U-shaped length of chain falling when one end is released

[NTesla]

Homework Statement

A chain of mass per unit length is hung from ceiling by both it's ends. At time t=0, one of it's end is allowed to fall freely. Find the force exerted at each moment of time, by the stationary portion of the chain to stop the elements of the falling portion of the chain, when the element of the falling portion of the chain comes to stop at the bottom of the stationary portion of the chain.

Relevant Equations

Force = $\frac {dP}{dt}$

In case of a straight chain falling on a surface, the force required to stop an element of length $\lambda dx$ of the chain is = $\frac{\lambda dxv}{dt}$ = $\lambda v^{2}$.

In case of U chain see fig. below, When the purely vertical portion(DE in pic below) falls a distance $y$, the points D and B fall a distance $\frac{y}{2}$. This means that the portion DE falls with velocity $v$ and portion DB falls with velocity $\frac{v}{2}$. Also, different element of portion DB have different downward velocity.

I'm not sure what would be the velocity of the element of the chain, which is just before point B, and is about to come to stop at point B. Will it be $\frac {v}{2}$ or $v$ ? I'm sure the portion DB has downward velocity of $\frac {v}{2}$, but it's also true that the velocity of different elements of portion DB is different.

 

[haruspex]

Homework Statement: A chain of mass per unit length is hung from ceiling by both it's ends. At time t=0, one of it's end is allowed to fall freely. Find the force exerted at each moment of time, by the stationary portion of the chain to stop the elements of the falling portion of the chain, when the element of the falling portion of the chain comes to stop at the bottom of the stationary portion of the chain.
Relevant Equations: Force = $\frac {dP}{dt}$

In case of a straight chain falling on a surface, the force required to stop an element of length $\lambda dx$ of the chain is = $\frac{\lambda dxv}{dt}$ = $\lambda v^{2}$.

In case of U chain see fig. below, When the purely vertical portion(DE in pic below) falls a distance $y$, the points D and B fall a distance $\frac{y}{2}$. This means that the portion DE falls with velocity $v$ and portion DB falls with velocity $\frac{v}{2}$. Also, different element of portion DB have different downward velocity.

I'm not sure what would be the velocity of the element of the chain, which is just before point B, and is about to come to stop at point B. Will it be $\frac {v}{2}$ or $v$ ? I'm sure the portion DB has downward velocity of $\frac {v}{2}$, but it's also true that the velocity of different elements of portion DB is different.
View attachment 369243

Based purely on the text description, I think you are supposed to treat O and A as the same point. Or was the diagram provided? The diagram is certainly inaccurate since it would start as a catenary.

 

[NTesla]

Based purely on the text description, I think you are supposed to treat O and A as the same point. Or was the diagram provided? The diagram is certainly inaccurate since it would start as a catenary.

Actually, the diagram is drawn for the situation at time t > 0, when the freely falling part has fallen a distance y in time t. At time t = 0, the point O and E, both were exactly at the same point i.e. at point O. The separation distance between point O and A tends to zero, and in that case, these 2 points can be assumed to be same point. However, for the sake of clarity, these 2 points, i.e. points O and A are drawn in the pic with some distance in between them. The chain is to be assumed to be infinitely flexible, therefore, the radius of curvature at the U bend can be assumed to be = 0. And, the chain's mass per unit length is $\lambda$.

 

[haruspex]

Actually, the diagram is drawn for the situation at time t > 0, when the freely falling part has fallen a distance y in time t. At time t = 0, the point O and E, both were exactly at the same point i.e. at point O. The separation distance between point O and A tends to zero, and in that case, these 2 points can be assumed to be same point. However, for the sake of clarity, these 2 points, i.e. points O and A are drawn in the pic with some distance in between them. The chain is to be assumed to be infinitely flexible, therefore, the radius of curvature at the U bend can be assumed to be = 0. And, the chain's mass per unit length is $\lambda$.

I believe this problem belongs to class that includes the chain fountain effect. In practice, it seems that when the link at the bottom gets rotated by the force from the static side its inertia leads to a significant pull down on the descending side.
E.g. consider a rod lying EW on a smooth surface. Given a northward kick at one end, the other will move somewhat south.
As a result, equations based on conservation of mechanical energy produce a more accurate answer than one might expect.

 

[NTesla]

equations based on conservation of mechanical energy produce a more accurate answer than one might expect.

Only one portion is falling, i.e. portion DE. When I apply conservation of energy, I'll have to take the speed of portion DE. I'm not sure how would that yield velocity of element of chain that is about to come to stop at point B, where point B belongs to the static portion of the chain ?

