Chain falling out of a horizontal tube onto a table

[jbriggs444]

I alluded to that in post #44, but I think it is ok because we can instead view it in terms a tension transmitted around the curve and write the separate equations for the accelerations of the two portions.
I believe this leads to the same equation.

I agree completely. The chain traverses a one dimensional fixed path. There is no net force nor freedom to move other than along the path. The chain moves along this path in lock step, all pieces keeping pace with all others. Accordingly, we are free to view the chain as if it were a single entity moving along a straight line. Though, obviously, only one part is subject to gravity. And we still have the issue with links dribbling off the one end.

Similarly, in pulley problems with two masses dangling from a single ideal pulley, it is valid to add the masses together to get a single effective mass for the combined system.

 

[NTesla]

I believe this leads to the same equation.

Which equation are you referring to ?

If we happen to write separate equations for accelerations for the two parts of the chain (horizontal and vertical), then we do reach the correct differential equation. I've been through that route and I understand how to solve this question using that route.

However, in my current attempt to solve this problem, I've taken the route of finding the change in momentum and equating it to the net external force, as may be deduced from my post #38 and 49.

I'm asking for help in writing the correct equations for $\vec {P_{i}}$ or $\vec {P_{f}}$ or the force applicable on the system. Kindly go through the equations that I've written in those aforementioned posts, and let me know which term is wrong or if any other term needs to be included in those expressions. After @haruspex's post in post#48, I tried writing the equation for force = rate of change of momentum of the system and this is the equation i've landed with:
$$ \vec {P_{f}} - \vec {P_{i}} = \lambda ((l - (x + \Delta x) - h)(v + \Delta v)- \lambda (l - x - h)v)\hat i + (h (v+\Delta v)-\Delta xv - hv)(-\hat {j})$$

$$\vec F_{ext} = \lambda hg(-\hat j) + \lambda \frac {\Delta x}{\Delta t}v \hat j$$

Now, by equating these two equations, i'm getting: $$((l - (x + \Delta x) - h)(v + \Delta v)- (l - x - h)v)\hat i + (h (v+\Delta v)-\Delta xv - hv)(-\hat {j}) = hg\Delta t(-\hat j) + {\Delta x}v \hat j $$

I don't know where to proceed from here.

 

[jbriggs444]

$$\vec F_{ext} = \lambda hg(-\hat j) + \lambda \frac {\Delta x}{\Delta t}v \hat j$$

The claim that $\vec F_\text{ext}$ is given by $\lambda hg(-\hat j) + \lambda \frac {\Delta x}{\Delta t}v \hat j$ is clearly incorrect.

We have the downward force from gravity given by $\lambda h g (-\hat j)$. That much is correct.

We have the upward momentum flow as downward moving chain links leave the system. That is indeed given by $\lambda \frac {\Delta x}{\Delta t}v$.

But we also have the force from the mouth of the tube which delivers a downward momentum flow due to the chain links emerging downward. The mouth is accelerating these links downward. It also delivers a leftward momentum flow due to the chain links entering the mouth from the left. It is bringing these links to a stop in the horizontal direction. It also delivers an upward force equal to the tension in the chain at the mouth. Finally it delivers a rightward force equal to the tension in the chain at the mouth.

It is way easier to make an argument and reduce the problem to one dimension.

 

[haruspex]

I don't know where to proceed from here.

Well, obviously, the usual steps of cancelling out the non-infinitesimal terms and discarding the second order infinitesimals. What does that give you?

 

[wrobel]

Most funny thing is that the system is Hamiltonian (see #18). Even the "energy" integral exists:
$$\frac{1}{2}\dot x^2+gh\ln(x+h)=\mathrm{const}$$

 

[NTesla]

Well, obviously, the usual steps of cancelling out the non-infinitesimal terms and discarding the second order infinitesimals. What does that give you?

Did you really thought that you were being helpful when you wrote that ?

When I wrote the last equation in post#52, it was obvious to me that even if I cancelled the 2nd order differentials and cancelled out the non-infinitesimals, the situation hasn't improved at all, in the direction of solving the question. But apparently, it wasn't obvious to you. I suppose it would have been obvious to you too, had you bothered to read the equation, instead of just glancing at the post. Atleast @jbriggs444 made that effort and came up with some reasoning why the last equation in post #52 wouldn't work, and I appreciate him for that.

