Slippery Chain!

Let $x(t)$ be the displacement of the chain at time $t$ . Then at any time $t$ the external force on the chain is $F = \sigma (h +x) g$ , where $\sigma$ is mass per length of the chain and $g$ is acceleration due to gravity. This force $F$ is used to accelerate the whole chain $F = \sigma l \dfrac {d^2x}{dt^2}$ . Therefore, we have:

$\begin{aligned} \sigma (h +x) g & = \sigma l \frac {d^2x}{dt^2} \\ \frac {d^2x}{dt^2} & = \frac gl x + \frac {gh}l \end{aligned}$

The general solution for the differential equation is as follows:

$\begin{aligned} x(t) & = c_1 e^{\sqrt{\frac gl}t} + c_2 e^{-\sqrt{\frac gl}t} - h & \small \color{#3D99F6}{\text{Note that }x(0) = 0} \\ x(0) & = c_1 + c_2 - h = 0 \\ \implies c_1 + c_2 & = h \quad ...(*) & \small \color{#3D99F6}{\text{Also that } \frac {dx}{dt}\big|_{t=0} = 0} \\ \frac {dx}{dt}\big|_{t=0} & = c_1 - c_2 = 0 \\ \implies c_1 & = c_2 = \color{#3D99F6}{\frac h2} & \small \color{#3D99F6}{(*): \ \ c_1 + c_2 = h} \end{aligned}$

$\begin{aligned} \implies x(t) & = \frac h2 \left(e^{\sqrt{\frac gl}t} + e^{-\sqrt{\frac gl}t} - 2\right) \end{aligned}$

When the chain slips off the table at $t_1$ , $x(t_1) = l - h$ , then,

$\begin{aligned} \frac h2 \left(e^{\sqrt{\frac gl}t_1} + e^{-\sqrt{\frac gl}t_1} - 2\right) & = l - h \\ \frac h2 \left(e^{\sqrt{\frac gl}t_1} + e^{-\sqrt{\frac gl}t_1} \right) & = l \\ e^{\sqrt{\frac gl}t_1} + e^{-\sqrt{\frac gl}t_1} & = \frac {2l}h & \small \color{#3D99F6}{\text{Multiply throughout by }e^{\sqrt{\frac gl}t_1}} \\ \left(e^{\sqrt{\frac gl}t_1}\right)^2 - \frac {2l}h e^{\sqrt{\frac gl}t_1} + 1& = 0 & \small \color{#3D99F6}{\text{A quadratic equation of }e^{\sqrt{\frac gl}t_1}} \end{aligned}$

$\begin{aligned} \implies e^{\sqrt{\frac gl}t_1} & = \frac lh \pm \sqrt{\left(\frac lh \right)^2-1} \\ t_1 & = \boxed{\sqrt{\dfrac lg} \ln \left[ \dfrac lh + \sqrt{\left(\dfrac lh \right)^2-1} \right]} \end{aligned}$

$\ln \left[ \dfrac lh - \sqrt{\left(\dfrac lh \right)^2-1} \right] < 0$ and it leads to negative time, which is unacceptable.

The differential equation can be obtained by conservation of energy (in the absence of friction, both internal and external, this is fine). The kinetic energy of the chain is $\tfrac12m\dot{x}^2$ , while the potential energy of the chain is $-\tfrac{x}{\ell}mg \times \tfrac12x \; = \; -\tfrac{mgx^2}{2\ell}$ and hence $\tfrac12 m \dot{x}^2 - \tfrac{mg}{2\ell}x^2 \; = \; -\tfrac{mg}{2\ell}h^2$ leading to $\dot{x}^2 \; = \; \tfrac{g}{\ell}(x^2 - h^2)$ After this, solving the differential equation proceeds normally.