Does $\dot{\theta}^2=-\frac{g}{l}\cos\theta\iff\ddot{\theta}=-\frac{g}{l}\sin\theta$ holds? (Simple pendulum problem)

Question

I solved a simple pendulum problem as follows (please see the above figure):

By the Newton's second law,

$$m\ddot{\vec{r}}=-ml\dot{\theta}^2\begin{pmatrix}\cos\theta \\ \sin\theta\end{pmatrix}+ml\ddot{\theta}\begin{pmatrix}-\sin\theta \\ \cos\theta\end{pmatrix}=mg\cos\theta\begin{pmatrix}\cos\theta \\ \sin\theta\end{pmatrix}-mg\sin\theta\begin{pmatrix}-\sin\theta \\ \cos\theta \end{pmatrix}.$$
From this equation, we get the following two differential equations:
$\dot{\theta}^2=-\frac{g}{l}\cos\theta.$
$\ddot{\theta}=-\frac{g}{l}\sin\theta.$

If $\theta$ satisfies $\dot{\theta}^2=-\frac{g}{l}\cos\theta$, then does $\theta$ satisfy $\ddot{\theta}=-\frac{g}{l}\sin\theta$?
If $\theta$ satisfies $\ddot{\theta}=-\frac{g}{l}\sin\theta$, then does $\theta$ satisfy $\dot{\theta}^2=-\frac{g}{l}\cos\theta$?

Answer 1

The derived equations do not account for the tension ${\bf T}$ constraining the bob. Conveniently, though, this means that we can describe our system correctly with a single degree of freedom.

If we denote the angular displacement by $\theta$, the displacement along the arc of radius $l$ centered at the top of the pendulum is $l \theta$, and so its tangential acceleration is $l \ddot\theta$. Drawing a right triangle shows that the tangential force is $-mg \sin \theta$ (and, incidentally, $T = mg \cos\theta$, though we don't need that fact to derive the equation of motion). Now, Newton's Second Law gives that $$m l \ddot \theta = -mg \sin \theta,$$ and canceling and rearranging gives the classical pendulum equation, $$\ddot\theta = -\frac{g}{l} \sin \theta .$$ To recover an equation in $\dot\theta$, we can multiply both sides by $\dot\theta$ and integrate, yielding $$\frac{1}{2} \dot\theta^2 = \frac{g}{l} (\cos \theta + C)$$ for some constant $C$ that we can express, e.g., in terms of the energy of the system or the maximum angular displacement.

Answer 2

Mathematically, not really.

I noticed that your formula for pendulum is not correct. You might mean these: $$\frac{\dot{\theta}^2}{2} = \frac{g}{l} \cos(\theta)$$ and $$\ddot{\theta} = -\frac{g}{l} \sin(\theta)$$

The trivial counterexample is $\theta \equiv \pi /2$, where $\dot{\theta}^2/2=\frac{g}{l}\cos\theta = 0$ but $\ddot{\theta}=0 \ne -\frac{g}{l}\sin\theta$.

In that case, you may assume that $\theta$ is not a constant.

By taking derivative, you get $\ddot{\theta}\dot{\theta}=\frac{g}{l}\sin\theta \dot{\theta}$ and you can cancel $\dot{\theta}$ because $\theta$ is not a constant, then you get $\ddot{\theta} = -\frac{g}{l}\sin\theta$. However, if you have $\ddot{\theta} = -\frac{g}{l}\sin\theta$, then you can't get $\frac{\dot{\theta}^2}{2} = \frac{g}{l} \cos(\theta)$, because actually $\frac{\dot{\theta}^2}{2} = \frac{g}{l} \cos(\theta) + C$.