We are given a standard diagram of a rotating cylinder with the parabolic shape in the rotation.


  • the distance between the vertex of parabola to the top of the parabola.
  • the original height of the water.

I have a really general question regarding this problem. From where the vertex of the parabola begins, up to the top, the pressure distribution varies with radius? And from the vertex of the parabola to the bottom of the cylinder, it only varies with height, since radius is fixed?

My question is: can I split the two different areas and integrate $P = \dfrac{\rho\omega^2}{2} -\gamma z$ twice? Once while holding the radius constant but the height a variable, and another while holding the height constant but integrating with respect to the radius.

I'm planning on using $\text{d}F= P\cdot\text{d}A$ when integrating each one.

A link to an image that looks similar to the one I have : http://demonstrations.wolfram.com/FluidRotatingInACylinder/

  • 1
    $\begingroup$ Are you looking for pressure at the wall, or pressure at any point inside they cylinder? That is, $P(z)$ or $P(z,r)$? $\endgroup$
    – Carlton
    Oct 25, 2015 at 13:44
  • $\begingroup$ Hello, I am looking for the pressure on the right half wall of the cylinder. $\endgroup$
    – user3496
    Oct 25, 2015 at 22:16

2 Answers 2


You don't need to split up the fluid domain into two parts. You can use the following equation for the pressure in a rotating fluid which I got from this PSU website:

$$p=p_a-\rho g z+ \frac{\rho r^2 \omega^2}{2}$$

$p \equiv$ pressure at location $(r,z)$

$p_a \equiv$ external (e.g. atmospheric) pressure

$\rho \equiv$ fluid density

$g \equiv$ gravity

$z \equiv$ height above the origin of the parabola (negative for the fluid below the parabola)

$r \equiv $ radial distance from the axis of rotation

$\omega \equiv$ angular velocity

Since you're only concerned with the pressure at the wall, $r$ is a constant and the only variable remaining is $z$. Just integrate this equation with respect to area over the height of the fluid (like you mentioned already) and that should give you total force on the container walls.


I would like to add to Carltons answer and go a little more in depth on how to approach problems like that in general.

Hydrostatics or Hydrodynamics

There are two possible solutions to the problem. You can either choose a reference frame that rotates with the fluids or that stays still and observes the system from the outside. This will determine if hydrostatics are allowed or not.

Rotating reference frame

The origin of the coordinate system (polar coordinates) is at the top of the cylinder, positive z axis direction is upward, against the gravity.

Since there is no movement in our system as we rotate the same way the fluid does, hydrostatics can be used.

$$\nabla p = \varrho \vec{f}$$ Due to the rotation of our reference frame we have to add the centrifugal force as an external force. $$\vec{f} = \begin{pmatrix} \Omega^2r\\ 0\\ -g \end{pmatrix}$$

Hence $$\begin{pmatrix}\frac{\partial p}{\partial r}\\ \frac{1}{r}\frac{\partial p}{\partial \varphi}\\ \frac{\partial p}{\partial z} \end{pmatrix} = \varrho \begin{pmatrix} \Omega^2r\\ 0\\ -g \end{pmatrix}$$

Successively Integration yields

$p(r,\varphi,z)=\varrho \Omega^2 \frac{r^2}{2} + C(\varphi,z) \tag{1}$

Inserting in the second row

$\frac{1}{r}\frac{\partial \varrho \Omega^2 \frac{r^2}{2} + C(\varphi,z)}{\partial \varphi} = 0$

$C'(\varphi,z) = 0$

Hence $C(z)$

Inserting in the third row yields

$C'(z) = \frac{\partial C(z)}{\partial z} = - \varrho g$

$C(z) = - \varrho g z + D$


$$p(r,z)=\varrho \Omega^2 \frac{r^2}{2} - \varrho g z + D$$

With the boundary condition that $p(r=0, z=0)=p_a$ it follows that

$$p(r,z)=p_a - \varrho g z +\varrho \frac{\Omega^2 r^2}{2} \tag{2} $$

This approach is more flexible as you can account for any changes in the problemset without relying on a formula that is just 1 solution for 1 specific problem.

Fixed reference frame

In this case we observe the system from the outside and we can no longer use hydrostatics for our problem.

Here Navier-Stokes comes in handy.

In polar coordinates for an inviscid, stationary flow the equation for the radial direction is

$u\frac{\partial u}{\partial r} + \frac{v}{r}\frac{\partial u}{\partial \varphi} + w\frac{\partial u}{\partial z} - \frac{v^2}{r} = -\frac{1}{\varrho} \frac{\partial p}{\partial r} + f_r$

with $\vec{v} = \begin{pmatrix} u\\ v\\ w \end{pmatrix}$

As there are now no external forces acting on the system $f_r=0$ and we omit all zero terms that follow from $u=0$ and $z=0$ as we only have a velocity in $v$ direction, it follows that.

$- \frac{v^2}{r} = -\frac{1}{\varrho} \frac{\partial p}{\partial r}$

with $\Omega = \frac{v}{r}$

$$\Omega^2 r = \frac{1}{\varrho} \frac{\partial p}{\partial r}$$

which equals (1).

Repeat with v and z direction and the same boundary condition you get to (2).

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I prefer general solutions over specific solutions that why I felt the need to add to this answer.

Calculating force

Now you can integrate over the walls of the cylinder.

$F = \int_A p(r=R,z) R d \varphi dz$


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