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My question arises from a classical mechanics problem from a Hong Kong physics training programme:

This is not a homework question as I am not asking about how to solve the problems in the image.

As an additional exercise, I tried to solve for $\alpha$ as a function of $t$.

From energy conservation,

$$\frac{m\dot x^2}2+\frac{m\dot y^2}2+mgy=mgy_0$$

By the substitution $x=L\cos\alpha, y=L\sin\alpha$, we have

$$\frac{L}{2g}\dot\alpha^2=\sin\alpha_0-\sin\alpha$$

Rearranging immediately gives $$\int^\alpha_{\alpha_0}\frac{d\theta}{\sqrt{\sin\alpha_0-\sin\theta}}=-\int^t_0\sqrt{\frac{2g}{L}}dt$$

(Negative square root is taken as $\dot\alpha<0$.)


If we naively substitute in $\alpha_0=\frac{\pi}2$, then the integral diverges. This makes sense as a vertical rod should stand still.


I guess that taking the limit $\alpha_0\to\frac\pi2^-$ would avoid the infinity problem, but I have no idea how can this be implemented.

How can I avoid the divergence and obtain $\alpha(t)$?

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    $\begingroup$ Hi Szeto. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Oct 13 '19 at 12:45
  • $\begingroup$ Unfortunately, the initial setup (i.e. bodies static and rod perfectly vertical) is a valid solution of the equations of motion for all times. Since the system is unstable, you can either add a small deviation from the $\frac{\pi}{2}$ angle or a small amount of initial kinetic energy, to branch out from the static solution and find a different one with the rod falling. $\endgroup$ – secavara Oct 14 '19 at 18:24
  • $\begingroup$ Related answer to this problem. $\endgroup$ – John Alexiou Oct 15 '19 at 21:53