Chain falling out of a horizontal tube onto a table

[NTesla]

Homework Statement

The question is from Irodov Q 1.184. See the Screenshot.

Relevant Equations

I'm trying to solve it using energy conservation. My thinking is that since all the surfaces are frictionless and only conservative gravitational force is acting, therefore, ME must be conserved.

My attempt:
Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE.
PE of hanging part = $\frac{1}{2} \frac{m}{l}gh^{2}$.
PE of part in the tube = $\frac{m}{l}(l - h)gh$.

Final ME = $\frac{1}{2}\frac{m}{l}gh^{2}$ + $\frac{1}{2}\frac{m}{l}hv^{2}$.

Since Initial ME = Final ME. Therefore, $\frac{1}{2}\frac{m}{l}hv^{2}$ = $\frac{m}{l}(l-h)gh$. Solving this gives: $v = \sqrt{2g(l-h)}$. But the answer in the book is: $v = \sqrt{2ghln(\frac{l}{h})}$.

 

[Steve4Physics]

Homework Statement: The question is from Irodov Q 1.184. See the Screenshot.

@NTesla, the last part of the question on the screenshot is missing. So here’s what (I believe) it should say:

"... At a certain moment, end-A of the chain is released. What is the speed of end-A when it reaches the right-hand end of the tube?"

Since Initial ME = Final ME

Are you sure about that? Suppose the whole chain, not just part of the chain, had reached the floor and come to rest. What would the final ME be? If it is different to the initial ME, how do you account for this difference?

 

[NTesla]

the last part of the question on the screenshot is missing

Yes, thanks. Edited. The missing screenshot has been added in the original post.

 

[NTesla]

Are you sure about that? Suppose the whole chain, not just part of the chain, had reached the floor and come to rest. What would the final ME be? If it is different to the initial ME, how do you account for this difference?

Yes, that's a valid point.

 

[jbriggs444]

There are a number of conundrums that deal with the unexpected behavior of falling ropes and chains. It seems clear that you are not intended to worry about the fine details involved with those.

The intent is that you pretend that the portion of the chain that reaches the table is quietly removed from existence as it gets there.

Spoiler

If a falling link strikes the table at an angle (perhaps nudged to an angle by the portion of the chain that had already fallen) then the strike will impart spin. That spin will pull down on the remainder of the falling chain. Almost paradoxically, the upward force from the table below has the net effect of a downward force on the falling chain.

https://en.wikipedia.org/wiki/Chain_fountain

 

[haruspex]

When writing "ln(x)” in LaTeX, you can make it clearer with "\ln": $\ln(x)$ v. $ln(x)$. This avoids confusion when $l$ or $n$ is also a variable.

 

[NTesla]

When writing "ln(x)” in LaTeX, you can make it clearer with "\ln": $\ln(x)$ v. $ln(x)$. This avoids confusion when $l$ or $n$ is also a variable.

All the equations that I had written have been deleted by the moderator. I don't understand why. Is that just because $\ln$ was not written as it should have been ?

 

[Steve4Physics]

All the equations that I had written have been deleted by the moderator. I don't understand why. Is that just because $\ln$ was not written as it should have been ?

Do you mean the equations from your Post #1? I can see them. Here's a copy, though LaTeX rendering may be a problem:

My attempt:
Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE.
PE of hanging part = $\frac{1}{2} \frac{m}{l}gh^{2}$.
PE of part in the tube = $\frac{m}{l}(l - h)gh$.

Final ME = $\frac{1}{2}\frac{m}{l}gh^{2}$ + $\frac{1}{2}\frac{m}{l}hv^{2}$.

Since Initial ME = Final ME. Therefore, $\frac{1}{2}\frac{m}{l}hv^{2}$ = $\frac{m}{l}(l-h)gh$. Solving this gives: $v = \sqrt{2g(l-h)}$. But the answer in the book is: $v = \sqrt{2ghln(\frac{l}{h})}$.

 

[kuruman]

And here is a screenshot.

 

[wrobel]

In principle one can introduce a tension force along the chain and write down the 2nd Newton law for each $\Delta m$. But where should one take boundary conditions for the tension force at the end of the tube?

 

[jbriggs444]

In principle one can introduce a tension force along the chain and write down the 2nd Newton law for each $\Delta m$. But where should one take boundary conditions for the tension force at the end of the tube?

I do not understand. There is no discontinuity in tension at the end of the tube where the chain turns downward. The obvious boundary conditions are at the left end of the chain in the tube (tension zero there) and at the bottom where the chain strikes the table (tension zero there as well).

