## Sunday, February 18, 2018

### Rotating Fluid

Suppose we have an infinitely tall cylinder of radius R, filled to a height H with an incompressible fluid. We then set the fluid rotating about the cylindrical axis at angular speed $\omega$. Suppose we take a differential chunk of fluid on the surface, a radius r from the axis. The resulting normal force will then be $N=F_c+W$. This normal force, as the name suggests, will be normal to the fluid surface. It follows by simple geometry, that $\frac{dy}{dr}=\frac{F_c}{W}=\frac{r \omega^2}{g}$ From which it follows that the height of the surface at any radius will be given by $y=\frac{r^2 \omega^2}{2g}+C$ Let us define $\omega_0=2\sqrt{gH}/R \\ u=\omega/\omega_0$ Given that the fluid is incompressible, we know that the total volume does not change. From this, we can determine that the height of the surface at any radius will be given by: $y(r)=2H\left ( ru/R \right )^2+\left\{\begin{matrix} H(1-u^2) \\ 2H(u-u^2) \end{matrix}\right. \, \, \, \, \, \, \, \begin{matrix} u \leq 1\\ u > 1 \end{matrix}$ The highest point on the liquid surface is then given by: $y_{\textrm{max}}=\left\{\begin{matrix} H(1+u^2)\\ 2Hu \end{matrix}\right. \, \, \, \, \, \, \, \begin{matrix} u \leq 1\\ u > 1 \end{matrix}$ If $u > 1$, the center of the base of the cylinder is not covered by fluid. There is a minimum radius at which fluid can be found. This minimum radius is given by: $r_{\textrm{min}}=R\sqrt{1-\frac{1}{u}}$ If the fluid is of uniform density and of total mass M, then the moment of inertia of the rotating fluid is given by $I=\left\{\begin{matrix} \frac{MR^2}{2}\left ( 1+\frac{u^2}{3} \right )\\ MR^2\left ( 1-\frac{1}{3u} \right ) \end{matrix}\right. \, \, \, \, \, \, \, \begin{matrix} u \leq 1\\ u > 1 \end{matrix}$ Note for each of these piecewise functions, the functions and their first derivatives are continuous.