Introductory Problems - Exercises

We can parameterize the coordinates of the particle by two variables, the angles \(\varphi\) and \(\theta\). This takes the form $$\begin{equation} \begin{cases} x = \ell \sin\theta \cos\varphi \\ y = \ell \sin\theta \sin\varphi \\ z = \ell \cos\theta \end{cases} \end{equation}$$ Thus, the time derivatives of each of these are $$\begin{equation} \begin{cases} \dot{x} = \ell(\dot{\theta}\cos\theta\cos\varphi - \dot{\varphi}\sin\theta\sin\varphi) \\ \dot{y} = \ell(\dot{\theta}\cos\theta\sin\varphi + \dot{\varphi}\sin\theta\cos\varphi) \\ \dot{z} = -\ell\dot{\theta}\sin\theta \end{cases} \end{equation}$$ Thus, the kinetic energy is given by $$\begin{align} T ={}& \frac{m}{2}(\dot{x}^2 + \dot{y}^2 + \dot{z}^2) \\ ={}& \frac{m}{2}\ell^2(\dot{\theta}^2\cos^2\theta\cos^2\varphi + \dot{\varphi}^2\sin^2\theta\sin^2\varphi + \dot{\theta}^2\cos^2\theta\sin^2\varphi + \dot{\varphi}^2\sin^2\theta\cos^2\varphi + \dot{\theta}^2\sin^2\theta) \\ ={}& \frac{m}{2}\ell^2(\dot{\theta}^2\cos^2\theta + \dot{\theta}^2\sin^2\theta + \dot{\varphi}^2\sin^2\theta) \\ ={}& \frac{1}{2}m\ell^2(\dot{\theta}^2 + \dot{\varphi}^2\sin^2\theta) \end{align}$$ The potential energy is \(V = mgz\), and since \(z = \ell \cos\theta\), we have that \(V = mg\ell\cos\theta\). Thus, the Lagrangian is given by \(\)\L = \frac{1}{2}m\ell^2(\dot{\theta}^2 + \dot{\varphi}^2\sin^2\theta) - mg\ell\cos\theta\(\)