Chain Challenge!

Let the length of chain below table at any arbitrary time 't' be $x$ . Net force acting on the hanging part of chain at that position is $\frac { M }{ L } (x)g$ where M is the mass of the chain and L is the length of the chain.

So $\frac { d(mv) }{ dt } =\frac { M }{ L } (x)g$

or $mv\frac { d(mv) }{ dt } =\frac { M }{ L } (x)g\times mv$

or $mv\frac { d(mv) }{ dt } =g{ \left( \frac { M }{ L } \right) }^{ 2 }{ x }^{ 2 }\frac { dx }{ dt\ }$

Let $s=mv$

So $\frac { { s }^{ 2 } }{ 2 } ={ \left( \frac { M }{ L } \right) }^{ 2 }\frac { { x }^{ 3 } }{ 3 } +c$

At t=0 v=0 and let $x={ x }_{ 0 }$

So $c=-{ \left( \frac { M }{ L } \right) }^{ 2 }\frac { { x }_{ 0 }^{ 3 } }{ 3 }$

Putting this value

$\frac { { v }^{ 2 } }{ 2 } =\frac { g }{ 3 } \left( x-\frac { { x }_{ 0 }^{ 3 } }{ { { x }^{ 2 } } } \right)$

or ${ v }^{ 2 }=\frac { 2g }{ 3 } \left( x-\frac { { x }_{ 0 }^{ 3 } }{ { { x }^{ 2 } } } \right)$

or $\int _{ { x }_{ 0 } }^{ x }{ { \sqrt { \frac { 2g }{ 3 } \left( x-\frac { { x }_{ 0 }^{ 3 } }{ { { x }^{ 2 } } } \right) } }^{ -1 } } dx=\int _{ 0 }^{ t }{ dt }$

or $t=\sqrt { \frac { 3 }{ 2g } } \int _{ { x }_{ 0 } }^{ x }{ \frac { xdx }{ \sqrt { { x }^{ 3 }-{ x }_{ 0 }^{ 3 } } } }$

So $\sqrt { \frac { 3 }{ 2g } } \int _{ 1/3 }^{ 1 }{ \frac { xdx }{ \sqrt { { x }^{ 3 }-{ \left( \frac { 1 }{ 3 } \right) }^{ 3 } } } } =t$

The value of integral comes out to be 1.130.

Please watch this to find the answer of the integration.

On putting this value the result comes out to be $t\simeq 0.44sec$

Please use wolframalpha for calculating the integral because that integral has no representation in elementary functions and one can only compute its numeric value for the given limits.