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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online