A very Edgy Problem!

Conserving Energy,

$\begin{aligned}mgR(1-\cos\theta_c) &= \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\\ \text{Since it is performing pure rolling, }\omega &= \dfrac{v}{R}\\ \Rightarrow mgR(1-\cos\theta_c) &= \dfrac{2mv^2+mv^2}{4}\\&=\dfrac{3mv^2}{4}\\ \Rightarrow v^2 &= \dfrac43gR(1-\cos\theta_c)\cdots(1)\end{aligned}$

Now, for performing rotation,

$mg\cos\theta_c - \mathbb{N} = \dfrac{mv^2}{R}$

When the contact is lost, $\mathbb{N} = 0$

$\Rightarrow gR\cos\theta_c = v^2\cdots(2)$

$\begin{aligned} (1) &= (2)\\ \Rightarrow gR\cos\theta_c &= \frac{4}{3}gR(1-\cos\theta_c)\end{aligned} \\ \Large\Rightarrow \boxed{\cos\theta_c = \dfrac{4}{7}}\\$

$\displaystyle\begin{aligned}v_{cm} &= v\\ &=\sqrt{gR\cos\theta_c}\\&=\sqrt{\dfrac{4gR}{7}}\\&=\sqrt{\dfrac{4\times10\times0.7}{7}}\\&= 2\end{aligned}\\ \Large\boxed{v_{cm} = 2}\\$

$\begin{aligned}KE_{tran\text{ when leaving}} &= \frac{1}{2}mv_cm^2\\&=\frac{1}{2}mgR\cos\theta_c\\&=\frac{2}{4}mgR\cos\theta_c\\ KE_{rot} &= \dfrac{1}{4}\dfrac{mR^2}{2}\left(\dfrac{v}{R}\right)^2\\ &=\frac{1}{4}mgR\cos\theta_c\\ \text{When the axis}&\text{ of the cylinder is in horizontal line of the edge, }\Delta h = R\cos\theta_c\\ \text{Hence } KE_{tran} &= KE_{0_{trans}} + mg\Delta h\\&= \frac{2}{4}mgR\cos\theta_c + mgR\cos\theta_c\\&=\frac{6}{4}mgR\cos\theta_c\\ \text{Hence }Ratio &= 6 \end{aligned}\\ \Large\boxed{c = 6}$

$\Huge\therefore\boxed{7\cos\theta_c + v_{cm} + c = 4+2+6 = 12}$