The position vector, velocity and acceleration of the particle are $\mathbf{r} = \binom{y^4}{-y} \hspace{1cm} \dot{\mathbf{r}} = \dot{y}\binom{4y^3}{-1} \hspace{1cm} \ddot{\mathbf{r}} = \ddot{y}\binom{4y^3}{-1} + 12\dot{y}^2\binom{y^2}{0}$ and so the equation of motion of the particle (while slipping occurs) is $m\ddot{y}\binom{4y^3}{-1} + 12m\dot{y}^2\binom{y^2}{0} \; = \; mR\binom{1}{4y^3} - \mu mR \binom{4y^3}{-1} + mg\binom{0}{-1}$ and hence $\begin{aligned} 12y^2\dot{y}^2 & = \; R(1 + 16y^6) - 4gy^3 \\ (1 + 16y^6)\ddot{y} + 48y^5\dot{y}^2 & = \; g - \mu R(1 + 16y^6) \end{aligned}$ Eliminating $R$ gives $\begin{aligned} (1 + 16y^6)\ddot{y} + (48y^5 + 12\mu y^2)\dot{y}^2 & = \; g(1 -4\mu y^3) \\ \frac{d}{dy}\left[\tfrac12(1 + 16y^6)e^{2\mu \tan^{-1}(4y^3)} \dot{y}^2\right] & = \; g(1 -4\mu y^3)e^{2\mu \tan^{-1}(4y^3)} \\ \tfrac12(1 + 16y^6)e^{2\mu \tan^{-1}(4y^3)}\dot{y}^2 &= \; gF(y) \end{aligned}$ where $F(y) \; = \; \int_0^y(1 - 4\mu u^3)e^{2\mu \tan^{-1}(4u^3)}\,du$ Thus the particle continues to move until it has dropped a depth $h$ , where $F(h) =0$ , and it takes time $T \; = \; \int_0^h \sqrt{\frac{1 + 16y^6}{2gF(y)}}e^{\mu \tan^{-1}(4y^3)}\,dy$ to do so.

The rest is just numerical slog. With $\mu = 0.064$ we obtain $h = 2.459409878$ and, with $g=10$ , we obtain $T = \boxed{13.4868}$ .