A Bead on a Circle (with Gravity)

If the radial line to the bead makes an angle $\theta$ with the downward vertical, the equations of motion of the bead are $mR\dot{\theta}^2 \; = \; N - mg\cos\theta \hspace{3cm} mR \ddot{\theta} \; = \; F - mg\sin\theta$ where $N$ and $F$ are the normal reaction and friction force acting on the bead. Since friction is limiting while motion occurs, $F = \mu N$ . Thus we obtain the equation $\begin{aligned} mR\ddot{\theta} + mg\sin\theta & = \; \mu\big(mR\dot{\theta}^2 + mg\cos\theta\big) \\ R\big(\ddot{\theta} - \mu\dot{\theta}^2\big) & = \; g\big(\mu\cos\theta - \sin\theta\big) \\ R\frac{d}{d\theta}\big[\tfrac12\dot{\theta}^2 e^{-2\mu \theta}\big] & = \; g(\mu\cos\theta - \sin\theta)e^{-2\mu\theta} \\ & = \; \frac{g}{1 + 4\mu^2}\frac{d}{d\theta}\left[\left((1 - 2\mu^2)\cos\theta + 3\mu\sin\theta\right)e^{-2\mu\theta}\right] \end{aligned}$ and hence, given the initial conditions $\dot{\theta}^2 \; = \; F_\mu(\theta) \; = \; \frac{2g}{R(1 + 4\mu^2)}\left[(1 - 2\mu^2)\cos\theta + 3\mu\sin\theta\right] + \left(\frac{v_0^2}{R^2} - \frac{6\mu g}{R(1 + 4\mu^2)}\right)e^{-\mu(\pi-2\theta)}$ Thus the time taken to reach the bottom is $T(\mu) \; = \; \int_0^{\frac12\pi} \frac{d\theta}{\sqrt{F_\mu(\theta)}}$ Numerical calculations give us that $t_1 \; = \; T(0) \; = \; 0.623225 \hspace{2cm} t_2 \; = \; T(0.4) \; = \; 0.849012$ and hence $\frac{t_2}{t_1} \; = \; \boxed{1.36229}$

This problem requires the numerical integration of

$\ddot{\alpha}= -\mu(\dot{\alpha}^2+\frac{g}{R} \sin \alpha)+\frac{g}{R} \cos \alpha$

with the initial condition of $\dot{\alpha}=v_0/R$ at $\alpha=0$ , and look for the time when $\alpha=\pi/2$ .