In an attempt to make the situation plain and simple to understand, and to take the discussion forward, I'll write down the equation after having cancelled out the non-infinitesimal terms and discarding the second order infinitesimals.:

In the $\hat i$ direction, we have: $l\Delta v - x\Delta v - \Delta x v - h\Delta v = 0$
and in the $\hat j$ direction, we have: $h\Delta v -\Delta x v = hg - \frac {\Delta x}{\Delta t}v$

From these 2 equations, we have: $l\Delta v -x\Delta v = hg - \frac {\Delta x}{\Delta t}v + 2 \Delta x v$


Any idea of what to do next after the above equations ? Or are you going to say "Ok, I'll give you that one" and then abandon the whole thing, again ?

 

[NTesla]

Most funny thing is that the system is Hamiltonian (see #18). Even the "energy" integral exists:
$$\frac{1}{2}\dot x^2+gh\ln(x+h)=\mathrm{const}$$

Could you kindly write for Newtonian mechanics. I have only studied Newtonian mechanics. And what is funny here, I don't understand.

 

[NTesla]

But we also have the force from the mouth of the tube which delivers a downward momentum flow due to the chain links emerging downward. The mouth is accelerating these links downward. It also delivers a leftward momentum flow due to the chain links entering the mouth from the left. It is bringing these links to a stop in the horizontal direction. It also delivers an upward force equal to the tension in the chain at the mouth. Finally it delivers a rightward force equal to the tension in the chain at the mouth.

I would argue that instead of a tube, if the chain was kept on a table, and everything else remained the same, even then the chain would bend and move down at the edge of the table. I don't see how mouth of the tube has anything to do with why the chain would go down.

 

[jbriggs444]

I would argue that instead of a tube, if the chain was kept on a table, and everything else remained the same, even then the chain would bend and move down at the edge of the table. I don't see how mouth of the tube has anything to do with why the chain would go down.

The drawing shows the chain descending vertically. Not in some sort of [evolving!] parabola. This implies that the chain is constrained not just by the bottom of the tube but also by its top and right where the tube turns downward at its mouth.

If the chain fell in a curve then the length of the falling portion would not be $h$.

 

[NTesla]

The drawing shows the chain descending vertically. Not in some sort of [evolving!] parabola. This implies that the chain is constrained not just by the bottom of the tube but also by its top and right where the tube turns downward at its mouth.

If the chain fell in a curve then the length of the falling portion would not be $h$.

Yes, that makes sense.

So, the rightmost part of the mouth of the tube would exert a leftward force on the piece of the chain which is right at the bend, and the magnitude of that force would be exactly equal to cancel the horizontal momentum carried by that piece. So, that piece would only have downward momentum from that point on.

 

[NTesla]

The drawing shows the chain descending vertically. Not in some sort of [evolving!] parabola. This implies that the chain is constrained not just by the bottom of the tube but also by its top and right where the tube turns downward at its mouth.

If the chain fell in a curve then the length of the falling portion would not be $h$.

I'm struggling to understand what you mentioned by "it's top" when you mentioned that "the chain is constrained not just by the bottom of the tube, but also by it's top and right."

Here's what I understood by your statement: The element of the chain which is just at the bend, would experience a leftward force (I agree on this point). However, your statement implies that the element would also experience a downward force by the top of the mouth. And the bottom of the tube is already supporting the weight of the horizontal part of the chain.

So, if we consider the hanging part of the chain, then it is experiencing 4 forces:
(1) Due to the weight of the hanging part of the chain: $\lambda hg(-\hat j)$
(2) Force exerted by the table on the element which is just touching it and is coming to rest: $\lambda \frac {\Delta x}{\Delta t}v (\hat j)$.
(3) Force exerted by the top of the mouth of the tube on the element which is just at the bend of the tube: $\lambda \frac {\Delta x}{\Delta t}v (-\hat j)$.
(4) Tension force due to the horizontal part of the chain = $\lambda h(g - \frac {\Delta v}{\Delta t})(-\hat j)$.

Is that correct ?

 

[jbriggs444]

Yes, that makes sense.

