Ex 1.9 Lagrange’s equations

Structure and Interpretation of Classical Mechanics

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Derive the Lagrange equations for the following systems, showing all of the intermediate steps as in the harmonic oscillator and orbital motion examples.

b. An ideal planar pendulum consists of a bob of mass \displaystyle{m} connected to a pivot by a massless rod of length \displaystyle{l} subject to uniform gravitational acceleration \displaystyle{g}. A Lagrangian is \displaystyle{L(t, \theta, \dot \theta) = \frac{1}{2} m l^2 \dot \theta^2 + mgl \cos \theta}. The formal parameters of \displaystyle{L} are \displaystyle{t}, \displaystyle{\theta}, and \displaystyle{\dot \theta}; \displaystyle{\theta} measures the angle of the pendulum rod to a plumb line and \displaystyle{\dot \theta} is the angular velocity of the rod.

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\displaystyle{ \begin{aligned}   L (t, \xi, \eta) &= \frac{1}{2} m l^2 \eta^2 + m g l \cos \xi \\  \end{aligned}}

\displaystyle{ \begin{aligned}   \partial_1 L (t, \xi, \eta) &= - m g l \sin \xi \\  \partial_2 L (t, \xi, \eta) &= m l^2 \eta  \\  \end{aligned}}

Put \displaystyle{q = \theta},

\displaystyle{ \begin{aligned}   \Gamma[q](t) &= (t; \theta(t); D\theta(t)) \\   \end{aligned}}

\displaystyle{ \begin{aligned}   \partial_1 L \circ \Gamma[q] (t) &= - m g l \sin \theta \\  \partial_2 L \circ \Gamma[q] (t) &= m l^2 D \theta  \\  \end{aligned}}

The Lagrange equation:

\displaystyle{ \begin{aligned}   D ( \partial_2 L \circ \Gamma[q]) - (\partial_1 L \circ \Gamma[q]) &= 0 \\   D (  m l^2 D \theta  ) - ( - m g l \sin \theta ) &= 0 \\   D^2 \theta + \frac{g}{l} \sin \theta &= 0 \\   \end{aligned}}

— Me@2020-09-28 05:40:42 PM

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2020.09.30 Wednesday (c) All rights reserved by ACHK