 

[wrobel]

The hypothesis about the shape of the falling chain is a little bit strange

 

[NTesla]

The hypothesis about the shape of the falling chain is a little bit strange

IMO, It's catenary as long as both it's end is fastened to the ceiling. It's not catenary from the moment one of it's end is released. But I would like to know your reasoning for the hypothesis being strange.

 

[wrobel]

I'm afraid it is up to the proposer to provide a rationale for their hypotheses.:)

 

[wrobel]

-

 

[NTesla]

I'm afraid it is up to the proposer to provide a rationale for their hypotheses.:)

I would like to understand which part did you thought was strange ? Only then, can i attempt to figure out the rationale.

 

[wrobel]

I would like to understand which part did you thought was strange ? Only then, can i attempt to figure out the rationale.

Why, for example, is it a semicircle? There are plenty of other curves out there.

 

[NTesla]

Why, for example, is it a semicircle? There are plenty of other curves out there.

hmm.. that's a valid point. However, in the limit that the distance between the two points O and A, tending to zero, and the assumption that the chain is infinitely flexible, as if it were a cloth ribbon, I suppose the shape of the bend can be assumed to be an infinitely small curve. In that situation, will the shape of the bend affect the velocity of the link that is about to come to stop at point B ?

 

[jbriggs444]

hmm.. that's a valid point. However, in the limit that the distance between the two points O and A, tending to zero, and the assumption that the chain is infinitely flexible, as if it were a cloth ribbon, I suppose the shape of the bend can be assumed to be an infinitely small curve. In that situation, will the shape of the bend affect the velocity of the link that is about to come to stop at point B ?

A good way to approach infinite quantities in physics is to model them as the limit of finite cases as some parameter is increased without bound (or decreased toward zero).

So we take the limit of a flexible ribbon and a tight curve as both flexibility and tightness increase without bound. That sounds like an indeterminate case to me.

We might imagine a limit where energy is conserved and the chain fountain effect dominates.
We might imagine a limit where energy is not conserved and the chain fountain effect does not manifest.

 

[wrobel]

At the point where the circle is tangent to the straight line, the acceleration of the chain links undergoes a discontinuity. Just a remark

 

[NTesla]

At the point where the circle is tangent to the straight line, the acceleration of the chain links undergoes a discontinuity. Just a remark

Just before point D in direction of point E, the acceleration is a. Just at point D, the downward acceleration is less than a, and it's overall acceleration is towards the centre of the semi-circle, and below point D, the acceleration of all the elements would be towards the centre of the circle, until point B. However, i wonder if that could be called as discontinuous acceleration. But even if we consider that the acceleration is discontinuous at point D, how's that going to help find the velocity of element which is just going to stop at point B ?

 

[wrobel]

Write the angular momentum change theorem about the suspension point of the chain; $(r>0)$

 

[haruspex]

Only one portion is falling, i.e. portion DE. When I apply conservation of energy, I'll have to take the speed of portion DE. I'm not sure how would that yield velocity of element of chain that is about to come to stop at point B, where point B belongs to the static portion of the chain ?

In the r=0 limit, there are only two velocities and one of them is zero.

 

[NTesla]

In the r=0 limit, there are only two velocities and one of them is zero.

In the limit $r=0$, the static portion's length increases by $\frac {v}{2}$, and the moving portion of the chain falls by velocity $v$. I'm not sure why you wrote that one of the velocity is zero. Which part's velocity are you referring to as zero velocity.

 

[jbriggs444]

In the limit $r=0$, the static portion's length increases by $\frac {v}{2}$, and the moving portion of the chain falls by velocity $v$. I'm not sure why you wrote that one of the velocity is zero. Which part's velocity are you referring to as zero velocity.

The static portion has velocity zero.
The falling part has velocity $v$.

 

[NTesla]

Write the angular momentum change theorem about the suspension point of the chain; $(r>0)$

Angular momentum of purely vertical falling portion of the chain = $\lambda (\frac {L}{2} - \frac {x}{2})2rv$.
since the velocity of the purely vertical portion, $v$.

However, for the semi-circular portion, the expression for angular momentum about the suspension point is somewhat lengthy one. I'm not understanding how finding the angular momentum is going to answer the question that i had posed in my original post.

 

[haruspex]

In the limit $r=0$, the static portion's length increases by $\frac {v}{2}$,

$\frac {v}{2}$ is not a length.

and the moving portion of the chain falls by velocity $v$.

Do you mean it falls with velocity $v$?

When the top has fallen by $y$, say, how much GPE has been lost?

 

[NTesla]

v2 is not a length.

Yes, that I understand. When I wrote that I meant that the length of the static portion increases by $\frac {v}{2}$ per unit of time. I should have clarified that in that previous post.

Do you mean it falls with velocity v?

Yes.

When the top has fallen by y, say, how much GPE has been lost?

