What a chain! 3

Let us denote the position of the end of the chain by $y$ . The mass of the moving segment of the chain is $m=\rho y$ , where $\rho$ is the mass per unit length. The velocity of the chain is $v=\dot y$ and the acceleration is $\ddot y$ . According to Newton's law

$mg= \frac{ d(mv)}{dt}= m \dot v+ v \dot m$

Since $\dot m= \rho \dot y$ , we can rewrite this as

$g= \ddot y + \frac{\dot y}{m} \dot m= \ddot y+ \frac{\dot y^2}{y}$

This is a second order ordinary nonlinear differential equation. It has a simple solution if the initial conditions are $y(0)=0$ and $\dot y(0)=0$ . The trial function that satisfies the initial conditions is $y=\frac{1}{2} at^2$ , where $a$ is a constant acceleration. With that, $\dot y=at$ . Inserting this to the equation we get

$g= a+\frac{2 (at)^2}{at^2}=3a$

and $a=g/3$ .

If the chain has a non-zero length hanging down when it is released this solution is not valid, but it is likely that the solution will always approach $a=g/3$ as the chain gets longer with time.

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The only external force on the chain is weight of the released chain

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Consider the system at time t....when dx length is about to join the moving part of chain .........Mass of the released portion is m and length L...

Impulse = d (mv)

mg = $\frac{m dv}{dt}$ + $\frac{v dm}{dt}$

mg = ma + $\frac{mv}{L}$ $\frac{dx}{dt}$

mg = ma + $\frac{mv^2}{L}$

$v^2 = (g - a) L$

comparing it with $v^2 = 2aL$

g - a = 2 a

$a = {\frac{g}{3}}$

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The forces acting on the falling part of chain are its weight and the tension which is equal to thrust force $mv^2/x$
So, $ma = mg - mv^2/x$ . Or, $vdv/dt=g-v^2/x$ Let $u=v^2$ Then $du/dt +2u/x =2g$ Multiplying by $x^2$ ,and putting $z=ux^2. , dz/dx = 2gx^2$ So $v^2x^2 = z = 2gx^3/3$ , considering ​the initial conditions $v=\sqrt(2gx/3)$