Why Does Water Taper When Falling?

[TFM]

Homework Statement

If a tap is turned on gently to give a continuous stream of water, the column of water tapers (i.e. the width of the column decreases) as is falls. Explain this observation. Show that the cross-sectional area, A, of the column at a point P below the mouth of the tap is inversely proportional to the velocity of the water at P.

Homework Equations

The Attempt at a Solution

For the first part, I have written:

"The water tapers because when the water leaves the tap it has a constant velocity, but as it falls down, gravity accelerates the water. This means that the water at the bottom is going faster than the water at the top. Since the amount of water is constant, the water increases its density and makes itself have a smaller volume as it falls down, meaning that the bottom is thinner then the top."

What I am not sure of is the second part:

Show that the cross-sectional area, A, of the column at a point P below the mouth of the tap is inversely proportional to the velocity of the water at P.

What I have written kind of agrees, the cross-sec. area at the top, where it is slowest, is greater then the cross-sec. area at the bottom, where it is fastest, but I am not sure how to prove it.

Any suggestions?

TFM

 

[nasu]

The density of water do not change (actually it is very hard to compress water).
It's much simpler. The rate of flow (m^3/s) is constant along the column of liquid. Remember that
rate of flow = v *A

 

[TFM]

so

$$rate of flow = v*A$$

Where v is the velocity, A is the area

since this is constant, then

$$v_1*A_1 = V_2*A_2$$

so if V1 is greater then V2, the area will have to decrease. But how do I show the cross-sectional area, A, of the column at a point P below the mouth of the tap is inversely proportional to the velocity of the water at P?

TFM

 

[nasu]

v1*A1=v2*A2=v*A = Q = constant
So A=Q/v for any point. This means "inversely proportional".

 

[TFM]

Ah, thanks

TFM