# Rotating Horizontal Tube filled with ideal liquid

1. Feb 12, 2015

### andyrk

A horizontally oriented tube is rotating with angular velocity ω about one end of the tube which is open. The tube has ideal liquid filled inside it as shown in the attachment. Why does there have to be a pressure difference between two points in the ideal liquid at the same horizontal level?
Doesn't pressure only change with depth and remains constant for all points on the same horiozontal level for a stationary liquid?

Does the pressure difference exist so as to create an unbalanced force which accounts for the centripetal force on the liquid? But centripetal force exists only when a particle is executing circular motion. And when the liquid would have just been poured in the tube, it would slided back towards the closed end of the tube due to centrifugal force. And in the process of doing so (sliding back), there wasn't any centripetal force. So why then all of a sudden does the liquid halt at the end of the tube and starts executing circular motion all of a sudden when it was not doing so anywhere before halting? Why doesn't the liquid keep moving backward because of the centrifugal force and burst open the closed end of the tube?

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Last edited: Feb 12, 2015
2. Feb 12, 2015

### Staff: Mentor

The fluid is not stationary. It is accelerating.
Yes. Exactly.
That's right. It was both sliding radially and rotating. The inability to apply a centripetal force allowed it to slide out radially. If a centripetal force could have been applied, it would not have slid radially.

What makes you think the closed end of the tube isn't strong enough to prevent bursting. Suppose the tube is very thick and strong.

Chet

3. Feb 12, 2015

### andyrk

So you mean to say that only when it is stopped by the closed end, does the liquid begin moving in circular motion? Before the stoppage, it was rotating but not executing circular motion. Is that correct?

But then, why did the liquid not experience any centripetal force when it was going towards the closed end? Why didn't it have any pressure difference when it was going to the closed end but hadn't reached there yet?

Why did the liquid had to wait all the way to get back at the end of the tube to start circular motion? Why couldn't it do the same thing before reaching the closed end? And if it starts circular motion at the closed end, then shouldn't the normal reaction from the end of the tube contribute to the centripetal force (because that is what appears to have opposed the centrifugal force) and not the pressure difference?

Last edited: Feb 12, 2015
4. Feb 12, 2015

### andyrk

Anyone there?

5. Feb 12, 2015

### Staff: Mentor

If you mean purely circular motion, then yes. But, before it gets to the closed end, it has a component of circular motion.
Not exactly. It's motion is a combination of radial (sliding) and circumferential (circular) motion.
Because the liquid did not block the entire cross section of the tube yet, so the gas outboard of the liquid was able to move inward as the liquid moved outward. The liquid did not experience any centripetal force because there was no resistance to push against. The centripetal force that would have been there if the fluid had been constrained is taken up by the radial movement.
Before it got to the end of the tube, it was moving both radially and circumferentially (circular). Once it got to the end of the tube, the radial motion was blocked, and it could only move circumferentially.
The normal reaction from the end of the tube is equal to the pressure at the closed end times the cross sectional area. So, it's really the same thing.

Chet

Last edited: Feb 15, 2015
6. Feb 12, 2015

### andyrk

Why couldn't it be blocked before it had reached the end of the tube? Couldn't an internal pressure difference among points at the same horizontal level have existed even though it hadn't reached the end? And if it did, why wasn't it enough to counteract the centrifugal force?
So you mean to say, the pressure at the end and the normal reaction, both contributed to stopping the liquid? Or does the normal reaction arise because of the pressing (pressure) of the liquid at the end of the tube? And that normal reaction led to stopping the liquid?

7. Feb 12, 2015

### Staff: Mentor

If gas were trapped between the liquid and the end of the tube, it could prevent the liquid from reaching the end by compressing enough to provide the centripetal force. But, if the gas is not trapped, this would not happen. And, I don't think it could be trapped in reality because the liquid is much denser and is going to try to segregate to the bottom of the tube cross section.
No. They're both the same thing. The pressure is the normal reaction force per unit area.
Yes. Exactly.

Chet

8. Feb 12, 2015

### andyrk

But is the trapping of gas a concern at all? The question is that why can't pressure difference exist within points at the same horizontal level even thought the liquid hasn't reached the end of the tube? And if it does exist, then why is it not enough to balance the centrifugal force, so that the liquid stops in between and doesn't need to reach the end to stop and start executing circular motion?

9. Feb 12, 2015

### andyrk

Wait. I also have another explanation which might be wrong but let's give it a try. Does the Normal Reaction from the end of tube create a wave/disturbance inside the liquid which travels through the liquid and hence creates pressure differences between on small elements of the liquid? Does the pressure difference arise because of the normal reaction imparted by the wall? And so, the centripetal force equals the sum of all these pressure differences on infinitesimally small liquid elements multiplied by the area of cross section of the tube?

