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Rope Through a Hole in a Frictionless Table

  1. Nov 11, 2014 #1
    1. The problem statement, all variables and given/known data
    A uniform rope of length L lies in a lump on a frictionless table. a
    very small piece at one end is held hanging down through a hole in the table just below the lump.
    This system is released, and the rope slides down through the hole. What is the
    speed of the other end of rope when it just leaves the table?

    2. Relevant equations


    3. The attempt at a solution
    according to me the loss in potential energy equals gain in kinetic energy
    thus mgL/2 = (1/2) mv^2
    v=sqrt(gL)

    this option doesnt exist in the options available in my book so pls anyone can help me out
    i have attached a file for the same
     

    Attached Files:

  2. jcsd
  3. Nov 11, 2014 #2
  4. Nov 11, 2014 #3

    collinsmark

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    There's a bit of energy lost by the rope holding itself together. So using conservation of energy, without considering the work done, isn't a valid approach for this particular problem. (Conservation of energy is a valid approach if you consider the work done though.)

    KE would equal PE if the "rope" fell through the hole like many tiny droplets, each droplet independent from all the others; the resultant "rope" would expand as a function of time. The difference in this problem is that the rope holds itself together.
     
  5. Nov 11, 2014 #4
    i think that this work done u r talking about is due to tension???? is it so???
    initially i was thining that since we consider gravitational force as internal force so here PE Loss shoud be equal to KE Gain.......what's ur say about that???
     
  6. Nov 12, 2014 #5

    collinsmark

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    The work done on the rope might be related to the rope's tension somehow or another. However, I wouldn't delve too much into the tension idea, unless you really want to get all crazy with it. It might just be better to think of the energy going into heat caused by the infinitesimal parts of the rope jumping from 0 to some non-zero velocity practically instantaneously.

    My point was that this problem differs from the simple KE = PE type of problem because the rope does not allow itself to stretch apart as it would if it were instead an infinite number of infinitesimal points acting independently of each other.

    ---------------

    More importantly though, I think there might be a mistake in your book's solution. I don't like the way your book treats this particular problem; it seems to me as if part of the book's solution treats the rope as extended across the table (in at least one part of the solution) instead of being piled up in a heap as is given in the problem statement. Whatever the case, I'm having a hard time following your book's solution.

    Anyway, I came up with the answer of
    [tex] v = \sqrt{\frac{2gL}{3}}. [/tex]
    Notice that the '2' is inside the square root; not a '4' like your book's solution. I went about solving the problem in a very different way than your book did. Still, either I'm wrong or the book is wrong. And I don't particularly like your book's treatment of this particular problem.
     
  7. Nov 12, 2014 #6

    collinsmark

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    I think I can safely say that there is a mistake in your book, somewhere along the way (regardless of whether my [itex] \sqrt{2gL/3} [/itex] answer is correct or not). If you were to simply drop the lump of rope altogether (or any simple object for that matter) a distance of L/2, without threading it through a hole or anything like that (instead just simply drop it) its final velocity would be only [itex] \sqrt{gL}[/itex]. The fact that the book's solution is greater than that indicates that your book's answer is simply wrong, otherwise it would lead to perpetual motion machines and the like.

    I found two things wrong with your book's solution
    (a) It treats the solution as if the rope is laid out flat on the frictionless table, such that all of the rope moves together; not just the part that hangs through the hole. And that's a fine problem to solve, by the way, but it is not the way it is worded in the problem statement as being in a "lump."
    (b) Just before it gives the final answer, it states [itex] \frac{MgL}{2} - \frac{MgL}{6} = \frac{1}{2}Mv^2 [/itex]. Let's just assume for a moment that that part is correct. Solving for [itex] v[/itex], the resulting answer is not [itex] v = \sqrt{\frac{4gL}{3}} [/itex]. So that's definitely a mistake there.
     
    Last edited: Nov 12, 2014
  8. Nov 12, 2014 #7

    haruspex

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    Assuming the rope is inelastic, I confirm your answer. Work is not conserved because as each small piece of the rope passes through the hole in the table it accelerates instantaneously from 0 to the current speed of the rope. That constitutes an inelastic 'impact'.
    If the rope is elastic then it gets a lot more complicated since there will be longitudinal oscillations in the rope. These will take up some of the KE without contributing to the rope's overall speed.
     
