What is the tension of the rope?

  • #1
Ebi
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Homework Statement:
Both ends of a rope with the mass of ρ kg/m and length of L are attached to a horizontal surface (see photo attached). At t=0, one end of the rope detaches from the surface and starts falling vertically. When this end falls the height X, what is the tension on the other end of the rope?
Relevant Equations:
F=ma
E=U+K
and any equation in classical physics.
I have attached two different attempts to solve this problem. They both look correct to me but they give two different answers! Which one is correct, which one is wrong and why?
 

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Answers and Replies

  • #2
haruspex
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Homework Statement:: Both ends of a rope with the mass of ρ kg/m and length of L are attached to a horizontal surface (see photo attached). At t=0, one end of the rope detaches from the surface and starts falling vertically. When this end falls the height X, what is the tension on the other end of the rope?
Relevant Equations:: F=ma
E=U+K
and any equation in classical physics.

I have attached two different attempts to solve this problem. They both look correct to me but they give two different answers! Which one is correct, which one is wrong and why?
The second attempt is correct. I am having difficulty following the first attempt, but I note that the answer you have there has a factor x, making the initial tension zero. There are also some unbounded terms in the working as x approaches L.
 
  • #3
Ebi
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The second attempt is correct. I am having difficulty following the first attempt, but I note that the answer you have there has a factor x, making the initial tension zero. There are also some unbounded terms in the working as x approaches L.
Hello,
Thanks for your response. Please note we are after tension at the fixed end which is referred to as R in both soloutions.
Thanks
 
  • #4
Delta2
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What exactly is ##T_1## and how you set it equal to that expression?
Also isn't there supposed to be some interaction between the moving part and the non moving part which will affect the velocity of the moving part which will no longer do free fall?
 
  • #5
haruspex
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Hello,
Thanks for your response. Please note we are after tension at the fixed end which is referred to as R in both soloutions.
Thanks
Ok, but in solution 1 there is still a divisor of L-x, which cannot be right.
 
  • #6
wrobel
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I do not understand why the total energy must conserve.
The equations of variable mass systems must be employed
 
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  • #7
Delta2
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Using my own approach I get the result $$R=\rho g(\frac{L}{2}+3x)$$

(No I don't have a typo in the final result).

I ll post my approach when I find the time.
 
  • #8
ergospherical
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Assume that the distance ##w## between the attachments to the horizontal surface satisfies ##w \ll L##. Write ##u = X/L##, then the distance ##\xi## of the centre of mass below the surface is given by ##\xi = \dfrac{L}{4} \left( 1 + 2u - u^2 \right)##. The free end of the rope is in free fall so ##\ddot{X} = g##, ##\dot{X} = gt## and ##X = \frac{1}{2}gt^2## hence ##\ddot{\xi} = \dfrac{g}{2} \left( 1 - 3 u \right)##.

The equation of motion satisfied by ##\xi## is ##\rho L g - T = \rho L \ddot{\xi}## which can be rearranged for ##T = \dfrac{\rho g L}{2} \left( 1 + 3 u \right)##.
 
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  • #9
Delta2
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Assume that the distance ##w## between the attachments to the horizontal surface satisfies ##w \ll L##. Write ##u = X/L##, then the distance ##\xi## of the centre of mass below the surface is given by ##\xi = \dfrac{L}{4} \left( 1 + 2u - u^2 \right)##.
This formula for ##\xi## looks like it "drops from the sky". Can you elaborate more on how you find it?
 
  • #10
ergospherical
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This formula for ##\xi## looks like it "drops from the sky". Can you elaborate more on how you find it?
From the definition ##\boldsymbol{R} = \dfrac{\sum_a m_a \mathbf{r}_a}{\sum_a m_a} ## write\begin{align*}

\xi = \dfrac{ X \left(\dfrac{X}{2} \right) + (L-X) \left( X + \dfrac{L-X}{4} \right)}{L} &= \dfrac{L}{4} + \dfrac{X}{2} - \dfrac{X^2}{4L} \\ \\

&= \dfrac{L}{4} (1 + 2u - u^2)

\end{align*}
 
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  • #12
TSny
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@Ebi
Your first method based on conservation of mechanical energy has some support in the literature. The falling part of the chain accelerates greater than ##g##.

