- #1

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## Homework Statement

I've pretty much solved it, but i'm unsure of my final integration

A uniform chain of length L and density /rho(kg/m) is initially stationary on a horizontal, frictionless table, with part of the chain (length y

_{o}) hanging over the edge. How much time passes before the entire chain has left the table?

## Homework Equations

arccosh(x) = log(sqrt(x

^{2}-1)+x)

## The Attempt at a Solution

I don't think i need to put all the work i've done.

my integral

[tex]\int \sqrt{(y^{2}-y^{2}_{o})g/l}^{-1/2}[/tex]

the answer i get is

[tex]arccosh( \sqrt{l/g}*y/y_{o})[/tex]

or

[tex]\sqrt{l/g}*log(2(\sqrt{y^{2}-y^{2}_{o}}+y))[/tex]

yet on wolfram and other websites they say that l/g should not be square rooted. Yet i don't see why.

thanks