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#55
Feb613, 04:27 PM

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#56
Feb613, 05:44 PM

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#57
Feb613, 07:57 PM

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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. 


#58
Feb613, 08:35 PM

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#59
Feb613, 09:20 PM

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#60
Feb613, 10:17 PM

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I just think we're missing somethingI don't know what.
It's as if we were doing a typical blockonfrictionlesssurface type problem, and the answer had a frictional loss in ita frictional loss that was the same for every block and surface. 


#61
Feb613, 10:55 PM

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#62
Feb613, 11:08 PM

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#63
Feb713, 12:58 AM

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When we say that the KE of the whole system is 1/2 (ρx)u^{2} we totally neglect the motion of the piece still on the ground. It has to move somehow, as the length of the chain is constant. And we do not know anything about that motion, but it contributes to KE.
I have shown an arrangement in #46 when the chain does not move on the table. In that case, the workenergy method gives the same result as the momentum method. The problem says that the chain is piled up, but it cannot be treated as a single point as it has length. As we do not know the details of the internal motion of the parts,the best thing is treating the chain as whole. See TSny's post #23. The acceleration of the CM is equal to the net force, including the lifting force, weight of the whole chain and the normal force from the table. ( the question: is the normal force really equal to the weight of the chain still on the ground?) The CM of the whole chain does accelerate while the lifted part moves with constant vertical speed. ehild 


#64
Feb713, 03:23 AM

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First let's look at my magical chain in post #17 where links magically appear out of nowhere and attach themselves to the chain. In that case I would agree with the answer implied by the text. All of the force needed to accelerate a newly created link to a velocity of u is supplied by the lifter. There's a problem here. Physics and magic don't mix well. Next let's look at what happens during the process of lifting a single link of the chain off the platform. Between the point in time when the link is lying flat on the platform and the point in time when the link is vertical and lifts off the platform, the lifter only applies a force equal to half of the link's weight. The platform bears the other half of the link's weight^{1}. It's only until the link is vertical and lifted off the platform that the lifter bears the entire weight of the link. By this time, the link has already been accelerated to an upward velocity of u. That the lifter bears only half of the weight during the process of lifting a link resolves the apparent discrepancy between a conservation of momentum and conservation of energy approach. Next, let's look at a number of different approaches to solving this problem. Use the work energy theorem and you'll find that the lifter exerts a force equal to ρgx+½ρu^{2}. Use Lagrangian mechanics and and you'll get a force equal to ρgx+½ρu^{2}. Use Hamiltonian mechanics and you'll get a force equal to ρgx+½ρu^{2}. Use conservation of momentum with the platform bearing half the weight during the process of lifting a single link and you'll get a force equal to ρgx+½ρu^{2}. They all agree. Finally, let's look at what happens if we use F=dp/dt. In a reference frame in which the platform is stationary you'll get a force equal to ρgx+ρu^{2}. In a reference frame whose origin is moving upward at a constant velocity u with respect to the platform you'll get a force equal to ρgx. The cost of picking up the chain is free in this frame per F=dp/dt. Use some other reference frame and you'll get yet another contradictory answer. ^{1}This assumes the link being lifted is directly below the lifter; i.e., that the suspended part of chain is vertical. The force needed to start lifting a link is a bit more than half the link's weight if the link is not directly below the lifter. That this is never truly the case does introduce a tiny bit of lossiness. There's lots of other little bits of lossiness that appear here and there. They don't add up to the erroneous answer given by F=dp/dt. 


#65
Feb713, 05:33 AM

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Suppose vertical slabs are placed on a frictionless horizontal surface as shown. All the slabs have a small hole through the center except the first slab on the left. A string is attached to the center of the first slab and the string is threaded through the holes in all the other slabs.
In the first scenario there is initially a small space between each slab. The string is then pulled to the right so that the first slab maintains a constant speed u. In this case, each slab attains a speed u before slamming into the next slab. The second scenario has all the slabs in contact before the string is pulled. In each case there will be a certain amount of work done to get all the slabs moving at speed u. The work required for the first scenario is approximately Mu^{2} while for the second scenario it is Mu^{2}/2, where M is the total mass of all the slabs. The reason for saying “approximately” Mu^{2} in the first case is that the work required to get the first slab moving is just mu^{2}/2 while the work for each succeeding slab is mu^{2}, where m is the mass of each slab. For a large number of slabs, the total work will be approximately Mu^{2}. The work is greater for the first scenario because internal energy is generated in the slabs due to the completely inelastic collisions. The workenergy theorem stated in the form W_{net} = ΔK is only applicable to a single point particle or to a system that can be treated as a single point particle. It is not generally true for a deformable system where parts of the system move relative to one another. 


#66
Feb713, 10:59 AM

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#67
Feb713, 11:50 AM

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It's best not to do physics in a world where magic occurs, but if you insist on doing so, you at least should get the mathematics right. 


#68
Feb713, 11:58 AM

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#69
Feb713, 12:51 PM

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Oh please !!
Why is everybody ignoring my previous posts in this thread. Please see my posts:20,21 and 48. I am just curious as to where I went wrong/or not. I (being a student) also did lot of such questions, but this one seem to bewilder me as well. 


#70
Feb713, 01:37 PM

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sankalpmittal, the correct answer is not (D), F=ρ(u^2gx). It makes no sense. Force decreases with increasing height? The force obviously must increase as x increases.



#71
Feb713, 01:47 PM

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#72
Feb713, 02:39 PM

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All that said, I think I see why it's such a divisive issue. What will happen in practice depends on details of the system. Some of the models I've offered, in particular the one with the rising chain of 'buckets', accentuate the coalescence aspect, so maximising the energy loss. This may well be appropriate for, for example, a fine gold chain. (There is something distinctive about how it feels when you lift a fine gold chain from one end. It feels heavier while lifting it than when held steady.) All of the models discussed will be lossy in some way, but some more than others. E.g. with your heavy chain lift, with large links, losses could be kept to a minimum by varying the lifting rate. As each link tilts, reduce the force, allowing the angular momentum of the link to assist in completing getting it to the vertical. By the time it becomes vertical, it is almost at rest. Now increase the force, initiating the lifting of the next link. I believe you can get arbitrarily close to lossless that way. But, as the link size tends to zero, the startstop technique remains incompatible with the notion of lifting at steady speed, so this is not applicable to the OP. 


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