Analyzing Velocity and Force of a Chain in Motion

[NTesla]

Homework Statement

A chain AB of length l is located in a smooth horizontal tube so that its fraction of length h hangs freely and touches the surface of a table with its end B . At certain moment the end A of the chain is set free. With what velocity will this end of the chain slip out of the tube ?

Relevant Equations

I've worked out the equation, but could somebody please let me know whether I've made any mistake or the equation is ok ? Also please let me know why the equation is wrong if it is wrong at any place.

My solution: For the horizontal portion of the chain: let at any instant the length of chain inside the tube is x, and at that instant the chain in the tube is having a velocity v. Then, at any instant:

$F = \frac{\mathrm{d} p}{\mathrm{d} t}$

$p$= mass of the chain in the tube at the instant t, multiplied by its velocity at that instant, i.e $v$.

mass of the total length $l$ of the chain = $m$.

mass of chain inside the tube = $\frac{m}{l}\times x$

velocity of the chain inside the tube = $-\frac{\mathrm{d} x}{\mathrm{d} t}$

Now $F= \frac{\mathrm{d} (mv)}{\mathrm{d} t} (General\;equation)$

$\therefore F= \frac{m\mathrm{d} v}{\mathrm{d} t} + \frac{v\mathrm{d} m}{\mathrm{d} t} (General\;equation)$

$\therefore$ $F = \frac{mx}{l}\frac{\mathrm{d} v}{\mathrm{d} t} + \frac{vm}{l}\frac{\mathrm{d} x}{\mathrm{d} t} (In\;the\;present\;question)$For the hanging part of the chain(which is of length h):

$\frac{mhg}{l} - F=\frac{mh}{l}\frac{\mathrm{d} v}{\mathrm{d} t}$

$\therefore \frac{mhg}{l}-\frac{mh}{l}\frac{\mathrm{d} v}{\mathrm{d} t}=\frac{m}{l}\left ( x\frac{\mathrm{d} v}{\mathrm{d} t}-v^2 \right )$

$\Rightarrow hg=x\frac{\mathrm{d} v}{\mathrm{d} t}+h\frac{\mathrm{d} v}{\mathrm{d} t}-v^2$

$\Rightarrow hg=-xv\frac{\mathrm{d} v}{\mathrm{d} x}-hv\frac{\mathrm{d} v}{\mathrm{d} x}-v^2$

Please let me know, if there is any mistake till here. If there is, then also let me know what is that mistake, and your reasoning.

 

[Lnewqban]

Interesting problem!
You only have gravity acceleration acting on $h$ length of chain of unknown linear density, but on a variable mass at the same time.
You are asked to calculate instantaneous velocity when mass reaches a minimum.

 

[haruspex]

Whenever I see $F=\frac{d(mv)}{dt}$ I get suspicious. The trouble is that mass is neither created nor destroyed, so a nonzero $\dot m$ implies mass is being added to or removed from the subsystem to which the equation applies. This raises the possibility that that mass change is also transferring momentum.
If F is the tension at the bend in the tube and the horizontal portion has mass $\frac{mx}l$ then surely $F=\frac{mx}l\frac{dv}{dt}$?

 

[NTesla]

In the equation:$F=\frac{mx}{l}\frac{\mathrm{d} v}{\mathrm{d} t} + \frac{vm}{l}\frac{\mathrm{d} x}{\mathrm{d} t}$, there must be solid grounds for rejecting the second term of the expression on the right hand side. In the 2nd term on the RHS, v is NOT equal to zero, $\frac{\mathrm{d} x}{\mathrm{d} t}$ is NOT equal to zero. Then, based on what reasoning should the 2nd term on the RHS be neglected/cancelled.

 

[haruspex]

In the equation:$F=\frac{mx}{l}\frac{\mathrm{d} v}{\mathrm{d} t} + \frac{vm}{l}\frac{\mathrm{d} x}{\mathrm{d} t}$, there must be solid grounds for rejecting the second term of the expression on the right hand side. In the 2nd term on the RHS, v is NOT equal to zero, $\frac{\mathrm{d} x}{\mathrm{d} t}$ is NOT equal to zero. Then, based on what reasoning should the 2nd term on the RHS be neglected/cancelled.

