# Pressure of a Rotating Bucket of Liquid

1. Aug 4, 2004

### e(ho0n3

A cylindrical bucket of liquid (density $\rho$) is rotated about its symmetry axis, which is vertical. If the angular velocity is $\omega$, show that the pressure at a distance r from the rotation axis is

$$P = P_0 + 1/2\rho \omega ^2 r^2$$

where $P_0$ is the pressure at r = 0.

Isn't depth supposed to play a role here!? Am I supposed to use Bernoulli's equation to solve this (although I don't see how)? It seems the presure increases as we move away from the rotation axis but this doesn't make any sense to me because the velocity of the fluid is (in my opinion) faster the farther away we are from the rotation axis and by Bernoulli's principle the pressure should be lower. Am I missing something?

Last edited: Aug 4, 2004
2. Aug 4, 2004

### arildno

Sure, depth will play a role here, if the effect of gravity is to be included
(Surfaces of constant pressure is paraboloids about the rotation axis in this case
The author's equation is correct by regarding P0 as a function of height)

In order to solve this problem, go into the rotating reference frame measured from which the velocity of the fluid is zero.
Look at the equations of motion gained.

Be careful how to interpret Bernoulli's equation!!
Given two points on a streamline at the same height, it is certainly true that at the point where we have the highest pressure, we have the lowest velocity.

But what is the streamlines in this case (on which Bernoulli is applicable)?
They are (plane) circles!!
But the velocities at different points lying on such a circle are the same constant, hence, the pressure at these points is the same as well, by Bernoulli.

What you were talking about can be stated:
How does the pressure vary across streamlines (not along, as in Bernoulli)?
(The answer is called Crocco's theorem, I believe)
The result is simply that the net pressure force must provide the required centripetal acceleration at a given point.
Since the centripetal acceleration increases outwards, so must the pressure.

EDIT: Actually, the main point is that the direction of the centripetal acceleration is inwards towards the axis at all points.
To see why this is true, consider a position at (r,z0), and regard the pressures, p(r+dr,z0), p(r-dr,z0).
Surely, in order to gain the required centripetal acceleration, we must have p(r+dr,z0)>p(r-dr,z0), that is $$\frac{\partial{p}}{\partial{r}}>0$$

Last edited: Aug 4, 2004
3. Aug 6, 2004

### e(ho0n3

But how is this supposed to help me determine pressure.

So you're saying that the pressure at these points on the circle do not depend on the value of r? Huh?

Now I'm just totally confused. The centripetal acceleration is v2/r and if v is constant (as you wrote above), then the centripetal acceleration should decrease outwards.

4. Aug 6, 2004

### arildno

1. I said that the velocities at different points on the SAME STREAMLINE are equal !!
It does not follow from this that the velocity at a point on a different streamline is equal to the same value (it is not)
Bernoulli's equation is only applicable along a streamline; since, on a given streamline the velocities at different points are the same, the pressures on these points must also be equal to each other (the circular streamline lies in some horizontal plane which means the gravitational potential at different points on the same streamline is the same)
Again, just because the pressure is constant along a given streamline, does not mean that the pressure cannot vary across streamlines (which means that the pressure may vary with the radius, which it does).

2. read my edit again.
The velocity on a given streamline is $$r\omega$$ (that is, constant along the circle; the direction is tangential to the circle)
The centripetal acceleration is therefore: $$r\omega^{2}$$ which increase with the radius (inwards radial direction).

3. I'll get back to the actual solution later

Last edited: Aug 6, 2004
5. Aug 6, 2004

### e(ho0n3

OK. I was misunderstanding the notion of a streamline. Now this makes much more sense.

OK, this makes perfect sense now. I just don't see how to calculate the pressure. I'm having trouble visualizing how the pressure is acting on the streamline.

