Exercise 5
The coordinates we use are defined in the following image:
Instead of using the cartesian coordinates \(x\) and \(y\), we make use of the generalized coordinates \(\varphi\) and \(l\). Since \(l\) is fixed, we have only one degree of freedom and are interested in only one equation of motion for \(\varphi\).
The equations for the coordinate transformation are as follows: \[x = l \sin \varphi\] \[y = - l \cos \varphi\]
\[\dot{x} = l \cos \varphi\, \dot{\varphi}\] \[\dot{y} = l \sin \varphi\, \dot{\varphi}\]
Kinetic energy \(T\): \[T=\frac{1}{2}m \left( \dot{x}^2 + \dot{y}^2 \right)\] \[= \frac{1}{2}ml^2 \dot{\varphi}^2 \cos^2 \varphi +\frac{1}{2}m l^2 \dot{\varphi}^2 \sin^2 \varphi\] \[= \frac{1}{2}m l^2 \dot{\varphi}^2\]
The potential energy \(V\) has the value 0 for \(y=-l\) (rest position) and has its biggest value for \(y=l\): \[V=mg\, (l+y) = mg \, (l -l \cos \varphi) = mg l \, (1-\cos \varphi)\]
Hence, the Lagrangian is: \[L = T-V =\frac{1}{2}m l^2 \dot{\varphi}^2 -mg l \, (1-\cos \varphi) \]
The Euler-Lagrange equation is:\[\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial L}{\partial \dot{\varphi}} - \frac{\partial L}{\partial \varphi} = 0\] \[ \frac{\partial L}{\partial \dot{\varphi}}= m l^2 \dot{\varphi}\] \[\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{\varphi}} = m l^2 \ddot{\varphi}\] \[\frac{\partial L}{\partial \varphi} = -mg \, l \sin \varphi\]
Equation of motion: \[ m l^2 \ddot{\varphi} + mg \, l \sin \varphi=0\] \[ l \ddot{\varphi} + g \sin \varphi=0\] \[\Rightarrow \ddot{\varphi} = - \frac{g}{l} \sin \varphi\]
Solving this differential equation is not easy, you can do it numerically or, for an analytical solution, you need Jacobi’s elliptical functions.
For small angles we can use the approximation \(\sin \varphi \approx \varphi\) (Exercise 5, Lecture 2). This way we get the nice and simple differential equation for the harmonic oscillator: \[\ddot{\varphi} = - \frac{g}{l} \varphi\] with well known solutions (Exercise 3, Lecture 4).