Exploring Newton's 2nd Law: Rope on a Frictionless Table

[Mr Davis 97]

Homework Statement

A rope of mass M and length $l$ lies on a frictionless table, with a short portion $l_0$, hanging through a hole. Initially the rope is at rest. Find the length of the rope through the hole as a function of time.

Homework Equations

The Attempt at a Solution

So I know what the solution is. First, you find the force of gravity on the small portion of the rope, which is $\displaystyle F = \frac{M}{l}x(t) g$. Then, the general equation of motion is $\displaystyle M \frac{dv}{dt} = \frac{M}{l} x(t) g$. I can easily solve this to get the general solution. My question is, why does Newton's 2nd law work in this scenario if the mass of the rope through the whole is constantly changing?

 

[Simon Bridge]

My question is, why does Newton's 2nd law work in this scenario if the mass of the rope through the whole is constantly changing?

Um. Because that's the law...

You mean as opposed to something like the rocket equation, where the maths is much harder?
Compare the two derivations and see... short answer: because there are two symmetrically changing masses that are linked together.
There are lots of ways that the mass can change that does not mess up the maths.

 

[haruspex]

Another way to look at it is to consider the tension acting on each part of the rope. Since the accelerations must be the same, the tensions must be proportional to the rope lengths.