# Rope falling in a hole

1. Dec 24, 2014

### geoffrey159

1. The problem statement, all variables and given/known data
A rope of mass M and length l lies on a friction less table, with a short portion, $l_0$ hanging through a hole. Initially the rope is at rest.

a. Find a general solution for x(t), the length of rope through the hole.

(Ans: $x=Ae^{\gamma t}+Be^{-\gamma t}$, where $\gamma^2=\frac{g}{l}$)

b. Evaluate the constants A and B so that the initial conditions are satisfied.

2. Relevant equations
Momentum

3. The attempt at a solution

I don't find this problem hard but I need an explanation.

The solution given solves the general equation $\ddot x = \gamma ^ 2 x$. If we multiply by rope mass on the right and on the left, we get

$\frac{M}{l}g x = M \ddot x$

Since the left part of this equation is equal to net vertical external force, right part is equal to $\frac{dP}{dt}$ in vertical direction. Then the vertical momentum is $P(t) = M \dot x$

This is where I have a problem. To me we should have $P(t) = m(t) \dot x$ where $m(t) = \frac{M}{l} x(t)$ is the hanging mass. The mass $M - m(t)$ is not moving vertically, so how could it have vertical speed ?

What do you think?

2. Dec 24, 2014

### Staff: Mentor

The hole provides the force to change the direction of motion, there is no need to take this into account within the scope of this problem. You simply assume the 90°-turn works, and the force of gravity has to accelerate the whole rope with the same acceleration.

Just using the hanging mass does not work, then nothing could accelerate the part on the table.

3. Dec 24, 2014

### rude man

I think you're neglecting the buildup in horizontal momentum as well as vertical, whereas the force is as you say just Mxg/l.

4. Dec 25, 2014

### geoffrey159

Oh thank you! I have neglected the fact that

$\frac{d}{dt}( (M-m(t)) \dot x ) = 0$.

So

$M \ddot x = \frac{d}{dt}(m(t) \dot x)= M \gamma^2x$,

which solves my problem.

Thanks to both of you, and merry Christmas !!

5. Dec 25, 2014

### rude man

Meery Christams to you too! Take some time off from physics! :-)