291 not necessarily innitesimal then m 1 2 using cos

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: + a0 , but this won’t be˙ important.) Consider the function ˙ x2 ≈ −2rθ1 + rθ2 =⇒ x2 ≈ −2rθ1 + rθ2 , y (t) = y0 (t) + f (t), where f (t) vanishes at the endpoints, but is otherwise arbitrary ˙ ˙ ˙ x3 ≈ −2rθ1 + 2rθ2 − rθ3 =⇒ x3 ≈ −2rθ1 + 2rθ2 − rθ3 . (291) (not necessarily infinitesimal). Then m θ1 2 Using cos θ ≈ 1 − θ /2, we have (up to additive constants in the y ’s) Sy = = 2 1 m y0 + f˙ − mg (y0 + f ) θ2 ˙ dt 1 2 y1 = r cos θ1 −→ −r , 2 1 2 my0 − mgy0 dt + my0 f˙2− mgf dt + ˙ θ2 2 y2 = 2r cos θ1 + r cos θ2 −→ −r θ1 + 2 , 1 ˙2 mf dt 2 2 1 = S 0 − m f (¨0 + g ) dt + m f˙2 dt, 2 θ2 y 2 y3 = 2r cosyθ1 + 2r cos θ2 + r cos θ3 −→ −r θ1 + θ2 + 3 2 2 . (292) Figure 17 (309) wherevalues are irrelevant, because their squares will belast line. The middle term here is The y we have integrated by parts to obtain the 4th order in the θ’s. The ˙ Lagrangian is therefore zero, because we are assuming that y0 makes the a...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online