Can with water rotates -- the water forms a paraboloid

[Karol]

Homework Statement

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.

Homework Equations

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

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"?

 

[haruspex]

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

 

[Karol]

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

 

[haruspex]

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

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?

 

[Karol]

I made a mistake in calculating h but now no better:

$$\pi R^2H=\frac{\pi\omega^2}{2g}R^4+\pi R^2h_1\;\rightarrow\; h_1=H-\frac{\omega^2R^2}{2g}$$
$$h=h_1+AR^2=H-\frac{\omega^2R^2}{2g}+\frac{\omega^2}{2g}R^2=H$$

 

[haruspex]

$h=h_1+AR^2=H−\frac{ω^2R^2}{2g}+\frac{ω^2}{2g}R^2=H$

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?

 

[Karol]

H cannot equal h.
I believe it should be h=H+ω2R22gh=H+\frac{ω^2R^2}{2g}. That is, H is half way between h and h1.

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 $V=\frac{\pi\omega^2}{2g}R^4$?

 

[haruspex]

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 $V=\frac{\pi\omega^2}{2g}R^4$?

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.

 

[Karol]

$$\pi R^2H=\frac{\pi\omega^2}{4g}R^4+\pi R^2h_1\;\rightarrow\; h_1=H-\frac{\omega^4R^2}{4g}$$
$$h=h_1+AR^2=H-\frac{\omega^2R^2}{4g}+\frac{\omega^2}{2g}R^2=H+\frac{\omega^2R^2}{4g}$$
The volume under a paraboloid is:
$$V=\frac{\pi\omega^2}{4g}R^4$$
I made a mistake earlier. i made a thread about it at the math forum:
https://www.physicsforums.com/threads/integral-area-of-a-revolved-parabola.845135/#post-5304042

 

[haruspex]

$$\pi R^2H=\frac{\pi\omega^2}{4g}R^4+\pi R^2h_1\;\rightarrow\; h_1=H-\frac{\omega^4R^2}{4g}$$
$$h=h_1+AR^2=H-\frac{\omega^4R^2}{4g}+\frac{\omega^2}{2g}R^2=H+\frac{\omega^4R^2}{4g}$$
The volume under a paraboloid is:
$$V=\frac{\pi\omega^2}{4g}R^4$$
I made a mistake earlier. i made a thread about it at the math forum:
https://www.physicsforums.com/threads/integral-area-of-a-revolved-parabola.845135/#post-5304042

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

 

[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:

No matter what R is, varying w will change h.

 

[Orodruin]

the only way i see is when r=1/ω

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.

 

[haruspex]

$$h=h_1+AR^2=H+\frac{\omega^2R^2}{4g}$$
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:

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?

 

[Karol]

Can you find an x such that y does not depend on omega?

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

 

[haruspex]

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

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?

 

[Karol]

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

 

[haruspex]

View attachment 93268 $$h=h_1+AR^2=H+\frac{\omega^2R^2}{4g}\;\rightarrow\; y_1=h_1+A_1x^2,\;y_2=h_2+A_2x^2$$
$$y_1=y_2:\;h_1+A_1x^2=h_2+A_2x^2\;\rightarrow\; x=\frac{\sqrt{2}}{2}$$
$$y_1=h_1+A_1\frac{1}{2}=H,\;y_2=H$$

Looks good.

 

[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: $I_0\omega_0=(I_W+I_B)\omega_1$
$$\frac{4}{3} \pi \rho HR^3 \sqrt{gH}=\left( \frac{7}{2} \pi \rho HR^4 \right) \frac{\sqrt{2}\sqrt{gH}}{R}$$
$$\rightarrow\; I_B=\left( \frac{8\sqrt{2}-7}{12} \right)\pi\rho HR^4$$

 

[haruspex]

$$\rightarrow\; I_B=\left( \frac{8\sqrt{2}-7}{12} \right)\pi\rho HR^4$$

I get the same I_B.

 

[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 ?

 

[Orodruin]

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 ?

Yup, it is a parabola. With the effective potential in the rotating frame, this is the shape of the equipotential surfaces.

 

[phinds]

Yup, it is a parabola. With the effective potential in the rotating frame, this is the shape of the equipotential surfaces.

OK, thanks.

 

[haruspex]

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 ?

Consider the FBD for a particle on the surface at radius x. No vertical acceleration so $N\cos(\theta)=mg$, where $\tan(\theta)=\frac{dy}{dx}$. Centripetal acceleration = $N\sin(\theta)=mx\omega^2$. Solve.

 

[Karol]

Thank you all. Haruspex and the rest