Find Force to Lift Chain of Length L and Mass ρ Up

[Dick]

It is very convincing 2 cents :) But what is the problem with the energy then? ehild

It's what DH noticed. Suppose the chain has length L. Then at the moment it all lifts off the platform it's center of mass is at L/2 so the potential energy is hmg=(L/2)(Lρ)g. It's kinetic energy is (1/2)mu^2=(1/2)(Lρ)u^2, but if you integrate F you get ρu^2L+(L/2)(Lρ)g. So you put more energy into lifting the chain then there is in the center of mass kinetic energy plus potential energy of the chain. Physically you can wave your hands and say it goes into sound and heat due to damping. If you idealize the chain to be silent and friction free, as DH suggested, and think of an otherwise realistic chain, then the answer would be that the links must be oscillating without any damping. So the extra kinetic energy must be in the wiggling chain. The equation doesn't tell you exactly where the extra energy went, but it does say that it has to happen.

 

[D H]

Regarding energy, I suspect that when each link in jerked off the platform, it's like an inelastic collision- causing internal energy to increase. The chain will get warmer!

No, it won't. At least not to the extent suggested by the erroneous F=dp/dt solution. To see why this is nonsense, suppose we reverse the process. Instead of raising the chain, we'll lower it. Pile the chain up in one big lump, carry it up en masse to some height, and lower the chain. The F=dp/dt solution yields an over unity energy gain. Free energy! Or maybe it's just a mistake.

 

[Dick]

No, it won't. At least not to the extent suggested by the erroneous F=dp/dt solution. To see why this is nonsense, suppose we reverse the process. Instead of raising the chain, we'll lower it. Pile the chain up in one big lump, carry it up en masse to some height, and lower the chain. The F=dp/dt solution yields an over unity energy gain. Free energy! Or maybe it's just a mistake.

I don't think the dp/dt solution is erroneous. It sounds like a reasonably accurate model of actually lifting a chain. If we unmagic your magical universe picture so we have a machine on the platform attaching links to the chain as it rises then act of attaching each link is basically an inelastic collision between the chain and link. Energy must be lost from the kinetic+potential sum.

 

[TSny]

But why? ehild

Suppose we have just 2 links each of mass m on a horizontal frictionless surface as shown. The right link is already in motion with speed u while the left link is still at rest. Let F Δt be the impulse that the right link gives to the left link to get the left link moving also at speed u. So, F Δt = mu. There will be a reaction impulse of equal magnitude acting to the left on the right link which we arrange to be exactly balanced by an external impulse (from the blue force) so that the right link maintains speed u. Then the work done by the blue force is F Δx = F(uΔt) =(F Δt) u = mu u = mu² = twice the increase in KE of the system. So, more work was done by the blue force than shows up as KE.

 

[D H]

It's what DH noticed. ... The equation doesn't tell you exactly where the extra energy went, but it does say that it has to happen.

You completely missed my point.

I'll be very explicit. Using an F=dp/dt approach and attributing all of that change in momentum to the action by the hoist yields a nonsense answer. This is one of many cases where naively using F=dp/dt is flat out wrong. It's F=ma, not F=dp/dt.

The heating implied by that F=dp/dt approach is ludicrous. Even more ludicrous is the cooling that would result from reversing the process were this approach correct. Even yet more ludicrous is the fact that this erroneous approach leads to an over unity device.

 

[Dick]

Suppose we have just 2 links each of mass m on a horizontal frictionless surface as shown. The right link is already in motion with speed u while the left link is still at rest. Left F Δt be the impulse that the right link gives to the left link to get the left link moving also at speed u. So, F Δt = mu. There will be a reaction impulse of equal magnitude acting to the left on the right link which we arrange to be exactly balanced by an external impulse (from the blue force) so that the right link maintains speed u. Then the work done by the blue force is F Δx = F(uΔt) =(F Δt) u = mu u = mu² = twice the increase in KE of the system. So, more work was done by the blue force than shows up as KE.

Now that I agree with. Picking up the links is an inelastic process.

