A chain AB of length $l$ is located in a smooth horizontal tube so that its fraction of length $h$ h

Question

A chain AB of length $l$ is located in a smooth horizontal tube so that its fraction of length $h$ hangs freely and touches the surface of the table with its end B. At a certain moment the end A of the chain is set free. With what velocity will this end of the chain slip out of the tube?

Professor PC Thomas Classes · Answer

Let $y$ be the length of the chain piled on the table. The moving mass is $m = \lambda(l-y)$, driven by the constant weight of the hanging part $F = \lambda h g$. Tension at the table is zero. By Newton's second law for a system losing mass at zero relative velocity: $m \frac{dv}{dt} = F \implies \lambda(l-y)v\frac{dv}{dy} = \lambda h g$. Integrating $\int_0^v v dv = \int_0^{l-h} \frac{hg}{l-y} dy$ gives $v^2/2 = hg \ln(l/h)$. Thus $v = \sqrt{2gh \ln(l/h)}$.