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Falling chain problem

  1. Oct 31, 2012 #1
    1. The problem statement, all variables and given/known data
    the question is about a very flexible strain falling on a rigid table and ask for expression of normal reaction of table at a certain instant


    2. Relevant equations
    when i was trying to solve problem , i resolve N(normal reaction) into N1, N2
    which N2 is the weight of chain already on table and N1 is the force by table to stop the chain(i consider it force to stop a small part of chain)


    3. The attempt at a solution
    then i use newtons 2nd law to find N2
    N2 = v*dm/dt + m*dv/dt ( v for speed of chain just before touch the table)
    i can find v*dm/dt = p*v^2 (p -linear density)
    but when i tried to find m*dv/dt ,
    i was confused about how to find the acceleration of that small part of chain with dt cannot be found
    can anyone help me in correcting my calculation/concept?
     
  2. jcsd
  3. Oct 31, 2012 #2
    sorry for the typing mistake
     
  4. Oct 31, 2012 #3
    anyone here can help me ?
     
  5. Oct 31, 2012 #4

    haruspex

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    In time dt, a portion of chain of mass pvdt comes to rest from speed v. Its change in momentum is therefore ....., and if that is brought about by force N1 acting for period dt then it is equal to .....
    I don't think there is a term m dv/dt here. What would m be for that?
     
  6. Oct 31, 2012 #5
    i know what you mean, you guide me to consider the motion of small portion of chain
    to obtain N1 = v * dm/dt
    but i still don't understand why the term m *dv/dt vanish even it should be in equation of Newton 2nd law
     
  7. Oct 31, 2012 #6

    sophiecentaur

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    There must be a 'rate of change of momentum' term involved,
    I need to rewrite this bu can't seem to delete it. I will be back.
     
    Last edited: Oct 31, 2012
  8. Oct 31, 2012 #7

    sophiecentaur

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    There must be a 'rate of change of momentum' term involved, which is m v/dt , where m (an infinitessimal) would be the mass of chain being brought to a halt in the time dt. That will relate to the speed just before impact and the linear density. Then there will also be the steadily increasing weight of the chain as it piles up. I think you can say the v will correspond to the 'free fall' speed after time t as every bit of the chain is under the same acceleration - before it hits the table.
    That should be enough for you to get an answer.

    I now see why you ask about m dv/dt. If you wanted to do it that way, as with many collision problems, you would need to know the actual acceleration of the chain as it stops - which is not known because the details of the table and chain materials aren't specified. Using change of momentum avoids having to do that.
     
  9. Oct 31, 2012 #8

    haruspex

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    You are right that the full expression would be d(mv)/dt, and if m and v both vary that expands to v dm/dt + m dv/dt. But you're trying to evaluate both of those in respect of the small piece of chain that comes to rest in dt. That's double-dipping: it is fully accounted for by the v dm/dt term.
    It helps to be clear about what the variables mean. If you define m to be the mass of the chain that has not yet reached the table, its velocity is constant. If you define it some other way then you will have trouble justifying your value for dm/dt. The m dv/dt term would be appropriate if the suspended part of the chain were also slowing down.
     
  10. Oct 31, 2012 #9

    sophiecentaur

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    I think the chain must be treated as a not allowing any compression forces to resist its fall (as with the ideal string, tension-only model, it must fold up with no backwards transmitted force) so it can't fall at uniform velocity. Each bit of it accelerates at g until it hits the table. In each interval dt there will be vdt length of chain hitting the table, which will have ρvdt mass. Its momentum will be this mass times v so the surely the retarding force from the table will be the momentum times dt. You then need to add the accumulated weight of the chain that ends up on the table.
    Why should Newton 2 be necessary?
     
  11. Oct 31, 2012 #10

    haruspex

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    Yes, of course you are right - the upper part of the chain may be accelerating. The OP does say 'falling', but for some reason I thought of it as 'being lowered'.
    However, that does not mean there is a dv/dt term in the force on the table. It just means the force varies over time.
     
  12. Nov 1, 2012 #11

    sophiecentaur

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    One way of finding the force on the table would be to look at the acceleration of each link as it hits the table (which would be where a dv/dt term could come in). This would involve knowing about the materials used and the shapes of the links and to study each impact over time. It would (should!) yield the same answer as the method involving change of momentum and impulse. The beauty of the impulse method (why it is what is frequently used in collsion problems) is that you don't need to know those details.
    'Cheeky shortcut' or 'intelligent strategy'?
     
  13. Nov 1, 2012 #12

    haruspex

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    That's where this thread started. The OP introduced a term v dm/dt and a term v dm/dt, both in respect of the chain links hitting the table. It appears logical on the basis of being the expansion of d(mv)/dt. The rest of the discussion has been explaining why there is no m dv/dt term.
    My view is that you have to decide what the m and v refer to. If m means 'that portion of the chain not yet in contact with the table', its only acceleration is due to gravity, and the table is not involved in that, so the force from the table is only v dm/dt. (but where v increase as the chain falls).
    If m means the whole chain then v dm/dt goes away and you have to evaluate m dv/dt instead. It's less obvious how to do that since the acceleration on an individual link is unknowable. In fact, with the whole chain view, it's better to think of the total momentum and not expand d(mv)/dt at all.
     
  14. Nov 1, 2012 #13

    sophiecentaur

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    We have a problem in deciding exactly what the scenario is. I am assuming the chain is infinitely flexible and doesn't transmit any thrust. To get a solution for that interpretation of the question, you don't need to know the details of the individual impacts but you just have a rate of change of momentum problem - just as if, instead of the chain, you dropped a string of ball bearings from a sideways moving vertical tube. The balls would hit the table in sequence, each one going slightly faster than the previous one.
    I have chosen the simplest option as it's the only one I could answer with so little initial information.
     
  15. Nov 1, 2012 #14

    haruspex

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    There's no disagreement on that.
     
  16. Nov 1, 2012 #15

    sophiecentaur

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    OK then. I think my falling balls equivalent model works and the solution is pretty straightforward and doesn't involve knowing any details of the structure of the links - except to assume they don't bounce!
     
  17. Nov 1, 2012 #16
    This is an Irodov problem. The problem initially asks to prove that if a chain is falling and if the part already lying on the surface exerts some normal force then the falling part exerts twice as much ( or was it half as much ? I forgot ) force. Bottom line, consider the part that has already fallen, its weight and the impulsive change of momentum as unit part of the chain comes crashing down.
     
  18. Nov 1, 2012 #17

    sophiecentaur

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    Yep - twice as much. The calc is very straightforward and I was surprised when the 'twice' came out of it.
     
  19. Nov 7, 2012 #18
    Well ,there are only two forces acting on the chain , namely the gravitational force and the reaction forces . Now , the reaction forces do not do any work on the chain , as at their points of application , the displacement is zero. Therefore the net mechanical energy of the whole chain is conserved. This should give you an expression of the velocity as a function of the length of the chain in the air.

    Now the rate of change of momentum of the chain is caused due to the resultant of the two forces , ie mg - N = m(t).dv/dt + v.dm(t)/dt , and dm/dt = -vdt*m/l
     
  20. Nov 7, 2012 #19

    haruspex

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    So the last angstrom of chain hits the table at relativistic speed? I don't think so.
     
  21. Nov 7, 2012 #20

    sophiecentaur

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    The last angstrom of chain hits the table at the speed you'd expect from its starting height. It's in free fall!
    No one said anything about an infinitely long chain - for a start, we're assuming acceleration of g, which can only be true for short (ish) chains.
     
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