 

[D H]

Now that I agree with. Picking up the links is an inelastic process.

Of course it is. The correct answer is that F\ge \rho(gx+\frac 1 2 u^2), with equality resulting only if the process is reversible (i.e., picking up the chain is an elastic process).

There's not one thing in the F=dp/dt approach that suggests an irreversible process. In fact, this erroneous approach yields a reversible process. Simply apply this exact same concept but with the chain being lowered rather than raised.

 

[tms]

This is one of many cases where naively using F=dp/dt is flat out wrong. It's F=ma, not F=dp/dt.

What!? F = dp/dt is the definition of force. F = ma only applies when the mass is constant. In this problem, since the speed is constant, there is no acceleration, and, according to F = ma, there is no force.

 

[TSny]

We can replace the linked chain by an “elastic” system. Two masses (each m) are connected by a spring as shown. Initially the spring is unstretched with the mass on left at rest and the mass on right moving to the right with speed u. Also we have arranged a variable force F acting on the right mass that will always have the same magnitude as the spring force. Thus, the right mass always moves with the same speed u. The force F acts between t = 0 and t = t_f where t_f is chosen to be the instant when the velocity of the center of mass (cm) of the two-mass system is u.

The total momentum at t_f will be $(M_{system})V_{cm} = (2m)u = 2mu$. Since we started at time t = 0 with total momentum $mu$, the change in momentum of the system between t = 0 and t = t_f is $mu$. This change in momentum must come from the impulse of the applied force:

$\int{Fdt} = mu$.

The energy of the system at t = 0 is $E_o =mu^2/2$.

The energy of the system at t = t_f is $E_f = KE_{system} + PE_{spring}$.

We can always write the KE of the system as KE due to motion of the cm plus KE relative to cm. The KE due to motion of the cm is

$(1/2)(2m)V_{cm}^2 = mV_{cm}^2 = mu^2$ at t_f.

The KE relative to cm is just the “vibrational” kinetic energy associated with the rate of separation of the two masses. We combine the vibrational KE with the spring PE into a total “internal energy”. So, our energy at time t_f is $E_f = mu^2 +E_{int}$.

Hence, the change in energy from t = 0 to t = t_f is

$\Delta E = mu^2/2 + E_{int}$.

The work done by the applied force between t = 0 and t = t_f is

$W = \int{Fdx} = \int{F\frac{dx}{dt}dt} = \int{Fudt} = u\int{Fdt} = u(mu) = mu^2$

Suppose for some reason we didn’t know about the “internal energy”. (Maybe the spring is hard to see and it’s very stiff so that the vibrational amplitudes are two small for us to notice.) Then we would only be aware of the energy due to overall motion of the center of mass. Thus, we would claim that the change in energy of the system between t = 0 and t = t_f would be just the term $mu^2/2$ in the expression for $\Delta E$.

So, it would appear that the work done was twice the increase in energy of the system in apparent violation of the work-energy theorem. But we see that in fact the other half of the work went into “hard-to-see” internal energy (i.e., vibrational KE and PE of the spring).

 

[Dick]

Of course it is. The correct answer is that F\ge \rho(gx+\frac 1 2 u^2), with equality resulting only if the process is reversible (i.e., picking up the chain is an elastic process).

There's not one thing in the F=dp/dt approach that suggests an irreversible process. In fact, this erroneous approach yields a reversible process. Simply apply this exact same concept but with the chain being lowered rather than raised.

When the chain is being lowered the force is not the same as when the chain is being raised. It's not reversible. This isn't even a uniquely confusing problem. Pouring sand onto a conveyor belt gives the same sort of result. Even the standard air table experiment with pucks and velcro shows it. As Tsny points out, you can replace the velcro with locking springs and get the same result. Then you can see where the kinetic energy went. It becomes internal to the two puck system instead of dissapated into friction. This is the same sort of problem.

 

[Saitama]

Thanks everyone for taking your time to explain the things so nicely. I am trying to understand each and every single post in this thread. This is probably going to take some time at my part. This thread is really interesting. Please continue your further discussions. I am following this thread, please don't think that I have left this thread. Its just that I am an average student and it takes me some (or a lot of) time to understand the things.

