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# m2 2 cos 2 and y 1 2 1f 318325 the condition

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Unformatted text preview: ve dV v2 mv 2 ˙ F =− = −mrθ2 = −mR = . y = f (x) =⇒ y = f x =⇒ y = f x + f x2 . ˙ ˙ ˙ dr R¨ R¨ (315) (322) (316) (323) 6.36. Atwood’s machine Plugging the x and y from the E-L equations into this, and solving for F , gives ¨ ¨ Let 1 and 2 be the lengths of string in the air, and 2 L = 1 + 2 . If η ≡ 1 + 2 − L, let mf x ˙ F= . (324) then the Lagrangian is ∂η /∂ y − f ∂ η /∂ x 1 1 L = m1 ˙2 ∂η m2 ˙2 + ∂η g 1 m2 is − angle (317) We must now determine + /∂ x and m1/∂ y .+If θ g 2theV (η ). the curve makes with 1 2 2 2 the x axis at a given point, then the slope there is f (x) = tan θ. Consider a point Usingx, y= near /dη ,curve also the deﬁnition of η , point is to the of motion are ( F ) −dV the and (we’ll assume that this the equations left of the curve, but the other side proceeds similarly). If∂η imagine varying only x, and then only y , you you dV m1 ¨1 = m1 g − the curve changes ¨1 = m1 g to can see that the distance to η · ∂ 1 =⇒ m1...
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## This note was uploaded on 01/19/2014 for the course PHYSICS 251 taught by Professor Brandenberger during the Fall '13 term at McGill.

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