Differential equation for describing a falling chain I'm trying to solve a problem which involves finding the final velocity of a chain as it falls off a table.
A chain of uniform mass is on a table with negligible friction. Length $b$ hangs off the table and length $a$ is on the table. Find the velocity as the last link leaves the table.
$$F = mg = \frac{dP}{dt}$$
$$\frac{\partial{m}}{\partial{t}} = \dot{m}v$$
$$\frac{\partial{m}}{\partial{t}} = ma$$
Therefore
$$mg = \dot{m}v + ma.$$
I assume that if I can solve this equation then that'll lead to the solution of this problem. However, the equation involves a first and second order derivative. I've never solved one of these before.
What is the right way to go about completing the problem?
 A: You can just use conservation of energy.  The length $b$ has moved downward by $a$ at the time the last link falls.  The length $a$ has moved downward by $\frac a2$ at the time the last link falls.  If we assume the linear density is $1$ (a different value will divide out) we have lost $g(ab+\frac {a^2}2)$ in potential energy.  We have gained $\frac 12v^2(a+b)$ in kinetic energy.  These must be equal, so 
$$g\left(ab+\frac {a^2}2\right)=\frac 12(a+b)v^2\\
v=\sqrt{\frac{2g\left(ab+\frac {a^2}2\right)}{a+b}}$$
A: Define $y$ positive down, and zero at the top of the table. I think you need to make some more careful distinctions. In Newton's Second Law, $F=\dot{p},$ the $p$ on the RHS is relevant, in your case, to the entire chain. Let $M$ be the mass of the entire chain, and let $\ell=a+b$ be the total length of the chain. Then $p=M\dot{y}.$ It follows that $\dot{p}=M\ddot{y}.$ Let $m$ be the mass of the vertical section, corresponding to $y$. Then the force on the chain is $mg$. Let $\lambda=M/\ell$ be the linear mass density. Newton's Second Law then tells us that 
\begin{align*}
mg&=M\ddot{y} \\
\lambda yg&=M\ddot{y} \\
\frac{M}{\ell}yg&=M\ddot{y} \\
\ddot{y}&=\frac{g}{\ell}\,y.
\end{align*}
This DE has the solution 
$$y=Ae^{t\sqrt{g/\ell}}+Be^{-t\sqrt{g/\ell}}. $$
The initial conditions are $y(0)=b$ and $\dot{y}(0)=0.$ Solving for $A$ and $B$ yields
\begin{align*}
y(t)&=\frac{b}{2}\,e^{t\sqrt{g/\ell}}+\frac{b}{2}e^{-t\sqrt{g/\ell}} \\
\dot{y}(t)&=\frac{b}{2}\,\sqrt{\frac{g}{\ell}}\,e^{t\sqrt{g/\ell}}-\frac{b}{2}\sqrt{\frac{g}{\ell}}\,e^{-t\sqrt{g/\ell}}.
\end{align*}
We can simplify these expressions by using hyperbolic trig functions:
\begin{align*}
y(t)&=b\cosh\left(t\sqrt{g/\ell}\right) \\
\dot{y}(t)&=b\sqrt{\frac{g}{\ell}}\,\sinh\left(t\sqrt{g/\ell}\right)
\end{align*}
Now then, we must solve for $t$ when $y=\ell,$ then plug that into $\dot{y}$. Can you finish from here?
Alternatively, you could use the hyperbolic trig identity
$$\cosh^2(\theta)-\sinh^2(\theta)=1$$
to eliminate the hyperbolic trig functions entirely (removing the parameter $t$ in the process), and solve for $\dot{y}$ directly. This would be a bit more elegant.
