Such a solution is usually a trivial solution but in

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Unformatted text preview: alf the sin(45 − 2 ), − cos(45 −the ,normal mode (θ, α) ∝ (1, 1) √ √ ◦ 0, we have the normal mode (θ , α) ∝ (1, −1) with ◦ with frequency ω=. And sif B1 = θ) + 2 sin α, − cos(45 + θ) − 2 cos α . (x, y )2 1 R in(45 + frequency ω2 . It makes sense that the frequency of the second mode is larger than the frequency of the first. a circle, so its speed is Rθ. Calculating x2 + y2 for the ˙ The left mass moves along ˙2 ˙2 right sliding on a the 6.45. Massmass (or using rimlaw of cosines), we have Let 2 be the2 horizontal coordinate2of M , and 2 θ √ the˙angle of ◦ along the hoop, x be 2 m 2 let ˙ ˙2 ˙ v1 = R2 θ , and v2 ˙ measured counterclockwise 2 = Rthe + 2R α of 2 2R θα cos(45 the − α). (344) from θ bottom + the hoop. Then + θ position and velocity of m are The Lagrangian is therefore (x, y )m = (x + R sin θ, R − R cos θ) √ 2 ◦ L = (m/2) (2Ry )˙2 + 2R2 αx + 2 θ 2R2θ˙,αRθ sin θ)+ θ − α) ˙2 ˙ =⇒ x, ˙ θ ˙ = ( ˙ + R ˙ cos...
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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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