Rod Rotating with a ring -part 1

If the bead is a distance $r$ from the centre of rotation when the rod makes and angle $\theta$ with its initial direction, then the angular momentum of the system about the centre of rotation is constant, and hence $\tfrac13m\ell^2 \dot{\theta} + mr \times r\dot{\theta}$ is constant. Thus $\begin{aligned} \tfrac13m(\ell^2 + 3r^2)\dot{\theta} & = \; \tfrac{7}{12} m\ell^2\omega \\ \dot{\theta} & = \; \frac{7 \ell^2 \omega}{4(\ell^2 + 3r^2)} \end{aligned}$ Since the rod is smooth, the only force acting on the bead is transverse, and so the radial acceleration of the bead is zero. Thus $\ddot{r} - r\dot{\theta}^2 = 0$ , and so $\begin{aligned} \ddot{r} & = \; r\dot{\theta}^2 \; = \; \frac{49 \ell^4 \omega^2 r}{16(\ell^2 + 3r^2)^2} \\ \tfrac12\dot{r}^2 & = \; \tfrac{7}{24}\ell^2\omega^2 - \frac{49\ell^4 \omega^2}{96(\ell^2 + 3r^2)} \; = \; \frac{7\ell^2 \omega^2}{96(\ell^2 + 3r^2)}\big[ 4(\ell^2 +3r^2) - 7\ell^2\big] \\ \dot{r}^2 & = \; \frac{7\ell^2 \omega^2(4r^2 - \ell^2)}{16(\ell^2 + 3r^2)} \end{aligned}$ Thus, when $r =\ell$ , $\dot{\theta} = \tfrac{7}{16}\omega$ and $\dot{r}^2 = \tfrac{21}{64}\ell^2\omega^2$ , and hence $v^2 \; = \; \dot{r}^2 + \ell^2\dot{\theta}^2 \; = \; \tfrac{21}{64}\ell^2\omega^2 + \tfrac{49}{256}\ell^2\omega^2 \; = \; \tfrac{133}{256}\ell^2\omega^2$ With $\ell = 168$ and $\omega = \tfrac{2}{21}$ we have $\ell \omega = 16$ , and this makes $v^2 = \boxed{133}$ .