Can with water rotates -- the water forms a paraboloid

1. Dec 2, 2015

Karol

1. The problem statement, all variables and given/known data
The angular velocity is ω, R is the radius of the vessel. at rest the water has depth H.
The face of the water form a paraboloid y=Ax2. find R for which the maximum height h of the water above the bottom doesn't depend on ω.

2. Relevant equations
Centripetal force: $F=m \omega^2 r$

3. The attempt at a solution
The function of the water face is:
$$y=\frac{\omega^2}{2g}r^2$$
The volume under the paraboloid is $V=\frac{\pi A}{2}R^4=\frac{\pi\omega^2}{2g}R^4$
The height of the highest point from the bottom, h, is derived from the difference between the volume at rest and the volume under the paraboloid:
$$\Delta V=\pi R^2 H-\frac{\pi\omega^2}{2g}R^4=\pi R^2 h\;\rightarrow\; h=H-\frac{\omega^2 R^2}{2g}R^2$$
If $R^2=\frac{1}{\omega^2}$ they cancel and $h=H-\frac{1}{2g}$ which doesn't include ω.
But that combination is for a specific ω, if ω changes R appears again, so what's the meaning of "R for which the maximum height h of the water above the bottom doesn't depend on ω"?

2. Dec 2, 2015

haruspex

The question makes no sense to me either. No matter what R is, varying w will change h.
Is this a translation?

3. Dec 2, 2015

Karol

yes, a translation, but i guess that's what they meant, i don't see anything else

4. Dec 2, 2015

haruspex

There is one interpretation I can think of that might work.
Consider the height of the surface as h=h(r), 0<r<R. Is there a value of r such that h(r) is independent of w?

5. Dec 4, 2015

Karol

I made a mistake in calculating h but now no better:
$$\pi R^2H=\frac{\pi\omega^2}{2g}R^4+\pi R^2h_1\;\rightarrow\; h_1=H-\frac{\omega^2R^2}{2g}$$
$$h=h_1+AR^2=H-\frac{\omega^2R^2}{2g}+\frac{\omega^2}{2g}R^2=H$$

6. Dec 5, 2015

haruspex

H cannot equal h.
I believe it should be $h=H+\frac{ω^2R^2}{2g}$. That is, H is half way between h and h1.
Did you try my interpretation in post #4?

7. Dec 5, 2015

Karol

Of course h≠H but how? i don't get $h=H+\frac{ω^2R^2}{2g}$, and if
$$\pi R^2H=\frac{\pi\omega^2}{2g}R^4+\pi R^2h_1\;\rightarrow\; h_1=H-\frac{\omega^2R^2}{2g}$$
Is correct then i don't know how to reach it, and if i don't get the right expression for h i have difficulty thinking further.
Is it true that the volume under a concave paraboloid is $V=\frac{\pi\omega^2}{2g}R^4$?

8. Dec 5, 2015

haruspex

In your post #5, how did you get the value of AR2? (I believe it is double the value you substituted.)
I agree with your expression for the volume under the paraboloid. Interestingly, it is also the volume above it, i.e. it is half the volume of the enclosing cylinder.

9. Dec 6, 2015

Karol

Last edited: Dec 6, 2015
10. Dec 6, 2015

haruspex

Ok, so where does that leave us with answering this question? Did you try my interpretation yet?

11. Dec 7, 2015

Karol

$$h=h_1+AR^2=H+\frac{\omega^2R^2}{4g}$$
I don't see how to interpret h(r) as independent of ω. the only way i see is when r=1/ω and they both cancel, but i told that before. you said:

12. Dec 7, 2015

Orodruin

Staff Emeritus
This absolutely cannot be true. Omega is an angular frequency and r is a length, they have strictly different physical dimensions. Anyway, this is not what haruspex is suggesting you to do.

13. Dec 7, 2015

haruspex

Maybe my use of h there is confusing you. Let's use x and y.
Let y=y(x) be the height of the surface at radius x. Can you find an x such that y does not depend on omega?

14. Dec 7, 2015

Karol

And at that x, if i change ω won't y change? i mean, is that a "point" x?

15. Dec 7, 2015

haruspex

It's a certain radius, yes. If you imagine the water profile as the rotation rate increases, it will dip in the middle and rise at the sides. So any two given profiles must intersect at some radius. The question is, is it the same radius all the time?

16. Dec 12, 2015

Karol

$$h=h_1+AR^2=H+\frac{\omega^2R^2}{4g}\;\rightarrow\; y_1=h_1+A_1x^2,\;y_2=h_2+A_2x^2$$
$$y_1=y_2:\;h_1+A_1x^2=h_2+A_2x^2\;\rightarrow\; x=\frac{\sqrt{2}}{2}$$
$$y_1=h_1+A_1\frac{1}{2}=H,\;y_2=H$$

17. Dec 12, 2015

haruspex

Looks good.

18. Dec 17, 2015

Karol

Continuation (or should i post a new thread?):
What's the angular velocity ω0 at which the lowest point of the water touches the bottom.
I equal volumes:
$$\pi R^2H=\frac{\omega^4}{4g}R^2\;\rightarrow\; \omega_0^2=\frac{4gH}{R^2}$$
At this ω0 the bottle is placed on a smooth table. there is friction between the water and the bottle. the final angular velocity of the bottle is ω1. express it using ω0. it's given that at this stage the height difference between the highest and lowest points on the water surface is H.
what's the moment of inertia of the bottle.
I found the moment of inertia of the water at ω0 to be:
$$I_0=\pi \frac{\rho \omega_0^2 R^6}{6g}=\frac{2}{3} \pi \rho HR^4$$
From $y=Ar^2$ i find ω1:
$$H=\frac{\omega_1^2}{2g}R^2\;\rightarrow\; \omega_1^2=\frac{2gH}{R^2}$$
I want to find h1. from the formula in post #9:
$$h_1=H-\frac{\omega_1^4R^2}{4g}=...=\frac{H}{2}$$
The moment of inertia of a solid cylinder: $I=\frac{1}{2}mr^2$
The moment of inertia of the paraboloid+the cylinder:
$$I_{W}=\frac{\pi\rho HR^4}{3}+\frac{1}{2}\pi\rho h_1R^4=...=\frac{7}{2}\pi\rho HR^4$$
Initial angular momentum:
$$I_0\omega_0=\frac{4}{3}\pi\rho HR^3\sqrt{gH}$$
Conservation of angular momentum: $I_0\omega_0=(I_W+I_B)\omega_1$
$$\frac{4}{3} \pi \rho HR^3 \sqrt{gH}=\left( \frac{7}{2} \pi \rho HR^4 \right) \frac{\sqrt{2}\sqrt{gH}}{R}$$
$$\rightarrow\; I_B=\left( \frac{8\sqrt{2}-7}{12} \right)\pi\rho HR^4$$

19. Dec 17, 2015

haruspex

I get the same IB.

20. Dec 17, 2015

phinds

Speaking apparently from a base of pure ignorance, I always thought that the cross section of the surface of water rotating like that was a catenary, like a free-hanging rope, not a parabola. I have no math or physics to back that up, it's just something that I heard (or made up?) MANY years ago and never had occasion to question it. I take it I'm wrong about that ?