Analytical Mech Homework Solutions 153

# Analytical Mech Homework Solutions 153 - V = mgy = mgr sin...

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Potential energy sin Vm g ym g r θ =− LTV Equation of constraint () ,0 fr r a =−= 1 f r = 0 f = 2 1 f x = Lagrange’s equations 0 ii i Ld L f xd tx x λ  ∂∂∂ −+   ± = x 2 ) 22 11 1 sin cos 0 d mx m r r dt θθ  −−− =  ± ±± ± 2 1 sin sin cos cos sin 0 d m r r r r r dt −− + + + = ± ± ± using constraint ; r a constant == 0 rr = = ± ( 2 1 2 12 sin cos m xa mm ) + + ± ( 1 ) θ ) 2 222 2 2 sin sin cos sin cos cos 0 r r r x rx r x r g r ++ + ± ± ± ± ± ± ± ± ± ± = using constraint ; r a constant 0 = = ± ) 2 sin cos 0 xg aa + = ( 2 ) r ) 2 2 1 cos sin sin sin 0 x r x g m + + + + = ± ± ± Apply constraint… 2 2 1 cos sin 0 g m += ± ( 3 ) Now, plug solution for 2 x (1) into equation (2): 2 1 sin cos sin cos 0 mg a + ± = 1s i n s i n c o s c o s g ff a ± 0 = where 1 1 m m f = + Now 2 1 2 dd d d d dt d dt d d = =  ± ± ± So – let 2 x = ± 2 2 1 sin 2 sin cos cos 0 dx g fx f da = Let 1 2 i n m yf sin 2 d dx dy g yx dda d 15
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