What Is the Incorrect Statement About the Dynamics of a Falling Chain?

[Saitama]

Homework Statement

One end of the chain falls through a hole in its support and pulls the remaining links after it in a steady flow. If the links which are initially at rest, acquire the velocity of the chain suddenly and without frictional resistance or interference from the support or from adjacent links. Choose the INCORRECT statement. (when x=0, then v=0) (length of chain is L and $\rho$ is the mass per unit length of the chain).

A) the velocity v of the chain as a function of x is $\sqrt{\frac{2gx}{3}}$.

B) the acceleration of a of the falling chain as a function of x is $\frac{g}{3}$.

C) the energy Q lost from the system as the last link leaves the platform is $\frac{\rho gL^2}{6}$.

D) tension at the middle point of the falling chain is $\frac{\rho gx}{3}$.

Homework Equations

The Attempt at a Solution

I am completely lost on this one. I could only write the following (from Newton's second law on falling part of chain):

$$\rho gx-T=\rho xa$$
where T is the tension acting on the part just getting out of the support. How do I write more equations?

Any help is appreciated. Thanks!

 

[voko]

I would use conservation of energy here.

 

[Saitama]

I would use conservation of energy here.

At what instants do I use conservation of energy? When the x length is out and the other when a dx part comes out, right?

 

[voko]

With x out, consider the potential/kinetic energy balance.

 

[Saitama]

With x out, consider the potential/kinetic energy balance.

What does "potential/kinetic energy balance" mean?

If you ask me to write down the energy when x length is out, this is what I get:

$$-\rho gx \frac{x}{2}+\frac{1}{2}\rho xv^2$$
where U=0 is at the support and v is the velocity of falling chain.

 

[voko]

Very well, but why did you not write the entire equation for conservation of energy? Do you think it is not conserved here?

 

[Saitama]

Very well, but why did you not write the entire equation for conservation of energy? Do you think it is not conserved here?

I did not understand what you meant by "potential/kinetic energy balance". I agree that energy is conserved.

Initial energy of system is zero. Equating it with what I wrote above, I get

$$v=\sqrt{gx}$$

which is wrong. The answer states D as the answer which means other options are correct.

 

[voko]

Hmm. It seems that with that assumption, more than one statement is incorrect. Yet I do not immediately see why energy would not be conserved here.

 

[Saitama]

Hmm. It seems that with that assumption, more than one statement is incorrect. Yet I do not immediately see why energy would not be conserved here.

I have the solutions but I don't want to look at them before I reach the answers. Should I open the solution booklet and post the solution if you find something wrong here?

 

[voko]

I think it is because links are supposed to acquire the velocity of the chain suddenly. That means there is an impulse accelerating them down, which does non-zero work.

Then conservation of energy is not usable here.

 

[Saitama]

I think it is because links are supposed to acquire the velocity of the chain suddenly. That means there is an impulse accelerating them down, which does non-zero work.

Then conservation of energy is not usable here.

So how should I proceed now?

 

[voko]

You could try treating that as a variable mass system, where mass is acquired, plus everything is accelerated by gravity.

 

[Saitama]

You could try treating that as a variable mass system, where mass is acquired, plus everything is accelerated by gravity.

At any instant, momentum of system is $\rho xv$.
$$\Rightarrow \rho gx=\rho \frac{d(xv)}{dt}$$
$$\Rightarrow gx=v\frac{dx}{dt}+x\frac{dv}{dt}$$
I can substitute $dx/dt=v$, but then how do I solve the D.E, I have three variables.

 

[D H]

The question asks for things such as velocity v of the chain as a function of x. Time is not mentioned. This suggests you want to find velocity and acceleration as functions of x rather than as functions of t.

 

[Saitama]

The question asks for things such as velocity v of the chain as a function of x. Time is not mentioned. This suggests you want to find velocity and acceleration as functions of x rather than as functions of t.

I am confused, can you please give a few more hints?

