Why do water streams get thinner as they fall?

[goleynik]

Homework Statement

Develop an equation that will allow you to determine the radius of the stream (r) at any distance (y) below the rim of the faucet.

It would seem that the initial velocity of the water coming out of the faucet and the acceleration due to gravity have something to do with the radius of the stream at any point.

Homework Equations

The only equations needed are :
Area of a circle,
Volume of a cylinder,
v=v_o + at
y=y_o + v_y0t+ 1/2 at²

The Attempt at a Solution

I solved for a faucet opening of .96 cm and determined that the water exits at 7.67cm³/sI don't even know if what I found is relevant but I am really stuck and any help would be greatly appreciated.

 

[Simon Bridge]

What are the forces on a small cylindrical section of a stream of falling water?

 

[Chestermiller]

In terms of the cross sectional area of the stream and the downward velocity of the water, what is the volume flow rate of the water? Is the volume flow rate out of the faucet constant? Does the volume flow rate change with distance below the faucet? If the downward velocity of the water is increasing with distance from the faucet, what is happening to the cross sectional area?

Chet

 

[astolfo]

As it fall a volume V of water gets thinner and longer due to gravity. This means that dV/dt is the same at t1 and t2 so until the water stream breaks up I got: dr/dt = g/2 *( dr**2/dh).

 

[haruspex]

As it fall a volume V of water gets thinner and longer due to gravity. This means that dV/dt is the same at t1 and t2 so until the water stream breaks up I got: dr/dt = g/2 *( dr**2/dh).

The question asks for r as a function of y. Time should not enter into it.
Presumably your answer refers, on the left, to how the radius of a given parcel of water varies with time as it descends. Is that also what the RHS refers to, with $h=h_0-y$?

 

[Ibix]

As it fall a volume V of water gets thinner and longer due to gravity. This means that dV/dt is the same at t1 and t2 so until the water stream breaks up I got: dr/dt = g/2 *( dr**2/dh).

I'm not sure this is helpful. Particularly as I'm not clear what dr**2/dh is supposed to be - you may wish to check the LaTeX Guide linked below the reply box to learn how to format maths here.

I think the key observation needed is to answer the third question posed by Chestermiller above - is the volume of water passing a point in any given time independent of height?

 

[haruspex]

I'm not clear what dr**2/dh is supposed to be

I think it is a fair guess that it means $\frac{dr^2}{dh}$, but since, for a given parcel of water, h is changing as a function of t it possibly should be $\frac{\partial r^2}{\partial h}$