I'm teaching a grade 9 math course that deals with volumes of pyramids, cones, prisms, and cylinders and I wanted to come up with an interesting situation that uses the 1/3 volume factor. I know that fluids are tricky but I was wondering if the logic I've used here is a valid simplification or not? It's fine if they learn more details should they take physics some other time, but I don't want to tell them anything wrong. Here's the example:
"Water flowing through a cylindrical hose flows through a cone-shaped spout at the end. If water is flowing through the hose at 3 cm/s, how fast is it flowing through the cone tip? You can assume that water cannot be compressed to take up less volume.
Answer: The volume of the spout is one-third of the volume of the hose. As the water gets to the end of the hose, the water is squished into a smaller space. So, the water must flow faster to compensate. Logically, the water must flow through the cone 3 times as fast. So, water should be passing through the cone at a rate of 9 cm/s.
This 9cm/s should be considered an average rate. In reality, the water coming out of the tiny tip would be moving faster, but water at the base of the cone where it is the widest would be traveling more slowly."
EDIT: Thanks so much for helping me think through this problem, everyone. I enjoyed reading all of your answers and am grateful you took the time to write at such length. I am going to shelve this problem as is for the grade 9s, but if I do get a chance to pose it with another age group then I'll be able to do it properly then.
Best Answer
That's not right, unfortunately. The principle governing this situation is the continuity equation, which says that the total flow rate past any given point is constant. Since the flow rate is given by the flow speed $v$ times the cross-sectional area of the hose $A$, one has that $$v_1 A_1 = v_2 A_2$$ for any two points along the flow. In particular, the velocity of the water coming out through the nozzle is $$v_{spout} = \frac{\big(3 \mathrm{cm/s} \big)A_{hose}}{A_{spout}}$$ where $A_{hose}$ is the cross-sectional area of the hose and $A_{spout}$ is the cross-sectional area of the aperature at the end of the nozzle.
Yes. There is a sense in which the 9 cm/s is a relevant figure. If you divide the water within the cone into many tiny little parcels, compute the speed of each parcel, and then average them all together, the average of all of those speeds approaches 9 cm/s as the area of the aperture at the end of the cone goes to zero. In particular it is not the velocity at the half-way point of the cone (in fact, it is the velocity at the point $1-1/\sqrt{3}\approx 42.26\%$ of the way from the base to the tip).
If that's what you're asking your students to compute, the volume of a cylinder vs. the volume of a cone with the same length and base area is a reasonable shortcut to the solution. However, for grade 9 students I fear that the question would be completely opaque.
Your intent appears to be to get them to divide the volume of a cylinder by the volume of a cone. It is in principle possible to devise a question to which that is the shortest path to a solution, but my personal opinion is that this route would simply confuse them, and that any attempt at explanation would ultimately lead you to "look, just divide these two volumes ... " which defeats the point of having a creative question.
But then again, you're their instructor, so that's a call you'd have to make.