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## 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?