Rotation

From a top down view we can derive the following relationships…

• $a_r=-\Omega r{\hat{e}}_r$
• $P\left(r,z\right)=\frac{1}{2}\rho r^2{\Omega }^2-\rho gz+C$

But what we really care about here is this relationship…
$gh=\frac{1}{2}{\Omega }^2{r}^2$
As well as this pressure relationship for fluids rotating on an axis…
$P_O=P_{atm}-\rho gh$

Solving Tips for Cylinder

• Find the volume of the open tank $V=\pi R^{2}h$
• Find $V_{spin}={\int }_0^{R}2\pi rh\left(r\right)dr$
• Now we can use our $P\left(r,z\right)$ formula to solve for our constant
• Now using our definition of C, we can solve for $z$ using our $P\left(r,z\right)$ formula
• Use $z$ in our $V_{spin}$ equation to solve for $\Omega$