The only thing I can think of is this:

$$gx\,\, dt=d(xv)$$
I am thinking of integrating the above and replacing $\int_0^t xdt$ with v. Does that sound good?

I am confused at one more point. I wrote
$$\rho gx=\rho\frac{d(xv)}{dt}$$

Shouldn't that be
$$\rho gx-T=\rho \frac{d(xv)}{dt}$$

 

[D H]

You need to get rid of the dependence on time. You're making it worse!

Hint: By the chain rule, $\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}$.

 

[Saitama]

You need to get rid of the dependence on time. You're making it worse!

Hint: By the chain rule, $\frac{dv}{dt} = \frac{dv}{dx}\frac{dx}{dt} = v\frac{dv}{dx}$.

Ah yes, I forgot about that. Can I blame that on my bad health?

So we have the equation,
$$gx=v^2+xv\frac{dv}{dx} \Rightarrow gx=v\left(v+x\frac{dv}{dx}\right)=v\frac{d(xv)}{dx}$$
$$\Rightarrow gxdx=vd(vx)$$
Multiplying by x on both sides,
$$\Rightarrow gx^2dx=vxd(xv) \Rightarrow \int_0^xgx^2dx=\int_0^{xv} d(xv)$$
$$\frac{gx^3}{3}=\frac{x^2v^2}{2}$$
$$v=\sqrt{\frac{2gx}{3}}$$

Does this look correct?

Can you please clear my doubt I posted in my previous reply?

 

[TSny]

So we have the equation,
$$gx=v^2+xv\frac{dv}{dx} \Rightarrow gx=v\left(v+x\frac{dv}{dx}\right)=v\frac{d(xv)}{dx}$$
$$\Rightarrow gxdx=vd(vx)$$
Multiplying by x on both sides,
$$\Rightarrow gx^2dx=vxd(xv) \Rightarrow \int_0^xgx^2dx=\int_0^{xv} d(xv)$$
$$\frac{gx^3}{3}=\frac{x^2v^2}{2}$$
$$v=\sqrt{\frac{2gx}{3}}$$

Does this look correct?

Yes. That looks good to me (except for a typo in the last integral where you left out part of the integrand)

I am confused at one more point. I wrote
$$\rho gx=\rho\frac{d(xv)}{dt}$$

Shouldn't that be
$$\rho gx-T=\rho \frac{d(xv)}{dt}$$

$T$ represents the tension at what point?

 

[haruspex]

So we have the equation,
$$gx=v^2+xv\frac{dv}{dx} \Rightarrow gx=v\left(v+x\frac{dv}{dx}\right)=v\frac{d(xv)}{dx}$$
$$\Rightarrow gxdx=vd(vx)$$
Multiplying by x on both sides,
$$\Rightarrow gx^2dx=vxd(xv) \Rightarrow \int_0^xgx^2dx=\int_0^{xv} d(xv)$$
$$\frac{gx^3}{3}=\frac{x^2v^2}{2}$$
$$v=\sqrt{\frac{2gx}{3}}$$

Yes, that looks fine.
This is very similar to https://www.physicsforums.com/showthread.php?t=669505, which you posted last Feb. It caused a schism between those who insisted work would be conserved, and those (including me) who declared it wasn't.
When you're given the form of the answer ($v=\sqrt{\frac{2gx}{3}}$), you can use shortcuts. Clearly that answer means x = Atⁿ, some A and n. So just plug that into your ODE and see what A and n satisfy it.

 

[D H]

Shouldn't that be
$$\rho gx-T=\rho \frac{d(xv)}{dt}$$

No, it shouldn't.

The reason is that tension is an internal force. By Newton's third law, the net internal force acting on an object is zero. It's only the external forces that count with regard to Newton's second law.

 

[Saitama]

$T$ represents the tension at what point?

T represents the tension acting on the part of chain just coming out of the support.

No, it shouldn't.