If one considers the gradient in tension across each $\Delta m$ then the gradient is constant within the tube and is also constant within the falling segment of chain. The gradient changes across the [negligibly small] curve at the mouth of the tunnel.

 

[wrobel]

There is no discontinuity in tension at the end of the tube where the chain turns downward.

why?

 

[jbriggs444]

why?

The tube is smooth (frictionless). The mouth of the tube acts like an ideal pulley. The direction of the tension force changes there. But the magnitude of the tension does not.

 

[wrobel]

The mouth of the tube acts like an ideal pulley.

In exercises as a rule pulleys are considered with massless strings.

But the magnitude of the tension does not.

This is a hypothesis it does not follow from equations of motion. In my opinion this hypothesis is not evident and thus it must be fixed in the statement of the problem.

 

[haruspex]

In exercises as a rule pulleys are considered with massless strings.

Whether it’s a track over a smooth shoulder or a massless pulley is clearly independent of whether the problem concerns a massless string or a massive one.

The mouth of the tube acts like an ideal pulley. The direction of the tension force changes there. But the magnitude of the tension does not.

This is a hypothesis it does not follow from equations of motion.

If there is a step change in the tension anywhere then there is a string element of infinitesimal mass subject to a non-infinitesimal net force in the tangential direction, no? Taking the string to be inextensible and following a fixed path, the magnitude of its tangential acceleration is the same everywhere.

 

[wrobel]

Taking the string to be inextensible and following a fixed path, the magnitude of its tangential acceleration is the same everywhere.

then it must be incompressible as well. Is the chain under consideration incompressible? Why?
If it is incompressible it stops when reaching the point B :)

 

[NTesla]

Here's how I'm trying to solve it, using Newton's 2nd law.
I'm considering the whole chain as the system. Let at any instant of time, t, the horizontal part of the chain is x distance from the left end. $v$ is the velocity of the chain at the instant t. $\lambda$ is the mass per unit length.

$\Delta\vec{P} = \vec{P_{f}} - \vec{P_{i}}$
$\vec{P_{i}} = \lambda (l - x - h)v\hat{i} - \lambda hv\hat{j}$
$\vec{P_{f}} = \lambda (l - (x + \Delta{x}) - h)(v + \Delta{v}) \hat{i} - \lambda h(v + \Delta{v}) \hat{j}$

This gives in the $\hat{i}$ direction: $\Delta\vec{P} = \lambda\left ( (l - x - h)\Delta{v} - \Delta{x}v \right )\hat{i}$. (ignoring $\Delta{x}\Delta{v}$).
Now, in the $\hat{i}$ direction, the force is due to the weight of the hanging part of the chain, which is constant, as long as x ranges from 0 to $(l-h)$ and that force is = $\lambda hg$.

From here, the differential equation in the limit $\Delta{t}$ tends to 0, gives: $$(l - x - h)\frac{dv}{dt} - v^{2} = hg$$.

But solving this differential equation is not giving the right answer. I'm dont know where I'm going wrong.

 

[wrobel]

my version is
$$\ddot x=-\frac{gh}{x+h},$$
where $x$ is a length of the horizontal tail

 

[NTesla]

....If $x_0$ is an initial length of the horizontal tail then
$$v=\sqrt{2gh\ln(1+x_0/h)}$$
is that the answer?

No. That's not the answer.

I've posted the correct answer in the original post.

 

[wrobel]

No. That's not the answer.

I've posted the correct answer in the original post.

look carefully :) My answer is the same as one in the original post

 

[NTesla]

look carefully :) My answer is the same as one in the original post

In your equation, you wrote, x is length of horizontal tail, that led me to think that by "tail" you meant the distance moved by the left end of the horizontal part of the chain. But now, I understand that by tail you wanted to mean length of the horizontal part of the chain. It would have been prudent to use the word "chain" instead of the word "tail" :) .

However, I still don't understand why my equations written in post#17 do not work.

 

[NTesla]

By reverse engineering the correct equation, I've reached a few conclusion: The $v^{2}$ term should not be there in the last equation in post#17. Also, the force on the horizontal part of the chain is not = $\lambda hg$, but is = $\lambda h(g - \frac{dv}{dt})$. But for eliminating the $v^{2}$ term in the differential equation, the term $\Delta {x}v\hat{i}$ must be added in the $\vec{P_{f}}$ that I've written. But I don't understand on what grounds should that be written in $\vec{P_{f}}$.

