What is the tension of the rope?

In summary: I was very surprised to see that the mechanical energy would be (largely) conserved. I would have worked it similarly to your second method where it is assumed that the falling part of the chain is in free fall, ##a = g##.
  • #36
Ok then :) let's develop the variable mass version of the story.

Equation from #11 takes the form
$$-2\frac{d}{dt}(\dot y y)=gy.$$ After a change ##y^2=z## we have
$$\ddot z=-g\sqrt z,$$
and let ##\dot z=u(z)##
then
$$\frac{1}{2}\frac{d}{dz}u^2=-g\sqrt z.$$
Eventually it follows that
$$\dot y=-\frac{1}{y}\sqrt{-\frac{1}{3}g\Big(y^3-\Big(\frac{L}{2}\Big)^3\Big)}.$$
it is clear ##\dot y\to-\infty## as
##y\to 0+##
 
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  • #37
wrobel said:
it is clear y˙→−∞ as
If I understand this correctly it means that the speed of the falling part approaches infinity when all of the rope falls. Isn't that a sign that something is wrong with your equations?
 
  • #38
Delta2 said:
it means that the speed of the falling part approaches infinity when all of the rope falls. Isn't that a sign that something is wrong with your equations?
just look at the picture carefully: y is a length of the loop it surely decreases. The velocity of the chain's end is ##\dot x=-2\dot y##
asymptotically one has
$$\dot y\sim -\frac{1}{\sqrt{c_1-c_2t}},\quad c_1,c_2>0$$
as ##t\to c_1/c_2-##
 
  • #39
Delta2 said:
If I understand this correctly it means that the speed of the falling part approaches infinity when all of the rope falls. Isn't that a sign that something is wrong with your equations?
It's an inevitable consequence of taking work to be conserved, since the KE therefore increases while becoming concentrated in a vanishing mass.
In the real world, there are losses. In the presence of an atmosphere, drag will increase dramatically, removing the infinity.
It would be interesting to conduct the experiment in a vacuum.
 
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  • #40
Sorry @wrobel, it seems after all that your equations are consistent with conservation of energy. @haruspex made me see that.
 
  • #41
Delta2 said:
it seems after all that your equations are consistent with conservation of energy.
I do not believe that. In my equations the total energy must decrease
You can check it by yourself all the formulas are brought
 
  • #42
There is another way of approaching all this that can be used, but it ain't easy. The rope can be modeled as an inextensible string, starting as a catenary at equilibrium and then cutting one end and modeling the subsequent motion as a function of time and position along the string. The initial placement of the two ends of the catenary can be relatively close together compared to the length of the string. I've formulated the equations, but they aren't pretty.
 
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  • #43
It seems to me for a "perfect" string, absent dissipative forces, this system will oscillate and part of the energy will be kinetic almost always. Or am I saying something obvious.
 
  • #44
I have to admit that my solution is completely incorrect. I am sorry
 
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  • #45
Delta2 said:
It seems that the correct answer to this problem involves breaking our basic assumption that the right half part of the rope does free fall. I haven't understand completely the papers provided by @TSny at posts #12,#13 but it seems that they prove that the acceleration of the falling part is time varying and larger than g
I agree with you that the falling part is not free falling. If tere is a non zero tension T at the bottom of the rope (where it bends between falling part and hung part), which I believe there is, then this force also pulling the falling part own, which means the acceleration is more than g.
But how do you say (I also found that that rate of change of momentum is ρv^2 and not 1/2ρv^2?
 
