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# 291 not necessarily innitesimal then m 1 2 using cos

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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...
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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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