Paraboloid Surface Dynamics

Parametrizing the position of the particle with $z$ and $\theta$ , the kinetic energy of the particle is $T \; =\; \tfrac12m\left(1 + \tfrac{1}{4z}\right)\dot{z}^2 + \tfrac12mz\dot{\theta}^2$ and the potential energy is $V \; = \; mgz \; =\; 10mz$ so the Lagrangian is $\mathcal{L} \; = \; \tfrac12m\left(1 + \tfrac{1}{4z}\right)\dot{z}^2 + \tfrac12mz\dot{\theta}^2 - 10mz$ Since $\mathcal{L}$ is independent of $\theta$ , we deduce that $\frac{\partial\mathcal{L}}{\partial \dot{\theta}}$ is constant, and hence that $z\dot{\theta} \; = \; 10$ This is conservation of angular momentum. The other equation of motion is $\begin{aligned} \frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{z}}\right) & = \; \frac{\partial \mathcal{L}}{\partial z} \\ \frac{d}{dt}\left[\left(1 + \tfrac{1}{4z}\right)\dot{z}\right] & = \; -\tfrac{1}{8z^2}\dot{z}^2 + \tfrac12\dot{\theta}^2 - 10 \\ \left(1 + \tfrac{1}{4z}\right)\ddot{z} - \tfrac{1}{8z^2}\dot{z}^2 & = \; \tfrac{50}{z^2} - 10 \\ \frac{d}{dz}\left[\tfrac12\left(1 + \tfrac{1}{4z}\right)\dot{z}^2\right] & = \; \tfrac{50}{z^2} - 10 \end{aligned}$ and hence $(1 + 4z)\dot{z}^2 \; = \; 500z - 400 - 80z^2 \; = \; 80(z - \alpha)(\beta - z)$ where $\alpha \; = \; \tfrac18(25 -\sqrt{305}) \hspace{1cm} \beta \; = \; \tfrac{1}{8}(25 + \sqrt{305})$ From this it is clear that the particle's height oscillates between $\alpha$ and $\beta$ , with half-period $T \; = \; \int_\alpha^\beta \sqrt{\frac{1 + 4z}{80(z-\alpha)(\beta-z)}}\,dz$ The substitution $z = \alpha\cos^2\phi + \beta\sin^2\phi$ is helpful here, giving $T \; =\; \frac{1}{2\sqrt{10}}\int_0^{\frac12\pi} \sqrt{27 - \sqrt{305}\cos2\phi}\,d\phi$ This can be evaluated explicitly in terms of complete elliptic integrals, but we shall simply note that $T = 1.25268$ .

It is also worth noting that, during each half-period, the particle moves through an angle $\Theta$ , where $\Theta \; =\; \int_\alpha^\beta \frac{10}{z \dot{z}}\,dz \; = \; \tfrac12\sqrt{5}\int_\alpha^\beta \sqrt{\frac{1 + 4z}{(z-\alpha)(\beta-z)}}\,\frac{dz}{z} \; = \; 4\sqrt{10}\int_0^{\frac12\pi} \frac{\sqrt{27 - \sqrt{305}\cos2\phi}}{25 - \sqrt{305}\cos2\phi}\,d\phi \;=\; 4.77483$

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To begin with, the particle needs to rise from its initial height of $z=1$ to $z=\beta$ . This takes time $T_0 \; = \; \int_1^\beta \sqrt{\frac{1 + 4z}{80(z-\alpha)(\beta-z)}}\,dz \; = \; \frac{1}{2\sqrt{10}}\int_u^{\frac12\pi} \sqrt{27 - \sqrt{305}\cos2\phi}\,d\phi$ where $u = \sin^{-1}\sqrt{\frac{1-\alpha}{\beta-\alpha}}$ , and hence $T_0 = 1.19581$ .

At time $t = T_0 + 21T = 27.5022$ , the particle has completed a total of $10\tfrac12$ full oscillations, and so at that time is at a height $z=\alpha$ . We need to find the height (during the next half-oscillation) that it has reached by time $t=28$ . In other words, we need to find the height $z$ such that $\int_\alpha^z \sqrt{\frac{1 + 4z}{80(z-\alpha)(\beta-z)}}\,dz \; =\; \frac{1}{2\sqrt{10}}\int_0^v \sqrt{27 - \sqrt{305}\cos2\phi}\,d\phi \; = \; 0.49781$ where $v = \sin^{-1}\sqrt{\frac{z-\alpha}{\beta-\alpha}}$ , and we deduce that the height at time $t=28$ is $\boxed{3.1770}$ .

Solution based on the Lagrangian approach. This solution does not take a deep dive into the salient features of this dynamical system, even though there are several insights to explore. Let the mass of the particle be unity. The kinetic energy of the system is:

$\mathcal{T} = \frac{1}{2} m \left(\dot{x}^2 + \dot{y}^2 +\dot{z}^2 \right)$ $\implies \mathcal{T} = 2\dot{r}^2 r^2 + \frac{\dot{r}^2}{2} + \frac{\dot{\theta}^2r^2}{2}$

Potential energy:

$\mathcal{V} = mgz =10r^2$

System Lagrangian: $\mathcal{L} = \mathcal{T}-\mathcal{V}$

Lagrange's equations read:

$\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{r}}\right) - \frac{\partial \mathcal{L}}{\partial r}=0$ $\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{\theta}}\right) - \frac{\partial \mathcal{L}}{\partial \theta}=0$

Finding the equations and numerically solving yields the required answer. The trajectory of the particle up to $t=100$ :