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A Paradox Which Must Be Resolved

  1. Mar 30, 2010 #1
    Thanks to the post by studiot, I can update my discussion with the calculations.

    Here's the problem:

    "Suppose that a flexible 4-ft rope starts with 3 ft of its length arranged in a heap right at the edge of a high horizontal table, with the remaining foot hanging (at rest) off the table. At time t = 0 the heap begins to unwind and the rope begins gradually to fall off the table, under the force of gravity pulling on the overhanging part. Under the assumption that frictional forces of all sorts are negligible, how long will it take for all the rope to fall off the table?"

    Note that many viewers will be familiar with a problem which is similar to this...but not the same! The rope on the table is arranged in a heap, which means that only the rope off the table is accelerating. The rope on the table remains at rest (it's not being dragged along).

    I didn't bring this up because I couldn't solve it--I could, but I got a different answer from the authors in my textbook. They used the definition F = d/dt [ma] as the basis of their model. I used the principle of conservation of energy (the system is, after all, conservative).

    It makes sense to set up the potential energy system so the rope on the table has a gravitational potential energy of 0 (and rope hanging off the table has negative gravitational potential). The following is my model:

    [tex]Energ{y_{sys}} = C = Energ{y_{gravitational\;potential\;rope}} + Energ{y_{kinetic\;rope}}\;[/tex]

    [tex]Energ{y_{grav}} = {m_{off}}gh = (wx)(g)( - x/2) = \frac{{ - wg{x^2}}}{2}[/tex]

    [tex]Energ{y_{kinetic}} = \frac{1}{2}{m_{off}}{\left( {\frac{{dx}}{{dt}}} \right)^2} = \frac{1}{2}wx{\left( {\frac{{dx}}{{dt}}} \right)^2}[/tex]​

    In these equations, m-off means mass off the table. w is the linear density (w is irrelevant, soon cancelled in the calculations). The goal now is to find a function t(x), which fits these equations. I have set up the x-axis with 0 at the table and pointing downward. Therefore, positive values of x correspond to negative potential energies.

    [tex]{E_{sys}} = {E_{grav}} + {E_{kin}}[/tex]​

    We can use the initial conditions to find the energy of the system. See below.

    [tex] - \frac{{wg}}{2} = - \frac{{wg{x^2}}}{2} + \frac{{wx}}{2}{\left( {\frac{{dx}}{{dt}}} \right)^2}[/tex]

    [tex]\sqrt {g\frac{{{x^2} - 1}}{x}} = \frac{{dx}}{{dt}}[/tex]​

    And variables are easily separated:

    [tex]\frac{{dt}}{{dx}} = \sqrt {\frac{1}{g}} \cdot \sqrt {\frac{x}{{{x^2} - 1}}} [/tex]

    [tex]t(4) = \sqrt {\frac{1}{g}} \int_1^4 {\sqrt {\frac{x}{{{x^2} - 1}}} } dx[/tex]​

    This integral is improper. A trigonometric substitution does the trick:

    [tex]\ell etting\;x = \sec \theta [/tex]

    [tex]t(4) = \sqrt {\frac{1}{g}} \int\limits_{{{\cos }^{ - 1}}(1)}^{{{\cos }^{ - 1}}(1/4)} {{{(\sec \theta )}^{3/2}}d\theta } [/tex]

    [tex]t(4) \approx .4876(s)\;using\;Simpson's\;Rule\;with\;n\;=\;100[/tex]​

    This is my answer. As you can see, I got it using a pretty straight-forward application of conservation of energy. If anyone doesn't understand the model or work, just ask. I'm pretty sure it's all right. Now, my textbook does NOT begin with conservation of energy. Instead, it begins with an application of F = d/dt [ma]. Here's the model:

    [tex]F = \frac{{dP}}{{dt}} = [mv];\quad m = wx;\quad F = mg = wxg[/tex]

    [tex]wxg = \frac{d}{{dt}}[wxv] = w\left( {x\frac{{dv}}{{dt}} + v\frac{{dx}}{{dt}}} \right)[/tex]

    [tex]xg = x\frac{{dv}}{{dt}} + v\frac{{dx}}{{dt}}[/tex]​

    Here, the author cleverly uses the chain rule to eliminate t and reduce the equation to two variables: v and x, where v is a function of the independent variable x.

    [tex]xg = x\frac{{dv}}{{dx}}\frac{{dx}}{{dt}} + v \cdot v[/tex]

    [tex]xg = xv\frac{{dv}}{{dx}} + {v^2}[/tex]​

    Now, the author sets up the equation in differential form to be solved as an exact differential equation:

    [tex]0 = \left( {xv} \right)\,dv\, + ({v^2}\, - xg)\,dx[/tex]​

    Multiplying both sides by the simple integrating factor x yields an exact equation:

    [tex]0 = ({x^2}v)\,dv + ({v^2}x - {x^2}g)dx = dF[/tex]​

    Given this differential form of F, dF, we can easily find F to be:

    [tex]F(x,v) = \frac{1}{2}{x^2}{v^2} - \frac{g}{3}{x^3} = C[/tex]​

    It's easy to check that

    [tex]\frac{{\delta F}}{{\delta v}} = {x^2}v\,;\,\frac{{\delta F}}{{\delta x}} = {v^2}x - {x^2}g[/tex]​

    And that F is a general solution to the differential equation. Now, we use the initial condition to find that

    [tex]C = - \frac{g}{3}[/tex]​

    We then replace v with dx/dt, then we separate variables and integrate.

