Analytical Mech Homework Solutions 90

Analytical Mech Homework Solutions 90 - (1 ) b) V ( x , y )...

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b) () 22 11 1 , 2 1 x y y xyx y α αα Vx + =−  −+ +  (7.4.13) ( ) [] ( ) 33 xx V x x −− + =+ Now x 0.5 at and 4 L 5 L also, each bracket term in the denominator equals 1 at , 4 L 5 L ( 1 0.5 0.5 1 0.5 V x − − + − +− − − ) 0.5 0.5 0.5 0.5 0 = ++− + 1 y Vy y y Again, the denominator in brackets equals 1 @ , 4 L 5 L So, 1 V y  ± + ± −±   3 2 333 3 0 222 2 ± ∓∓ Thus ˆˆ ,0 VV Vxy i j xy ∂∂ + ∇= n of moment : at L , . 4 5 L 7.14 Conservatio um 4 pp mv ′′ D GG G 45 o v’ φ v’ p cos 45 4 cos p vv v φ D D 4c o s 2 p v D v o 0 sin 45 4 sin p D 4s i n 2 p v v = 16 2 v + DD 2 Conservation of energy: 11 1 4 22 2 p 2 p D 16 4 4 p D v
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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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