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The impact of a rope versus the stagnation pressure of a fluid

  1. Jul 14, 2014 #1
    Let [itex]\lambda[/itex] 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
    [itex]\frac{d p}{d t} = v\frac{d m}{d t} = \lambda v^2[/itex]

    Where as the stagnation pressure from stopping a column of water in excess of static pressure is

    [itex]\frac{1}{2}\rho v^2[/itex]

    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 [itex]\frac{1}{2}[/itex] factor would remain different.

    So what is the explanation for this relative factor of [itex]\frac{1}{2}[/itex]? I have tossed around a few ideas but I'm curious what you may think.
     
  2. jcsd
  3. Jul 14, 2014 #2
    For water, the velocity of the water after hitting the tube is sideways, so thus only the kinetic energy of the forward flow is measured.

    For the rope, are you assuming an elastic collision with the rope rebounding with the velocity -v.
     
  4. Jul 15, 2014 #3
    I assume by sideways you mean radially outward, orthogonal to the original flow. Why do you think that is the case? What would happen when the "sideways" water hit the sides of the pipe? Why would that mean kinetic energy is measured?

    I don't agree.

    It appears specifically we have found the force required to stop the rope from moving and nothing more. We did not assume the rope was to rebound with velocity -v. We assumed our force had to take a certain amount of momentum per second and bring it to a stop, nothing more.
     
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