 

[jbriggs444]

then it must be incompressible as well. Is the chain under consideration incompressible? Why?
If it is incompressible it stops when reaching the point B :)

Chains buckle under compression. Like cables, ropes, cords, strings and threads.

The intent of the problem is clearly that the chain crumples into the table below without passing the resulting contact force back up the chain, either as compression or as tension.

It is clear that the chain is under tension throughout its moving length, from point B, up through the mouth of the tube and left through the tube all the way to the chain's tail end in the tube.

 

[wrobel]

NTesla:
Consider the vertical part of the chain. Let $N(x)$ be a tension force that acts in the section that lies at the height $x$. For an infinitesimal chain's interval of length $ds$ write down the 2nd Newton:
$$\rho ds a=-g\rho ds+N(x+ds)-N(x)=-g\rho ds+N'(x)ds+o(ds),$$ here $\rho$ is a density of the chain; $a$ is an acceleration.
Thus we have
$$\rho a=N'(x)-g\rho.$$
Take the integral $\int_0^h dx$ from the both sides of this equality
$$\rho a h=N(h)-N(0)-g\rho h,\quad N(0)=0.$$
Do the same for the horizontal part of the chain.

 

[NTesla]

NTesla:
Consider the vertical part of the chain. Let $N(x)$ be a tension force that acts in the section that lies at the height $x$. For an infinitesimal chain's interval of length $ds$ write down the 2nd Newton:
$$\rho ds a=-g\rho ds+N(x+ds)-N(x)=-g\rho ds+N'(x)ds+o(ds),$$ here $\rho$ is a density of the chain; $a$ is an acceleration.
Thus we have
$$\rho a=N'(x)-g\rho.$$
Take the integral $\int_0^h dx$ from the both sides of this equality
$$\rho a h=N(h)-N(0)-g\rho h,\quad N(0)=0.$$
Do the same for the horizontal part of the chain.

I'm trying to understand your solution. Could you kindly tell me what does the last term $o(ds)$ mean, and where did that come from. I understood till the term $N'(x)ds$.

Secondly, by simplifying the equation: $$\rho a=N'(x)-g\rho.$$, we get:
$$\rho vdv = dN - g\rho dx$$. LHS of this can be integrated for some values of v, but you've mentioned to use $\int_0^h dx$ on both sides of this equation. This part I dont understand. Kindly explain.

 

[Philip Wood]

Have you now appreciated that there WOULD usually be a loss of mechanical energy, because of the inelastic collision between the chain and the floor? You mustn't equate the total loss of gravitational PE, $\frac {m}{l} g(l-h)$, to the KE of the length, $h$, still in motion at the end-time.

Now suppose that when the lower part of the chain meets the deck it is guided round a smooth quarter-circle ramp in a vertical plane, so it doesn't lose KE but carries on moving, but across the deck. All parts of the chain will be moving at the same speed. In that case the speed at the end time is found by equating the loss in gravity PE to the KE of the whole length, $l$, of chain.

In my original post I claimed he end-speed found for the ramp set-up will also be the speed at which the chain hits the deck inelastically (and forms a stationary heap – of negligible height!). I'd argued that the upward force on the vertical part of the chain will be the same in both cases because vertical momentum is destroyed at the same rate). But in the ramp case I'd neglected the negative tension in the chain at the bottom bend where it passes round the ramp; the negative tension is needed to give the part of the chain moving across the deck its acceleration – the same |acceleration| as that of the rest of the chain.

Here, instead, is a force-based argument for the case where the chain is brought to rest on hitting the deck. Suppose that when length $s$ has hit the deck, the rest of the chain is moving at speed $w$. Write $\mu=\frac m l$, the mass per unit length of chain.

The rate of loss of downward momentum of the chain when length $s$ has hit the deck is $\mu w^2$. This is therefore the upward force on the falling chain. Applying $F=ma$ to the moving part of the chain we have
$$\mu h g - \mu w^2 = \mu (l-s) \frac{dw}{dt}$$

[Strictly the part of the chain in still in the tube and the vertical parts ought to be treated separately, as they are accelerating in different directions, but the above equation is a valid short-cut.]
We now use some mathematical trickery to re-express $\frac{dw}{dt}$...

$$\frac{dw}{dt}= \frac{ds}{dt}.\frac {dw}{ds}= w\frac {dw}{ds}= \frac 1 2 \times \frac{d{w^2}{ds}$$

So our $F=ma$ equation becomes

$$\mu h g - \mu w^2 = \mu (l-s) \frac 1 2 \times \frac{d{w^2}{ds}$$

Imposing the initial condition $w=0$ when $s=0$, we find that

$$w^2=g h \left(1-\frac{(l-s)^2}{l^2}.$$

So the speed, $v$ at the end point, when $s=(l-h)##, is given by

$$v^2=g h \left(1-\frac{h^2}{l^2}\right).$$

 

[jbriggs444]

Now suppose that when the lower part of the chain meets the deck it is guided round a smooth quarter-circle ramp in a vertical plane, so it doesn't lose KE but carries on moving, but across the deck.

