# Volumetric flow of a overflowing cone

• jackkk_gatz
In summary, the professor is asking if when the cone overflows, the volumetric flow is constant or increases. The student solves for the volumetric flow as a function of ω and then confirms that the volumetric flow is indeed constant at a higher ω.

#### jackkk_gatz

Homework Statement
A cone (with height H and radius R) has a total volume V, contains a fluid of density ρ and volume mV, where m is a fraction (0<m<1). Such cone is rotated about its axis of symmetry and accelerates very slowly. The magnitude of gravity is g. Find the volumetric flow rate at which the liquid overflows from the container.
Relevant Equations
p1-p2=pw^2(r1^2-r2^2)/2-pg(z1-z2)
I solved the case where m=0.99999. Then the height at which it overflows can be obtained with the equation, when points on the liquid surface are chosen. Then the cross-sectional area is given by the circumference of the circle times the height that the parabola reaches, that cross-sectional area multiplied by the linear velocity, which is obtained from v=wR, gives the volumetric flow.
I don't know if my approach is correct, but I want to solve for all cases, not just one. And I'm not quite sure how to do that, If m is small, it is troublesome to find the new hight, the one that starts at H so the liquid overflows. Also the fact that it accelerates very slowly, makes me think a minimum velocity is reached, then How big has to be the velocity for the cone to overflow? I can obtain the angular velocity before it overflows, but not while it overflows. I could find it if I knew the acceleration.

Please explain the posted equation (what is the context and what are all those variables?) and please post your solution to the case you solved (I'm not sure I understand what you did).
Even the question is hard to interpret. Won't the flow rate change over time, as the cone speeds up?

haruspex said:
Please explain the posted equation (what is the context and what are all those variables?) and please post your solution to the case you solved (I'm not sure I understand what you did).
Even the question is hard to interpret. Won't the flow rate change over time, as the cone speeds up?
I too am having difficulty understanding the question, because it is quite a few things that change. It makes sense and I think it is the right thing to think, that the volumetric flow is not constant, I don't understand what my professor meant at all. The only thing he told me when I asked him my doubts was, imagine what would happen, a pretty bad answer to be honest.

The equation is

##p-p_0= \frac {ρw^2(r^2 -r_0^2)} {2}-ρg(z-z_0)##

It is on cylindrical coordinates. The context is, i think is analogous to the bernoulli equation, except it is for rotating bodies. Where the term ##ρg(z-z_0)## works the same as hidrostatic pressure, where you choose two points. It works the same for the term ##\frac {ρw^2(r^2 -r_0^2)} {2}## except, since we are on cylindrical coordinates instead of height, it is the radius measured from the origin of our coordinate system to the points chosen. And ##w## is the angular velocity

My solution is, since a parabola is formed when the liquid starts rotating. I chose two points on the surface, the one at the lowest height (but still at the surface) and the other at the highest point.

Since both are at the surface, the must have the same pressure, the atmospheric pressure so:
##p_1=p_2=P_{atm}##
So I have

##0= \frac {ρw^2(r^2 -r_0^2)} {2}-ρg(z-z_0)##

I chose my coordinate system on the bottom of the container. My firs point is aligned with the z axis, so ##r_1=0## and my second point is away R ##r_2=R##.

##0= \frac {ρw^2(R^2)} {2}-ρg(z_2-z_1)##
##ρg(z_2-z_1)= \frac {ρw^2(R^2)} {2}##

If my container is almost full (m=0,99999) I can say that ##z_2-z_1## is the height of the cross-sectional area. I substitue ##z_2-z_1=h##
##ρg(h)= \frac {ρw^2(R^2)} {2}##
##h= \frac {ρw^2(R^2)} {2ρg}##
##h= \frac {w^2(R^2)} {2g}##

Using this on ##VA=Q##

##(wR)(2\pi R)(\frac {w^2(R^2)} {2g})=Q##

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jackkk_gatz said:
I too am having difficulty understanding the question
Are you sure he did not ask what the rotation rate will be when it overflows?

haruspex said:
Are you sure he did not ask what the rotation rate will be when it overflows?
the problem ask two things, the first one is asking for the angular velocity for the cone to overflow, I already solved that, it was not a problem. The second one is the one that asks for the volumetric flow, which is the one that I wrote, the one I'm struggling with

jackkk_gatz said:
the problem ask two things, the first one is asking for the angular velocity for the cone to overflow, I already solved that, it was not a problem. The second one is the one that asks for the volumetric flow, which is the one that I wrote, the one I'm struggling with
Ok, so assume it is rotating at some rate ω greater than that threshold value.
The centripetal force, plus gravity, gives a potential. Bernoulli then gives the linear flow rate at the lip.
From the total volume of fluid and the surface shape, you can calculate the thickness of the fluid layer at the lip of the cone, and thus get a volumetric flow rate as a function of ω.
Does that work?

haruspex said:
Ok, so assume it is rotating at some rate ω greater than that threshold value.
The centripetal force, plus gravity, gives a potential. Bernoulli then gives the linear flow rate at the lip.
From the total volume of fluid and the surface shape, you can calculate the thickness of the fluid layer at the lip of the cone, and thus get a volumetric flow rate as a function of ω.
Does that work?
what do you mean by gives a potential?