Question

A tube filled with water and closed at both ends uniformly rotates in a horizontal plane about the $O{O}^{{}^{\prime }}$ axis. The manometers fixed in the tube wall at distances $r_1$ and $r_2$ from the
rotational axis indicate pressures $p_1$ and $p_2$ respectively (Fig)
Determine the angular velocity $\omega$ of rotational of the tube, assuming that the density ${\rho }_w$ of water is known

Answer 1

Let us consider the condition of equilibrium for the mass of water contained between cross section separated by $x$ and $x+\Delta x$ from the rotational axis relative to the tube. This part of the liquid, whose mass i s${\rho }_{w}S\Delta x$, uniformly rotates at an angular velocity $\omega$ under the action of the forces of pressure on its lateral surface. Denoting the pressure in the section $x$ by $p\left(x\right),$ we obtained
$\left[p\right(x+\Delta x)-p(x\left)\right]S=\rho S\Delta x{\omega }^2(x+\frac{\Delta x}{2})$
Making $\Delta x$ tend to zero, we obtain the following equation:
$\frac{dp}{dx}={\rho }_w{\omega }^{2}x$,
whence
$p\left(x\right)={\rho }_w{\omega }^2\left(\frac{x^2}{2}\right)+p_0$
Using the condition of the problem,
$p_1=p\left(r_1\right){\rho }_w{\omega }^2\left(\frac{r_1^2}{2}\right)+p_0$
$p_2=p\left(r_2\right){\rho }_w{\omega }^2\left(\frac{r_2^2}{2}\right)+p_0$
we obtain the angular velocity of the tube:
$\omega =\sqrt{\frac{2}{\rho w}\frac{p_2-p_1}{r_2^2-r_1^2}}$.