Slipping shoelace

We consider the potential energy $U$ of the lace. In doing so, we divide the shoelace into the sub-masses $m - m_x$ and $m_x = \frac{x}{l} m$ , which respectively lie on the table top or overhang at the edge. While the mass $m - m_x$ is at the height $h_0$ (height of the table), the height of the mass $m_x$ can be determined by the position of its center of gravity at $h_x = h_0 - \frac{1}{2} x$ . The potential is thus $\begin{aligned} U(x) &= (m - m_x) g h_0 + m_x g h_x \\ &= m g \left[ \frac{l - x}{l} h_0 + \frac{x}{l} \left(h_0 - \frac{x}{2} \right) \right] \\ &= - \frac{m g}{2 l} x^2 + \text{const} \end{aligned}$ The force on the shoelace results $m \ddot x = F = - U'(x) = \frac{m g}{l} x$ This differential equation can be solved with an expotential function $\begin{aligned} x(t) &= A \cdot e^{+ \lambda t} + B \cdot e^{- \lambda t} \\ \Rightarrow \quad \ddot x(t) &= \lambda^2 A \cdot e^{+ \lambda t} + \lambda^2 B \cdot e^{- \lambda t} = \lambda^2 \cdot x(t) \end{aligned}$ with the exponent $\lambda = \sqrt{g/l}$ and two coefficients $A$ and $B$ , that are fixed by the initial conditions: $\begin{aligned} x(t = 0) &= A + B \stackrel{!}{=} x_0 \\ \dot x(t= 0) &= \lambda A - \lambda B \stackrel{!}{=} 0 \end{aligned}$ Therefore, $A = B = \frac{1}{2}x_0$ and we get the result $x(t) = \frac{x_0}{2} \left[e^{+ \lambda t} + e^{- \lambda t} \right] = x_0 \cosh(\lambda t)$ Finally, the equation $x(T) = l$ can be solved by the time $T$ : $T = \sqrt{\frac{l}{g}} \cdot \text{arccosh}\left(\frac{l}{x_0}\right) \approx 0.8 \,\text{s}$