When the top has fallen by y, then the change in GPE is given by:
$d(P.E) = - \frac{\lambda g}{2}(-\frac{y^{2}}{2} + Ly)$

 

[haruspex]

Yes, that I understand. When I wrote that I meant that the length of the static portion increases by $\frac {v}{2}$ per unit of time. I should have clarified that in that previous post.

Yes.

When the top has fallen by y, then the change in GPE is given by:
$d(P.E) = - \frac{\lambda g}{2}(-\frac{y^{2}}{2} + Ly)$

Ok, so what velocity does that imply, assuming mechanical work is conserved?

 

[NTesla]

Yes, that I understand. When I wrote that I meant that the length of the static portion increases by $\frac {v}{2}$ per unit of time. I should have clarified that in that previous post.

Yes.

When the top has fallen by y, then the change in GPE is given by:
$d(P.E) = - \frac{\lambda g}{2}(-\frac{y^{2}}{2} + Ly)$

When I equate this d(P.E) with negative of d(K.E), I get an expression for velocity in terms of y: $v^{2} = gy\frac {2L - y}{L - y}$.

Now, when I use the Newton's 2nd law on the entire chain($\frac {dP}{dt} = F_{ext}$), then I get the expression for tension at the ceiling. From that I get the tension at the bottom of the static portion of the chain, $T_{bottom} = \frac {\lambda v^{2}}{4}$.

However, when I take just the falling portion of the chain as a system, then since this system is a variable mass system, the form of Newton's 2nd law applicable to this system will be: $\frac {dP}{dt} = F_{ext} + u\frac {dm}{dt}$, where P is the momentum of the main mass in the system, u is the velocity of dm w.r.t ground frame of reference.
For this variable mass system, we have: m = $\lambda (\frac {L}{2} - \frac {x}{2})$, therefore, $\frac {dm}{dt} = -\frac {\lambda v}{2}$, $u = v$, $F_{ext} =\lambda (\frac {L}{2} - \frac {x}{2})g + T_{bottom}$. After putting these values in the eq, I get the value of Tension, $T_{bottom}$ = $\frac {\lambda v^{2}}{4}$.

In case of Energy non-conservation, the acceleration of falling chain = $g$, and therefore, the value of $T_{bottom}$ = $\frac {\lambda v^{2}}{2}$.

My questions are:

(1) Since the speed of all the elements in the semi-circular curve is $\frac {v}{2}$, then why do we need to take $u = v$ to get to the correct expression for tension at the bottom of the falling portion of the chain. Taking $u = \frac {v}{2}$ will give wrong answer but why ?

(2) The force required to stop an element of straight falling chain on a surface is = $\lambda v^{2}$. But in case of U chain, in Energy conserving approach, the force required to stop an element of the falling portion of the chain = $\frac {\lambda v^{2}}{4}$ and in case of Energy non-conserving approach, this force =$\frac {\lambda v^{2}}{2}$ . Why is this force $\frac {1}{4}$th or $\frac {1}{2}$ (depending upon the approach) of the force in case of straight falling chain. What could be the qualitative/intuitive explanation for this ?

 

[haruspex]

When I equate this d(P.E) with negative of d(K.E), I get an expression for velocity in terms of y: $v^{2} = gy\frac {2L - y}{L - y}$.

Now, when I use the Newton's 2nd law on the entire chain($\frac {dP}{dt} = F_{ext}$), then I get the expression for tension at the ceiling. From that I get the tension at the bottom of the static portion of the chain, $T_{bottom} = \frac {\lambda v^{2}}{4}$.

However, when I take just the falling portion of the chain as a system, then since this system is a variable mass system, the form of Newton's 2nd law applicable to this system will be: $\frac {dP}{dt} = F_{ext} + u\frac {dm}{dt}$, where P is the momentum of the main mass in the system, u is the velocity of dm w.r.t ground frame of reference.
For this variable mass system, we have: m = $\lambda (\frac {L}{2} - \frac {x}{2})$, therefore, $\frac {dm}{dt} = -\frac {\lambda v}{2}$, $u = v$, $F_{ext} =\lambda (\frac {L}{2} - \frac {x}{2})g + T_{bottom}$. After putting these values in the eq, I get the value of Tension, $T_{bottom}$ = $\frac {\lambda v^{2}}{4}$.

In case of Energy non-conservation, the acceleration of falling chain = $g$, and therefore, the value of $T_{bottom}$ = $\frac {\lambda v^{2}}{2}$.

My questions are:

(1) Since the speed of all the elements in the semi-circular curve is $\frac {v}{2}$, then why do we need to take $u = v$ to get to the correct expression for tension at the bottom of the falling portion of the chain. Taking $u = \frac {v}{2}$ will give wrong answer but why ?

