More Paraboloid Surface Dynamics

Working in cylindrical polar coordinates, the position vector of the particle can be written as $\mathbf{r} \; = \; \left(\begin{array}{c} 2\sqrt{az}\cos\theta \\ 2\sqrt{az}\sin\theta \\ z \end{array}\right)$ Differentiating with respect to $t$ we obtain the kinetic energy of the particle $T \; = \; \tfrac12m\left(1 + \tfrac{a}{z}\right)\dot{z}^2 + 2maz\dot{\theta}^2$ Since the potential energy of the particle is $V = mgz$ we obtain the Lagrangian $\mathcal{L} \; = \; \tfrac12m\left(1 + \tfrac{a}{z}\right)\dot{z}^2 + 2maz\dot{\theta}^2 - mgz$ Initially we know that $r=2a$ , $z = a$ , $\dot{r} = \dot{z} = 0$ and $\dot{\theta} = \omega$ . Since $\mathcal{L}$ is independent of $\theta$ , we deduce that $\frac{\partial \mathcal{L}}{\partial \dot{\theta}}$ is constant, and we deduce that $z\dot{\theta} = a\omega$ . The second equation of motion is $\begin{aligned} 0 & = \; \displaystyle \frac{d}{dt}\Big(\frac{\partial \mathcal{L}}{\partial \dot{z}}\Big) - \frac{\partial \mathcal{L}}{\partial z} \; = \; \frac{d}{dt}\Big[m\left(1 + \tfrac{a}{z}\right)\dot{z}\Big] + \frac{ma}{2z^2}\dot{z}^2 - 2ma\dot{\theta}^2 + mg \\ & = \; \displaystyle m\left(1 + \tfrac{a}{z}\right)\ddot{z} - \frac{ma}{2z^2}\dot{z}^2 - \frac{2ma^3\omega^2}{z^2} + mg \; = \; \frac{d}{dz}\Big[\tfrac12m\left(1 + \tfrac{a}{z}\right)\dot{z}^2 + \frac{mg\alpha a^2}{z} + mgz \Big] \end{aligned}$ so that $\tfrac12m\left(1 + \tfrac{a}{z}\right)\dot{z}^2 + \frac{mg\alpha a^2}{z}+ mgz \; = \; mga(\alpha + 1)$ and hence $\dot{z}^2 \; = \; \frac{2g(\alpha a - z)(z - a)}{a + z}$ Thus the particle oscillates between height $z=a$ and $z=\alpha a$ with period $T \; = \; 2\int_a^{\alpha a}\sqrt{\frac{a+z}{2g(\alpha a - z)(z - a)}}\,dz$ Using the substitution $z = a(\cos^2\theta + \alpha\sin^2\theta)$ , we obtain $\begin{aligned} T & = \; \displaystyle 2\int_0^{\frac12\pi} \sqrt{\frac{a + a(\cos^2\theta + \alpha \sin^2\theta)}{2g}}\,2d\theta \; = \; 2\sqrt{\tfrac{2a}{g}}\int_0^{\frac12\pi}\sqrt{1 + \cos^2\theta + \alpha\sin^2\theta}\,d\theta \\[2ex] & = \; \displaystyle 2\sqrt{\tfrac{2a}{g}} \int_0^{\frac12\pi} \sqrt{2 + (a-1)\sin^2\theta}\,d\theta \; = \; 4\sqrt{\tfrac{a}{g}}\mathsf{E}\left(-\tfrac{\alpha-1}{2}\right) \end{aligned}$ where $\mathsf{E}$ is the complete elliptic integral.

During a half period the particle moves through an angle $\Theta(\alpha)$ , where $\begin{aligned} \Theta(\alpha) & = \; \int_a^{\alpha a}\frac{d\theta}{dz}\,dz \; = \; \int_a^{\alpha a} \frac{\dot{\theta}}{\dot{z}}\,dz \; = \; a\omega \int_a^{\alpha a}\frac{dz}{z \dot{z}} \; = \; a\omega \int_a^{\alpha a} \sqrt{\frac{a+z}{2g(\alpha a - z)(z - a)}}\,\frac{dz}{z} \\ & = \; \omega\sqrt{\tfrac{2a}{g}}\int_0^{\frac12\pi} \frac{\sqrt{1 + \cos^2\theta + \alpha \sin^2\theta}}{\cos^2\theta + \alpha\sin^2\theta}\,d\theta \; = \; \sqrt{\alpha}\int_0^{\frac12\pi} \frac{\sqrt{2 + (\alpha-1)\sin^2\theta}}{1 + (\alpha-1)\sin^2\theta}\,d\theta \\ & = \; \sqrt{\tfrac{\alpha}{2}}\left[\mathsf{K}\left(-\tfrac{\alpha-1}{2}\right) + \mathsf{\Pi}\left(-(\alpha-1),-\tfrac{\alpha-1}{2}\right)\right] \end{aligned}$ where $\mathsf{K}$ is the complete elliptic integral of the first kind, while $\mathsf{\Pi}$ is the complete elliptic integral of the third kind.

We want to find the value of $\alpha > 1$ for which $\Theta(\alpha) = \pi$ . Now $\Theta(\alpha) \; > \; \sqrt{\alpha}\int_0^{\frac12\pi} \frac{d\theta}{\sqrt{1 + (\alpha-1)\sin^2\theta}} \; = \; \mathsf{K}\left(\tfrac{\alpha-1}{\alpha}\right)$ and since $\mathsf{K}$ is an increasing function on $(0,1)$ , with $\mathsf{K}\left(\tfrac{39}{40}\right) > \pi$ , we deduce that $\Theta > \pi$ for all $\alpha > 40$ . Inspection of the graph of $\Theta$ for $1 < \alpha < 40$ shows that there is a unique value $\alpha_0$ for which $\Theta(\alpha_0) = \pi$ . Numerical calculation gives us $\alpha_0 = 11.9389732...$ , and hence $\lfloor 10^5 \alpha_0 \rfloor = \boxed{1193897}$ .