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Lifting up a chain

by Pranav-Arora
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Pranav-Arora
#19
Feb5-13, 08:41 AM
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Quote Quote by D H View Post
OK, let's do that.

Suppose that at some point in time [itex]t[/itex] the length of the chain being held off the platform is [itex]x(t)[/itex]. This length of chain is moving upwards at a constant velocity [itex]u[/itex] with respect to the platform. Given that the mass per unit length of the chain is [itex]\rho[/itex], this means the momentum of the chain with respect to the platform is [itex]p(t)=\rho x(t) u[/itex], directed upward. Some very short time [itex]\Delta t[/itex] later, the length of the chain moving upward is [itex]x(t)+u\Delta t[/itex], making the momentum [itex]p(t+\Delta t)=\rho (x(t) + u\Delta t)u[/itex]. The change in momentum is [itex]\Delta p = \rho u^2 \Delta t[/itex]. Applying [itex]F_{\text{net}}=\lim_{\Delta t \to 0} \Delta p / \Delta t[/itex] yields [itex]F_{\text{net}}=\rho u^2[/itex]. Adding the weight of the chain yields the total force needed to keep the chain moving at a constant velocity, [itex]F_{\text{tot}} = \rho gx + \rho u^2[/itex]. That's answer (B). Done!

Or maybe we're not done. It's always good to do a sanity check.

Let's see how much work is done by this force and compare this to the change in kinetic energy. The net force is [itex]F_{\text{net}}=\rho u^2[/itex], a constant. Calculating the work performed by this constant net force yields [itex]W=\int_0^x F\,dl = \rho x u^2[/itex]. The change in kinetic energy is half this amount. At this point we can do one of two things:
(a) Attribute this discrepancy to energy that is somehow lost.
(b) Figure out where we went wrong.

The right option is (b), figure out where we went wrong. Energy is not somehow lost. It's a conserved quantity. Where we went wrong was in attributing all of this Fnet to the hoist that is lifting the chain. We weren't solving the chain-lifted-off-a-platform problem. We were instead solving this problem:
The end of a chain of length x and mass per unit length ρ is lifted vertically with a constant velocity u by a variable force F. At any point in time, mass magically appears out of nowhere at a rate [itex]dm/dt = \rho u[/itex] with zero velocity with respect to the ground and attaches itself to the end of the chain. Find F as a function of height x of the end above platform.
It's best not to solve problems in universes where magic occurs.
You explained it nicely. Thanks!

But do you mean that the question given is wrong? If I was asked to calculate the energy lost, how can I find it?

Can you give links to some good resources where I can learn more about these variable mass systems?
sankalpmittal
#20
Feb5-13, 11:41 AM
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Quote Quote by Pranav-Arora View Post
You explained it nicely. Thanks!

But do you mean that the question given is wrong? If I was asked to calculate the energy lost, how can I find it?

Can you give links to some good resources where I can learn more about these variable mass systems?
This question did surprise me too. I completed H.C. Verma work and energy and there was 57th question in which we had a chain just touching the ground and it was released from rest. It was asked that out of chain of length L , x length of it strike the floor. No heap formed. We had to calculate force exerted by chain on floor as a function of displacement x. I tried and got the wrong answer until I realized that weight of chain also acts in addition to change in momentum. In this case also , normal reaction acts. By using this logic , I get the same answer as DH. Mechanical energy is not conserved as I suspect, may be some non conservative forces be acting in chain+earth system.

But again, work energy theorem,i.e. Wnet=ΔK.E. still applies. Did you try applying it ? If work energy theorem fails, then I am sure that this question is somewhat a magic. Work energy theorem applies to every inertial earth system, as per H.C. Verma.
sankalpmittal
#21
Feb5-13, 12:15 PM
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Quote Quote by D H View Post
OK, let's do that.

