What is the tension of the rope?

• Ebi
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##.

Ebi

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?

Attachments

• Problem.GIF
20.1 KB · Views: 171
• attempt 1.jpg
36.9 KB · Views: 157
• attempt 2.png
36.1 KB · Views: 164
Delta2
Ebi said:
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.

haruspex said:
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

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?

Ebi said:
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.

Ebi
I do not understand why the total energy must conserve.
The equations of variable mass systems must be employed

Ebi and Delta2
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.

Ebi
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)##.

Last edited:
Ebi Rogha and Ebi
ergospherical said:
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?

Ebi
Delta2 said:
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*}

Ebi Rogha and Ebi
Something like that I guess

Last edited:
@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)

https://www.researchgate.net/publication/226367296_On_the_Paradox_of_the_Free_Falling_Folded_Chain 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##.

Last edited:
Ebi Rogha and Delta2
The references given in the previous post are all for chains rather than ropes. Ropes are mentioned in the introduction of https://www.researchgate.net/publication/230932068_Dynamics_of_ropes_and_chains_I_The_fall_of_the_folded_chain. (Login required to see the full paper.) Apparently, ropes are harder to analyze than chains!

Last edited:
TSny said:
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.

Last edited:
wrobel said:
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##.

TSny said:
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

TSny
TSny said:
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.

Lnewqban, Ebi Rogha and TSny
TSny said:
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

TSny said:
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

TSny said:
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

hutchphd
Delta2 said:
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?

Delta2 said:
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

Attachments

• IMG_20210626_202752.jpg
60 KB · Views: 87
Delta2
Ebi Rogha said:
Sorry for the late response. The equation you askef is drived in the reference attached.
Thanks

Attachments

• IMG_20210626_202847.jpg
93 KB · Views: 99
Delta2
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.

Delta2
Chestermiller said:
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.

Ebi Rogha
Delta2
Chestermiller said:
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.

Ebi Rogha and Chestermiller
haruspex said:
Neither is exactly correct. It all hinges on how much the KE is conserved.
I think one needs to be a bit more definitive here. The idealized solution for a perfectly flexible massive rope is exactly soluble and energy is conserved. See reference. Of course such systems are never perfect descriptors of the real world, but they are idealized with purpose.

Chestermiller
haruspex said:
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.
Good answer. Of course, it is always possible to have the same mass per unit length of rope in the limit where the moment of inertia of the cross section is zero. I guess in that case, the tension on the falling portion of the rope would be zero, and there would be a sudden change in tension at the reversal.

hutchphd said:
The idealized solution for a perfectly flexible massive rope is exactly soluble and energy is conserved.
If any specific idealizations are required to arrive at the intended answer then they should be stated.
In the present case, perfect flexibility doesn't seem to be enough. It needs to have a nonzero flexural modulus too.
Chestermiller said:
in the limit where the moment of inertia of the cross section is zero
Need to be careful not to take the limit too soon. The correct approach is to establish (an approximation to) the general result and then take the limit.

haruspex said:
Need to be careful not to take the limit too soon. The correct approach is to establish (an approximation to) the general result and then take the limit.
Sorry to disagree, but I see nothing wrong with solving this for a idealized rope having zero flexural rigidity and zero rotational inertia.

Chestermiller said:
Sorry to disagree, but I see nothing wrong with solving this for a idealized rope having zero flexural rigidity and zero rotational inertia.
A solution for an idealization is only valid if it is the limit behaviour as the parameters approach the idealized values. In post #27, you did not mention zero flexural modulus, only MoI.
My interpretation of your post is to take the rope as composed of progressively shorter rigid links. In the limit, each has zero MoI. But maybe if we consider the tension on the falling section before taking the limit we might find that does not disappear. Progressively shorter links means more of them for the same length of rope.
Now, I'm not saying the effect won't disappear, just that one does need to be careful.

haruspex said:
A solution for an idealization is only valid if it is the limit behaviour as the parameters approach the idealized values. In post #27, you did not mention zero flexural modulus, only MoI.
My interpretation of your post is to take the rope as composed of progressively shorter rigid links. In the limit, each has zero MoI. But maybe if we consider the tension on the falling section before taking the limit we might find that does not disappear. Progressively shorter links means more of them for the same length of rope.
Now, I'm not saying the effect won't disappear, just that one does need to be careful.
I wasn't assuming that the rope is composed of rigid links at all. I was assuming that the rope is continuous and has zero bending rigidity and zero rotational moment of inertia. It's like how we solve a catenary.

Chestermiller said:
I wasn't assuming that the rope is composed of rigid links at all. I was assuming that the rope is continuous and has zero bending rigidity and zero rotational moment of inertia.
Ah, that it inherently has no rotational moment of inertia, not merely none in the limit of thickness and/or element length.
Seems to violate some basic physical principle. Even if I represent it as rigid links with the mass of each concentrated at its centre, it still has MoI about its endpoints, and that is enough to cause the yank down.

After writing my previous post, it dawned on me that I have overlooked another model: links connected by side-to-side springs (don't know if there is technical term for these). That might be accurate for a rope and could be more significant than MoI.

haruspex said:
Ah, that it inherently has no rotational moment of inertia, not merely none in the limit of thickness and/or element length.
Seems to violate some basic physical principle. Even if I represent it as rigid links with the mass of each concentrated at its centre, it still has MoI about its endpoints, and that is enough to cause the yank down.

After writing my previous post, it dawned on me that I have overlooked another model: links connected by side-to-side springs (don't know if there is technical term for these). That might be accurate for a rope and could be more significant than MoI.
The model you are thinking of is balls connected by springs. Now, replace the springs by inextensible massless strings.

Chestermiller said:
The model you are thinking of is balls connected by springs. Now, replace the springs by inextensible massless strings.
Yes, that makes it like the 'sliding slats' model I mentioned in post #17. But your version is more realistic.

Can you not put a very small massive pulley on the bottom and let its mass go to zero at the end of the calculation? Maybe I'll take a look tomorrow...not now!

Delta2