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Classical Mechanics (Marion)

  1. Dec 19, 2013 #1
    From (Marion 5th ed. Problem 9-15)
    A smooth rope is placed above a hole in a table. One end of the rope falls through the hole at t = 0, pulling steadily on the remainder of the rope. Find the velocity and acceleration of the rope as a function of the distance to the end of the rope x.

    Ignore all friction. The total length of the rope is L.


    I attempt to solve it by lagrangian L = T - U
    T = 1/2 (λx) v^2
    U = -(λx)g(x/2)

    where λ is mass density
    But someone said it is not correct. He claimed that if T depends on generalized coordinate explicitly(in this case x), then it violates Newton's Second Law.
    So, lagrangian mechanics cannot be used in open system ?



    Marion's Solution
    F = dp/dt = m(dv/dt) + v(dm/dt) (see Fig)

    I learnt that we cannot differentiate m (i.e. dm/dt) if the system is open.


    So, which one is correct ??
    Can anyone explain to me in detail ?
     

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  3. Dec 19, 2013 #2

    vanhees71

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    I don't see what's wrong with the given Lagrangian. The Euler-Lagrange equation give the same equations of motion as given in the book.
     
  4. Dec 19, 2013 #3
    Bad news. The solution in Merion is incorrect. The entire rope is accelerating, and not just the part that has passed through the hole. This needs to be taken into account. Let F be the tension in the rope at the hole. Then the force balance on the part of the rope below the hole at any time is:
    [tex]σgx-F=\frac{d(σxv)}{dt}[/tex]
    The force balance on the part of the rope above the hole is:
    [tex]F=\frac{d(σ(L-x)v)}{dt}[/tex]
    If we add these two equations to eliminate F, we get:
    [tex]gx=L\frac{dv}{dt}[/tex]
     
  5. Dec 20, 2013 #4

    vanhees71

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    Well, then I misunderstood the drawing. I thought the rope is curled up on the table and thus only the part gone through the hole is accelerating. If you have the other case, of course you get another equation of motion.
     
  6. Dec 20, 2013 #5

    D H

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    No, it doesn't. You can't use a Lagrangian formulation for this problem because energy is not a conserved quantity in this problem.

    No, it's not. The rope is coiled just above the hole. The only part of the rope that is accelerating is the part that is falling.

    Your understanding is correct. The rope is coiled up just above the hole so that the only part that is accelerating is the part of the rope that has already passed through the hole.
     
  7. Dec 20, 2013 #6

    vela

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    Why wouldn't energy be conserved in this problem? The only force doing work is gravity, which is conservative.
     
  8. Dec 20, 2013 #7

    vela

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    You've omitted the normal force exerted by the table on the coiled part of the rope.
     
  9. Dec 20, 2013 #8

    vela

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    That's not right. For example, the kinetic energy in polar coordinates, ##T = \frac{1}{2}m[\dot{r}^2 + (r\dot{\theta})^2]##, depends explicitly on the coordinate ##r##. Changing from Cartesian to polar coordinates certainly doesn't result in a violation of Newton's second law.
     
  10. Dec 20, 2013 #9
    This was the horizontal force balance on the part of the rope on the table.
     
  11. Dec 20, 2013 #10
    Even if it's coiled, part of it is still accelerating, although not nearly as much as if it is straight. But, actually, I didn't notice that it was coiled.
     
  12. Dec 20, 2013 #11

    vela

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    Ah, okay. I think the intent of the problem is that the rope on the table remains at rest, and only a small piece of length v dt is accelerated from rest to speed v at time t.
     
  13. Dec 20, 2013 #12

    D H

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    Energy is not conserved because the part of the rope that is falling is constantly bringing new parts of non-falling rope up to speed. One way to look at this is that this acts as a drag on the falling part of the rope. Drag is almost inevitably a non-conservative force.


    Hint to HAMJOOP:

    Suppose that at some time t the length the rope x(t) that is already falling with velocity v(t) is about to pull a small length of rope Δx up to speed. At some small time tt later, the length of rope x(t)+Δx will be moving at some speed v(t)+Δv. During this short span of time, the only force acting on this segment of rope is gravity. Assume there is no interaction (yet) between the rest of the coiled up rope and that small segment that is about to be brought up to speed. (You might want to justify this assumption.)

    I'll leave the rest up to HAMJOOP.
     
