# Sum of forces with liquid in rotation

It's not homework (I'm teacher). I would like to compute sum of forces on this study : The shape is symmetrical like that I'm sure the center of gravity is in the center of the shape. I compute forces on axis $X$ only. No external gravity. Air container are fixed on big container. But air container has very low pressure, I don't want to compute it, consider it like vacuum.

I know force of pressure is like :

$\frac{1}{2}{\rho}h_{\epsilon}^2w^2R^2$ with R the distance to container from center of gravity of container, $h_{\epsilon}^2$ is a small area, true ?

I compute forces on water, depth is d, it's:

$$d\omega^2\rho*((\int_{0}^{h1}(r1^2+y^2)cos(atan(\frac{y}{r1}))dy)-(\int_{0}^{h1}(r2^2+y^2)cos(atan(\frac{y}{r2}))dy)+(\int_{0}^{h2}(r3^2+y^2)cos(atan(\frac{y}{r3}))dy))$$

so, it's:

$$d\omega^2\rho*((\int_{0}^{h1}\sqrt{(r1^2+y^2)}r1dy)-(\int_{0}^{h1}\sqrt{(r2^2+y^2)}r2dy)+(\int_{0}^{h2}\sqrt{(r3^2+y^2)}r3dy))$$

Is it ok ?

Now, if I want to compute forces on rectangle solid, I need to integrate 2 times ? Like that :

$$\frac{1}{2}d\rho\omega^2*(\int_{-h2}^{h2}\int_{0}^{r3}(x^2+y^2)^{0.5}cos(atan(\frac{y}{x}))dxdy-(\int_{-h1}^{h1}\int_{r1}^{r2}(x^2+y^2)^{0.5}cos(atan(\frac{y}{x}))dxdy)$$

So it's:

$$\frac{1}{2}d\rho\omega^2*(\int_{-h2}^{h2}\int_{0}^{r3}(x^2+y^2)^{0.25}xdxdy-\int_{-h1}^{h1}\int_{r1}^{r2}(x^2+y^2)^{0.25}xdxdy)$$

How to add mass inside integrales ?

I have a problem, I don't find $0$ for the sum of forces.

There are 2 additional forces $F_a$ that decrease forces from liquid: But the shape can be like that: So in this case forces $F_a = 0$. In last case, $F_a$ are compensated by what forces ?

If I take a macroscopic model, with compressible balls for understand where F forces are compensated, consider the center of gravity like red point (balls have more density than solid, just enough for compensated): And there is another force to take in account it's the red force : This red force is apply all along the surface of container ?

The question is confusing for several reasons.

First, it doesn't specify if the air is fixed in place somehow, or free to move. The air container would experience a force in the direction of the axis of rotation the magnitude of which would depend upon the density of the liquid, the density of the air, the rate or rotation, and distance of each point of the container from the axis.

Second, it introduces a variable or constant "k" without any explanation.

Third, it's not clear what force is being considered. If the water is not moving toward or away from the axis, there is no net force on the water. There will be a pressure at each point in the water that depends upon distance from the axis. The pressure of the water exerts a varying force on each infinitesimal area of the water container, and the container exerts an equal and opposite force.

The pressure at any point in the water (assume in air is in a fixed container) would be $\rho \omega ^2 r^2/2$, where "r" is distance from the axis of rotation.

• $dM=\rho Adx$ where $dM$ is the mass of a cross section and $A$ is the area of the cross section. You haven't told us a 3-dimension shape for the container, so we can not say what $A$ is in terms of the variables provided. – DavePhD Apr 21 '14 at 13:40