6. Aug 7, 2004

### arildno

The pressure acts on a fluid element (representable as a point) which at a given time happens to coincide with some point on some streamline

Now, how to solve it:
You are seeking a rigid body motion solution for the fluid, or, equivalently, if you go into the rotating reference frame (angular velocity $$\omega\vec{k}$$), the fluid exhibits no relative motion in this frame (hence, no reklative accelerations either; the appropriate acceleration terms appears on the other side of the equation as pseudo-forces)

The equations of motion reads:
$$\vec{0}=-\frac{1}{\rho}\nabla{p}-g\vec{k}+\omega^{2}r\vec{i}_{r}$$
Here, $$\omega^{2}r\vec{i}_{r}$$ is the centrifugal force pr. unit mass; its direction is radially outward (away from) the rotation axis (i.e, the vertical, anti-parallell to the direction of the gravity force pr. unit mass, $$-g\vec{k}$$)

Essentially, therefore, the most convenient set of coordinates is cylindrical coordinates.
We assume a constant density $$\rho$$

Note that $$\omega^{2}r\vec{i}_{r}=\vec{i}_{r}\frac{\partial}{\partial{r}}\frac{1}{2}\omega^{2}r^{2}=\nabla(\frac{1}{2}\omega^{2}r^{2})$$

I'm using here the form of the gradient operator appropriate to cylindrical coordinates:
$$\nabla=\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta}\frac{\partial}{r\partial{\theta}}+\vec{k}\frac{\partial}{\partial{z}}$$

Clearly, we also have: $$-\frac{1}{\rho}\nabla{p}=\nabla(-\frac{p}{\rho})$$
And:$$-g\vec{k}=\nabla(-gz)$$

Hence, the equations of motion may be rewritten as:
$$\nabla(\frac{p}{\rho}+gz-\frac{1}{2}\omega^{2}r^{2})=\vec{0}$$

Clearly, a function whose derivative with respect to all its variables is zero everywhere in the region, must be equal to some constant.
(This is just a multivariable version of if f'(x)=0 for all x, then f(x)=C, where C is some constant)

Hence, we have, in the fluid region:
$$p(r,z)=C-\rho{gz}+\frac{\rho}{2}\omega^{2}r^{2}$$
Note that $$P0=C-\rho{gz}$$ yields a hydrostatic pressure distribution along the rotation axis.

Last edited: Aug 7, 2004
7. Aug 7, 2004

### e(ho0n3

Where does $-\frac{1}{\rho}\nabla{p}$ come from and why is the equation set to $\vec{0}$.

Last edited: Aug 7, 2004
8. Aug 7, 2004

### e(ho0n3

OK, i just realized why you set it to $\vec{0}$ (because of the frame of reference you chose).

9. Aug 7, 2004

### arildno

This is how the pressure force pr. unit mass occurs in Newton's 2.law of motion for a fluid.

10. Aug 7, 2004

### e(ho0n3

I'm not familiar with Newton's 2. law as applied to a fluid. Would you explicate?

11. Aug 8, 2004

### arildno

Sure enough.
For starters, let's just assume we're in a reference frame where the fluid is at rest.

Then, for any arbitrary region of the fluid, the (vector) sum of forces aciting upon that region must be equal to zero.
The forces may act on the surface on the region (in our case, pressure force), or acting on all fluid particles contained in our region (body forces, like gravity or pseudo-forces if we're in a non-inertial reference frame)

The pressure force $$\vec{P}$$ acts along the local vector normal on the surface, so we have:
$$\vec{P}=-\int_{S}p\vec{n}dS$$
Here, S is the surface of our region V, the scalar function p is the (isotropic) pressure, and $$\vec{n}$$ is the unit outward normal on the surface.
The total body force may in our case be written as:
$$\vec{B}=\int_{V}-\rho{g}\vec{k}+\rho\omega^{2}r\vec{i}_{r}dV$$

Or, we have that for the region V, Newton 2.law states:
$$\vec{B}+\vec{P}=\vec{0}$$

We transform the surface integral into a volume integral through the relation for a function f:
$$\int_{S}f\vec{n}dS=\int_{V}\nabla{f}dV$$

Hence we have:
$$\int_{V}-\nabla{p}-\rho{g}\vec{k}+\rho\omega^{2}r\vec{i}_{r}dV=\vec{0}$$

Since V is arbitrary, the integrand must be identically zero, which is equivalent to the form of Newton's 2.law I posted first

Last edited: Aug 8, 2004
12. Aug 8, 2004

### e(ho0n3

Thanks for the explanation. It makes total sense now. The level of maths. used is beyond the treatment given in my physics book (where I got the problem from). Not that I don't understand (as I said, it makes total sense), but I believe there must be a simpler (mathematically speaking) derivation.

13. Aug 8, 2004

### arildno

Has the book covered/assumed the knowledge of partial differentiation?
In particular, has it covered the use of cylindrical coordinates?

