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# D arclength is therefore d l in terms of the

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Unformatted text preview: ˙ r , ∂ 2 L θ˙32= r . L β +2 ββ + β . ∂ x2 ∂ x∂ x ˙ ∂ x2 ˙ ∂x ˙ (295) ˙˙ ββ   (310) 6.29. Cycloidal pendulum 6.34. x(a) The given parametrization of the cycloid yields ¨ dependence We havey ) = R(θ − sin θ, −1+cos θ) =⇒ (dx, dy ) = R dθ(1 − cos θ, − sin θ). (296) (x, dS ∂L ∂L ˙ ∂L ¨ = β+ β+ β dt. Therefore, da ∂x ∂x ˙ ∂x ¨ (311) We can integrate the= dx = 1 − cosby = tan(θ/2) usual. θIntegrating the last term by middle term θ parts as =⇒ tan α = 2α. (297) |dy | sin θ parts twice gives (b) In terms of θ, the diﬀerential arclength along the cycloid is given by ∂L ¨ β = ∂L ˙ β− d ∂L ˙ β ∂x ¨ ∂¨ dt =¨ ds2 = dx2 + dy 2 = R2x θ2 (2 − 2 cos θ) ∂ x4R2 dθ2 sin2 (θ/2). d ∂ arclength is therefore d ∂L In terms of α, the diﬀerentialL β − ˙ = β− ∂x ¨ dt ∂x ¨ ds = 2R sin(θ/2) dθ = 4R sin α dα. 2 d dt2 ∂L β ∂x ¨ (298) . (312) (299)   74 CHAPTER 6. THE LAGRANGIAN METHOD Put...
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