    [tex]{v^2} = \frac{{2g}}{3}\left( {\frac{{{x^3} - 1}}{{{x^2}}}} \right)[/tex]

    [tex]\frac{{dx}}{{dt}} = \sqrt {\frac{{2g}}{3}} \cdot \sqrt {\frac{{{x^3} - 1}}{{{x^2}}}} [/tex]

    [tex]t = \sqrt {\frac{3}{{2g}}} \int {\sqrt {\frac{{{x^2}}}{{{x^3} - 1}}} } dx[/tex]

    [tex]t(4) = \sqrt {\frac{3}{{2g}}} \int\limits_1^4 {\sqrt {\frac{{{x^2}}}{{{x^3} - 1}}} } dx[/tex]​

    Which, like in the first example, is improper. Trig sub!

    [tex]\ell etting\;{x^3} = {\sec ^2}\theta [/tex]

    [tex]t(4) = \sqrt {\frac{3}{{2g}}} \int\limits_0^{{{\cos }^{ - 1}}(1/8)} {{{(\sec \theta )}^{4/3}}d\theta } [/tex]

    [tex]t(4) \approx 0.5411(s)[/tex]​

    Now, here's an easy exercise. Show that this model does NOT obey conservation of energy. How is this possible? In the first model, using conservation of energy from the start, we get that the rope falls off in 0.4876 seconds. In the second model, using F = dP/dt, we get that the rope falls off in 0.5411 seconds. What's going on here? The calculations are right, so is there a problem with one of the models? Obviously, there must be. Thanks in advance for any thoughtful responses.


    All credit for this problem and the second solution go to the authors C.H. Edwards, Jr. and David E. Penney of the book Elementary Differential Equations With Boundary Value Problems, 2nd Edition.
  2. jcsd
  3. Mar 31, 2010 #2

    Doc Al

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    No it isn't. The coiled sections of rope must be jerked to the full speed of the falling sections, as they are dragged along, bit by bit. Mechanical energy isn't conserved; some is "lost" to internal energy.
  4. Mar 31, 2010 #3
    That makes some sense...Now, I'm confused as to how exactly this "internal energy" manifests itself. The whole thing is frictionless, so I don't want to say heat...

    I guess what I'm saying is this: What does it mean for a system to lose energy to "internal energy" if there is no heat output? Or is there a heat output? How can I predict (in future problems) whether a system will or will not experience this phenomenon?
  5. Mar 31, 2010 #4

    Doc Al

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    The rope will "heat up", so loosely speaking energy does convert to heat. But it's better to say internal energy, since "heat" really means a transfer of energy due to a temperature difference.

    The rope does warm up.
    Hard to say. Look for things equivalent to inelastic collisions. When it doubt, solve the problem multiple ways. Barring calculational errors, if you get different answers at least one of your solutions must be based on an incorrect assumption.
  6. Mar 31, 2010 #5
    That was very helpful Doc. Thanks. I've never seen this happen before. I'm glad that at least the calculations were correct though :cool:

    It makes sense that, doing it my way, the rope would slide down in less time, since ALL of it's energy was converted to kinetic in my model, rather than (realistically) only MOST of it being converted.

    The question's settled then.
  7. Mar 31, 2010 #6


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    In any real situation where the rope is coiled up, energy will indeed be lost.
    But in the way the problem is posed, the assumption that the rope is coiled is not used at all. The model more appropriately applies to a flexible, nonstretchable, rope that lies stretched out initially. Energy would be conserved in this case.

    I disagree with the method in the book. Variable mass problems in Newtonian mechanics are often handled incorrectly. You should never use the equation [tex]F = v\frac{dm}{dt}+ma[/tex], it is simply wrong (it violates Galilean relativity). Mass is conserved in classical physics and the correct way to handle it (like in the rocket or falling waterdrop kind of problems) is to treat the mass as being accumulated from elsewhere. (See http://articles.adsabs.harvard.edu//full/1992CeMDA..53..227P/0000227.000.html" for a treatise on this).

    Your method is correct, but you did make a small mistake in the equation for the kinetic energy:
    Since the entire rope moves with the same velocity, the kinetic energy is [tex]\frac{1}{2}m(\frac{dx}{dt})^2[/tex], where [tex]m[/tex] is the mass of the entire rope, not just the part that's hanging off.
    If you redo the calculation you'll get: t(4) = 0.727 s.

    It's actually simply to solve for the equation of motion from Newtons law. The force on the rope depends on x and is simply F(x)=wxg. With the initial conditions this leads to:
    You can then solve x(t)=4 numerically. It's also easy to verify that energy is conserved.
    Last edited by a moderator: Apr 24, 2017
  8. Apr 2, 2010 #7
    That's actually a different problem. Don't forget, it states that the rope is "arranged in a heap" at the edge of the table. Take another look. You and Doc Al agree that in the case of this problem, energy should and will be lost. Take another look.
  9. Apr 2, 2010 #8

    Doc Al

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    Sorry that I didn't see this post earlier.
    Sure it is. That assumption prevents you from modeling it as a stretched out rope. When the rope is 'arranged in a heap' you cannot model all parts of the rope as having the same speed, as you can when the rope is stretched out on the table.
    Yes, energy is conserved in that case, but that's a different problem, as Mazerakham says.
  10. Apr 4, 2010 #9


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    Sheesh, I just read right over that part, even though it was mentioned explicitly.
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