## **Problem**

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)} $.