Analytical Mech Homework Solutions 137

# Analytical Mech Homework Solutions 137 - dV d = cos ( Lz Lz...

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()() 2 2 3 cos cos sin cos sin 0 sin zz z LL L dV mgl dI θθ θ ′′ = −− + =− D DD D D = let cos γ D Then and solving for 22 cos sin sin 0 z Lm g l I γθ −+ 4 = D 1 2 2 2 sin 4 cos 11 2cos z z g l I L       D now is large since z L ψ ± is large and the precession rate is small, so we can expand the term in square root above and use the (-) solution since must be positive … 2 2 sin 2 cos z z g l I L ≈− + D 2 sin cos z mglI L D D From equations 9.7.2, 9.7.5 and 9.7.7 … () 00 sin cos cos zs LI I 0 φ θφ =++ ±± ± 0 and … cos ( cos )cos φθ =+ ± ± so … 2 sin cos cos cos cos ss s s II I I I θθφ + − D D D 2 sin sin sin cos z s mglI mglI I L I =≈ = + D D ± ± ± and since 0 >> ± , we can ignore the ± term in the denominator and we have … s mgl I ± ± Hence, if ± large,
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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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