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

Click For Summary
SUMMARY

The discussion focuses on the comparison between the force exerted by a moving rope and the stagnation pressure of a fluid, specifically water. The force due to the rope's linear density, represented as d p/d t = \lambda v^2, contrasts with the stagnation pressure formula 1/2 \rho v^2. The key point of contention is the factor of 1/2, which arises from the differing dynamics of the rope and fluid interactions upon impact. The conversation emphasizes that the force calculated for the rope does not assume elastic rebound, but rather the momentum required to stop the rope.

PREREQUISITES
  • Understanding of linear density in physics
  • Familiarity with fluid dynamics principles
  • Knowledge of momentum and kinetic energy concepts
  • Basic grasp of collision types (elastic vs inelastic)
NEXT STEPS
  • Research the principles of linear momentum in fluid dynamics
  • Study the derivation and implications of stagnation pressure in fluid mechanics
  • Explore the differences between elastic and inelastic collisions
  • Investigate the effects of fluid velocity on pressure measurements in pipes
USEFUL FOR

Physicists, engineers, and students studying fluid dynamics and mechanics, particularly those interested in the interactions between moving objects and fluids.

MisterX
Messages
758
Reaction score
71
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.
 
Physics news on Phys.org
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.
 
256bits said:
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.
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?

256bits said:
For the rope, are you assuming an elastic collision with the rope rebounding with the velocity -v.

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.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 67 ·
3
Replies
67
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
11
Views
3K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 40 ·
2
Replies
40
Views
5K