Let $\lambda$ be a linear density of a rope which is moving into a scale at velocity v. The additional force on the scale due to the collision is given as $\frac{d p}{d t} = v\frac{d m}{d t} = \lambda v^2$.

For an incompressible fluid, the stagnation pressure from stopping a column of water in excess of static pressure is

$$\frac{1}{2}\rho v^2$$

We can easily compare the forms by, for example multiplying by the width of the column to obtain a linear density of the fluid, or consider hitting the scale with a continuum of infinitesimal ropes. It seems the $\frac{1}{2}$ factor would remain different.

So what is the explanation for this relative factor of $\frac{1}{2}$? I have tossed around a few ideas but I'm curious what you may think.

  • $\begingroup$ I don't see why the mass of your rope changes? $\endgroup$
    – Bernhard
    Jul 16 '14 at 6:19
  • $\begingroup$ Can you provide a link for your equation $\tfrac{1}{2}\rho v^2$? $\endgroup$ Jul 16 '14 at 6:27

Your derivation of the additional force on the scale for the falling rope is wrong, both cases yield the same results.

If I understand correctly you are comparing the effect of a rope falling on a scale, to the similar fall of a liquid, something like this schematically:

The problem of your reasoning is related to a misconception of the stagnation pressure, which takes no part in this problem. You have used a term of the Bernoulli equation usually called dynamic pressure. The stagnation pressure is the pressure inside any point of a liquid in static conditions, and it will be the maximum pressure attainable inside an incompressible liquid (see Princeton site). Here you can argue that the liquid already fallen and inside the vessel has this pressure, but in any case none of these pressures are relevant to your problem: the falling of the liquid is completely analogous to the falling of the rope, and your derivation should be the same, using $\rho$ instead of $\lambda$.

Although the expressions are deceivingly similar, they refer to different scenarios.

Now on deriving your expression, the fundamental hint indicating that the above is not true, is that if you find the work invested in stopping the rope you get $mv_o^2$ ($v_o$ is the initial velocity of the rope) which is twice the total energy (just kinetic) that the rope had initially.

The thing is you are assuming that each portion of rope (of mass $\lambda$) stops immediately, which means that there is an infinite magnitude force. $$F_{\lambda}=-\frac{dp}{dt}= A \delta (t-t_o)$$ where $A$ is a constant unknown, $t_o$ is the time in which the force $F_{\lambda}$ acts, and the negative sign is because the force stopping the rope is opposed to the movement. We can find this constant by equating the impulse involved in stopping a rope element: $$ J_{\lambda} = \int_0^{\infty} F_{\lambda} dt = A$$ and we know that this impulse has to equal the momentum loss $\Delta p_{\lambda} = \lambda \Delta x v_o$, where we the portion of mass of the rope fragment is put as $\Delta m = \lambda \Delta x$. Therefore we have: $$ J_{\lambda} = \int_0^{\infty} F_{\lambda} dt = A\int_0^{\infty} \delta (t-t_o) dt = A = \lambda \Delta x v_o $$ So the proper expression of the element of force on the scale (opposite to the one on rope) is:

$$F_{\lambda}= \Delta m v_o \delta(t-t_o)$$

Some important notes:

1) Here we must not mistake $\Delta x$ as related to $v_o$, since it is the length of rope corresponding to the rope mass $\Delta m$. So $\Delta t = \frac{\Delta x}{v_o}$ is not the time in which the force acts (unless we assume so, and the problem is different).

2) The dimensions are correct, bearing in mind that $\delta (t - t_o)$ has to have dimensions of $T^{-1}$ since $\int_0^{\infty} \delta (t-t_o) dt = 1$

3) There is no mechanical work performed here $\int W dx = 0$ since we have assumed no displacement of the scale. But in fact this would also imply no deformation of the rope, and no redistribution of the water inside the glass to fill it. So in reality the loss of kinetic energy has been invested as the work involved in deforming the rope (and redistributing water) which, although not evident to calculate, should ultimately yield $\frac {m v_o^2}{2}$ the total initial kinetic energy.

  • $\begingroup$ Would you be able to derive the correct expression for the falling rope? $\endgroup$
    – WhatIAm
    Jul 20 '14 at 22:35
  • $\begingroup$ Sorry I forgot to add. It is now. $\endgroup$
    – rmhleo
    Jul 24 '14 at 8:44
  • $\begingroup$ Isn't $\int F dt$ impulse, not work? $\endgroup$
    – WhatIAm
    Jul 29 '14 at 20:49
  • $\begingroup$ Indeed, and the analysis should be done with impulse instead, the dimensions are wrong therefore. Thanks, I will edit it. $\endgroup$
    – rmhleo
    Jul 29 '14 at 21:24

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