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Problem

Chain

A uniform chain of mass $ m $, and length $ L $ is held onto a horizontal frictionless table. The table has a large hole in which a segment of length $ D $ is hanging over the hole. The chain is then released from rest, and the chain begins to descend. How long will it take the chain to completely slide through the hole?

Hint

$ \int \frac{dx}{\sqrt{x^2 - a^2}} = \ln \left(x + \sqrt{x^2 -a^2} \right) + C $


Solution

Define $ \sigma $ as the constant linear mass density of the chain. Let $ x $ represent the variable length hanging over the hole.

$ {F}_{net} = ma $
$ \sigma x g = \sigma L \frac{dv}{dt} $

Since $ \frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v \frac{dv}{dx} $, the equation for force become the differential equation

$ v \frac{dv}{dx} = \frac{gx}{L} $.

Since the chain was initially at rest, and the initial length hanging over the hole is $ D $, the solution for velocity is

$ v = \sqrt{\frac{g}{L}} \sqrt{{x}^{2} - {D}^{2}} $.

Since $ v = \frac{dx}{dt} $, the above equation can be written as

$ \frac{dx}{dt} = \sqrt{\frac{g}{L}} \sqrt{{x}^{2} - {D}^{2}} $.

Integrate the above differential equation with the integrating bounds $ x = [D, L] $ and $ t = [0, T] $.

$ \int_{D}^{L} \frac{dx}{\sqrt{{x}^{2} - {D}^{2}}} = \int_{0}^{T} \sqrt{\frac{g}{L}} dt $.

The solution for time is

$ T = \sqrt{\frac{L}{g}} \ln{\left(\frac{L + \sqrt{{L}^{2} - {D}^{2}}}{D}\right)} $.
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