Problem source

One end of a thin uniform inextensible, but perfectly flexible, string of length $l$ and uniform mass per unit length is held at a point on a smooth table a distance $d(< l)$ away from a small vertical hole in the surface of the table. The string passes through the hole so that a length $l-d$ of the string hangs vertically. The string is released from rest. Assuming that the height of the table is greater than $l$, find the time taken for the end of the string to reach the top of the hole.

Solution source

Consider some point once the string is moving, there will be $x$ above the table and $l - x$ hanging in the air.

For the hanging string we must have $(l-x)mg - T = -(l-x)m\ddot{x}$. For the string on the table we must have that $T = -xm \ddot{x}$. Eliminating T, we have $(l-x)g = -l \ddot{x}$

Solving the differential equation, we must have $x = A \cosh \sqrt \frac{g}{l}t+B \sinh\sqrt \frac{g}{l}t+l$,

Since $x(0) = d, \dot{x}(0) = 0 \Rightarrow B = 0, A = (-d)$. Therefore $x = l-(l-d) \cosh \sqrt \frac{g}{l} t  \Rightarrow t =\sqrt \frac{l}{g} \cosh^{-1} \l \frac{l-x}{l-d} \r$ and we go over the edge when $x = 0$, ie $\sqrt \frac{l}{g} \cosh^{-1} \l \frac{l}{l-d} \r$