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Analytical Mech Homework Solutions 166

# Analytical Mech Homework Solutions 166 - 3 V= k 4l 2 4 y 2...

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3 ( ) 2 2 2 2 2 4 4 8 4 2 k V l y l l y l m = + + + gy ( ) 2 2 2 2 2 2 2 V k l y l l y mg = + + 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 ( ) 2 2 V y k y l l y mgy = + ( ) 1 2 2 2 ( ) 2 2 2 dV y k y yl l y mg dy = + At equilibrium, the above expression is zero, so … 2 2 4 4 0 kly ky mg l y + = 2 2 4 4 kly ky mg l y = + 2 2 2 2 2 2 2 2 2 16 16 8 k l y k y kmgy m g l y + = + ( ) 2 4 3 2 2 2 2 2 2 2 2 2 2 2 16 8 16 16 8 0 k y kmgy k l m g k l y kl mgy l m g + + + = 4 2 2 2 3 2 4 4 2 4 2 2 2 0 2 16 2 16 y mg m g mg m g y y y l kl k l kl k l + + 2 = letting y u l = and 4 mg a kl = 4 3 2 2 2 2 2 u au a u au a +
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