So, the rightmost part of the mouth of the tube would exert a leftward force on the piece of the chain which is right at the bend, and the magnitude of that force would be exactly equal to cancel the horizontal momentum carried by that piece. So, that piece would only have downward momentum from that point on.

A leftward force to bring the chain to a stop horizontally.
And a downward force to bring the chain to a matching rate of motion vertically.

 

[jbriggs444]

I'm struggling to understand what you mentioned by "it's top" when you mentioned that "the chain is constrained not just by the bottom of the tube, but also by it's top and right."

The "it" in that passage is the tube.

The top of the tube exerts a downward force to impart downward momentum to the chain.
The right of the tube exerts a leftward force to impart leftward momentum to the chain.

At some times during the scenario it is possible that the tension in the chain will be adequate to supply the requisite downward and leftward momentum. But not at all times. When the fall is nearly complete, the tension drops toward zero.

Here's what I understood by your statement: The element of the chain which is just at the bend, would experience a leftward force (I agree on this point). However, your statement implies that the element would also experience a downward force by the top of the mouth.

Yes. The required force on the incremental portion of chain in the mouth supplied by the resultant of the tensions and the mouth of the guide tube. The guide tube is a constraint. The resultant force is whatever it has to be to keep the chain on its prescribed trajectory. The trajectory is around a 90 degree bend with unchanged speed. So the required resultant force is at 45 degrees down and to the left.

The mouth is effectively a pulley. A pulley with a cover so that centrifugal force cannot fling the chain off of the pulley.

And the bottom of the tube is already supporting the weight of the horizontal part of the chain.

So, if we consider the hanging part of the chain, then it is experiencing 4 forces:
(1) Due to the weight of the hanging part of the chain: $\lambda hg(-\hat j)$

Yes.

(2) Force exerted by the table on the element which is just touching it and is coming to rest: $\lambda \frac {\Delta x}{\Delta t}v (\hat j)$.

Yes.

(3) Force exerted by the top of the mouth of the tube on the element which is just at the bend of the tube: $\lambda \frac {\Delta x}{\Delta t}v (-\hat j)$.

Yes. Nicely cancelling with the force from the table.

(4) Tension force due to the horizontal part of the chain = $\lambda h(g - \frac {\Delta v}{\Delta t})(-\hat j)$.

The formula seems qualitatively correct. Though it would be nice to see a justification.

Apparently you are working a force balance on the hanging portion of the chain. The acceleration of that piece is given by $a = \frac{g-t}{\lambda h}$. The interactions at the table and at the mouth of the guide tube cancel each other out. If we solve for $t$ we get $t = (a - g) \lambda h$ which matches your formula.

So yes, I agree.

 

[TSny]

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.

As pointed out by @Steve4Physics, mechanical energy is not conserved. However, you can still solve it using energy concepts.

(1) Find an expression for the KE of the chain at the instant the upper end of the chain has moved a distance $x$ from point $A$. Express in terms of $\lambda$, $x$, $l$, and the speed $v$ at that instant.

(2) Find an expression for the potential energy $U$ of the chain at the same instant. Express in terms of $\lambda$, $x$, $g$, $h,$ and $U_0$, where $U_0$ is the initial potential energy of the chain when it was released.

(3) Find an expression for the rate at which mechanical energy is lost at point $B$. Assume that each link makes a completely inelastic collision with the surface at $B$. Express in terms of $\lambda$ and $v$.

Consider how these three expressions are related.

 

[haruspex]

Did you really thought that you were being helpful when you wrote that ?

Obviously you had done those steps, so I expect you to post where you actually got to, not several equations back, making me redundantly go through those steps too.

From these 2 equations, we have: $l\Delta v -x\Delta v = hg - \frac {\Delta x}{\Delta t}v + 2 \Delta x v$

which is clearly wrong because the non-infinitesimals have not cancelled. This strongly suggests missing forces: the ones @jbriggs444 mentions in post #53 and I mentioned in post #44.

 

[NTesla]

I expect you to post where you actually got to, not several equations back, making me redundantly go through those steps too.

