Spyrals!

We can represent mass as a funtion of distance covered $x$ , $m(x) = M\left(1-\dfrac{x}{L}\right)$

Now, since $R\ll L$ , we can neglect Upward velocity and Potential Energy,

$\displaystyle\begin{aligned}\frac{1}{2}Mv_0^2 + \frac{1}{2}MR^2\left(\dfrac{v_0}{R}\right)^2 &=\frac{1}{2}m(x)v^2(x) + \frac{1}{2}m(x)r^2\left(\dfrac{v(x)}{r}\right)^2\\ \Rightarrow v^2(x)&=\dfrac{mv_0^2}{m(x)}\\\Rightarrow v(x) &= \dfrac{v_0}{\sqrt{1-\dfrac{x}{L}}}\\ \Rightarrow \dfrac{1}{v_0}\int\limits_0^L \sqrt{1-\dfrac{x}{L}}\mathrm{d}x&=\int\mathrm{d}t = T\\ \Rightarrow T&=\dfrac{L}{v_0}\int\limits_0^1\sqrt{1-u}\mathrm{d}u\\&=\dfrac{2L}{3v_0}\end{aligned}$

$\displaystyle\Large\therefore\boxed{T = \dfrac{2L}{3v_0}}\approx 9.523809$