Chain falling out of a horizontal tube onto a table

[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.

 

[NTesla]

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.

If we consider an element of length $\Delta x$ at point B, which is just about to strike the table, then it's K.E at an instant $t$ = $\frac {1}{2}\lambda \Delta x v^{2}$. Let's suppose that at time $t + \Delta t$, it comes to a stop. Then in time $\Delta t$, $\Delta K.E = -\frac {1}{2}\lambda \Delta x v^{2}$.
Rate of change of it's K.E in the limit $\Delta t \rightarrow 0$=$-\frac {1}{2}\lambda v^{3}$.

But how much time does it take for $dK$ to be "destroyed" (i.e. converted to heat), I dont think I fully understand how to tackle this. If I'm already considering that in time $dt$, the K.E of this element is going to become zero, then how do I find this dt ? Will I have to use some other method to find out what this dt is going to be ?

Also, regarding finding P.E, I could write an expression of the element's P.E if I had considered it at a distance x units above the table, but since this element is already near the point B, will the expression derived in case of it being at x units above the table, be valid even though it is considered very near to point B ?

But, for the sake of argument, let's say this element is at $x$ units above the table, then it's P.E = $\lambda dx gx$.

Then $\frac {d}{dt} PE$ = $\lambda g \frac {d}{dt} xdx$ = $\lambda g (-xv -vdx)$.

Now, $\frac {d}{dt} (M.E)$ = $-\frac {1}{2}\lambda v^{3} - \lambda g xv - \lambda g vdx$. I don't understand how to remove the $dx$ term in the last term in RHS.

Now, considering the chain, it's K.E at any instant t = $\frac {1}{2}\lambda (l - x)v^{2}$.
Let at time $t$, the leftmost point of the horizontal portion of the chain is at $x$ units from it's starting position.
It's Potential energy U = $U_{0}$ - $\lambda ghx$.
Therefore, total M.E = $\frac {1}{2}\lambda (l - x)v^{2}$ + $U_{0}$ - $\lambda ghx$.
$\frac {d(M.E)}{dt} = - \lambda ghv + \frac {1}{2} \lambda (l2v\frac {dv}{dt} - x2v\frac {dv}{dt} - v^{3})$
= $- \lambda ghv + \lambda lv\frac {dv}{dt} - \lambda xv\frac {dv}{dt} - \frac {1}{2}\lambda v^{3}$.

Now, if i equate this $\frac {d(M.E)}{dt}$ with the $\frac {d(M.E)}{dt}$ derived for that element $dx$, I get $-gv(x + dx) = -ghv + lv\frac {dv}{dt} - xv\frac {dv}{dt}$, which clearly shows that something has gone terribly wrong somewhere or that I've misunderstood/miscalculated the rate at which M.E is lost at point B.
Help !

 

[haruspex]

But how much time does it take for dK to be "destroyed

You may have misunderstood what @TSny was asking, which was to write an expression for $dt$ in terms of $dK$. And you have that, in effect.
Besides, it was only a step to finding the rate of loss of KE, which you have also done.

Then $\frac {d}{dt} PE$ = $\lambda g \frac {d}{dt} xdx$ = $\lambda g (-xv -vdx)$.

Careful, you have lost a "d". You can’t turn $\frac d{dt}dx$ into $\frac {dx}{dt}$. It has resulted in an equation which combines infinitesimals with non-cancelling non-infinitesimals. As I mentioned, that shows you have gone wrong.

Instead of differentiating an expression for the energy of dx, find the total energy at an instant and differentiate that wrt t.

Edit:
Wrt $\frac d{dt}dx$, $dx$ is the length of an element at offset $x$. Over time, $x$ varies but $dx$ is constant (the chain being inextensible). So $\frac d{dt}dx=0$.

 

[TSny]

If we consider an element of length $\Delta x$ at point B, which is just about to strike the table, then it's K.E at an instant $t$ = $\frac {1}{2}\lambda \Delta x v^{2}$. Let's suppose that at time $t + \Delta t$, it comes to a stop. Then in time $\Delta t$, $\Delta K.E = -\frac {1}{2}\lambda \Delta x v^{2}$.
Rate of change of it's K.E in the limit $\Delta t \rightarrow 0$=$-\frac {1}{2}\lambda v^{3}$.

