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Problem Set 7 - Triangular Pendulum dene as the pendulums...

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Triangular Pendulum define θ as the pendulum’s angle of rotation. We can find the kinetic energy of the pendulum from its velocity. Since both masses are moving in polar coordinates at constant radius l , their velocity is simply l ˙ θ and T = 1 2 m ( l 2 ˙ θ 2 + l 2 ˙ θ 2 ) = ml 2 ˙ θ 2 The potential energy can be found from the heigt of both masses. - mgl (cos( π/ 6 - θ ) + cos( π/ 6 + θ ) and the Lagrangian is L = ml 2 ˙ θ 2 + mgl (cos( π/ 6 - θ ) + cos( π/ 6 + θ ) The equation of motion is can then be found ∂L ∂θ = - mgl (sin( π/ 6 + θ - sin( π/ 6 - θ )) and d dt ∂L ˙ θ = d dt 2 ml 2 ˙ θ = 2 ml 2 ¨ θ Setting these two equal, we see ¨ θ = - g 2 l (sin( π/ 6 + θ - sin( π/ 6 - θ )) for small values of θ , we can use the Taylor expansion of the sine functions sin( π/ 6 + θ ) sin( π/ 6) + θ cos( π/ 6) = 1 / 2 + 3 θ/ 2 and we have ¨ θ = - g l ( 3 2 θ ) For small oscillatons, we’ll assume θ ( t ) = θ o e iωt ¨ θ ( t ) = - ω 2 θ ( t ) Dividing by the common factor of θ , we have ω 2 = - 3 2 g l Bead on a Helix
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