Roller coaster loop top speed For a roller coaster loop, if it were perfectly circular, we would have a minimum speed of $v_{min} = \sqrt{gR}$ at the top of the loop where $g=9.8 m/s^2$ and $R$ is the radius of the 'circle'. However, most roller coaster loops are actually not circular but more elliptical. I've been looking for ways to calculate the min. speed at the top for an elliptical loop, but so far I haven't been able to. How could I go about that? 
 A: As it turns out, you actually can use that same formula $ v_{min} = \sqrt{gR} $. However $R$ is the radius of curvature at the top of the loop, which is simply equal to the radius in the case of a circle. See here for more information on finding the curvature of an ellipse.
In general, the curvature of a plane curve given by $ (x(t),y(t)) $ can be found using the formula $ \kappa = \frac{|\dot x \ddot y - \dot y \ddot x|}{(\dot x^2 + \dot y^2)^\frac 3 2} $ (or one of many other formulae) where $ R = \frac 1 \kappa $
A: The ellipse equation in polar coordinate is:
$\begin{bmatrix}
              x \\
              y \\
            \end{bmatrix}=\left[ \begin {array}{c} r\cos \left( \varphi  \right) 
\\ r\sin \left( \varphi  \right) \end {array}
 \right] 
$
$r={\frac {ba}{\sqrt {{a}^{2} \left( \sin \left( \varphi  \right) 
 \right) ^{2}+{b}^{2} \left( \cos \left( \varphi  \right)  \right) ^{2
}}}}
$
where  $2a$ is the major axis and  $2b$ the minor axis  
Kinetic energy  
$T=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)$
with 
$\begin{bmatrix}
              \dot{x} \\
              \dot{y} \\
            \end{bmatrix}
=\left[ \begin {array}{c} -{\frac {b{a}^{3}\sin \left( \varphi 
 \right) \varphi p}{ \left( {b}^{2} \left( \cos \left( \varphi 
 \right)  \right) ^{2}+{a}^{2}-{a}^{2} \left( \cos \left( \varphi 
 \right)  \right) ^{2} \right) ^{3/2}}}\\ {\frac {{b
}^{3}a\cos \left( \varphi  \right) \varphi p}{ \left( {b}^{2} \left( 
\cos \left( \varphi  \right)  \right) ^{2}+{a}^{2}-{a}^{2} \left( \cos
 \left( \varphi  \right)  \right) ^{2} \right) ^{3/2}}}\end {array}
 \right] 
$
and $\varphi p=\dot{\varphi}$
Potential Enegry
$V=m\,g\,y$
we solve the equation $T=V$ for $\dot{\varphi}$
and get for $v_{max}=r\,\dot{\varphi}|_{(\varphi=\frac{\pi}{2})}$
$v_{max}=\sqrt{2\,b\,g} $
For a circle ($b=a=R$) with radius $R$  we get:
$v_{max}=\sqrt{2\,R\,g} $
Simulation result