Your result $-\frac 1 2 \lambda v^3$ is correct for the rate at which KE is lost to heat at point B. I’m not sure of your thought process in going from $\Delta KE = - \frac 1 2 \lambda \Delta x v^3$ to the result for the rate $\frac{dKE}{dt}$. But it might be fine.

But how much time does it take for $dK$ to be "destroyed" (i.e. converted to heat), I dont think I fully understand how to tackle this. If I'm already considering that in time $dt$, the K.E of this element is going to become zero, then how do I find this dt ? Will I have to use some other method to find out what this dt is going to be ?

I was just trying to guide you a little with a way to get $- \frac 1 2 \lambda v^3.$ The time $dt$ that it takes an infinitesimal length $dx$ of the chain at B to strike the table is $dt = \text{distance}/\text{speed} = \frac{dx}{v}$. The KE in the infinitesimal segment is $dK =\frac 1 2 \lambda dx v^2$. So, $\frac{dK}{dt} = \frac {\frac 1 2 \lambda dx v^2}{\frac{dx}{v}} = \frac 1 2 \lambda v^3$. This is the rate at which KE is converted to heat at B.

Also, regarding finding P.E, I could write an expression of the element's P.E if I had considered it at a distance x units above the table, but since this element is already near the point B, will the expression derived in case of it being at x units above the table, be valid even though it is considered very near to point B ?

But, for the sake of argument, let's say this element is at $x$ units above the table, then it's P.E = $\lambda dx gx$.

Then $\frac {d}{dt} PE$ = $\lambda g \frac {d}{dt} xdx$ = $\lambda g (-xv -vdx)$.

I’m not following this. The only type of mechanical energy that is being converted to heat at B is kinetic energy. When an element $dx$ arrives at B, it essentially has zero potential energy. More accurately, it has a potential energy $dU = (\lambda dx)g(\frac {dx}{2})$, where $\frac{dx}{2}$ is the height of the center of mass of the element above the table. Since $dU$ is second order in $dx$, it can be neglected. So, the rate at which mechanical energy is being lost at B is $-\frac 1 2 \lambda v^3$, not $-\frac 1 2 \lambda v^3 -\lambda ghv$. [Edit: The last term $-\lambda ghv$ appeared in your post #76 at an earlier time. It no longer appears there, possibly due to an edit of yours. Anyway, the rate of loss of ME to heat at B is due to the conversion of the flux of KE at B into heat. PE does not contribute here.]

The rest of your post appears to be okay to me. See what you get if you make the above correction to the rate of loss of ME to heat.

 

[NTesla]

Your result $-\frac 1 2 \lambda v^3$ is correct for the rate at which KE is lost to heat at point B. I’m not sure of your thought process in going from $\Delta KE = - \frac 1 2 \lambda \Delta x v^3$ to the result for the rate $\frac{dKE}{dt}$. But it might be fine.


I was just trying to guide you a little with a way to get $- \frac 1 2 \lambda v^3.$ The time $dt$ that it takes an infinitesimal length $dx$ of the chain at B to strike the table is $dt = \text{distance}/\text{speed} = \frac{dx}{v}$. The KE in the infinitesimal segment is $dK =\frac 1 2 \lambda dx v^2$. So, $\frac{dK}{dt} = \frac {\frac 1 2 \lambda dx v^2}{\frac{dx}{v}} = \frac 1 2 \lambda v^3$. This is the rate at which KE is converted to heat at B.


I’m not following this. The only type of mechanical energy that is being converted to heat at B is kinetic energy. When an element $dx$ arrives at B, it essentially has zero potential energy. More accurately, it has a potential energy $dU = (\lambda dx)g(\frac {dx}{2})$, where $\frac{dx}{2}$ is the height of the center of mass of the element above the table. Since $dU$ is second order in $dx$, it can be neglected. So, the rate at which mechanical energy is being lost at B is $-\frac 1 2 \lambda v^3$, not $-\frac 1 2 \lambda v^3 -\lambda ghv$. [Edit: The last term $-\lambda ghv$ appeared in your post #76 at an earlier time. It no longer appears there, possibly due to an edit of yours. Anyway, the rate of loss of ME to heat at B is due to the conversion of the flux of KE at B into heat. PE does not contribute here.]

