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# D y y using r r r r 0 the rst of these equations

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Unformatted text preview: ting it all together gives t2 d2 dt2 ∂ L ˙ t2 β. ∂ x t1 ¨ t1 t1 (313) The boundary term involving β is zero, because β is assumed to vanish at the ˙ endpoints. But the boundary term involving β is not necessarily zero, because the derivative is not assumed to be zero at the endpoints. The proposed result is therefore not valid. dS = da β ∂L d − ∂x dt ∂L ∂x ˙ + ∂L ∂x ¨ dt + β ∂L d − ∂x ˙ dt t2 ∂L ∂x ¨ + 6.35. Constraint on a circle Let Constraint on a curve 6.38. the constraining potential be V (r). Then the Lagrangian is 75   The true Lagrangian is L = (m/2 x22 + y 2 ) − V (η ), where η is the distance from 2)( ˙ ˙ ˙ 1 L = m(r + r θ2 ) − V (r). ˙ (314) the curve. The equations of 2 motion are The equations of motion are dV ∂η ∂η mx = − ¨ · =⇒ mx = F ¨ , dη ∂ x ∂x d dV d 2˙ ˙2 , (mr) = − ˙ + mrθ ∂η and (mr θ) = 0. dV ∂η dt dr dt my = − ¨ · =⇒ my = F ¨ . dη ∂ y ∂y Using r = R =⇒ r = r = 0, the ﬁrst of these equations gives ˙¨ For a point on the curve, we ha...
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