Sliding Up an Elliptical Loop

The position, velocity and acceleration vectors of a particle touching the ellipse can be given by $\mathbf{r} \; = \; \binom{3\cos\theta}{2\sin\theta} \hspace{1cm} \mathbf{\dot{r}} \; = \; \binom{-3\sin\theta}{2\cos\theta}\dot{\theta} \hspace{1cm} \mathbf{\ddot{r}} \; = \; \binom{-3\sin\theta}{2\cos\theta}\ddot{\theta} - \binom{3\cos\theta}{2\sin\theta}\dot{\theta}^2$ The normal reaction between the particle and the ellipse will be of the form $\mathbf{R} \; = \; -m\lambda\binom{2\cos\theta}{3\sin\theta}$ where $\lambda > 0$ , and the equation of motion of the particle is $m\mathbf{\ddot{r}} \; = \; \mathbf{R} + \binom{0}{-mg} \hspace{3cm} (\star)$ Taking the scalar product of $(\star)$ with $\binom{-3\sin\theta}{2\cos\theta}$ , we obtain $\begin{aligned} (9\sin^2\theta + 4\cos^2\theta)\ddot{\theta} + 5\sin\theta\cos\theta \dot{\theta}^2 & = \; -2g\cos\theta \\ \frac{d}{d\theta}\left[\tfrac12(9\sin^2\theta + 4\cos^2\theta)\dot{\theta}^2 + 2g\sin\theta\right] & = \; 0 \\ \tfrac12(9\sin^2\theta + 4\cos^2\theta)\dot{\theta}^2 + 2g\sin\theta & = \; \tfrac12v^2 \\ \dot{\theta}^2 & = \; \frac{v^2 - 4g\sin\theta}{4 + 5\sin^2\theta} \end{aligned}$ Taking the scalar product of $(\star)$ with $\binom{2\cos\theta}{3\sin\theta}$ , we obtain that $\lambda \; = \; \frac{3(2\dot{\theta}^2 - g\sin\theta)}{4 + 5\sin^2\theta} \; = \; \frac{3}{(4 + 5\sin^2\theta)^2}\big[ 2v^2 - g\sin\theta(12 + 5\sin^2\theta)\big]$ and hence, to reach the top of the ellipse, we must have $\lambda \ge 0$ throughout the motion, and so $2v^2 \; \ge \; g\sin\theta(12 + 5\sin^2\theta) \hspace{1cm} 0 \le \theta \le \tfrac12\pi$ The RHS of this inequality is an increasing function of $\theta$ , so we want $2v^2 \ge 17g$ , and hence $v \ge \boxed{\sqrt{9.5g} = 9.219544457} \mathrm{m\,s}^{-1}$ .

$\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$ $\text{Taking derivative, }\dfrac{2x}{9}dx+\dfrac{2y}{4}dy=0$ $\dfrac{dy}{dx}=- \dfrac{4x}{9y} \cdots (\star)$

At the highest point, for the particle, $m \textbf{g}+ \textbf{N}_{\text{by loop}} =m \dfrac{d\textbf{v}}{dt}.$ For the particle to reach the top without losing contact with the loop with minimum initial velocity, at the highest point, the value of $\textbf{N}$ should be just negligibly greater than $0$ . So, $\dfrac{d\textbf{v}}{dt}=\textbf{g}.$ Here's how we can find acceleration at the point $(0,2)$ : $dV_x=v(\cos d\theta -1) \approx 0, dV_y =v \sin d\theta \approx v d \theta$ So $\dfrac{d\textbf{v}}{dt}= v \dfrac{d\theta}{dt} =\dfrac{-2v^2}{9}=-g. \text{ (Since } dx/dt=v)$ So $v^2=\dfrac{9g}{2}\implies \boxed{v_0^2=\dfrac{17g}{2}}.$