Analytical Mech Homework Solutions 172

Analytical Mech Homework Solutions 172 - 9 11.14 At time t...

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9 11.14 At time t = 0 and short times thereafter … cos 1 t ∆ ≈ and sin 0 t ∆ ≈ . Thus, … 10 2 cos 0 x A t and x ω ≈≈ . This situation occurs again when 2 t π ∆ = . 11 22 21 12 kk k mm ωω  +   ∆= =    2 mk  +   for , ′<< 1 2 2 1 1 1 ... 1 2 k k ′′ + −≈+ + −= 1 2 1 2 1 2 1 2 2 k T k ππ == = 2 1 2 k TT k = 11.15 () 2 1 2 Tm l θ =+ ±± 1 cos 1 cos k Vm g l r θθ = −+ 1 02 sin sin rr l l = +− 1 1 2 1 cos sin sin sin kl V mgl rl l 1 =− 2 1 23 2 1 021 sin 2 cos cos sin sin sin sin kl kl V mgl l l 1 1 2 l r 11 0 2 Vk km g r l l 2 2 3 2c o sc o s sin sin
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