Bead Sliding on Disk

@Mark Hennings – Well, OK, the moment of inertia of the disc about the point of rotation is $\tfrac{15}{2}mR^2 = \alpha mR^2$ where $\alpha = \tfrac{15}{2} > 4$ . The fact that $\alpha > 4$ is enough to ensure that $\dot{\psi}$ is never zero, and hence we can always take the positive square root. If the intial value of $\dot{\phi}$ is $\omega$ , then we obtain $\begin{aligned} \dot{\psi} & = \; \omega\sqrt{\frac{\alpha-2 + 2\cos\psi}{\alpha + \sin^2\psi}} \\ \dot{\theta} & = \; \frac{\omega}{\alpha+2+2\cos\psi}\left[2 - (1 + \cos\psi)\sqrt{\frac{\alpha-2+2\cos\psi}{\alpha+\sin^2\psi}}\right] \end{aligned}$ so that $\frac{d\theta}{d\psi} \; = \; \frac{1}{\alpha+2+2\cos\psi}\left[2\sqrt{\frac{\alpha+\sin^2\psi}{\alpha - 2 + 2\cos\psi}} - 1 - \cos\psi\right]$ and hence the desired angle is $\int_0^{2\pi} \frac{1}{\alpha+2+2\cos\psi}\left[2\sqrt{\frac{\alpha+\sin^2\psi}{\alpha - 2 + 2\cos\psi}} - 1 - \cos\psi\right]\,d\psi$ which (reassuringly) evaluates to $63.2^\circ$ .

@Mark Hennings – I love Lagrange, because I don't have to worry about anything. I just write out the geometry and then mindlessly crunch out some derivatives.