± ± or ( ) 2 2co s co s θθ =− D ± () 1 0 2 4 s s d T =   D D Time for pendulum to swing from 0 = to = D is 4 T Now—substitute sin 2 sin 2 sin φ = D so 2 π = at = D and use the identity 2 cos 1 2sin 2 1 0 2 22 4 4s i n s i n d T =    D D and after some algebra … 11 2 1s i n i n s i n 2 dd =   D or (a) 2 1 2 0 2 4 i n d T α = where 2 sin 2 = D (b) 1 2 2 13 1 sin 1 sin sin 28 αφ −≈ ++ 4 + 2 4 0 41s i n s i n Td φα =+ + +  2 9 21 46 4 T πα =+ + + (c) 2 32 2 sin 4 8 4 =≈ + DD D …∼ D . . . 2 16 T + D ------------------------------------------------------------------------------------------------------
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