Sliding on a Rough Quarter-Pipe - On Ice!

At time $t$ , let $X$ be the distance travelled by the quarter-pipe, and let $\theta$ be the angle $PCA$ , where $C$ is the centre of the circle of the quarter-pipe surface. The the position vector of the particle (relative to $O$ , the initial position of the bottom-left corner of the quarter-pipe) is $\mathbf{r} \; = \; \left(\begin{array}{c} -X + a - a\cos\theta \\ a - a\sin\theta \end{array}\right)$ While it is sliding, the particle is subject to a normal reaction force $mR$ and a friction force $\mu mR$ , as well as the force of gravity. Thus

$-m\ddot{X}\left(\begin{array}{c} 1 \\ 0 \end{array}\right) + ma\ddot{\theta}\left(\begin{array}{c} \sin\theta \\ -\cos\theta \end{array}\right) + ma\dot{\theta}^2\left(\begin{array}{c}\cos\theta \\ \sin\theta \end{array}\right) \; = \; mR\left(\begin{array}{c} \cos\theta \\ \sin\theta \end{array}\right) + \mu mR\left(\begin{array}{c}-\sin\theta \\ \cos\theta \end{array}\right) + mg\left(\begin{array}{c}0 \\ -1 \end{array}\right)$ and hence $\begin{aligned} -\ddot{X}\cos\theta + a\dot{\theta}^2 & = \; R - g\sin\theta \\ -\ddot{X}\sin\theta + a\ddot{\theta} & = \; -\mu R + g\cos\theta \end{aligned}$ The only external forces acting on the quarter-pipe and the particle are gravity and the normal reaction from the ice, all of which are vertical. Thus the horizontal component of the center of mass of the particle/quarter-pipe system stays constant throughout the motion. Hence $-MX + m(-X + a - a\cos\theta)$ is constant, and so $(M+m)\dot{X} \; = \; ma\dot\theta\sin\theta$ Eliminating both $R$ and $X$ from these equations, we obtain the formidable differential equation $\big(2M + m + m\cos2\theta - \mu m\sin2\theta\big)\ddot{\theta} + \big(\mu(2M+m) - m\sin2\theta - \mu m \cos2\theta\big)\dot{\theta}^2 \; = \; \frac{2(M+m)g}{a}(\cos\theta - \mu\sin\theta)$ Remarkably, we can find the first integral of this differential equation. To find the required integrating factor, we need to perform the integral $\int \frac{2\big(\mu(2M+m) - m\sin2\theta - \mu m \cos2\theta\big)}{2M + m + m\cos2\theta - \mu m\sin2\theta}\,d\theta \;=\; \ln\big(2M+m+m\cos2\theta-\mu m\sin2\theta\big) + \int \frac{2\mu(2M+m)}{2M+m+m\cos2\theta-\mu m\sin2\theta}\,d\theta$ This can be handled using a $t = \tan\theta$ substitution. Omitting the details, if we define the function $F(\theta) \; = \; \mathrm{exp}\,\left[\frac{2\mu(2M+m)}{\sqrt{4M^2 + 4Mm - \mu^2m^2}}\tan^{-1}\left(\frac{2M\tan\theta - \mu m}{\sqrt{4M^2 + 4Mm - \mu^2m^2}}\right)\right]$ then $\frac{d}{d\theta}\Big[\left[2M+m + m\cos2\theta - \mu m \sin2\theta\right]F(\theta)\Big] \; = \; 2\big[\mu(2M+m) - m\sin2\theta - \mu m\cos2\theta\big]F(\theta)$ so the differential equation becomes $\begin{aligned} \frac{d}{d\theta}\Big[\tfrac12\big(2M+m + m\cos2\theta - \mu m \sin2\theta\big)F(\theta)\dot{\theta}^2\Big] & = \; \frac{2(M+m)g}{a}(\cos\theta - g\sin\theta)F(\theta) \\ \big(2M+m + m\cos2\theta - \mu m \sin2\theta)F(\theta)\dot{\theta}^2 & = \; \frac{4(M+m)g}{a} G(\theta) \end{aligned}$ where $G(\theta) \; = \; \int_0^\theta (\cos\varphi - \mu \sin\varphi)F(\varphi)\,d\varphi$ Now $\big|m\cos2\theta - \mu m \sin2\theta\big| \; \le \; m\sqrt{1 + \mu^2} \; < \; m\sqrt{2} \; < \; 2M+m$ and hence $2M+m + m\cos2\theta - \mu m\sin2\theta > 0$ for all $0 < \theta < \tfrac12\pi$ . We can also show that $R \; = \; \frac{2M(a\dot{\theta}^2 + g\sin\theta)}{2M + m + m\cos2\theta - \mu m\sin2\theta} \; > \; 0$

Thus the particle never leaves the surface of the quarter-pipe before reaching $B$ , and so either the particle reaches the bottom $B$ of the quarter-pipe, where $\theta = \tfrac12\pi$ , or else the particle comes to rest at some point with a value of $\theta$ less than $\tfrac12\pi$ . The function $G(\theta)$ is increasing for $0 < \theta < \cot^{-1}\mu$ , and decreasing for $\cot^{-1}\mu < \theta < \tfrac12\pi$ . Thus the particle will have come to rest before reaching $B$ if $G(\tfrac12\pi) < 0$ , and will pass $B$ and move onto the ice if $G(\tfrac12\pi) > 0$ . Thus the critical value $\mu_0$ is the one for which $G(\tfrac12\pi) = 0$ . When $M = 10$ , $m = 1$ , plotting $G(\tfrac12\pi)$ against $\mu$ shows that there is unique value $\mu_0$ for which $G(\tfrac12\pi) = 0$ . Numerical calculations show that $\mu_0 = 0.5971498673...$ , and hence $\lfloor 10^6 \mu_0 \rfloor \; = \; \boxed{597149}$ .