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# 2 2 2 357 for small angles using sin the equations

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Unformatted text preview: θ ˙ cos(45 . (351) m √ 2 ˙ +mg R cos(45◦ −˙ 2 + Rxθ cos ◦ R θ 2R second (345) The square of the speed is then vm = xθ)+ 2R cos(45θ + θ)2+2 . To cos α . order, we ˙˙ may set cos θ ≈ 1 here. So the Lagrangian is ˙ L = (M/2)x2 + (m/2)(x2 + 2Rxθ + R2 θ2 ) + mgR cos θ. ˙ ˙ ˙˙ (352) The equations of motion are ¨ (M + m)¨ + mRθ = 0, x and ¨ x + R θ ≈ −g θ , ¨ (353) where we have used sin θ ≈ θ. To ﬁnd the normal modes, we can use the determinant method. Or we can just solve for x in the ﬁrst equation and plug into the second, ¨ which gives M +m ¨ θ=− M g θ =⇒ θ(t) = A cos(ω t + φ), R where ω = M +m M g . R (354)   79 The corresponding x is x(t) = − MR MR θ=− A cos(ω t + φ). M +m M +m (355) In this mode, the two masses move in opposite directions. The other solution to Eq. (354) is θ = 0 identically (this solution would pop out of the determinant method). Such a solution is usually a trivial solution, but in this problem both equations of motion give x = 0 =⇒ x(t) = Dt + E . So in this mode, ¨ the wheel moves at a constant rate, with the mass always at th...
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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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