Suppose that at some point in time [itex]t[/itex] the length of the chain being held off the platform is [itex]x(t)[/itex]. This length of chain is moving upwards at a constant velocity [itex]u[/itex] with respect to the platform. Given that the mass per unit length of the chain is [itex]\rho[/itex], this means the momentum of the chain with respect to the platform is [itex]p(t)=\rho x(t) u[/itex], directed upward. Some very short time [itex]\Delta t[/itex] later, the length of the chain moving upward is [itex]x(t)+u\Delta t[/itex], making the momentum [itex]p(t+\Delta t)=\rho (x(t) + u\Delta t)u[/itex]. The change in momentum is [itex]\Delta p = \rho u^2 \Delta t[/itex]. Applying [itex]F_{\text{net}}=\lim_{\Delta t \to 0} \Delta p / \Delta t[/itex] yields [itex]F_{\text{net}}=\rho u^2[/itex]. Adding the weight of the chain yields the total force needed to keep the chain moving at a constant velocity, [itex]F_{\text{tot}} = \rho gx + \rho u^2[/itex]. That's answer (B). Done!

Or maybe we're not done. It's always good to do a sanity check.

Let's see how much work is done by this force and compare this to the change in kinetic energy. The net force is [itex]F_{\text{net}}=\rho u^2[/itex], a constant. Calculating the work performed by this constant net force yields [itex]W=\int_0^x F\,dl = \rho x u^2[/itex]. The change in kinetic energy is half this amount. At this point we can do one of two things:
(a) Attribute this discrepancy to energy that is somehow lost.
(b) Figure out where we went wrong.

The right option is (b), figure out where we went wrong. Energy is not somehow lost. It's a conserved quantity. Where we went wrong was in attributing all of this Fnet to the hoist that is lifting the chain. We weren't solving the chain-lifted-off-a-platform problem. We were instead solving this problem:
The end of a chain of length x and mass per unit length ρ is lifted vertically with a constant velocity u by a variable force F. At any point in time, mass magically appears out of nowhere at a rate [itex]dm/dt = \rho u[/itex] with zero velocity with respect to the ground and attaches itself to the end of the chain. Find F as a function of height x of the end above platform.
It's best not to solve problems in universes where magic occurs.
Wait , I may be stupid , but lets see. Wnet=ΔKE

Fnet = ρu2-ρgx

Now as x=u2/2g

Fnet = ρu2-ρu2/2
F net =ρu2/2

Integrating net F with respect to dx from 0 to x,

W=ρu2x/2

So W=ρu2x/2 which is equal to ΔKE. As length x is in air , its incorrect to consider normal reaction. I think this is where you blundered.

Edit: As per me the correct answer is D. Then only this question is justified by work energy theorem.
D H
#22
Feb5-13, 01:03 PM
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Quote Quote by Pranav-Arora View Post
But do you mean that the question given is wrong?
Yep. In a multiple choice type question, the question itself is wrong when the correct answer is not one of listed answers. That's the case here.

This is not the first time this has happened. Textbooks can be erroneous. Multiple choice: The problem apparently is particularly bad
(a) In far too many online automated physics homework problem systems.
(b) In far too many Indian physics texts.
(c) Both of the above are true.

Apparently India has rather weak enforcement of its copyright laws, which leads Indian textbook authors to copy problems from one another. It's much easier to copy some other author's already-worked problems rather than to create new ones. That's fine (other than copyright issues) if the problems are well written and the solutions are correct. It's not so fine if the question is poorly written, if the correct answer isn't present in the multiple choice list, or if the solution in the answer guide / worked example is incorrect. We've chased down a number of problems with Indian physics texts at this site, occasionally seeing the exact same bad question replicated in dozens of different texts.

If I was asked to calculate the energy lost, how can I find it?
The correct answer is that energy is not lost, at least not ideally. Work is being done against gravity in this problem Gravity is a conservative force. There are no losses in working against gravity.

In reality, there will be some energy loss due to entropy. (Note: the energy isn't really "lost". It's just converted to unusable energy: Heat.) A cable or chain may heat up a bit due to non-conservative interactions as the cable/chain grows taut, links shift, etc. We can ignore all that messiness by assuming an ideal chain. Even a terribly constructed, non-ideal chain will not heat up to the extent implied by the wrong answer to this problem.

