# 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.

~~~ \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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