There's no reason to go on resolving the equations further if the 1st equation itself seems incomplete or wrong to begin with. I expect you to atleast go through the equations once, before you jump to the conclusion that I've fallen into the classic trap or any other conclusion for that matter. I also expect you to clarify further when I point out that your own two statements made in connection with the same question are not in congruence with each other, instead of just saying: "Ok, I'll give you this one", and move on as if it doesn't warrant clarification and leave the other person bewildered.

 

[NTesla]

The "it" in that passage is the tube.

The top of the tube exerts a downward force to impart downward momentum to the chain.
The right of the tube exerts a leftward force to impart leftward momentum to the chain.

At some times during the scenario it is possible that the tension in the chain will be adequate to supply the requisite downward and leftward momentum. But not at all times. When the fall is nearly complete, the tension drops toward zero.


Yes. The required force on the incremental portion of chain in the mouth supplied by the resultant of the tensions and the mouth of the guide tube. The guide tube is a constraint. The resultant force is whatever it has to be to keep the chain on its prescribed trajectory. The trajectory is around a 90 degree bend with unchanged speed. So the required resultant force is at 45 degrees down and to the left.

The mouth is effectively a pulley. A pulley with a cover so that centrifugal force cannot fling the chain off of the pulley.

Yes.

Yes.

Yes. Nicely cancelling with the force from the table.

The formula seems qualitatively correct. Though it would be nice to see a justification.

Apparently you are working a force balance on the hanging portion of the chain. The acceleration of that piece is given by $a = \frac{g-t}{\lambda h}$. The interactions at the table and at the mouth of the guide tube cancel each other out. If we solve for $t$ we get $t = (a - g) \lambda h$ which matches your formula.

So yes, I agree.

Much appreciated. Thank you.

However, now the equation that I'm writing for the hanging part of the chain is cancelling out all together.
Here's the calculation:
in the $\hat j$ direction: $\vec {\Delta P} = \lambda h(v + \Delta v)(-\hat j) - \lambda hv(-\hat j)$
= $\lambda h\Delta v(-\hat j)$.

Force in the $\hat j$ direction = $(-\lambda hg + \lambda \frac{\Delta x}{\Delta t}v + \lambda h(g - \frac{dv}{dt}) - \lambda \frac {\Delta x}{\Delta t}v)\hat j$ = $-\lambda h\frac {dv}{dt}\hat j$

Now, equating change in momentum to force, we get: $\lambda h \frac {dv}{dt} = \lambda h \frac {dv}{dt}$

Am I missing something here ?

I have derived the correct differential equation for the horizontal part of the chain. But when Im trying to do that same for the vertical part of the chain, as shown above, it's coming to an impasse. I do understand that solving for the horizontal part of the chain solves the question, but I'm thinking that one should be able to solve the question by focusing on the hanging part of the chain as well.

 

[NTesla]

As pointed out by @Steve4Physics, mechanical energy is not conserved. However, you can still solve it using energy concepts.

(1) Find an expression for the KE of the chain at the instant the upper end of the chain has moved a distance $x$ from point $A$. Express in terms of $\lambda$, $x$, $l$, and the speed $v$ at that instant.

(2) Find an expression for the potential energy $U$ of the chain at the same instant. Express in terms of $\lambda$, $x$, $g$, $h,$ and $U_0$, where $U_0$ is the initial potential energy of the chain when it was released.

(3) Find an expression for the rate at which mechanical energy is lost at point $B$. Assume that each link makes a completely inelastic collision with the surface at $B$. Express in terms of $\lambda$ and $v$.

Consider how these three expressions are related.

Yes, thank you. That seems plausible.

Here's my attempt using energy method:
K.E of the chain, at the instant the upper end of the chain has moved a distance $x$ from point $A$ = $\frac {1}{2} \lambda (l - x)v^{2}$.
Potential Energy $U$ of the chain at the same instant = $\lambda (l - h - x)gh + \frac {1}{2}\lambda gh^{2}$.