The reason is that tension is an internal force. By Newton's third law, the net internal force acting on an object is zero. It's only the external forces that count with regard to Newton's second law.

I don't get how this is an internal force. Do I apply F=dp/dt on the part hanging out of the support or the part hanging plus the part still above the support?

 

[haruspex]

Shouldn't that be
$$\rho gx-T=\rho \frac{d(xv)}{dt}$$

That would effectively count T twice over.
In time dt, an element of chain mass ρ v dt is accelerated from 0 to speed v. If we regard T as the force responsible, we get T = ρ v².
Now look at the derivative of the momentum:
$\rho \frac{d(xv)}{dt} = \rho v^2 + \rho x\dot v$
The RHS consists of the force to accelerate the new element plus the force to accelerate the existing vertical part of the chain.

 

[D H]

T represents the tension acting on the part of chain just coming out of the support.

That is zero. There is no tension at the top of the chain. From the opening post, "the links which are initially at rest, acquire the velocity of the chain suddenly and without frictional resistance or interference from the support or from adjacent links."

This is a non-physical problem.

 

[Saitama]

That would effectively count T twice over.
In time dt, an element of chain mass ρ v dt is accelerated from 0 to speed v. If we regard T as the force responsible, we get T = ρ v².
Now look at the derivative of the momentum:
$\rho \frac{d(xv)}{dt} = \rho v^2 + \rho x\dot v$
The RHS consists of the force to accelerate the new element plus the force to accelerate the existing vertical part of the chain.

Thanks haruspex! I understand it through the equations but how D H found that tension is zero using the problem statement? How does those statements imply that tension is zero?

That is zero. There is no tension at the top of the chain. From the opening post, "the links which are initially at rest, acquire the velocity of the chain suddenly and without frictional resistance or interference from the support or from adjacent links."

 

[haruspex]

Thanks haruspex! I understand it through the equations but how D H found that tension is zero using the problem statement? How does those statements imply that tension is zero?

I didn't follow D H's reasoning either.

 

[D H]

Thanks haruspex! I understand it through the equations but how D H found that tension is zero using the problem statement? How does those statements imply that tension is zero?

This response is late, but this question has come up again, this time in the context of a question from Classical Dynamics of Particles and Systems (Marion).

Whether the tension at the top of the falling chain is zero or non-zero depends on what is meant by "the top of the falling chain". In other words, it depends on where one draws the system boundary. With dynamic mass questions where one draws the system boundary is of extreme importance.

Suppose that at some point in time t the n links that are already falling are just about to pull the n+1^st through the hole. If you draw the system boundary so that it encloses only those n links that are already falling then there is a non-zero tension T at the top of the chain is non-zero, and the net force in the system is mg-T. The tension T is the force needed to bring the new link up to speed.

On the other hand, if you draw the system boundary so that it encloses the n links that are already falling plus that new link that the falling chain is about to pull through the hole, then the tension at the top of the chain is zero. That new link will pull on the next link a tiny bit of time later, but for now that next link might as well not exist. With this system boundary, the only external force is gravity, but now one has to calculate what is internally needed to bring that extra link up to speed.

Either approach will yield $v=\sqrt{2gx/3}$ -- assuming infinitesimally small links, that is. This becomes a nasty difference equation as opposed to a nice differential equation if the links are not infinitesimally small.

 

[Arun Sharma]

Sir is it not that the tension T is an external force for the system of hanging part of the chain?

 

[Arun Sharma]

If we put tension zero then it becomes the case of free fall which it can't.

 

[haruspex]

If we put tension zero then it becomes the case of free fall which it can't.

As D H wrote, it depends exactly what you mean by the top of the chain.
Think of the chain as being in three parts: the hanging part descending at some current speed, the reservoir resting on the table, and, connecting the two, a tiny section currently being accelerated rapidly from 0 to the speed of the descending section.
The tension between the second and third is zero, but the tension between the first and third is not.