Analytical Mech Homework Solutions 94

Analytical Mech Homework Solutions 94 - vcm 1 m1 = =...

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1 12 1 1 1 cm v m vmm α == ++ Equation 7.6.11 Thus () 2 22 1 11 r γ αα =−+ + 1 2 1 2 1 1 v r v αγ + Simplifying 1 2 21 0 rr γα  −+   = Let 2 x r = and solving the resulting quadratic for x 1 2 1 1 x  =+  Squaring 1 2 2 2 2 2 1 2 1 rx +− +  + 1 Now 1 1 2 2 1 1 2 1 2 1 1 T r T =−=− + + + And, after a little algebra, we get the desired solution 1 2 1 1 1 1 T T γγ =− + + 7.20 From Equation 7.6.15 … 1 1 2 2 1 mv m v vm m m mm v 1 Now we solve for 1 1 v v 1 2 1 1 2 T v m = and now solving for 1 v starting with Equation 7.6.9 … vv µµ Q and using 2 1 m mm ′ = + 1 v v we get … 2 1 T Q Q = 2 1 2 1 2 2 T vQ
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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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