10. Feb 13, 2015

### Staff: Mentor

Good questions. If you examine the kinematics of the motion, you find that the radial component of acceleration for each fluid particle in the tube is given by:
$$a_r=\frac{d^2r}{dt^2}-\omega ^2 r$$
(see if you can derive this)
The $\omega ^2 r$ term represents the centripetal portion of the acceleration and the $\frac{d^2r}{dt^2}$ represents the contribution of the radial sliding. Before the fluid reaches the end wall, there are no external forces acting on it in the radial direction, such that the radial acceleration is zero, and the sliding motion is described by
$$\frac{d^2r}{dt^2}=\omega ^2 r$$
The external forces that are absent if the above relationship is satisfied are pressure gradients across the fluid. These cannot develop until the fluid encounters the end wall such that
$$\frac{d^2r}{dt^2}=0$$
The situation is very much analogous to what would happen if you had a vertical pipe closed at the bottom, and you introduced a small stream of water into the top of the pipe and allowed the water to fall under gravity. The pressure within the water stream would not change until it started encountering the base of the pipe.

Chet

11. Feb 13, 2015

### Staff: Mentor

This is pretty close to what happens.

Chet

12. Feb 13, 2015

### sophiecentaur

I have to ask whether this question was intended also to apply to a solid, free-moving piston (with mass) placed in the tube or is there something about the fact that the liquid can sag towards the lower part of the tube.

13. Feb 13, 2015

### Staff: Mentor

If there's vacuum on both sides of the piston, then the problem is pretty much the same. But, if there's gas on both sides that can't get past the piston, then the ability of the fluid to sag makes a difference because it lets the gas get by (and not be trapped).

Chet

Last edited: Feb 13, 2015
14. Feb 13, 2015

### sophiecentaur

I realise that you know the possibilities of the situation but I was trying to get to what the OP was actually asking - rather than doing the PF thing and reinventing the question. ( have done that so many times and I can often sense OPs backing away from the overkill answers.)
As it stands, the question is not answerable imo, because the liquid can flop to the bottom and run out to the the far end - unless we have a very small diameter tube with surface tension operating significantly.
It may be much better to consider a solid piston first as it is a very possible scenario.

15. Feb 15, 2015

### andyrk

Sorry. But I don't agree with this. $a_r$ is 0 only when the liquid has reached the end. Anytime before reaching the end, it is not 0. And so its not $a_r$ that becomes 0 but infact $\omega ^2 r$ that becomes 0 making $a_r=\frac{d^2r}{dt^2}$. Rest of the explanation is pretty fine with me. :)

16. Feb 15, 2015

### Staff: Mentor

You seem to be thinking that ar is equal to d2r/dt2, but it's not. It is the sum of the two terms, one being the centripetal acceleration.

$\omega ^2 r$ is never zero in this system. The tube is forcing the fluid to rotate, irrespective whether the fluid it is also moving radially.

Chet

17. Feb 15, 2015

### andyrk

$\omega ^2 r$ is the centripetal acceleration. And d2r/dt2 is the centrifugal force. At the end, both of them are equal. But before reaching the end,$\omega ^2 r$ has to be 0 because there is no force that would cause this centripetal acceleration, since the force that causes a centripetal acceleration would come into play only when it has reached the end because of pressure difference caused by the normal reaction. So it makes all the more sense to say that $\omega ^2 r$ is 0 since there doesn't exist any such force until the liquid has reached the end.

And also, d2r/dt2 means the centripetal acceleration on a body. But that comes into play when the body has started to execute circular motion. Before it, it is 0. So, I need to correct myself a little bit. Its not $\omega ^2 r$ that is 0 until the end (because this is the centrifugal force) but infact it is d2r/dt2 that is 0 until the end( since this represents the centripetal acceleration). And so ar is not 0, but rather it is equal to $-\omega ^2 r$, the minus sign indicating that it is in a direction opposite to the radially inward direction.

Last edited: Feb 15, 2015
18. Feb 15, 2015

### Staff: Mentor

$\omega ^2 r$ is strictly kinematic, irrespective of the forces involved. Is the tube rotating with angular velocity ω? Is the fluid rotating with the tube? Is the radius r zero? Then $\omega ^2 r$ is not zero both before and after the fluid reaches the end of the tube.

Sorry. None of this makes any sense to me. You can call d2r/dt2 whatever you want, but it is not the radial acceleration (unless the fluid is not rotating) and it is certainly not the centripetal acceleration. Kinematically, the radial acceleration is given by the first equation I presented in post #10.

If you want some second opinions on this, I can bring some additional Mentors into the discussion.

Chet