The rest of your post appears to be okay to me. See what you get if you make the above correction to the rate of loss of ME to heat.

It did solve the problem that I was facing. I had given up on solving the question using energy considerations, until your intervention. It allowed me to see the entire thing in a new way. So, Thank you very much. And the resulting differential equation did solve the question.

 

[NTesla]

You may have misunderstood what @TSny was asking, which was to write an expression for $dt$ in terms of $dK$. And you have that, in effect.
Besides, it was only a step to finding the rate of loss of KE, which you have also done.


Careful, you have lost a "d". You can’t turn $\frac d{dt}dx$ into $\frac {dx}{dt}$. It has resulted in an equation which combines infinitesimals with non-cancelling non-infinitesimals. As I mentioned, that shows you have gone wrong.

Instead of differentiating an expression for the energy of dx, find the total energy at an instant and differentiate that wrt t.

Edit:
Wrt $\frac d{dt}dx$, $dx$ is the length of an element at offset $x$. Over time, $x$ varies but $dx$ is constant (the chain being inextensible). So $\frac d{dt}dx=0$.

This post did clarify a crucial error in my calcualtion, especially the part written in edit. Much appreciated. Thank you.

 

[NTesla]

Is that one worth revisiting now or did someone else manage to sort it out to your satisfaction?

No, that particular issue in that question still remains unresolved. No one picked up that particular point.

 

[TSny]

It did solve the problem that I was facing. I had given up on solving the question using energy considerations, until your intervention. It allowed me to see the entire thing in a new way. So, Thank you very much. And the resulting differential equation did solve the question.

Very good. I'm glad I could help.

 

[NTesla]

@jbriggs444 , Kindly see post No. 67 and respond if and when possible.

 

[NTesla]

I still have a question.
When I am trying to solve this question by equating rate of change of momentum to the force applicable, then there is a slight conundrum that I'm facing.
Here's the details:
For the horizontal portion of the chain, $\frac {dp}{dt}$ = $\lambda ((l - x - h)\frac {dv}{dt} - v^{2})\hat i$.
And the force applicable on this horizontal portion of the chain is given by: Force due to tension + force provided by the mouth of the tube in the leftward direction.

Here's the calculation for deriving the force due to tension:
For the horizontal portion of the chain: T = $\lambda (l - h - x)a$
For the vertical portion of the chain, $\lambda hg - T = \lambda h a$.
From these equations, we get: T = $\lambda h(g - \frac {dv}{dt})\hat i$
Therefore, force due to tension = $\lambda h(g - \frac {dv}{dt})\hat i$ ........................(i)
and force provided by the tube's mouth in the leftward direction = $\lambda v^{2} (-\hat i)$.

Therefore, $\lambda ((l - x - h)\frac {dv}{dt} - v^{2})\hat i$ = $\lambda h(g - \frac {dv}{dt})\hat i$ + $\lambda v^{2} (-\hat i)$. ..................................................(ii)

However, now my question is that for deriving the force due to tension (eq. (i) above), we did not consider the force provided by the tube's mouth. But when we are writing eq.(ii) we did consider the force provided by tube's mouth. Why's that ?

Here's my argument and I'm wondering if I'm correct in my argument: While deriving the tension on the horizontal portion of the chain(eq (i)), we did not consider the force provided by the tube's mouth, because this force (the one by tube's mouth) is not getting transferred to the entire horizontal chain. The impulse due to the force(by tube's mouth) is only cancelling out the momentum of the infinitesimal element of the chain. So when we are writing equation (ii) above, wherein we are equating the rate of change of the momentum to the force applicable on the horizontal portion of the chain, then we must include the force provided by the tube's mouth since it is affecting the momentum of the chain, and therefore is affecting the rate of change of momentum(even though it's only affecting the momentum of the infinitesimal element of the chain).