(2) The force required to stop an element of straight falling chain on a surface is = $\lambda v^{2}$. But in case of U chain, in Energy conserving approach, the force required to stop an element of the falling portion of the chain = $\frac {\lambda v^{2}}{4}$ and in case of Energy non-conserving approach, this force =$\frac {\lambda v^{2}}{2}$ . Why is this force $\frac {1}{4}$th or $\frac {1}{2}$ (depending upon the approach) of the force in case of straight falling chain. What could be the qualitative/intuitive explanation for this ?

To explain it, we have to come up with a model for how more work is being conserved than we might guess.
Consider a rod hinged to a fixed point at one end and subject to a perpendicular force $\vec F$ at the other. No gravity.
The reaction from the hinge normal to the rod is $\vec F/2$.
This shows how tension can be transferred around lowest part of the chain.

Turning that into analysis that predicts approximately how much work is lost looks to be a challenge.

 

[NTesla]

To explain it, we have to come up with a model for how more work is being conserved than we might guess.
Consider a rod hinged to a fixed point at one end and subject to a perpendicular force $\vec F$ at the other. No gravity.
The reaction from the hinge normal to the rod is $\vec F/2$.
This shows how tension can be transferred around lowest part of the chain.

Turning that into analysis that predicts approximately how much work is lost looks to be a challenge.

Even if we consider that Energy is not conserved, then in that case also, the value of tension at the bottom of both, the static portion as well as the falling portion of the chain = $\frac {\lambda v^{2}}{2}$. Why is this force $\frac {1}{2}$ of the force required to stop a straight falling chain.

 

[jbriggs444]

Even if we consider that Energy is not conserved, then in that case also, the value of tension at the bottom of both, the static portion as well as the falling portion of the chain = $\frac {\lambda v^{2}}{2}$. Why is this force $\frac {1}{2}$ of the force required to stop a straight falling chain.

Let me be sure I understand. For this calculation we are abstracting away the interaction zone between the falling portion and the static portion and making blind assertions about the behavior there.

Your particular blind assertions are that energy is not conserved and that the tensions from the two chain segments at the interfaces to the interaction zone are equal.

The reason for factor of $\frac{1}{2}$ is obvious. Because the momentum flow from chain going in and out of the interaction zone is $\frac{\lambda v^2}{2}$. In the limit as the size of the interaction zone goes to zero, there is zero momentum in the interaction zone. Accordingly, the total momentum flow from the tension forces is equal and opposite. By your blind assertion, each tension force contributes half.

By contrast, with energy conservation we would be forced to distribute the total supporting force of $\lambda v^2$ differently.

 

[NTesla]

Let me be sure I understand. For this calculation we are abstracting away the interaction zone between the falling portion and the static portion and making blind assertions about the behavior there.

Your particular blind assertions are that energy is not conserved and that the tensions from the two chain segments at the interfaces to the interaction zone are equal.

The reason for factor of $\frac{1}{2}$ is obvious. Because the momentum flow from chain going in and out of the interaction zone is $\frac{\lambda v^2}{2}$. In the limit as the size of the interaction zone goes to zero, there is zero momentum in the interaction zone. Accordingly, the total momentum flow from the tension forces is equal and opposite. By your blind assertion, each tension force contributes half.

By contrast, with energy conservation we would be forced to distribute the total supporting force of $\lambda v^2$ differently.

[Mentor Note: this post has been heavily edited to remove numerous insults.]

You mentioned that my blind assertion is that Energy is not conserved: In my post#24, I have considered both the cases: Energy being conserved and other case of Energy not being conserved. I have never solely said that Energy is definitely conserved or that Energy is definitely not conserved. When haruspex mentioned in post#25 a model for work being conserved, thereby commenting about the Energy conserving case, I wrote my post#26. It was inherent in that post that even if we for the sake of discussion consider a model where Energy is not conserved, the magnitude of force required to stop the element of the falling part of the chain is $\frac {1}{2}$ of the force required to stop a straight falling chain. And if we consider the model where Energy is conserved, then this force = $\frac {1}{4}$ of the force required to stop a straight falling chain.

You further wrote that my another blind assertion is that the tensions from the two chain segments at the interfaces to the interaction zones are equal: Actually, this is quite easy to prove. The tension at the bottom of the static side and the falling side would be approximately equal, and the difference between them tends to zero as the radius of the semi-circle tends to zero. This can be proved easily.

You further wrote that the factor of $\frac {1}{2}$ is obvious and it is because the momentum flow from chain going in and out of the interaction zone is $\frac {\lambda v^{2}}{2}$. I had put forth in post#24 -- In Question 2 of that post, I had asked for a qualitative/intuitive explanation for this.

 

[haruspex]

@NTesla, that does it! Look for help elsewhere.

 

[berkeman]

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