Analytical Mech Homework Solutions 97

Analytical Mech Homework Solutions 97 - 7.24 From eqn 7.7.6...

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7.24 From eqn. 7.7.6, , where m and v refer to the portion of the chain hanging over the edge of the table. mg mv vm =+ ±± z m λ = and v where λ is the mass per unit length of chain z = ± mz = ± ± and vz = ± 2 1() 2 dz dz dz dz d z zz dt dz dt dz dz ==⋅= = ± ± ± () 2 2 dz zg z z z dz λλ    ± 22 2 z dz z == Because of the initial condition 0 zb = D , a normal power series solution to this differential equation (…as in Prob. 7.22) does not work. Instead, we use the Method of Frobenius … 2 0 ns n n za z + = = ± 2 1 n n ansz dz + − ± 2 1 n n z az z +− = ± 11 1 2 nn n s a z + −+ = ∑∑
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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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