Rope Falling

1. Apr 17, 2005

shraps

I seem to be having trouble conceptualizing this question that I was asked to solve.

In the problem, I have a rope of length L held so that the bottom of the rope barely touches the surface below it. At this point, the rope is released and falls to the surface. The question involves the mass of the rope that is still in the air at any given time.

After analyzing the situation, I have decided that obviously the normal force applied to the rope by the surface after parts of the rope touch the surface effects the way in which the rope falls. It is for this reason that the rope does not fall in a straight line, it is effected by the rope already on the surface.

However, I have found no equations which help me to relate the mass of the rope above the surface with the height of the top of the rope, original mass, original length, and force of gravity.

I am fairly sure that this requires the use of Newton's Second Law for Continuously Changing Masses, but I still can't quite grasp how to utilize this, and my book does an extremely poor job of explaning it.

Thanks for the help guys,
shraps

2. Apr 17, 2005

Pengwuino

Not sure why some of the rope being on the ground already has anthing to do with it. All of the rope should afll at the same rate and in the same way if air resistance is negligable.

3. Apr 17, 2005

shraps

Well, I could be wrong, but using my reasoning abilities, I think something must be effecting this.

This is the first part of a larger problem, and it asks for my answer for the mass m of the rope above the surface in terms of M, L, y (height of top of rope above surface), and g. If nothing else were effecting the rope wouldn't the mass just be m = y * M/L, because the rope falls in a straight line?

Thanks,
shraps

4. Apr 17, 2005

OlderDan

The rope on the surface has no effect on the rope above the surface. Every part of the rope experiences the same acceleration due to gravity. If the rope were to suddenly break into millions of disconnected pieces at the moment it is released, those pieces would fall exactly the way they do when they are connected. You are correct about the mass of the rope still above the surface. Now you need to think about how y depends on the time.

There may be something coming that has to do with changing mass, but in this problem the force on the rope taken as a whole is proportional to the remaining mass above the surface. The fact that the total force on the rope is getting smaller as the rope falls means that the acceleration of the center of mass of the rope is less than g, but this is an unnecessarily complicated way of looking at this problem. Perhaps the problem wants you to come up with an answer a simpler way to prove that there is an equivalent way when considering a changing mass.

5. Apr 17, 2005

shraps

You guys are correct, thank you for the help. After understanding that the acceleration is in fact constant, the rest of the problem goes relatively easy.

Thanks,
shraps