Sliding Along a Rotating Rod

Here is the energy / Lagrange way. We will operate in polar coordinates $(r,\theta)$

Bead kinetic energy:

$E_{bead} = \frac{1}{2} m (r^2 \dot{\theta}^2 + \dot{r}^2)$

Rod kinetic energy: $E_{rod} = \frac{1}{2} \frac{m L^2}{3} \dot{\theta}^2$

Bead potential energy:

$U_{bead} = mgr \, sin \theta$

Rod potential energy:

$U_{rod} = \frac{1}{2} mgL \, sin \theta$

Lagrangian:

$L = E_{total} - U_{total} \\ L = \frac{1}{2} m (r^2 \dot{\theta}^2 + \dot{r}^2) + \frac{1}{2} \frac{m L^2}{3} \dot{\theta}^2 - mgr \, sin \theta - \frac{1}{2} mgL \, sin \theta$

Equations of motion:

$\frac{d}{dt} \frac{\partial{L}}{\partial{\dot{r}}} = \frac{\partial{L}}{\partial{r}} \\ \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{\theta}}} = \frac{\partial{L}}{\partial{\theta}}$

There is nothing special about the math from here on, so I'll just post the results. They match what @Mark Hennings derived via Newtonian mechanics. Having expressions for the "double-dot" terms for the two variables, we can readily solve numerically.

$\ddot{r} = r \dot{\theta}^2 - g \, sin{\theta} \\ \ddot{\theta} = \frac{-2 r \dot{r} \dot{\theta} - gr \, cos{\theta} - \frac{1}{2} g L \, cos{\theta}}{r^2 + \frac{L^2}{3}}$

Here are some interesting plots of the behavior (time is on the horizontal axis). We see that when $\theta$ is positive, the bead slides toward the origin as expected. At this time, the angle is decreasing, also as expected. When the angle turns negative, gravity begins to pull the bead toward the end of the rod, causing the radius to increase once more. It eventually slides off at around $t = 0.650$ .

If the rod makes an angle $\theta$ above the horizontal, and the bead is a distance $x$ from the hinge, then the position vector of the bead is $\mathbf{r} \; = \; x\binom{\cos\theta}{\sin\theta}$ and so the equation of motion of the bead is $(\ddot{x} - x\dot{\theta}^2)\binom{\cos\theta}{\sin\theta} + (x\ddot{\theta} + 2\dot{x}\dot{\theta})\binom{-\sin\theta}{\cos\theta} \; = \; \binom{0}{-10} + N\binom{-\sin\theta}{\cos\theta}$ where $N$ is the magnitude of the normal reaction between the bead and the rod. Taking radial and tangential components, we obtain the equations $\begin{aligned} \ddot{x} - x\dot{\theta}^2 & = -10\sin\theta \\ x\ddot{\theta} + 2\dot{x}\dot{\theta} & = N - 10\cos\theta \end{aligned}$ On the other hand, the equation of motion of the rod is $\tfrac13\ddot{\theta} \; = \; -(Nx + 5\cos\theta)$ Eliminating $N$ from these equations, we obtain the simultaneous differential equations $\begin{aligned} \ddot{x} - x\dot{\theta}^2 & = -10\sin\theta \\ x^2\ddot{\theta} + 2x\dot{x}\dot{\theta} + 10x\cos\theta & = -\tfrac13\ddot{\theta} - 5\cos\theta \end{aligned}$ together with the initial conditions $x(0) = 1$ , $\dot{x}(0) = 0$ , $\theta(0) = \tfrac13\pi$ , $\dot\theta(0) = 0$ .

Solving these equations numerically, I obtain that $x=1$ for the first time after $t=0$ at time $t = \boxed{0.650123}$ .