Here are a few references:

Falling Chains by Wong and Yasui 2006 (This paper was also published in the American Journal of Physics). Your result for the support force, ##R##, in your first method agrees with equation (34) of this paper where ##\hat x \equiv x/L##.

The dynamics of a falling chain:1 by Calkin and March 1989 (Login required to see full paper)

On the paradox of the free falling folded chain by Schagerl, et al. 1997 (Login required to see full paper)

A falling, folded string by K T McDonald 2017 (updated 2020)

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##.
 
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  • #13
TSny
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The references given in the previous post are all for chains rather than ropes. Ropes are mentioned in the introduction of this paper. (Login required to see the full paper.) Apparently, ropes are harder to analyze than chains!
 
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  • #14
wrobel
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Here are a few references:

Falling Chains by Wong and Yasui 2006 (This paper was also published in the American Journal of Physics). Your result for the support force, , in your first method agrees with equation (34) of this paper where .

The one-dimensional fall of a folded chain with one end suspended from a rigid support and achain falling from a resting heap on a table is studied. Because their Lagrangians contain no explicittime dependence, the falling chains are conservative systems.

Funny argument little bit. Why do that guys consider this system to be Lagrangian? Assume that the energy is conserved and prove that it is consered.


Actually at the folding point the links of the chain lose their velocities instantly from some nonzero values to the zero value. It is like a nonelastic collision.

If the energy would be conserved then the chain would have never come at the rest.

If we want to consider the Hamiltonian statement of the problem that is ok but then we have to use the true equations of chain's motion in ##\mathbb{R}^2## but they are damned crazy complicated.
 
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  • #15
TSny
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The one-dimensional fall of a folded chain with one end suspended from a rigid support and achain falling from a resting heap on a table is studied. Because their Lagrangians contain no explicittime dependence, the falling chains are conservative systems.

Funny argument little bit. Why do that guys consider this system to be Lagrangian? Assume that the energy is conserved and prove that it is consered.

Yes. They are essentially assuming conservation of mechanical energy when writing the Lagrangian. What caught my attention was the reference to experiments that show that the side of the chain that is falling has a downward acceleration greater than ##g## and that the supporting force was measured to approach approximately 25 times the weight of the chain as ##x## approached ##L##. On the other hand, the model that assumes that the acceleration of the falling portion equals ##g## predicts that the maximum support force is only twice the weight of the chain.

Actually at the folding point the links of the chain lose their velocities instantly from some nonzero values to the zero value. It is like a nonelastic collision.
That's what I would have thought. But, it seems that experiments show that the chain acts like it's approximately conserving mechanical energy for most of the fall.

If the energy would be conserved then the chain would have never come at the rest.
Yes, the real chain will eventually transform the loss of gravitational potential energy into internal energy (heat). So, the Lagrangian (energy-conserving) model must break down. For most of the fall, experiments seem to show that mechanical energy is largely conserved. So, the breakdown of the energy-conserving model apparently occurs during the last part of the fall (or perhaps after the fall is over and the chain swings in some fashion and eventually comes to rest through dissipation of energy). The energy-conserving solution predicts that the velocity of the falling tip of the string and the support force both approach infinity as ##x## approaches ##L##. Clearly, this is not correct. But the fact that the experiment measured the support force to approach a whopping 25 times the weight of the chain seems to show that the Lagrangian (energy conserving) model does a better job than the model where the acceleration is assumed to equal ##g##.
 