I had considered offering this as a supposed equivalence. But it is not quite right.

The portions of the chain that are falling through the air or being pulled from the tube share the same acceleration. In magnitude at least. The portion of the chain that is sliding across the table will not be accelerating. This leads to a problem. The chain must buckle at point B due to the mismatched accelerations.

If the chain is going to buckle at point B anyway, we should go ahead and let it just pile up on the table.

 

[Philip Wood]

"The portion of the chain that is sliding across the table will not be accelerating." Why can't it have the same |acceleration| as the rest of the chain? I should have extended the ramp in a straight line across the deck. But I now agree that I was wrong to argue that the end-speed found for the ramp set-up will also be the speed at which the chain hits the deck inelastically (and forms a stationary heap – of negligible height!).

 

[NTesla]

Have you now appreciated that there WOULD usually be a loss of mechanical energy, because of the inelastic collision between the chain and the floor? You mustn't equate the total loss of gravitational PE, $\frac {m}{l} g(l-h)$, to the KE of the length, $h$, still in motion at the end-time.

Now suppose that when the lower part of the chain meets the deck it is guided round a smooth quarter-circle ramp in a vertical plane, so it doesn't lose KE but carries on moving, but across the deck. All parts of the chain will be moving at the same speed. In that case the speed at the end time is found by equating the loss in gravity PE to the KE of the whole length, $l$, of chain.

Now here's the subtle bit (well I didn't spot it for a while)... The end-speed found for the ramp set-up will also be the speed at which the chain hits the deck inelastically (and forms a stationary heap – of negligible height!). This is because the upward force on the vertical part of the chain will be the same in both cases because vertical momentum is destroyed at the same rate).

How do I make the latex delimiters spring into action?

Appreciate your input. However, I've already understood that mechanical energy will not be conserved in the given situation. That's why I've now tried to solve the question using Newton's 2nd law. Kindly see the post#17 and 22 above, wherein I've written the equations. I don't understand why the equations that I've written in post#17 don't work. If you have insights regarding that, it will be very much welcome.

 

[haruspex]

"The portion of the chain that is sliding across the table will not be accelerating." Why can't it have the same |acceleration| as the rest of the chain? Perhaps I should have extended the ramp in a straight line across the deck.

Since the problem statement does not specify what happens to the chain on reaching the table we have to make a reasonable guess about the intent.
Judging by the diagram, there is no ramp that would redirect it across the table.
A real chain would pile up haphazardly, coming to rest. It would also stack up, reducing the drop height, but we've not enough information about it to process that and it would be unreasonably complicated.
As noted by @jbriggs444 in post #5, chains (and perhaps ropes of any stiffness?) can have a significant effect from the rotation of the segments stacking. I do not know whether that depends on the segment length, or, if it does, whether the effect disappears in the limit as that length tends to zero. That would be an interesting study, but still likely way beyond the intent of the question.
This leaves us with the model that the chain annihilates on contact with the table.

 

[Steve4Physics]

@NTesla, maybe this is one source of your difficulty...

Now, in the $\hat{i}$ direction, the force is due to the weight of the hanging part of the chain,

That's not entirely correct. The horizontal section of the chain is accelerated by tension,T, acting at its right-most end:

We have to treat the falling vertical chain section as if there are only two forces acting on it*. The forces are its weight, $\lambda h g$ downwards and $T$ acting upwards at its top end:

The acceleration, $a$, is the same for both sections, so you can (using N2L) construct 2 equations and eliminate T. This gives an equation relating $x$ and $a$.

Hint: When you get this equation that remember $a = v \frac {dv}{dx}$.

*The absence of a force from the ground, which affects the motion of the vertical section of the chain, has been discussed by @jbriggs444 and @haruspex.

 

[jbriggs444]

"The portion of the chain that is sliding across the table will not be accelerating." Why can't it have the same |acceleration| as the rest of the chain?

It could if, for instance, we let the chain ride in a frictionless tube on the table. Then the section of chain in the tube would be accelerating (we assume without evidence that the links can sustain a compressive load as long as they are forced to glide in a straight line). But then the compressive load from this would feed back into the falling chain. So then we would have to extend the guide tube upward from the table. With this in place the whole chain would count in our equation for acceleration.