You need to define what m refers to in $\frac{d(mv)}{dt}$.
$F=\frac{d(mv)}{dt}$ applies to a specified subsystem acted on by a force F. You cannot apply it with m=m(t) defined as the mass of the horizontal portion of chain at time t since that is not a fixed subsystem.
It is ok with m taken as the horizontal portion at a single instant of time. In the next dt, that portion accelerates from $v$ by $\frac{F} mdt$, but $vdt$ of that length is now past the bend, so the momentum of the horizontal portion goes from $mv$ to $(m+dm)(v+dv)=mv+v.dm+m.dv=mv+v.dm+F.dt$.
So if p=p(t) represents the momentum of the horizontal portion, $\frac{dp}{dt}=v\dot m+F$, whence $F=m\dot v$.

Section 6 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-momentum-impacts/ makes the same point.
Also, at https://en.m.wikipedia.org/wiki/Newton%27s_laws_of_motion#Newton's_second_law, note where it states "Newton's second law is valid only for constant-mass systems".

 

[NTesla]

You need to define what m refers to in $\frac{d(mv)}{dt}$.
$F=\frac{d(mv)}{dt}$ applies to a specified subsystem acted on by a force F. You cannot apply it with m=m(t) defined as the mass of the horizontal portion of chain at time t since that is not a fixed subsystem.
It is ok with m taken as the horizontal portion at a single instant of time. In the next dt, that portion accelerates from $v$ by $\frac{F} mdt$, but $vdt$ of that length is now past the bend, so the momentum of the horizontal portion goes from $mv$ to $(m+dm)(v+dv)=mv+v.dm+m.dv=mv+v.dm+F.dt$.
So if p=p(t) represents the momentum of the horizontal portion, $\frac{dp}{dt}=v\dot m+F$, whence $F=m\dot v$.

Section 6 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-momentum-impacts/ makes the same point.

$F= \frac{\mathrm{d} (mv)}{\mathrm{d} t}$ is used in deriving rocket propulsion equations, in which mass varies with time, which is NOT a fixed system.
The part of the chain inside the tube is almost like a rocket(except having an engine of itself), being propelled by a constant force F(which is provided by the weight of the hanging part of the chain), and the mass of the chain inside the tube keeps on decreasing with time. So $F= \frac{\mathrm{d} (mv)}{\mathrm{d} t}$ is the most appropriate form of equation to be applied, in my opinion.
I still don't understand the reasoning behind not using this equation to begin with.

 

[haruspex]

$F= \frac{\mathrm{d} (mv)}{\mathrm{d} t}$ is used in deriving rocket propulsion equations, in which mass varies with time, which is NOT a fixed system.
The part of the chain inside the tube is almost like a rocket(except having an engine of itself), being propelled by a constant force F(which is provided by the weight of the hanging part of the chain), and the mass of the chain inside the tube keeps on decreasing with time. So $F= \frac{\mathrm{d} (mv)}{\mathrm{d} t}$ is the most appropriate form of equation to be applied, in my opinion.
I still don't understand the reasoning behind not using this equation to begin with.

Yes, it can be used in rocket equations, but note e.g. at https://en.m.wikipedia.org/wiki/Tsiolkovsky_rocket_equation#Most_popular_derivation:
"Newton's second law of motion relates external forces (

) to the change in linear momentum of the whole system (including rocket and exhaust)".
I.e. it takes rocket+unburnt fuel+exhausted fuel as a closed system, which therefore has no change in momentum. Thus, the change in momentum of a portion of fuel as it is exhausted is equal and opposite to that of the remaining vessel and fuel.
The flaw in your analysis is that you do not consider the momentum carried away by the 'exhaust', i.e. the mass that has moved around the bend.

 

[TSny]

I still don't understand the reasoning behind not using this equation to begin with.

Consider a fairly trivial example. Suppose you have a rod that slides through a pipe that is fixed on a horizontal table. Assume no friction anywhere. No external forces are applied to the rod other than the force of gravity and the normal force. So the rod slides with constant speed and there is no tension at any point in the rod.

