Question

A vertical U-tube has a liquid up to a height of h. If the tube is slowly rotated to an angular speed $\omega =\sqrt{{g}/{h}\;}$, find the
(i) heights of the liquid column,
(ii) pressure at the points 1,2 and 3 in the limbs in steady state.

Answer 1

As the tube is slowly rotated to an angular speed, so, we will consider a mass element at some distance from the axis of rotation. Then, we will consider the equation of the circular motion for this element in terms of the pressure, area, mass and angular frequency. After integrating this equation, we will get the required results.
Formula used:
$P=\dfrac{F}{A}$

Complete answer:
Take a mass element at a distance x from the axis of rotation.

The equation of the circular motion for this element is,

& (p+dp)A-pA=dm{{\omega }^{2}}x \\\ & dpA=(\rho .Adx){{\omega }^{2}}x \\\ & pA=\rho {{\omega }^{2}}xdx \\\ \end{aligned}$$ Here p represents the pressure, A represents the area, dp represents the pressure of the smaller area considered, dx represents the small element, m is the mass and $$\omega $$is the angular frequency. Integrating both sides, we get, $$\int\limits_{{{p}_{1}}}^{{{p}_{3}}}{dp}=\rho {{\omega }^{2}}\int\limits_{-h}^{2h}{x\,dx}$$ Substitute the values of the limits after integrating the above equation. $$\begin{aligned} & {{p}_{3}}-{{p}_{1}}=\rho {{\omega }^{2}}\left[ \dfrac{{{x}^{2}}}{2} \right]=\dfrac{\rho {{\omega }^{2}}}{2}\left[ 4{{h}^{2}}-{{h}^{2}} \right] \\\ & \rho g({{h}_{3}}-{{h}_{1}})=\dfrac{3\rho {{\omega }^{2}}{{h}^{2}}}{2}=\dfrac{3\rho }{2}{{h}^{2}}\left( \dfrac{g}{h} \right) \\\ & ({{h}_{3}}-{{h}_{1}})=\dfrac{3}{2}h \\\ \end{aligned}$$ Thus, the difference between the heights at levels 1 and 3 is $$\dfrac{3}{2}h$$. But $${{h}_{3}}+{{h}_{1}}=2h$$ After solving the above 2 equations we get the values of the heights at levels 1 and 3 as follows. $${{h}_{3}}=\dfrac{7h}{4}$$and $${{h}_{1}}=\dfrac{h}{4}$$ Now substitute the expressions of the heights at the levels 1 and 3 in the equations of the pressures at the levels 1 and 3 respectively. So, we get, $${{p}_{3}}={{p}_{0}}+\rho g\left( \dfrac{7h}{4} \right)$$ and$${{p}_{1}}={{p}_{0}}+\rho g\left( \dfrac{h}{4} \right)$$ Therefore, the pressure at the points 1 and 3 in the limbs in steady state are as above. and $${{p}_{3}}={{p}_{0}}=\dfrac{\rho {{w}^{2}}}{2}\left[ {{x}^{2}} \right]_{0}^{2h}=\dfrac{\rho {{w}^{2}}}{2}{{\left( 4h \right)}^{2}}$$ As we have obtained the expressions of the pressure values at the levels 1 and 3, so, now we will compute the expression of the pressure value at the level 2. So, we have, $$\begin{aligned} & {{p}_{3}}-{{p}_{2}}=\dfrac{\rho }{2}\left( \dfrac{g}{h} \right).4{{h}^{2}}=2\rho gh \\\ & {{p}_{0}}+\rho g\left( \dfrac{7h}{4} \right)-{{p}_{2}}=2\rho gh \\\ & \therefore {{p}_{2}}={{p}_{0}}-\dfrac{\rho gh}{4} \\\ \end{aligned}$$ Therefore, the pressure at the point 2 in the limbs in steady state are as above. $$\therefore $$ (i) heights of the liquid column, $${{h}_{3}}=\dfrac{7h}{4}$$and $${{h}_{1}}=\dfrac{h}{4}$$ (ii) pressure at the points 1,2 and 3 in the limbs in steady state.$${{p}_{2}}={{p}_{0}}-\dfrac{\rho gh}{4}$$, $${{p}_{3}}={{p}_{0}}+\rho g\left( \dfrac{7h}{4} \right)$$ and$${{p}_{1}}={{p}_{0}}+\rho g\left( \dfrac{h}{4} \right)$$. **Note:** The formula that relates the parameters pressure, force and area and the formula that relates the parameters pressure, density of the fluid and the height of the fluid are important to solve this problem.