  9. Nov 12, 2014 #8
    @Jazz hey can you have a look at this one too and add what's your say about this question.....
     
  10. Nov 12, 2014 #9
    please can you explain how did you get this answer......
     
  11. Nov 12, 2014 #10
    i'm also in agreement to the fact that it should not be equal to sqrt(gL) but don't have any other ideas of solving it.....
     
  12. Nov 12, 2014 #11
    i agree with your point of instantaneous acceleration i.e. here energy is lost due to abrupt motion of the rope in lump where it instantaneously changes from zero to a non-zero velocity.......and thus we can't apply conservation of energy here in this problem.....

    please help me out with the solution to it........
     
  13. Nov 12, 2014 #12
    I would like to add something useful to what has been said, but since I'm just beginning with Calculus, I can't point out something about the math, which is what matters. However, these were my first thoughts about it:

    - Is it supposed to be a frictionless rope? If it's as a lump then, when falling, friction in the rope will do a non negligible work. And so it should be either a frictionless rope or it's lying on the table in a stretched manner.

    - I started to think about it in the same way that the problem of a train leaving the station with an acceleration ##a## is dealt. The velocity of the end of the rope when leaving the station would be ##\sqrt{2gL}##. Then I realized that this approach is not correct here, because the acceleration ##g## is assumed to be applied in a fixed mass ##m##, which is not the case. ##g## is fixed, but ##m## varies as the length ##x## of the rope falls, which in turn make the downward force to vary. Here is where I think Calculus comes into play |:

    - Then the answer states that the center of mass falls a distance ##\frac{L}{2}##, which can be understood that after falling that height, there is no acceleration. Therefore the velocity of the end of the rope would be ##\sqrt{gL}##. A velocity greater than this would mean a ##g## greater than that on the Earth's surface or mass was added to the rope from somewhere else.

    - Since ##\sqrt{2gL}## is the wrong answer, ##\sqrt{gL}## is not among the options, ##(b) > \sqrt{gL}##, which can't be right, ##(d)## is true if there is no pull by the rest of the rope that has fallen (which we know isn't so); then the answer by discarding is ##(c)## :)

    I'm not sure whether the distance ##\frac{L}{2}## is something that must be indicated in the problem's statement or it was obtained after the integration showed in the answer.
     
  14. Nov 12, 2014 #13

    haruspex

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    We've agreed, I hope, that we cannot use conservation of work energy. What other conservation laws might be relevant?
     
  15. Nov 12, 2014 #14
    sorry to say but i'm unable to understand what you want to say in the above reply......:confused:
     
  16. Nov 12, 2014 #15
    can you please explain its solution for this as i'm facing difficulty in it..........
     
  17. Nov 12, 2014 #16

    haruspex

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    What do you know about momentum?
     
  18. Nov 12, 2014 #17

    collinsmark

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    << Mentor's note: The full solution should not have been posted, since no correct solution was available. The advocated approach here would have been to try to guide the OP to the answer. If ever considering providing a complete solution for any reason, please contact a mentor for approval before posting. This post has not been deleted as the OP has already read it and the continued discussion in the thread refers to it. >>

    Normally I wouldn't give a detailed answer in this sub-forum on PF, but since the topic this thread is about solution given to you in the book, and the book's solutions have some obvious mistakes anyway [Edit: not to mention its approach to the solution doesn't even match the problem statement], I do not think that it is inappropriate to introduce an alternate solution in this case.

    I may skip a few details in the steps though for you to work through yourself.

    Anyhoo, my solution didn't use conservation laws. It's all just kinematics.

    Define the mass, [itex] m [/itex] as being only the mass of the part of the rope that has fallen though the hole. Since the rest of the rope is sitting there quietly, piled up in its sleepy lump, minding its own business, we can ignore it for most of this derivation.

