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Unformatted text preview: 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 ) =...
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This note was uploaded on 01/22/2012 for the course PHYS 3201 taught by Professor Deheer during the Spring '10 term at Georgia Institute of Technology.
 Spring '10
 deHeer
 mechanics, Energy, Kinetic Energy, Mass

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