Analytical Mech Homework Solutions 151

Analytical Mech Homework Solutions 151 - 2g 1 = 2 = 1 1 1 1...

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1 12 3 2 111 1 4 g mm m λ =   ++     2 3 4 g m = As the check, let and 23 m == 1 2 = . Thus, there is no acceleration and 1 2 mg = 10.18 (See Example 5.3.3) (a) ˆ r rr e = ˆˆ r er e θ =+ ± ± ± Constraint: () 0 ft θθ ω =− = so … = ± and 0 = ±± 22 2 1 2 Tm r r ± ± L = 0 LdL rd tr ∂∂ ± = and 0 f dt ∂∂∂ += ± 2 = 2 20 mrr mr −−+ ± ± = tt rA e B e 0 r 0 = apply constraint e =− ± 0 rl = ± so … 0 AB 2 mrr = ± ABl ωω −= 2 Al = 2 l A = 2 l B = − thus . . 2 l e re sinh cosh t t = = ± 2s i n h c o s h ml t t = now at so = tT = 1 10 sinh 1 T . 8 8 (b) There are 2 ways to calculate F … (i) See Example 5.3.3 … 2 Fm x = ± ( 2 t l xe e ) t and 2 l e ± 2 2c o s h l t = (ii)
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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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