Now, rate of loss of Mechanical energy = rate of work done by the force that the table is exerting on the element of the chain that is coming to rest after having fallen on the table.
Force exerted by the table on the element which is coming to rest on it = $\lambda \frac {\Delta x}{\Delta t} v$.
Rate of work done by this force = $\frac {d}{dt} (\lambda v^{2}dx)$ = $\lambda (v^{2} \frac {dx}{dt} + dx2v\frac {dv}{dt})$

Now, when I'm equating $\frac {d}{dt} K.E$ + $\frac {d}{dt} P.E$ = $\frac {d}{dt} (\lambda \frac {\Delta x}{\Delta t} v)$, I'm getting:

$\lambda (v^{2} \frac {dx}{dt} + dx2v\frac {dv}{dt})$ = $\lambda (v^{2}(v + \frac {dv}{dt} + vgx))$.

Now, I don't see this equation going any further. The 2nd term on the LHS is causing trouble. Even if I write $dx2v\frac {dv}{dt}$ as $2v^{2}dv$, the dv term remains. How to solve this.?

 

[TSny]

Here's my attempt using energy method:
K.E of the chain, at the instant the upper end of the chain has moved a distance $x$ from point $A$ = $\frac {1}{2} \lambda (l - x)v^{2}$.

Good.

Potential Energy $U$ of the chain at the same instant = $\lambda (l - h - x)gh + \frac {1}{2}\lambda gh^{2}$.

Ok. This is less of an eyesore if you write it as $U = U_0 -\lambda ghx$, where $U_0$ is the total potential energy at the moment of release ($x = 0$).

Now, rate of loss of Mechanical energy = rate of work done by the force that the table is exerting on the element of the chain that is coming to rest after having fallen on the table.

The force that the table exerts on the chain does not do any mechanical work since the force doesn't move the chain through any distance.

Consider a small element $dx$ of the chain at point $B$, which is just about to strike the table. Find an expression for the kinetic energy $dK$ contained in this element. If the speed of the chain at this instant is $v$, how much time $dt$ does it take for $dK$ to be "destroyed" (i.e., converted to heat). Write an expression for the rate at which mechanical energy is converted to heat at point B.

 

[Philip Wood]

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.

Thank you for your suggestion. I'm not writing multiple lines of LaTex between double dollar and double dollar. I am enclosing between double hash and double hash stuff that I want to appear in lines of text. I haven't omitted any $$ or ## delimiters, nor nested any. I'm completely baffled, especially since the first half of my post rendered as it should.

I've now discovered my mistakes. The LaTex error messages were totally misleading. I still can't find the Preview button; what, please, do you mean by 'the controls'? Are they local to my post, or higher up the page, or what?

 

[Steve4Physics]

$\lambda (v^{2} \frac {dx}{dt} + dx2v\frac {dv}{dt})$ = $\lambda (v^{2}(v + \frac {dv}{dt} + vgx))$.

FWIW, on the right-hand side you are adding quantities with different dimensions: $v$ has dimensions $LT^{-1}$, $\frac {dv}{dt}$ has dimensions $LT^{-2}$ and $vgx$ has dimensions $L^3 T^{-3}$.

 

[Philip Wood]

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

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

Sorry: I had problems using the site. Please see amended post.

 

[haruspex]

There's no reason to go on resolving the equations further if the 1st equation itself seems incomplete or wrong to begin with.

I do not see how it was clear that the last equation in post #52 was wrong without doing the simplification. Maybe you are better at mental manipulation than I am.

I expect you to atleast go through the equations once, before you jump to the conclusion that I've fallen into the classic trap or any other conclusion for that matter.

But you had. You had ignored the force from the ground bringing the chain links to rest. That is the same as ignoring the force that brings the leaking water to rest in the cart example.
It might not be all that was wrong, but it was certainly a crucial error and an important point to learn.

I also expect you to clarify further when I point out that your own two statements made in connection with the same question are not in congruence with each other, instead of just saying: "Ok, I'll give you this one", and move on as if it doesn't warrant clarification and leave the other person bewildered.

Sorry, I thought it was clear… I was saying I was confused somewhere and was unsure how to resolve it. Had hoped to get back to it but ran out of time. Is that one worth revisiting now or did someone else manage to sort it out to your satisfaction?

 

[Philip Wood]

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.

View attachment 367117
View attachment 367122
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: $v = \sqrt{2ghln(\frac{l}{h})}$.

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

Sorry – I had technical problems. Please see amended post.

 

[Philip Wood]

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.

Problems solved. Thank you so much for your help.