Kindly let me know if the above argument is right or wrong.

 

[haruspex]

This post did clarify a crucial error in my calcualtion, especially the part written in edit. Much appreciated. Thank you.

Ok, good. But I do wonder what happens if it's an extensible string. Such cases can be confusing.

 

[jbriggs444]

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: $\vec {\Delta P} = \lambda h(v + \Delta v)(-\hat j) - \lambda hv(-\hat j)$
= $\lambda h\Delta v(-\hat j)$.

Let us reverse engineer that equation. It would have been nice to have you provide the justification for your own equations.

You are calculating the change in momentum ($\vec{\Delta P}$) for the falling portion of chain for a brief interval. Let us take it as a given that we are considering the vertical momentum only. We can work with scalars (positive up) and drop the $\hat j$.

The first term is $\lambda h (v + \Delta v)$. That is the obviously just the final momentum.
The second term is $\lambda h (v)$. That is obviously just the initial momentum.

So this equation is simply asserting that the change in momentum is equal to the final momentum minus the initial momentum.

Manifestly correct.

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

This one is trickier. Where have we drawn the lines around our system of interest?

I know where I would draw them. Right under the mouth and right above the table. In that case we would have four identifiable momentum flows in to (or out from) the system of interest. Those four being the mass flow out of the mouth of the tube, the downward force of gravity, the upward force from tension and the mass flow onto the table.

Let us see if your equation has those four contributions.

I see $-\lambda h g$. That is gravity. A force
I see $\lambda \frac{\Delta x}{\Delta t}v$. That is the table. A rate of change of momentum.
I see $-\lambda \frac{\Delta x}{\Delta t}v$. That is the mouth of the tube. A rate of change of momentum.
I see $\lambda h(g-\frac{dv}{dt})$. That is a way of phrasing the force of tension. That awkward phrasing is probably why you got a tautology out.

Yes, that all seems correct.

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

Yes, that is a tautology that you can derive.

I would have been be working toward an equation like $a = \frac{g-t}{\lambda h}$ for the hanging chain section and a similar equation for the horizontal chain section. Set the two accelerations and the two tensions equal and solve.

 

[haruspex]

And the force applicable on this horizontal portion of the chain is given by: Force due to tension + force provided by the mouth of the tube in the leftward direction.

It might help to treat the mouth in more detail. Say it is a smooth quadrant of radius $r$. We are interested in the limit as $r\rightarrow 0$.

At any instant, in that curve is a section of chain of mass $\lambda\pi r/2$ rotating at rate $v/r$ about the centre of curvature, O.
Since it is smooth, the tube exerts no tangential force on that section. The radial forces exert no torque about O. If the tensions at the ends are $T_h, T_v$ then:
$(T_v-T_h)r =( \lambda\pi r/2)(r^2)(\dot v/r)$
$T_v-T_h= \lambda\pi r\dot v/2\rightarrow 0$.

This allows us to model the movement of the chain without getting tangled in the details of the normal forces the mouth exerts on the chain. Those forces will be responsible for:

• slowing the horizontal speed of the elements entering the mouth
• increasing the vertical speed of the elements leaving the mouth
• translating between the vertical force exerted by $T_v$ and the horizontal one exerted by $T_h$.

Note that although the first two affect the magnitude of the third (the faster the flow, the less the tension) they do not change their equivalence in the limit. Note also that the force slowing the elements entering the curve is not an additional force on the horizontal section; it appears as a reduction in the tension compared to what it would be were the chain stationary.

Exactly how these normal forces are distributed around the arc would be a challenge to compute.

Next, consider the purely horizontal portion of the chain over time dt. Whether we consider the whole $l-x-h$ at time t or just the leftmost $l-x-h-v\cdot dt$ makes no material difference to the tension. The tension at its right hand end is only infinitesimally different from that at $v\cdot dt$ to the left of that point. The effect of that difference on the motion over time dt is therefore a second order infinitesimal.

Likewise at the top and bottom of the vertical portion.