 

[Dick]

Thanks everyone for taking your time to explain the things so nicely. I am trying to understand each and every single post in this thread. This is probably going to take some time at my part. This thread is really interesting. Please continue your further discussions. I am following this thread, please don't think that I have left this thread. Its just that I am an average student and it takes me some (or a lot of) time to understand the things.

That's good news! If people with a lot of experience with stuff like this can disagree, no reason it shouldn't take you time to digest it. Feel free to chip in.

 

[Saitama]

One more question.

@D H: Then what should be the answer to the second part of the question mentioned in the pdf you posted on the first page. Please don't take me wrong but the pdf you posted is hosted under IIT Kanpur's (a reputed institution here) website and its very unlikely for them to do such mistakes (or maybe they did this time). Were you able to find the solution key of that problem sheet?

 

[haruspex]

Only just found this intriguing thread. Can't resist chipping in.
Yes indeed, TSny, it's the inelasticity. (It could be made even harder to comprehend by dealing with an inelastic string instead of a chain of links, but the same applies.) In time δt, a length uδt, mass ρuδt, accelerates from 0 to u. Note that this is an unbounded acceleration, so we're dealing with impulses; energy cannot be taken to be conserved. Momentum is conserved, so the force is (ignoring gravity - this would also work in free fall) ρu². The KE gained in δt = ρu³δt/2, while the work done, as the force moves distance uδt, is twice that.
With an elastic string, each uδt would take time to reach speed u, and would thereafter oscillate. Now that would make one helluva spring problem!

 

[Dick]

Only just found this intriguing thread. Can't resist chipping in.
Yes indeed, TSny, it's the inelasticity. (It could be made even harder to comprehend by dealing with an inelastic string instead of a chain of links, but the same applies.) In time δt, a length uδt, mass ρuδt, accelerates from 0 to u. Note that this is an unbounded acceleration, so we're dealing with impulses; energy cannot be taken to be conserved. Momentum is conserved, so the force is (ignoring gravity - this would also work in free fall) ρu². The KE gained in δt = ρu³δt/2, while the work done, as the force moves distance uδt, is twice that.
With an elastic string, each uδt would take time to reach speed u, and would thereafter oscillate. Now that would make one helluva spring problem!

Right. But energy is always conserved. What's not conserved is the center of mass kinetic energy of a composite system. It can hit something and absorb it into internal vibration. That was the wiggling chain I was referring to earlier.

 

[ehild]

I got an idea while sleeping . We assumed the situation on the left in the pic, that the chain stays at the same place while lifted, as if it was in a tube. But that involves that the parts on the floor have to move to that tube somehow, which involves some motion.

Imagine that the chain is straight on the floor and the person who lifts the chain moves his hand always above the end of the horizontal part. That means, the vertical part moves not only up but also sideways. The length of the chain is constant, so both components of the velocity are the same u. That means
KE=1/2 m (v²(vertical)+v²(horizontal))=1/2 (ρx)(2u²)
E = PE+KE=1/2 ρx²g+ρxu², dW=F(lift)dx=(ρxg+ρu²[/B])dx. F(lift)=(ρxg+ρu²) from work energy considerations
Anyway, it can not be assumed that the chain does not have any horizontal component of velocity. Either that part which is on the floor, or that part which is in the air or both. The KE has to include both terms. But the time derivative of the vertical component of momentum is equal to the vertical force. ehild

 

[D H]

OPlease don't take me wrong but the pdf you posted is hosted under IIT Kanpur's (a reputed institution here) website and its very unlikely for them to do such mistakes (or maybe they did this time). Were you able to find the solution key of that problem sheet?

No, I wasn't able to find the solution key.

Don't take something written by reputed institutions as sacrosant. MIT makes mistakes. So does Caltech. So do highly reputable publishers. That's why they all write erratum sheets. Those mistakes typically don't propagate in countries with strong copyright laws. There's a problem in India. India either has weak copyright law or has rather weak enforcement of them. Here's a google book search for the exact phrase "A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration" that illustrates what happens in India: http://www.google.com/search?q=A+pa...ius+r+such+that+its+centripetal+acceleration"

Aha! Pranav-Arora, it turns out you were involved in the thread that instigated this search: [thread]616190[/thread].