  • #46
Delta2 said:
It seems that the correct answer to this problem involves breaking our basic assumption that the right half part of the rope does free fall. I haven't understand completely the papers provided by @TSny at posts #12,#13 but it seems that they prove that the acceleration of the falling part is time varying and larger than g.
I agree with you on the falling part has acceleration greater than g. So using free falling equation for velocity of free falling object (V^2=2gx) in approach 2, is not correct. My reason for that is since there is a non-zero tension at the bottom of the rope (where falling part comes to rest at hung part), so the falling part is under a tension, therefore its acceleration is greater than g.
However, can you please explain why you believe "The rate of change of momentum for the free fall part entering the hung part is just ρv^2 ρv2 and not 1/2ρv^2)
 
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  • #47
Ebi Rogha said:
However, can you please explain why you believe "The rate of change of momentum for the free fall part entering the hung part is just ρv^2 ρv2 and not 1/2ρv^2)
Well it is $$\frac{\rho dx (v-0)}{dt}=\rho v^2$$
 
  • #48
haruspex said:
It's an inevitable consequence of taking work to be conserved, since the KE therefore increases while becoming concentrated in a vanishing mass.
This.

haruspex said:
It would be interesting to conduct the experiment in a vacuum.
I predict that the string will get a bit warmer.
 
  • #49
pbuk said:
I predict that the string will get a bit warmer.
All the lost GPE ends up as heat anyway. The question is the mechanism.
But my suggestion for an experiment in a vacuum was to find what speed and tension it could get up to.
 
  • #50
haruspex said:
But my suggestion for an experiment in a vacuum was to find what speed and tension it could get up to.
My guess is that losses due to non-ideal flexibility and extensibility of the rope are much greater than to air resistance and so it wouldn't make a lot of difference. Or are you thinking that a vacuum would make it possible to use something really thin and flexible like an aramid or Vectran fibre? OK, this would be interesting :cool:
 
  • #51
pbuk said:
My guess is that losses due to non-ideal flexibility and extensibility of the rope are much greater than to air resistance and so it wouldn't make a lot of difference.

Again one needs to be precise here. Neither lack of flexibility nor extensibility is necessarily a "loss" mechanism. This same problem with a good spring ( or a mass on a spring ) will self-evidently produce a system that will oscillate forever. This system will do the same.
The details of the loss mechanism depend upon which other degrees off freedom are coupled...for me that is not particularly interesting
 
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  • #52
hutchphd said:
Again one needs to be precise here. Neither lack of flexibility nor extensibility is necessarily a "loss" mechanism. This same problem with a good spring ( or a mass on a spring ) will self-evidently produce a system that will oscillate forever. This system will do the same.
The details of the loss mechanism depend upon which other degrees off freedom are coupled...for me that is not particularly interesting
I totally agree. I am formulating a catenary version of this model in which the rope is extensible, but has no bending rigidity (negligible moment of inertia). Anyone interested in participating and discussing such a model (which is non-dissipative)?
 
  • #53
hutchphd said:
Again one needs to be precise here. Neither lack of flexibility nor extensibility is necessarily a "loss" mechanism.
OK, I should have said:

My guess is that the practical limitation in peak kinetic energy due to non-ideal flexibility and extensibility of the rope is much greater than due to air resistance and so it wouldn't make a lot of difference.​

hutchphd said:
This same problem with a good spring ( or a mass on a spring ) will self-evidently produce a system that will oscillate forever. This system will do the same.
With an ideal (inextensible) rope what is there to oscillate?
 
  • #54
Where do I find an inextensible rope?
Seriously in the limit of unobtainable materials there will be an infinite tension at the bottom and the tip of the rope will snap back up to the original position and the cycle begins again. In the real world this is of course fabulously unlikely to happen. Absent other "frictional" degrees of freedom this thing will just keep wiggling.
If the rope is a perfect spring it is easy to see that this oscillation is correct.
 
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  • #55
pbuk said:
My guess is that losses due to non-ideal flexibility and extensibility of the rope are much greater than to air resistance and so it wouldn't make a lot of difference. Or are you thinking that a vacuum would make it possible to use something really thin and flexible like an aramid or Vectran fibre? OK, this would be interesting :cool:
The energy-conserving equation says the initial GPE ends up concentrated in a vanishingly small element of rope.
Air resistance rises with speed, quadratically even. That makes it very effective at capping the speed. Non-ideal flexibility should be speed-independent, I would think, so only absorbs a certain amount of energy.
 

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