  • #16
hutchphd
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A falling, folded string by K T McDonald 2017 (updated 2020)

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.
This is a very nice treatment and the eqns (4), (5) seem perfectly obvious (after about 20 minutes of internal debate in your head). I am obviously surprised. Nice
 
  • #17
haruspex
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Yes, the real chain will eventually transform the loss of gravitational potential energy into internal energy (heat). So, the Lagrangian (energy-conserving) model must break down.
For a chain with low friction joints, I can understand that as each link reaches the bottom it is yanked around, converting its linear velocity to rotational. Its moment of inertia results in that exerting a downward force on the next link, so that rotational energy is, at least in part, translated into linear KE of the descending chain.
When the last part of the chain reaches the bottom, there is nothing to arrest its rotation, so the accumulated KE goes into whipping it around and partly back up the other side.

Another system could be rods or slats that slide against each other but do not rotate. In this system, it is clear that energy is not conserved.

A rope could be like the chain, though perhaps a slightly rusty one, or more like the slats. I feel it will depend on the detailed properties of the rope.
 
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  • #18
wrobel
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Clearly, this is not correct. But the fact that the experiment measured the support force to approach a whopping 25 times the weight of the chain seems to show that the Lagrangian (energy conserving) model does a better job than the model where the acceleration is assumed to equal g.
sure it is just incorrect to apply the standard Newton law to a variable mass system

What caught my attention was the reference to experiments that show that the side of the chain that is falling has a downward acceleration greater than g
it is the acceleration of the center of mass who can not exceed g
 
  • #19
wrobel
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The energy-conserving solution predicts that the velocity of the falling tip of the string and the support force both approach infinity as approaches .
for the velocity the variable mass equation gives the same ,
I did not check for the force
 
  • #20
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What exactly is ##T_1## and how you set it equal to that expression?
Also isn't there supposed to be some interaction between the moving part and the non moving part which will affect the velocity of the moving part which will no longer do free fall?

What exactly is ##T_1## and how you set it equal to that expression?
Also isn't there supposed to be some interaction between the moving part and the non moving part which will affect the velocity of the moving part which will no longer do free fall?
Sorry for the late response. The equation you askef is drived in the reference attached.
Thanks
 

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  • #21
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Sorry for the late response. The equation you askef is drived in the reference attached.
Thanks
 

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  • #22
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So what exactly is the correct answer to this problem? My answer agrees with that of @Delta2 in post #7. The rate of change of momentum for the free fall part entering the hung part is just ##\rho v^2##, where v is the free fall velocity. This is the added tension on the hung part.
 
  • #23
Delta2
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So what exactly is the correct answer to this problem? My answer agrees with that of @Delta2 in post #7. The rate of change of momentum for the free fall part entering the hung part is just ##\rho v^2##, where v is the free fall velocity. This is the added tension on the hung part.
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.

Under the basic assumption that the falling half of the rope does free fall, my analysis seems to agree completely with you (I also found that that rate of change of momentum is ##\rho v^2## and not ##\frac{1}{2}\rho v^2## as the OP suggests. However I can't find a mistake in the approach of @ergospherical which agrees with the result of the OP method 2.
 
  • #25
haruspex
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So what exactly is the correct answer to this problem? My answer agrees with that of @Delta2 in post #7. The rate of change of momentum for the free fall part entering the hung part is just ##\rho v^2##, where v is the free fall velocity. This is the added tension on the hung part.
Neither is exactly correct. It all hinges on how much the KE is conserved.

The natural assumption for most of us seems to be that it is not conserved at all, that as each small element reaches the bottom it is simply and linearly yanked to a halt, with no consequence for the next element. But as described in various places in this thread and some links, it need not be that way. Treating each element as a bar of some rotational inertia, it is whipped around, exerting a downward force on the next element.
The question then is what happens in practice? Links in this thread describe experiments indicating the KE loss is relatively small, but it will depend on the material used.

No doubt the "chain fountain" works similarly, though whether that requires an actual chain, not just a string, I do not know.

This leaves me wondering what answer was expected by the problem setter.
 
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