Does the equation $F = \frac{dp}{dt}$ apply if $p$ denotes the momentum within the pipe?

 

[etotheipi]

That is really quite an excellent example @TSny! I hope I don't spoil the fun for the OP, but the key is that the momentum inside some defined boundary can change either due to the action of a force, or mass crossing through that boundary! It is nicely described by applying the Reynolds' transport theorem,$$\frac{d\vec{p}_{\mathrm{sys}}}{dt} = \frac{d\vec{p}_{\mathrm{cv}}}{dt} + \frac{d\vec{p}_{\mathrm{in}}}{dt} - \frac{d\vec{p}_{\mathrm{out}}}{dt}$$Since $\frac{d\vec{p}_{\mathrm{sys}}}{dt} = \vec{F}_{\text{ext}}$, and $\frac{d\vec{p}_{\text{in}}}{dt} = \vec{0}$, then$$\vec{F}_{\text{ext}} =\frac{d\vec{p}_{\mathrm{cv}}}{dt} - \frac{d\vec{p}_{\mathrm{out}}}{dt} = \frac{d(m\vec{v})}{dt} - \frac{dm}{dt}\vec{v} = m\frac{d\vec{v}}{dt}$$which in this case, is consistent with $\frac{d\vec{v}}{dt} = \vec{0}$ and $\vec{F}_{\text{ext}} = \vec{0}$.

 

[haruspex]

@NTesla , wrt to solving the chain problem, I recommend considering the whole moving part of the chain. It is easy to write down its mass and the force on it.

 

[NTesla]

Consider a fairly trivial example. Suppose you have a rod that slides through a pipe that is fixed on a horizontal table. Assume no friction anywhere. No external forces are applied to the rod other than the force of gravity and the normal force. So the rod slides with constant speed and there is no tension at any point in the rod.

Does the equation $F = \frac{dp}{dt}$ apply if $p$ denotes the momentum within the pipe?

View attachment 270096

The equation $F = \frac{dp}{dt}$ does apply, in the case you've mentioned, but since there is no external force in horizontal direction, and since there is nothing to slow down the velocity or to change its direction, therefore, p is constant. Thus, the equation $F = \frac{dp}{dt}$ is satisfied even if we take only the part of the chain inside the tube or even when we take the whole chain.

 

[TSny]

The equation $F = \frac{dp}{dt}$ does apply, in the case you've mentioned, but since there is no external force in horizontal direction, and since there is nothing to slow down the velocity or to change its direction, therefore, p is constant.

If $p$ represents the momentum contained within the pipe, then $p$ is decreasing since the amount of rod in the pipe is decreasing.

 

[NTesla]

If $p$ represents the momentum contained within the pipe, then $p$ is decreasing since the amount of rod in the pipe is decreasing.

Oh yes.. my mistake..

 

[NTesla]

@NTesla , wrt to solving the chain problem, I recommend considering the whole moving part of the chain. It is easy to write down its mass and the force on it.

Why only the moving part of the chain. The part which has come to rest on the table carried off some momentum which it imparted to the table and then it came to rest due to the impulsive force provided by the table. Should this part(which had carried off some momentum) not be considered while writing the equation. In all the answers to this question that I've seen online, all the solution have taken only the moving part of the chain into their equation. So, is there anything wrong in considering the whole chain (including the momentum carried off by the chain which is now at rest on the table) ?

 

[haruspex]

The part which has come to rest on the table carried off some momentum which it imparted to the table and then it came to rest due to the impulsive force provided by the table.

Yes, but I'm not suggesting working in terms of momentum. Just think of that mass and the force on it and calculate its acceleration.
If it helps your intuition, imagine that the chain is cut just above the table. Since the chain on the table has no effect on the rest of it, this cut doesn't alter how the moving part accelerates.

 

[NTesla]

@haruspex The links provided by you in your earlier comments to the original question have been extremely helpful. Thank you so much..

But, I'm still wondering if the parts which carried off some momentum before coming to rest on the table should be included in the equations..