Can you give links to some good resources where I can learn more about these variable mass systems?
The best thing to do in an introductory physics class is to try to pose the question without worrying about variable mass. It's messy, and easy to mess up. IMO, it's not a subject introductory physics classes should delve around in much (other than perhaps rockets as an exemplar). This particular problem can easily be answered without looking at variable mass. Simply look at the chain of fixed length L as a whole.
TSny
#23
Feb5-13, 03:17 PM
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I’ll add my 2 cents to the confusion! For any system ##F_{net}^{ext} = M a_{cm}## where M is the total mass of the system.

Let the system be the entire chain alone. The net external force is ##F_{net}^{ext} = F_{lift} – Mg + F_N## where ##F_{lift}## is the lifting force and ##F_N## is the normal force acting on the part of the chain still resting on the platform.

When amount ##x## of chain is off the table, the center of mass of the section that is off the table is at ##x/2## above the platform and has mass ##\rho x##. So, relative to the platform, the location of the center of mass of the whole chain is

##x_{cm} = \frac{(\rho x)(x/2)}{M} = \frac{\rho}{2M}x^2##.

Taking the second time derivative of this yields ##a_{cm} = \frac{\rho}{M}(\dot{x}^2 + x\ddot{x}) = \frac{\rho}{M}u^2## since ##\ddot{x} = 0##

So, we have ##F_{lift} – Mg + F_N = \rho u^2## or

##F_{lift}= \rho u^2 + Mg - F_N##

If we take the normal force to equal the weight of the part of the chain resting on the table, then ##F_N = (M-\rho x)g## and we get

##F_{lift}= \rho u^2 + \rho g x = \rho (u^2+g x)##
ehild
#24
Feb5-13, 05:01 PM
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Quote Quote by TSny View Post
Iíll add my 2 cents to the confusion!
##F_{lift}= \rho u^2 + \rho g x = \rho (u^2+g x)##
It is very convincing 2 cents :) But what is the problem with the energy then?


ehild
tms
#25
Feb5-13, 05:15 PM
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Quote Quote by ehild View Post
It is very convincing 2 cents :) But what is the problem with the energy then?
I think the problem is with the integration [itex]W = \int F\,dx[/itex]. For any particular bit of the chain that is correct, but each bit of the chain goes a different distance so the force acts on each bit for a different distance, thus doing a different amount of work. I haven't yet figured out the correct integral, though. I know I'm not being abundantly clear, but I think the answer is in this direction.

At any rate, if the energy is right, what is wrong with the F = dp/dt solution? Since that is basically the definition of force, it, too, must be right. Picking the energy to be right because dm/dt is "magic" is not really good enough.
ehild
#26
Feb5-13, 05:21 PM
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F acts at the lifted end of the rod and its work is ∫Fdx. We do that work on the whole system, and that work increases the energy. Unless something else non-conservative force does also work.

ehild
TSny
#27
Feb5-13, 05:27 PM
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Regarding energy, I suspect that when each link is jerked off the platform, it's like an inelastic collision- causing internal energy to increase. The chain will get warmer!
ehild
#28
Feb5-13, 05:34 PM
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Quote Quote by TSny View Post
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!
But why?

ehild
tms
#29
Feb5-13, 05:38 PM
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Quote Quote by ehild View Post
F acts at the lifted end of the rod and its work is ∫Fdx. We do that work on the whole system, and that work increases the energy. Unless something else non-conservative force does also work.
I'm confused, too. I still don't see anything wrong with the F = dp/dt solution.
tms
#30
Feb5-13, 05:42 PM
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Quote Quote by TSny View Post
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!
But there is nothing in the problem or solution(s) about the internal working of the "chain". The solution(s) apply just the same to the anchor chain of an aircraft carrier or to a piece of string.
Dick
#31
Feb5-13, 05:57 PM
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Quote Quote by ehild View Post
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
#32
Feb5-13, 06:02 PM
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Quote Quote by TSny View Post
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
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Feb5-13, 06:15 PM
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Quote Quote by D H View Post
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
#34
Feb5-13, 06:16 PM
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Quote Quote by ehild View Post
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.
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D H
#35
Feb5-13, 06:16 PM
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Quote Quote by Dick View Post
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
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Feb5-13, 06:28 PM
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Quote Quote by TSny View Post
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.


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