Arguably, that would make the problem both too easy to solve and too messy to pose. We would regain mechanical energy conservation. The solution proposed in the OP would apply.

 

[NTesla]

Equations that you wrote are not legible. Kindly see the attached pic.

 

[NTesla]

@NTesla, maybe this is one source of your difficulty...


That's not entirely correct. The horizontal section of the chain is accelerated by tension,T, acting at its right-most end:
View attachment 367233
We have to treat the falling vertical chain section as if there are only two forces acting on it*. The forces are its weight, $\lambda h g$ downwards and $T$ acting upwards at its top end:
View attachment 367231
The acceleration, $a$, is the same for both sections, so you can (using N2L) construct 2 equations and eliminate T. This gives an equation relating $x$ and $a$.

Hint: When you get this equation that remember $a = v \frac {dv}{dx}$.

*The absence of a force from the ground, which affects the motion of the vertical section of the chain, has been discussed by @jbriggs444 and @haruspex.

Yes, and I've already mentioned that in post#22.

But there's another problem that I've mentioned in post#22, that is the cause of much problem.

 

[Nigel Wood]

I know this is the wrong place to ask, but I need help with this site and don't know where to find it... (a) I can't find the 'edit' button on my longish post above with red error messages. (b) I can't understand the error messages (c) I can't find the latex guide, the one that uses a quadratic equation as an example.

 

[haruspex]

But there's another problem that I've mentioned in post#22, that is the cause of much problem.

Please show the details of your differentiation of the horizontal momentum. Looks like you missed a term.

 

[haruspex]

I know this is the wrong place to ask, but I need help with this site and don't know where to find it... (a) I can't find the 'edit' button on my longish post above with red error messages. (b) I can't understand the error messages (c) I can't find the latex guide, the one that uses a quadratic equation as an example.

Are you writing multiple lines of LaTeX inside one pair of $ $ brackets? Try not doing that.
And please use the preview button (magnifying glass at right hand end of controls) before posting.

 

[NTesla]

Please show the details of your differentiation of the horizontal momentum. Looks like you missed a term.

Here's all the steps:
$\Delta\vec{P} = \vec{P_{f}} - \vec{P_{i}}$
$\vec{P_{i}} = \lambda (l - x - h)v\hat{i} - \lambda hv\hat{j}$
$\vec{P_{f}} = \lambda (l - (x + \Delta{x}) - h)(v + \Delta{v}) \hat{i} - \lambda h(v + \Delta{v}) \hat{j}$

In the $\hat{i}$ direction: $\Delta P\hat{i} = \lambda(l - (x + \Delta x) - h)(v + \Delta v)\hat i - \lambda (l - x - h)v\hat i$
= $\lambda((l - x - h)\Delta v - \Delta x v)\hat i$ (ignoring $\Delta x \Delta v$).

Dividing both sides by $\Delta t$, we get:
$\frac{\Delta P}{\Delta t} = \lambda((l - x - h)\frac{\Delta v}{\Delta t} - \frac{\Delta x}{\Delta t} v)$

Now, in the limit, $\Delta t\rightarrow 0$ the equation becomes:
in the $\hat i$ direction: $$\frac {dP}{dt} = \lambda((l - x - h)\frac {dv}{dt} - \frac {dx}{dt}v)$$

$\Rightarrow$ For the horizontal part of the chain: $\frac {dP}{dt} = \lambda((l - x - h)\frac{dv}{dt} - v^{2})$

Now, by comparing the force applicable on the horizontal and vertical part of the chains, for same acceleration throughout, we get: Force applicable on the horizontal part of the chain, $F_{ext}$ = $\lambda h(g - \frac{\mathrm{d} v}{\mathrm{d} t})$.

$\therefore$ $\frac {dP}{dt} = \lambda((l - x - h)\frac{dv}{dt} - v^{2})$ = $\lambda h(g - \frac{\mathrm{d} v}{\mathrm{d} t})$.

But, this is not the right differential equation which leads to the correct answer. As I had mentioned in post#22, for reaching the correct differential equation, the $v^{2}$ term needs to vanish. For that to happen the term $( - \Delta x v\hat i)$ terms needs to vanish. That can only happen if there's an additional term $+ \Delta x v\hat i$ in the expression for $\vec {P_{f}}$. However, I do not see how or why there could be an additional term $( + \Delta x v\hat i)$ in the expression for $\vec {P_{f}}$. This is where I'm facing the problem.