1 problem found
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: 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 \(\sqrt \frac{l}{g} \cosh^{-1} \l \frac{l}{l-d} \r\)