Analytical Mech Homework Solutions 150

Analytical Mech Homework Solutions 150 - For L m = k x l m...

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() 2 Lm kx l m g x  =− + +   32 mm mx k x l m ′′ += + + ±± g For 2 gm yxl m k =−− + , 0 3 m my k y ++ = The block oscillates about the point 2 xl m k =+ + with a period … 2 3 2 m m k π ω T + == D 10.17 Note: 4 objects move – their coordinates are labeled i x : The coordinate of the movable, massless pulley is labeled p x . Two equations of constraint: ( ) ( ) 11 1 ,0 pp fxx x lx = −− = ( ) ( ) ( ) 223 2 3 ,, 2 0 fxxx x x x l = +− += 22 2 1 1 2 3 33 111 222 LTV m xm g x m g x m g x =−= + + ±± ± 3 0 j j j ii i f LdL qd tq q λ ∂∂ Thus: (1) 1 1 0 mg −− += (2) 2 2 0 mg mx (3) 3 2 0 Now – apply Lagrange’s equations to the movable pulley – note 0 p m So: (4) 12 0 ff xx λλ or 20 = Now - p x can be eliminated between
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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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