Will the hose of jet water placed on a frictionless horizontal ground eventually become a straight line?

Question

Will the hose of jet water placed on a frictionless horizontal ground eventually become a straight line?

This is a top view of a hose for jet water placed on the ground. One end is vertically fixed to the wall and the other end is free. The ground is horizontal and frictionless. The flow of water is stable.

If the reaction force is only caused by the bending of the hose, and the object must have a centripetal force to move in a curve, and the ground is frictionless, the hose of jet water will become a straight line perpendicular to the wall. But in fact, the hose of jet water doesn't become a straight line. It keeps swinging back and forth. Why? Some people explain that this is the result of water gaining momentum. I doubt this explanation, because the momentum of water is obtained at the pump, which is generally considered not to be obtained in the hose, so it should not affect the hose at the outlet. How do I understand this phenomenon?

Answer 1

This is called the garden hose instability. The behavior depends on several factors, including the stiffness of the pipe and the flow rate. For low flow rates you can get a steady situation, but for higher flow rates the system becomes unstable and “flutters”. In principle, if you lined the hose up perfectly straight, the flow would be straight, but it is unstable and any slight perturbation will cause the hose to buckle and the fluttering motion to begin as the nozzle turns and causes the hose to accelerate laterally.

Here is a nice video explaining in more depth:

https://www.coursera.org/lecture/fluid-solid-interaction/6-1-the-garden-hose-instability-fvRjz

And here is a slow motion video of the flutter

https://youtu.be/KMxyy5NrZ-o

As you can see, it does not stay straight.

Answer 2

An actual straight hose is clearly in equilibrium, but I'd argue it's unstable.

Imagine an almost straight section of the hose, with slight curvature. The water passing around the curve is changing direction, which means it is accelerating, which means a force is being applied to it. The force is "inwards" to the curve, and so the equal-and-opposite force the water applies to the hose is "outwards". So, a curved section is pushed "out".

Where I want to disagree with the other answers is that if the tip of the hose is free to move, and the hose is in-extensible, then the curve will move outwards and the tip will be dragged along behind it. This dragging of the tip behind a moving section will make the angle of curvature greater, at least at first.

However, because our curve is moving outwards but one end is attached to the wall, a new curve must be created, curving the other way, to accommodate its new position. This will be pushed outwards in the other direction.

I think the net result is the the curves in the hose travel towards the loose end like waves, each one inducing an opposing curve at the wall end, and the result is a tip that whips back and forth. I am less confident of this last paragraph than the preceding ones.

Answer 3

Consider a constant cross section hose that follows the curve $\mathbf{x}(s)$ where $s$ is the distance along it. The tangent vector is $\mathbf{T}(s)=d\mathbf{x}/ds$ and the normal vector is $\mathbf{N}(s)=(d\mathbf{T}/ds)/||(d\mathbf{T}/ds)||$. Then $\frac{d\mathbf{T}}{ds} = \kappa(s) \mathbf{N},$ where $\kappa(s)$ is the curvature.

When there is a bend the force will be proportional to $\kappa(s)$ and along $\mathbf{N}$. So if we assume constant density 1 and so on, the force will be $\mathbf{F} = \kappa(s) \mathbf{N}$. So we get a differential equation $$\frac{d^2\mathbf{N}}{dt^2} = \kappa(s) \mathbf{N}.$$

To get the full dynamics we need to add constraints on $\kappa(s)$ and the link to $\mathbf{T}(s)$ to keep the hose length constant; this looks messy. But it is enough to look at the equation to see that this isn't stable: if $\kappa(s)=0$ the hose remains straight, but if it is nonzero then you get exponential growth of $\mathbf{N}$: any deviation will be amplified until the constraints start to counteract it, producing the swinging motions.

With friction there will be a $-f (d\mathbf{N}/dt)$ term in the equation making sufficiently low-curvature sections of the hose stay still.

The whole system has an energy flow through it (in the form of the kinetic energy of the literal flow of water) that can drive non-equilibrium dynamics.

Answer 4

For a hanging pipe, there are two characteristic frequencies that can reinforce each other, the pendulum frequency $f_p$, and the flow length frequency $f_{uL}$, where $u$ is fluid velocity:

$$f_p= \frac{\sqrt{g}}{2\pi \sqrt{L}} \text{ , } f_{uL}=\frac{u}{L}$$

To drive oscillation: $$ \sqrt{gL}=2\pi u \text{ , eg: } L=1m, u=\tfrac{1}{2}\tfrac{m}{s}$$

However, our problem has no restoring force and will reach equilibrium

Assume as given in the comments that the the flow of water is stable, which is a big assumption, but no other assumptions:

The tube will settle down to an equilibrium that is nearly straight, and perfectly straight if the tube is perfectly symmetrical in dimension and material.

Before equilibrium, the forces will move and bend (mostly unbend) the tube. A curved tube with water flowing through it has a net force for the ongoing momentum transfer, in the direction radially outward (essentially centrifugal). This refers to the radius of the bend, not of the tube. This is what makes the tube move in some situations. (For example, in power plants, large elbows of cooling water must be braced to maintain a lot of lateral force far exceeding the weight of the pipe plus water, and changing weight during changing flows, particularly during startup and shutdown.) This force also tends to straighten the tube.