Analytical Mech Homework Solutions 166

Analytical Mech Homework Solutions 166 - 3 V= ( ) k 4l 2 +...

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3 ( ) 22 2 44 8 4 2 k Vl y l l y l m =+ −+ + g y ( ) 2 Vk ly l lym g + y The first term, , is an additive constant to the potential energy, so with appropriate adjustment of the zero reference point … 2 4 kl ( ) 2 () 2 2 Vy ky ll y m g y =−+ () 12 22 2 dV y ky y l l y m g dy  =−+ −   At equilibrium, the above expression is zero, so … 4 40 kly ky mg ly −− + = 4 4 kly ky mg −= + 16 16 8 kly km g y mg −+= + 24 3 2 2 2 22 16 8 16 16 8 0 k y kmgy k l m g k l y kl mgy l m g −++ − += 42 2 2 32 44 2 4 2 2 2 0 21 6 21 6 ym g m g m g m g yy y lk l k l k l k l 2 = letting y u l = and 4 mg a kl = 43 2 2 2 ua u a u a −+−+ = 0 11.5 ( )
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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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