Ex 1.9 Lagrange’s equations

Structure and Interpretation of Classical Mechanics


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.


\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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