The stream of water from a faucet decreases in diameter as it falls. Derive an equation for the diameter of the stream as a function of the distance y below the faucet, given that the water has speed v_{0} when it leaves the faucet, whose diameter is D. The water molecules in contact with the air are being slowed down because they are colliding with the air molecules (essentially friction). These water molecules in turn slow down other water molecules in nearby water layers and so on. This is much is clear. However, I don't know how to model this situation mathematically. Can anybody give me a start-push?
I strongly doubt that the author of this question intends you to include air resistance in your calculation. Consider each particle of water as an indepent particle falling freely from the faucet. The speed gained by a particle falling is easily calculated via the conservation of energy: [tex]\frac{1}{2} m v^2 = m g h[/tex] [tex]v = \sqrt{2 g h}[/tex] Add the initial velocity [itex]v_0[/itex] to that expression and you're done. - Warren
But that would mean all the water molecules in the stream have the same velocity. And anyways, I'm not looking for the velocity, I'm looking for the diameter of the stream as a function of y.
No, it means all the water particles at a particular distance h below the faucet have the same velocity -- and they do. If you're looking for the diameter of the stream, just use the continuity equation: the same volume of water must pass through any cross-section of the stream in a given unit of time. If the water is moving twice as fast at some distance below the faucet, the stream must have half the area. You can calculate the change in diameter easily. - Warren
BTW, I apologize for slightly misreading the question; the variable you called y I called h, and didn't notice the ultimate point of the problem. - Warren
That is what I meant. However, this doesn't really explain why the diameter of the stream decreases as it falls. I'm not really familiar with the continuity equation. I'll have to look into it. Thanks.
Continuity equations are simple -- this one just means that the volume per unit time must be constant everywhere along the stream. If more water passes per unit second at the bottom than at the top, for example, then water must be being created somewhere! Do you understand how to proceed with this problem? - Warren
Using conservation of mechanical energy on a water molecule of mass m yields [tex]\frac{1}{2}mv_0^2 = \frac{1}{2}mv(y)^2 - mgy[/tex] Simplifying: [itex]v(y) = \sqrt{v_0^2 +2gy}[/itex]. Assuming the horizontal cross-sections are completely circular, the continuity equation yields [tex]\pi r(y)^2 v(y) = v_0 \pi (D/2)^2[/tex] Solving for r(y) gives [tex]r(y) = \frac{D}{2}\sqrt{\frac{v_0}{v(y)}}[/tex] The diameter of the stream at y is just 2r(y). I'm somewhat insecure though. According to my book, the continuity equation applies to an incompressible fluid flowing through an enclosed tube. Since the stream I'm dealing with in this problem isn't flowing in an enclosed tube, does the continuity equation apply?
Sure it does. All you're saying is that the volume rate of water flowing through a given cross section is equal to the rate through the next one. If this were not true, you would either have to have a source or drain between the two cross sections or allow a change in density. The continuity relation is very intuitive. Think about it a little and you'll see why it's true.
You may be thinking only in terms of cross section and how molecules next to each other should stay next to each other. Think about the length of the stream. Since the further the water falls, the faster it falls, the front end of the stream will accelerate away from the back end and the stream will stretch. And, just like with a rubber band, when you stretch it, it will decrease in cross section.
The enclosed tube needn't be an actual physical tube. An imaginary one works just as well. The water's surface tension holds it together as a continuous "tube" of flow.