What!? F = dp/dt is the definition of force. F = ma only applies when the mass is constant.

Where did you get that idea? It's certainly not anywhere in Newton's Principia. (Then again, neither is F=ma. Newton's Principia is almost entirely calculus-free.)

There's a big problem with F defined via F=dp/dt: The \dot m v term makes force a frame dependent quantity in the case of variable mass systems. On the other hand, defining force defined via F=ma means that force is the same quantity in all reference frames since both mass and acceleration are invariant quantities in Newtonian mechanics. This is why aerospace engineers almost inevitably choose F=ma rather than F=dp/dt.

In this problem, since the speed is constant, there is no acceleration, and, according to F = ma, there is no force.

Speed isn't constant. The links at rest on the platform change velocity by u as they are picked up. There is a problem is with F=dp/dt, however. The momentum of the rising part of the chain is identically zero from the perspective of a reference frame moving upwards at a constant speed u with respect to the platform. In this frame, F as calculated with dp/dt is zero if you are looking only at the momentum of the rising part of the chain.

There is a way around this mess, and that's not to use variable mass. Look at what happens to the chain as a whole. Now F=dp/dt and F=ma yield the same answer because \dot m is zero. Unfortunately that will not solve the problem. The problem is that this is a kinematic view of force rather than a dynamic view of force. We need to attribute this kinematical net force to root causes to convert this kinematical POV to a dynamical POV. Naively attributing all of this force to the hoist, or whatever is lifting the chain, is incorrect.

To see that it must be incorrect, simply reverse the process: Lower the chain rather than raising it. To keep the chain moving downward at a constant velocity we'll let connect the upper end of the chain to some physical device and let the lowering chain do work on that device. How much work is performed? If we attribute all of the change in momentum to work done by that the device we get an over unity system. More work will be performed than is allowed by the conservation laws. That doesn't make sense.

One way out of this morass (and the way out of the raising morass as well) is to attribute some of the change in the momentum of the chain to the platform. For example, one could look at the transients as a link is picked up from the platform. Yech. That's getting into finite element analysis, something that is far beyond the scope of an introductory physics course.

The easiest way out of this morass is to look to the work energy theorem. This yields a nice simple answer (one that is not listed in the provided choices) if all forces are conservative. All bets are off if the system is not conservative. You aren't going to get a nice simple answer in the case of nonconservative forces. Nonconservative forces almost inevitably yield ugly solutions.

 

[sankalpmittal]

Pranav, from wherever you posted the question, what answer it gives ? I got (D) as my answer and then only can the work energy theorem be justified. Else , it appears magic.

If this is not the correct answer, please look at posts 20 and 21 and tell, where I went wrong.

@DH: As far as the other problem https://www.physicsforums.com/showthread.php?t=616190 is concerned that textbook answer which OP there posted is likely wrong. My textbook too has the question but its answer is way different...

The easiest way out of this morass is to look to the work energy theorem. This yields a nice simple answer (one that is not listed in the provided choices) if all forces are conservative. All bets are off if the system is not conservative. You aren't going to get a nice simple answer in the case of nonconservative forces. Nonconservative forces almost inevitably yield ugly solutions.

Well, I always use this equation if non conservative forces are involved.

W_nc=E_f-E_i

W_nc:Work done by non conservative forces.
E_f: Final Mechanical Energy.
E_i: Initial Mechanical energy.

 

[haruspex]

The easiest way out of this morass is to look to the work energy theorem. This yields a nice simple answer (one that is not listed in the provided choices)

... and which is wrong. The correct answer is achievable by consideration of conservation of momentum and can be done without involving the dp/dt formulation. Conservation of momentum and conservation of work energy (as opposed to thermal energy) lead to different answers - one must be flawed.

