0 Solver Challenge (Level 6 Mechanics)

If $\theta$ is the angle the radial line to the particle makes with the vertical, the tangential equation of motion is $mR\ddot{\theta} \; = \; -F - mg\sin\theta \; = \; -\mu N - mg\sin\theta$ while the radial equation of motion is $mR\dot{\theta}^2 \; = \; N - mg\cos\theta$ Putting these equations together, we see that $\begin{aligned} mR\ddot{\theta} + mg\sin\theta + \mu(mR\dot{\theta}^2 + mg\cos\theta) & = \; 0 \\ R(\ddot{\theta} + \mu\dot{\theta}^2) + g(\sin\theta + \mu\cos\theta) & = \; 0 \\ \frac{d}{d\theta}\big[\tfrac12R\dot{\theta}^2e^{2\mu\theta}\big] + g(\sin\theta + \mu\cos\theta)e^{2\mu\theta} & = \; 0 \\ \tfrac12R\dot{\theta}^2 e^{2\mu\theta} + \frac{g}{4\mu^2+1}\big[(2\mu^2-1)\cos\theta + 3\mu\sin\theta\big]e^{2\mu\theta} & = \; \frac{v_0^2}{2R} + \frac{g(2\mu^2-1)}{4\mu^2+1} \end{aligned}$ and hence $\begin{aligned} R\dot{\theta}^2 & = \; \left(\frac{v_0^2}{R} + \frac{2g(2\mu^2-1)}{4\mu^2+1}\right)e^{-2\mu\theta} - \frac{2g}{4\mu^2+1}\big[(2\mu^2-1)\cos\theta + 3\mu\sin\theta\big] \\ \tfrac{1}{m}N & = \; \left(\frac{v_0^2}{R} + \frac{2g(2\mu^2-1)}{4\mu^2+1}\right)e^{-2\mu\theta} + \frac{3g}{4\mu^2+1}\big[\cos\theta - 2\mu\sin\theta\big] \end{aligned}$ The least velocity that will enable the particle to reach the top is the one for which $N = 0$ for the first time at $\theta = \pi$ . For $N$ to be $0$ when $\theta = \pi$ we require that $\begin{aligned} \left(\frac{v_0^2}{R} + \frac{2g(2\mu^2-1)}{4\mu^2+1}\right)e^{-2\mu\pi} & = \; \frac{3g}{4\mu^2+1} \\ \frac{v_0^2}{R} & = \; g\left(e^{2\mu\pi} - \frac{2(2\mu^2-1)}{4\mu^2+1}(e^{2\mu\pi}+1)\right) \\ v_0 & = \; \sqrt{Rg\left(e^{2\mu\pi} - \frac{2(2\mu^2-1)}{4\mu^2+1}(e^{2\mu\pi}+1)\right)} \end{aligned}$ Notice that, for this value of $v_0$ , we have $\begin{aligned} N & = \; \frac{3mg}{4\mu^2+1}\left[e^{2\mu(\pi-\theta)} + \cos\theta - 2\mu\sin\theta\right] \\ \frac{dN}{d\theta} & = \; -\frac{3mg}{4\mu^2+1}\left[2\mu\big(e^{2\mu(\pi-\theta)} + \cos\theta\big) + \sin\theta\right] \end{aligned}$ and hence it is clear that $\frac{dN}{d\theta} \le 0$ for all $0 < \theta < \pi$ , so we deduce that $N \ge 0$ for all $0 \le \theta \le \pi$ . Thus the smallest value of $v_0$ for which the particle gets to the top is $v_0 \; = \; \sqrt{Rg\left(e^{2\mu\pi} - \frac{2(2\mu^2-1)}{4\mu^2+1}(e^{2\mu\pi}+1)\right)}$ which makes the answer $1+2+2+2+2+4+1+2+1+2 = \boxed{19}$ .