    The only force acting on this part of the rope is the force due to gravity, [itex] mg [/itex]. This gravitational force is equal to that part of the rope's rate of change in momentum.
    [tex] \vec F = \frac{d \left( m \vec v \right)}{dt} [/tex]
    If the mass of this part of the rope was constant, the result would be equal to the familiar [itex] \vec F = m \vec a [/itex]. But that's not the case here since that part of the rope's mass changes as a function of time. So we need to use the chain rule.
    [tex] \vec F = \left( m \right) \left( \frac{d \vec v}{dt} \right) + \left( \frac{d m}{dt} \right) \left( \vec v \right) [/tex]

    Now let's make our [itex] mg [/itex] substitution. And while we're at it, lets switch over to differential equation notation with the dots indicating time-based derivatives.
    [tex] mg = m \ddot x + \dot m \dot x [/tex]
    The next step is to make appropriate substitutions to represent the differential equation with the variables given in the problem statement. Or at least change them to make them jive with the defintions given by your book's solution, of the rope has length L, mass M, and the length of the rope hanging through the hole is x.
    [tex] m = M\frac{x}{L} [/tex]
    [tex] \dot m = M\frac{ \dot x}{L} [/tex]
    which gives, after a little simplification,
    [tex] xg = x \ddot x + \left( \dot x \right)^2 [/tex]
    The solution to which is
    [tex] x = \frac{1}{2} \left( \frac{g}{3} \right) t^2 [/tex]
    I'll skip the differential equation solution process, But you can check it yourself to make sure it fits. Take the derivative to find v and another derivative to find a.
    [tex] \dot x = v = \frac{g}{3}t [/tex]
    [tex] \ddot x = a = \frac{g}{3} [/tex]
    Plug those functions into the differential equation and you can see that the solution is valid. [Edit: and you can check that it satisfies our initial conditions of x = 0 and v = 0 when t = 0.]

    The next step is to solve for the time T it takes for the rope to fall through a distance L (substitute in L for x and T for t and solve for T).
    [tex] T = \sqrt{\frac{6L}{g}} [/tex]

    Plop that into the velocity equation and we finish with
    [tex] v = \sqrt{\frac{2gL}{3}} [/tex]

    And there ya' have it.

    This derivation did not use any conservation laws. I'd be interested to hear the conservation of momentum idea. Perhaps model the energy loss coming from an infinite number of infinitesimal inelastic collisions (causing each small section of rope to jump from 0 to a finite speed almost instantaneously, like what happens with inelastic collisions).

    (Again, normally such a detailed solution [by someone other than the OP] as given above is in violation of the forum rules. But in this case the OP's question was about a solution given in a textbook, and that given solution was incorrect on several counts. I don't think it's inappropriate in this case to provide an alternative.)
     
    Last edited by a moderator: Nov 13, 2014
  19. Nov 12, 2014 #18

    haruspex

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    I believe relying on this
    ##\vec F = \frac{d \left( m \vec v \right)}{dt}##
    as a way of finding the velocity effectively assumes conservation of momentum. If momentum were not conserved then you could not rely on the final momentum matching the total momentum imparted by the applied force.
     
  20. Nov 13, 2014 #19

    collinsmark

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    I was treating it merely as an extension of Newton's second law of motion. My initial assumption was that the force of gravity was the only external force acting on the system. But now that you mention it, I wonder if the greats, including physicists such as Joseph-Louis Lagrange, Jean le Rond d'Alembert and contemporaries weren't contemplating the same sort of things in their day. <collinsmark scratches chin. "hmmm." contemplates.> Perhaps so! :)

    [Edit: btw, as much as I like the thought process, the [itex] \vec F = \frac{d \left( m \vec v \right)}{dt} [/itex] doesn't say anything about the momentum value itself until boundary conditions are applied. (The boundary conditions were applied in solving the differential equation such that v = 0 and x = 0 when t = 0). It only says something about the rate of change in momentum, not the final or total value of the momentum in particular. That said, I still like your line of thinking. (And it may very well been the same sort of thought process, in part, that brought about more advanced physics.) So hats off to you there!]
     
    Last edited: Nov 13, 2014
  21. Nov 13, 2014 #20
    @collinsmark thanks for this solution it really helped me.......:):):):):)
     
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