Analytical Mech Homework Solutions 96

Analytical Mech Homework Solutions 96 - 7.23 4 m = r3 3 m =...

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7.23 3 4 3 mr π ρ = 2 4 r = ±± where vz 2 rz ± = ± k a constant of proportionality rk z = ± ± rrk z =+ D From eqn. 7.7.6, mg mv vm () 33 2 44 4 rg r k z z πρ ± ± 2 3 kz gz r ± 2 3 z zg r z k =− + D ± For 0 r = , D 2 3 z z ± A series solution is used for this differential equation: 2 0 n n n za = = ± z 2 1() 2 dz dz dz dz d z zz dt dz dt dz dz ==⋅= = ± ± ± 2 1 n n n dz anz dz = ± 2 1 n n n z az z = ± 11 1 3 2 nn n z ga z −− == ∑∑ For 1 n = : 1 3 2 aga 1 2 7 ag = For 1 n
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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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