The acceleration of a falling rope. At the climbing wall this week, after pulling the rope from its anchor at the top of the wall and letting it fall freely, I wondered about how the rope accelerates.
I believe that the rope just falls at $g\,ms^{-2}$ (ignoring any air resistance). But I have a line of logic that leads to something else and I do not know where the flaw in my logic lies! Could you please help me to find it? 
Here's the setup:

The rope has total mass $M$ and full length $L$. Its mass is uniform along its length. 
One end of it is held to a height of $x_0<L$ and released at time $t=0$. The rope forms a pile on the ground of negligible height. I'm ignoring all resistive forces.
As time passes, the tip of the rope is at distance $x(t)$ from the ground. The mass of the portion of the rope still falling is $\frac{x(t)}{L}M=:m(t)$ and the centre of mass of the falling portion of rope is at $\frac{1}{2}x(t)$.
The force of gravity on the falling portion of rope is $-m(t)g$. This force will act at the rope's centre of mass. So, using Newton's Second Law, at the centre of mass:
$$\begin{align}
-m(t)g &= \frac{d}{dt}\Big(m(t)\frac{d}{dt}\big(\frac{1}{2}x(t)\big)\Big) \\ \implies -\frac{M}{L}x(t)g &= \frac{M}{2L}\frac{d}{dt}\Big(x(t)\dot{x}(t)\Big) \\
\implies -gx(t) &= \frac{1}{2}\big(\dot{x}(t)^2 + x(t)\ddot{x}(t)\big)
\end{align}$$
... and so I have the differential equation $2gx(t)+\dot{x}(t)^2 + x(t)\ddot{x}(t)=0$, that I do not know how to solve but which definitely does not lead to $\ddot{x}(t)=-g$.
Thanks for any help!
 A: You confuse the top of the rope and the center of mass. They have different equations of motions. If you talk about center of mass, there are two forces acting on the rope. Gravity is acting on the entire rope, and the reaction from the ground is acting only on the part that is not still in the air. So if you have two forces, there is no reason to say that the center of mass has acceleration $g$. 
You can solve the problem much easier. The piece of rope not on the ground is falling with acceleration $g$, including the top end of the rope. Then the position of that point is giving you the length of the rope above ground:
$$l(t)=x_0-\frac 12 gt^2$$
Then the center of mass is at $$y_{CM}(t)=\frac{\frac{Ml(t)}{L}\frac{l(t)}2+\frac{M(L-l(t))}{L}0}{M}=\frac{l^2(t)}{2L}$$
Here $Ml(t)/L$ is the mass of the rope in the air, and $M(L-l(t))/L$ is the mass of the rope already on the ground (height $0$).
A: Each bit of the rope falls under gravity.  There is no tension in the rope because it all accelerates the same way.  When a bit hits the floor it decelerates to $0$.  The center of mass of the rope changes in position on the rope, which is what makes your equations complicated.
