A vertical U tube rotates about the vertical axis -- Find omega to cause this height offset

[Aurelius120]

Homework Statement

A vertical U tubes having a liquid rotation about the axis as shown. Then the angular velocity is

Relevant Equations

Force due to liquid column=$V\rho g$
Centripetal force on horizontal liquid =$m\omega^2r$
Area=A

I thought I could use force balance as
$$V\rho g=m\omega^2r$$
But that gives two different values and neither correct
$$(Ab)\rho g=(Ab\rho)\omega^2 (b) \text{ on the left}$$
$$(A×3b)\rho g=(A×2b×\rho)\omega^2(2b) \text{ on the right}$$

Just in case, the answer in the book in B

 

[haruspex]

The water in a horizontal section is not all at the same distance from the axis.
You seem to have assumed pressure at the axis is atmospheric.

 

[Aurelius120]

The water in a horizontal section is not all at the same distance from the axis.
You seem to have assumed pressure at the axis is atmospheric.

So I have to integrate or something?
Isn't the pressure atmospheric everywhere , so it doesn't affect the net results anywhere?

 

[haruspex]

So I have to integrate or something?

Yes.

Isn't the pressure atmospheric everywhere , so it doesn't affect the net results anywhere?

It can't be assumed atmospheric anywhere inside the water-filled sections.
You may not realise your equations assume that it is atmospheric in the water at the rotational axis.

 

[Aurelius120]

Yes.

It can't be assumed atmospheric anywhere inside the water-filled sections.
You may not realise your equations assume that it is atmospheric in the water at the rotational axis.

So it will be
$$\int{\omega^2r dm}= V\rho g$$
On either side with limits $0\text{ to }b$ and $0\text{ to }2b$ and $V$ changed according to height of liquid column

 

[haruspex]

So it will be
$$\int{\omega^2r dm}= V\rho g$$
On either side with limits $0\text{ to }b$ and $0\text{ to }2b$ and $V$ changed according to height of liquid column

Better, but you still are not allowing an arbitrary pressure in the water at the axis.

 

[Aurelius120]

Better, but you still are not allowing an arbitrary pressure in the water at the axis.

So how do I account for that? Its arbitrary so how to know its value

 

[haruspex]

So how do I account for that? Its arbitrary so how to know its value

There will be enough equations. Remember, in post #1 you had conflicting equations.

 

[Lnewqban]

I thought I could use force balance as
$$V\rho g=m\omega^2r$$
But that gives two different values and neither correct

Please, consider that at radius b, the shape of the curve must be symmetrical all around the vertical axis of rotation.
Therefore, "on the left" is similar to "on the right" anywhere between the axis of rotation and the radius b (concentric rotation).

There is balance within that range of radii.
There is nothing to balance the rotating mass located between the radius b and 2b (excentric rotation).

Perhaps "force balance" is not the way to the solution.
Which shown option satisfies that the height is triple (3b) when the radius is double (2b)?

 

[haruspex]

Perhaps "force balance" is not the way to the solution.

It is a perfectly reasonable way.

Which shown option satisfies that the height is triple (3b) when the radius is double (2b)?

Since one option is "none of the above" you can't use that shortcut.

 

[Aurelius120]

It is a perfectly reasonable way.

Since one option is "none of the above" you can't use that shortcut.

I think I got it
$$\int^b_0{\omega^2 A\rho rdr}-PA=A(b)\rho g$$
$$\int^{2b}_0{\omega^2A\rho rdr}-PA=A(3b)\rho g$$
Therefore
Subtracting equation 1 from 2
$$\frac{3A\rho\omega^2b^2}{2}=2Ab\rho g$$

 

[Aurelius120]

Please, consider that at radius b, the shape of the curve must be symmetrical all around the vertical axis of rotation.
Therefore, "on the left" is similar to "on the right" anywhere between the axis of rotation and the radius b (concentric rotation).

There is balance within that range of radii.
There is nothing to balance the rotating mass located between the radius b and 2b (excentric rotation).

Perhaps "force balance" is not the way to the solution.
Which shown option satisfies that the height is triple (3b) when the radius is double (2b)?

I am not sure I understand. The water can't be treated as rectangular/cylindrical blocks with forces $V\rho g$ on either side and arbitrary force due pressure in-between️

 

[haruspex]

I think I got it
$$\int^b_0{\omega^2 A\rho rdr}-PA=A(b)\rho g$$
$$\int^{2b}_0{\omega^2A\rho rdr}-PA=A(3b)\rho g$$
Therefore
Subtracting equation 1 from 2
$$\frac{3A\rho\omega^2b^2}{2}=2Ab\rho g$$

Yes, except that the balance is of pressures, not forces. You don’t need A in there. In fact, if the cross-sectional area of the horizontal tube were different from that of the vertical tubes then having those two different areas in your equation would lead to an incorrect result.

 

[Aurelius120]

Yes, except that the balance is of pressures, not forces. You don’t need A in there. In fact, if the cross-sectional area of the horizontal tube were different from that of the vertical tubes then having those two different areas in your equation would lead to an incorrect result.

But isn't it equating the force required for rotation to force exerted? I don't know if there is something like Centripetal pressure that is necessary for rotation

 

[haruspex]

But isn't it equating the force required for rotation to force exerted? I don't know if there is something like Centripetal pressure that is necessary for rotation

It is just like gravity, except that it is horizontal and varies along the length of the tube.
Suppose the horizontal tube were to vary in cross section, starting wide at the axis then narrowing. Where the sides of the tube go inwards, they would take some of the total centrifugal force. What is reliably transmitted to the next bit of fluid is the pressure.

 

[Lnewqban]

ωI am not sure I understand. The water can't be treated as rectangular/cylindrical blocks with forces $V\rho g$ on either side and arbitrary force due pressure in-between️

Apologies for not being able to explain myself in a clearer way, as well as for late response.
Those forces acting on the block must be associated with its cross-section areas and with the static pressure acting on each side.

Besides the correct Math solution, did you get a full understanding of the physics behind the different levels reached by the liquid in each branch due to rotation with angular velocity ω?

Please, see:
https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/14-1-fluids-density-and-pressure/