Discussion
The Lagrangian of a system in terms of polar coordinates \((r, \theta)\) is given by
\[ \begin{aligned} L = {} &\frac{1}{2} m \dot{r}^{2} + \frac{1}{2} m (\dot{r}^{2} + r^{2}\dot{\theta}^{2}) \\ &- m g r (1 - \cos(\theta)), \end{aligned} \]
where \(m\) is the mass, \(g\) is the acceleration due to gravity and \(\dot{s}\) denotes the derivative of \(s\) with respect to time. Then the equations of motion are
Answer: A
The equations of motion are \(2\ddot{r} = r\dot{\theta}^{2} - g(1 - \cos\theta)\) and \(\frac{d}{dt}(r^{2}\dot{\theta}) = -g r \sin\theta\).
First simplify: \(L = m\dot{r}^{2} + \frac{1}{2}m r^{2}\dot{\theta}^{2} - m g r (1 - \cos\theta)\). Apply Lagrange's equations \(\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} = \frac{\partial L}{\partial q}\).
\[ \begin{aligned} r:\ \tfrac{d}{dt}(2m\dot{r}) &= m r \dot{\theta}^{2} - m g (1 - \cos\theta) \\ \theta:\ \tfrac{d}{dt}(m r^{2}\dot{\theta}) &= -m g r \sin\theta \end{aligned} \]
Dividing by \(m\) gives the first option. The other options change the sign of the gravity term in one equation or the other: the potential \(m g r(1 - \cos\theta)\) enters \(L\) with a minus sign, so both generalised forces carry a minus sign.
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