  14. Dec 25, 2013 #13
    So, there is a non conservative force in this problem.

    if I am able to formulate the new "potential", is it still possible to use Lagrangian formulation ?
     
  15. Dec 25, 2013 #14

    vanhees71

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    Indeed, now that I checked the problem more carefully, I also see that this can not be treated via the Lagrange formalism. The reason seems to be that in the motion always new material of the rope is added to the moving part and thus that we have not a closed system.

    The correct way is to use Newton's 2nd Law directly:
    [tex]\mathrm{d}_t p=m g \; \Rightarrow \; x \ddot{x}+\dot{x}^2=m g.[/tex]
     
  16. Dec 27, 2013 #15
  17. Dec 27, 2013 #16

    vanhees71

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    That's true, but in this case the mass elements entering the moving part of the rope have 0 momentum and thus does not contribute an extra momentum. The formula by Marion (quoted in my previous posting) is thus correct. I just checked this in my favorite mechanics book (the good old Sommerfeld Lectures on Theoretical Physics, Vol. 1) :-)).
     
  18. Dec 27, 2013 #17

    D H

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    Actually, it isn't correct. Look at the units!

    Yes, Fext=d(mv)/dt can get you in trouble sometimes. But sometimes wikipedia's alternative expression can also be wrong.

    This problem can be solved without resorting to variable mass. Look at what is happening to the rope as a whole. Since the rope's mass isn't changing, it is perfectly safe to use Fext=dP/dt. What is the net external force on the rope? What is the rope's total momentum, and what is it's time derivative?
     
  19. Dec 28, 2013 #18

    vanhees71

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    But not the whole rope is moving, i.e., not the total mass is to be put into the momentum. I try to get the equation right, again: The moving mass is [itex]m=\lambda x[/itex] and thus
    [tex]\lambda \frac{\mathrm{d}}{\mathrm{d} t} (x \dot{x})=\lambda x g \; \Rightarrow \; x \ddot{x}+\dot{x}^2=g x.[/tex]
    This equation is correct, because the material entering the motion is at rest before and thus carries no momentum.

    Newton's Law
    [tex]\frac{\mathrm{d} \vec{p}}{\mathrm{d} t} = \vec{F}[/tex]
    only holds for the center of mass of a closed system. An example is the rocket equation. Here a gas with relative velocity [itex]u[/itex] (wrt. to the restframe of the rocket) is expelled (mass per unit time [itex]\mu[/itex]). The closed system is "rocket + gas", and we can write (for a rocket close to the earth)
    [tex]\frac{\mathrm{d} m v}{\mathrm{d} t}+\mu (v-u)=-m g,[/tex]
    where [itex]v[/itex] is the velocity of the rocket, and [itex]m=m_0-\mu t[/itex] is the mass still contained in the rocket.

    This leads to the Tsiolkovsky equation (with gravitational force in approximation close to earth),
    [tex]-\mu v+(m-\mu t) \dot{v} + \mu(v-u)=-m g[/tex]
    or
    [tex](m-\mu t) \dot{v}=\mu u - (m-\mu t) g.[/tex]
    This leads to
    [tex]\dot{v} = \frac{\mu u}{m-\mu t} -g[/tex]
    and can be simply integrated twice to get [itex]x(t)[/itex].
     
  20. Dec 28, 2013 #19

    D H

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    I'll address the first sentence later. With regard to the equation of motion, that is the correct equation. The left hand and right hand sides now agree. They didn't agree in what you wrote previously.

    Regarding "But not the whole rope is moving, i.e., not the total mass is to be put into the momentum": So? Newtonian mechanics (and it's rewrite as Lagrangian mechanics and Hamiltonian mechanics) can handle a system of particles, each moving with different speeds and each subject to different forces. That's the whole point of the latter half of chapter 2 in Marion, and in just about any other classical mechanics text.

    One way to escape the variable mass imbroglio is not to use variable mass. Treating the entire rope as a system, the external force acting on the rope is ##F_{\text{ext}} = F_{\text{grav}}+F_{\text{normal}} = \rho L g - \rho(L-x)g = \rho g x##. The momentum is just that of the part that is falling: ##p=\rho x v = \rho x \dot x##. Being a constant mass system, there's no problem with using F=dp/dt, yielding ##gx = \dot x^2 + x \ddot x##. Same equations as yours, but without the variable mass stuff.
     
  21. Dec 29, 2013 #20

    vanhees71

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    That's a good point, because it makes the whole business much clearer! Thanks.
     
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