Rope in the hole

Let us denote the length of the rope hanging from the table by $x$ and the instantaneous velocity of the rope by $v(x)$ . Let the table be at reference height zero for computation of the gravitational potential energy. Then using conservation of energy we have $\frac{1}{2}(\lambda L)v^2(x)=\frac{1}{2}(\lambda g)x^2-\frac{1}{2}(\lambda g)(\delta l)^2$ i.e., $\frac{dx}{dt}=\sqrt{\frac{g}{L}}\sqrt{x^2-(\delta l)^2}$ Hence, $t_c=\sqrt{\frac{L}{g}}\int_{\delta l}^{L}\frac{dx}{\sqrt{x^2-(\delta l)^2}}=5.349 \hspace{3pt}\text{sec}$

Let $l$ denote the length of rope hanging through the hole. By drawing the free body diagram (or by applying the Lagrangian formulation), we get the following differential equation:

$\ddot{l} = \frac{g}{L} l$

The general solution of the differential equation is $l = c_{1} \exp\left (t\sqrt{\frac{g}{l} }\right) + c_{2} \exp\left(-t\sqrt{\frac{g}{l} } \right)$

We can differentiate both sides to get velocity.

$v = \sqrt{\frac{g}{l}} \left( c_{1} \exp\left (t\sqrt{\frac{g}{l} }\right) - c_{2} \exp\left(-t\sqrt{\frac{g}{l} } \right)\right)$

Note that at $t = 0$ , we have the initial conditions: $l = \delta l$ and $v = 0$ . Substituting these values in the above equations gives $c_{1} = c_{2} = \frac{1}{2} \delta l$ .

We get the particular solution: $l = \frac{\delta l}{2} \left(\exp\left (t\sqrt{\frac{g}{l} }\right) + \exp\left(-t\sqrt{\frac{g}{l} } \right)\right)$ .
This can also be written as

$l = \delta l \cosh\left(t\sqrt{\frac{g}{l}}\right)$

We make $t$ the subject of the formula:

$t = \sqrt{\frac{l}{g}} \cosh^{-1} \left( \frac{l}{\delta l}\right)$

When the end of the rope passes through the hole, we have $l = L = 10 \text{ m},$ $g = 9.81 \text{ m/s}^2,$ $\delta l = 0.1 \text{ m},$ $t = t_{c}$ . We substitute these values and obtain

$t_{c} = \sqrt{\frac{10}{9.81}} \times \cosh^{-1} \left( \frac{10}{0.1}\right)\approx \boxed{5.349} \ _{\square}$