What we are left with is $\lambda (l-x-h)\dot v=T$, $\lambda h\dot v=\lambda hg-T$.
$\lambda (l-x-h)\dot v=-\lambda h\dot v+\lambda hg$.
$(l-x)\dot v= hg$.

This is the same equation as obtained with @TSny’s energy method.

 

[NTesla]

Next, consider the purely horizontal portion of the chain over time dt. Whether we consider the whole l−x−h at time t or just the leftmost l−x−h−v⋅dt makes no material difference to the tension. The tension at its right hand end is only infinitesimally different from that at v⋅dt to the left of that point. The effect of that difference on the motion over time dt is therefore a second order infinitesimal.

(1) Although I agree with your assertion that tension at it's right hand end is only infinitesimally different from that at $v. dt$ to the left of that point, but I'm having trouble understanding what you meant when you mentioned that: "the effect of that difference on the motion over time $dt$ is therefore a second order infinitesimal". It's the mention of "second order infinitesimal", that I'm failing to understand.

Here's how I'm approaching it:

For the purely horizontal portion, I've calculated that tension at any point in the chain can be given by the equation, $T_{any point}$ = $\frac {mT}{l - h - x}$,
where $m$ is the distance of the point where Tension is required to be calculated, measured from the leftmost point of the horizontal chain itself. And $T$ is the tension at the rightmost portion of the purely horizontal chain.

When $m = (l - h - x ) - v.dt$, then Tension at that point, $T_{1}$ = $T - \frac {v.dt}{l - h - x} T$.

So, difference of Tension = $T - T_{1}$ = $\frac {v.dt}{l - h - x} T$ = $\frac {dx}{l - h - x} T$.
But hereafter, how do I calculate the effect of this difference in Tension on the motion over time $dt$ ?

If I start with $T_{1} = \lambda m \frac {dv}{dt}$, then when I substitute $T_{1} = (1 - \frac {dx}{l - h - x})T$, and $T = \lambda (l - h - x)\frac {dv}{dt}$ then immediately all terms cancel out.


(2)

Note also that the force slowing the elements entering the curve is not an additional force on the horizontal section; it appears as a reduction in the tension compared to what it would be were the chain stationary.

Regarding this paragraph that you've written: At an intuitive level, I understand that the Tension in the chain when it is moving would be less than the tension when the system has not yet begun to move.
But when the chain is in motion, I don't understand how the force which is slowing the element's horizontal motion, only "appears" to be doing something. The way I'm seeing it, this force is actually acting on the horizontal portion of the chain. So, I'm not able to find reason to say that it's not an additional force.

Although when we write the two force equations: $T = \lambda (l - h - x)a$ and $\lambda hg - T = \lambda ha$ we have definitely not taken that aforementioned force on the element into account, and these 2 equations do solve the question. But still I feel that I don't understand why the aforementioned force on the element is not an additional force on the horizontal portion of the chain.

 

[haruspex]

It's the mention of "second order infinitesimal", that I'm failing to understand.

The effect of a force F on a mass m in time $dt$ is to accelerate it by $dv=F\cdot dt/m$. If instead the force is $F-dF$ then the velocity increase is only $(F-dF)\cdot dt/m$, a decrease of $dF\cdot dt/m$. That’s what I mean by a second order infinitesimal.
Writing out the full equation for the acceleration of the horizontal portion, there are non-cancelling first order infinitesimals. That means we can discard the second order ones.

The way I'm seeing it, this force is actually acting on the horizontal portion of the chain.

The reason I broke it into three parts - horizontal, vertical and quadrant - was so that the element being retarded is not included in the horizontal portion. The retarding force necessarily acts on that element. The horizontal portion does not feel that force directly. What the horizontal portion feels is less tension than there would have been if there were no retardation.

… when we write the two force equations: $T = \lambda (l - h - x)a$ and $\lambda hg - T = \lambda ha$ we have definitely not taken that aforementioned force on the element into account,

Yes we have. It is incorporated in $T$ as a reduction in its value. This is not directly visible in the equations because the same reduction occurs in the horizontal and vertical instances of $T$.