 

[ehild]

Any comment about my post#46?

ehild

 

[haruspex]

Any comment about my post#46?

ehild

I would have thought that the chain/string could be arbitrarily thin. The thinner it is, the less it will matter if it's all piled up.

 

[ehild]

It has constant length. It can not happen that the lifted part moves vertically upward and the horizontal part does not move at all. So the KE can not be calculated from u, the vertical component alone.

ehild

 

[haruspex]

It has constant length. It can not happen that the lifted part moves vertically upward and the horizontal part does not move at all. So the KE can not be calculated from u, the vertical component alone.

ehild

If you lay it out straight, sure, but in the OP it was piled up, and as you make it progressively thinner that matters less and less.

 

[Dick]

To see that it must be incorrect, simply reverse the process: Lower the chain rather than raising it. To keep the chain moving downward at a constant velocity we'll let connect the upper end of the chain to some physical device and let the lowering chain do work on that device. How much work is performed? If we attribute all of the change in momentum to work done by that the device we get an over unity system. More work will be performed than is allowed by the conservation laws. That doesn't make sense.

It's no more reversible than any other inelastic process. Hence there is no energy paradox. But I think I've said that before.

 

[tms]

... and which is wrong. The correct answer is achievable by consideration of conservation of momentum and can be done without involving the dp/dt formulation.

Why is momentum conserved? That implies no net force.

Conservation of momentum and conservation of work energy (as opposed to thermal energy) lead to different answers - one must be flawed.

Or both.

 

[haruspex]

Why is momentum conserved? That implies no net force.

No, it only requires that all forces have been taken into account. Using the work energy theorem requires all losses have been taken into account, and there's a good reason to suspect there are losses: an elastic string would have been accelerated on take up and overshot equilibrium, leading to oscillations. The energy that would have gone into these must have been dissipated somehow.

 

[tms]

No, it only requires that all forces have been taken into account. Using the work energy theorem requires all losses have been taken into account, and there's a good reason to suspect there are losses: an elastic string would have been accelerated on take up and overshot equilibrium, leading to oscillations. The energy that would have gone into these must have been dissipated somehow.

But the problem involves idealized object, the way most intro physics problems do. There is no information about possible sources of energy loss. Since a supposed loss is calculated without such information, it implies that every situation involves the same loss. I find it very hard to believe that pulling up an anchor chain (with links 2 feet across) and pulling up a shoestring both create exactly the same loss.

Using F = dp/dt, on the other hand, presupposes only that you know the change in momentum in order to get the net force, and I don't see how that could be wrong.

 

[Dick]

But the problem involves idealized object, the way most intro physics problems do. There is no information about possible sources of energy loss. Since a supposed loss is calculated without such information, it implies that every situation involves the same loss. I find it very hard to believe that pulling up an anchor chain (with links 2 feet across) and pulling up a shoestring both create exactly the same loss.

Using F = dp/dt, on the other hand, presupposes only that you know the change in momentum in order to get the net force, and I don't see how that could be wrong.

I think haruspex would agree that F=dp/dt is the right thing to do, and would also agree the loss doesn't depend on the detailed dynamics of how the energy was lost. Was just saying you can infer by thinking about possible ways the loss can occur that you can conclude it must happen. You don't disagree. DH is the one who disagrees.

 

[haruspex]

I think haruspex would agree that F=dp/dt is the right thing to do, and would also agree the loss doesn't depend on the detailed dynamics of how the energy was lost. Was just saying you can infer by thinking about possible ways the loss can occur that you can conclude it must happen. You don't disagree. DH is the one who disagrees.

Thanks Dick, well put.

I find it very hard to believe that pulling up an anchor chain (with links 2 feet across) and pulling up a shoestring both create exactly the same loss.

The mechanism of the loss may be rather different; I picture the anchor chain links as swinging side to side briefly as they leave the ground, rather than bouncing up and down. But it would appear that about half the energy is lost either way.

 

[tms]

I just think we're missing something---I don't know what.

It's as if we were doing a typical block-on-frictionless-surface type problem, and the answer had a frictional loss in it---a frictional loss that was the same for every block and surface.