14. Aug 8, 2004

### e(ho0n3

The book assumes that you have this knowlege I guess (especially for the harder problems since most of them seem to involve calculus), although I still haven't encountered a problem that uses partial differentiation, cylindrical coordinates, or the divergence theorem. It's a pretty dumbed down book.

15. Aug 8, 2004

### arildno

What about the distinction of a field description (Eulerian) and a particle description( Lagrangian)?
(The reason why I ask this, is that I'm trying to decipher the author's intended approach (assuming he didn't intend a flawed approach))

16. Aug 9, 2004

### e(ho0n3

I have no idea what you mean by that and there is no mention of it in my book.

17. Aug 9, 2004

### Gokul43201

Staff Emeritus
I think there may a simpler approach - hopefully, it's not the flawed approach that arildno's talking about.

Let the axis of the cylinder be the z-axis. Our frame is the rotating frame.
Consider a point in the liquid given by P(r,z), where r is the radial distance from the axis.

Consider a fluid element of mass m, at this point. The forces on this element are mg downwards and $$m\omega^2r$$ outwards. The x and y components of $$m\omega^2r$$ are respectively $$m\omega^2x$$ and $$m\omega^2y$$.

Darn, I need partial derivatives now...but I'll try without. Let the elementary mass be a cubical box (since we're working in rectangular co-ords) of volume, dV = dX dY dZ, with faces parallel to the co-orsinate planes. The force on the element in the x-direction means there's more pressure on one of the YZ planes than on the other. Let this difference in pressure be dP. Clearly, dP.dYdZ = F_x = dV (dP/dX)

So, $$~\frac{dP}{dX} = \frac{m}{dV}\omega^2x = \rho\omega^2x$$

No, this is too much sleight of hand....and I'll need to use partial derivatives to go anywhere. But, if you do, it's pretty straightforward. You just get,

$$P = \frac {\rho\omega^2x^2}{2} + \frac {\rho\omega^2y^2}{2} + P_0$$

and you can do it all in rectangular co-ordinates.

Last edited: Aug 9, 2004
18. Aug 9, 2004

### Gokul43201

Staff Emeritus
Here's a possibly flawed approach : Work in the rotating frame. Let the tangent to the surface at some P(r,z) be at an angle $$\theta$$. The resultant force at an element on the surface is due to mg downwards and $$m\omega^2r$$ outwards. We expect this resultant to be normal to the surface, so that gives us,
$$tan\theta = \frac {m\omega^2r}{mg} = \frac {\omega^2r}{g} = \frac {dz}{dr}$$

From this you get the shape of the surface, $$z(r) = \frac {\omega^2r^2}{2g} + z_0$$

So, $$P(r) = \rho g z(r) = \frac {\rho \omega^2r^2}{2} + P_0$$

19. Aug 9, 2004

### arildno

1."We expect this resultant to be normal to the surface"
Oh, really? Why should we expect that?
Let's first look at the rationale behind this idea, as clarified by looking at the differential form of Newton's 2.law (rewritten slightly):
$$-g\vec{k}+\omega^{2}\vec{i}_{r}=\frac{1}{\rho}\nabla{p}$$
Here we see the justification for the assertion:
At any isobar level (surface with constant pressure), the remaining force is parallell to the normal of the isobar, and in the direction of increasing pressure value.

Is this rather subtle argument, which along with knowledge of the differential equations also requires familiarity with the isobar concept really simple, and difficult to misunderstand?
For example, since the argument is invalid for all other types of surfaces than isobars, should one really expect a student to hit upon this strategy?
2.
How you conjured forth this particular formula is beyond me, here's my version:
a) Given an isobar i, we have:
$$P(r,z_{i}(r))=P_{i}$$
Where $$P_{i}$$ is the pressure value on isobar i

b) Since there are no vertical acceleration or velocity, we may assume that the pressure's dependency on the vertical is hydrostatic; hence, the pressure may in general be written on the form:
$$P(r,z)=f(r)-\rho{gz}$$
Using the result from a), we have, on an arbitrary isobar:
$$P_{i}=f(r)-\rho{gz_{i}(r)}$$
by which the form of f(r) may be elucidated.

I don't really see this method as particularly simple; however, your argument had a certain elegance in shortness..

Last edited: Aug 10, 2004