Rope falling in a hole

[geoffrey159]

Homework Statement

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.

Homework Equations

Momentum

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?

 

[mfb]

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.

 

[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.

 

[geoffrey159]

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

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 !

 

[rude man]

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 !

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