Chapter 2, Problem 2.159P

To determine

(a)

The formula for change in height of the tube.

Answer to Problem 2.159P

The formula for change in height of the tube is $\Delta h_{centre}=-\frac{1}{3}\frac{{\Omega }^2{R}^2}{g}$.

Explanation of Solution

Given information:

The depth of water is $20\text{cm}$, the angular velocity is $120\text{r}/\text{min}$.

Write the expression for total displacement between outer and center menisci.

$h_{total}=\frac{v^2}{2g}$ ……. (I)

Here, total displacement is $h_{total}$, velocity is $v$ and the acceleration due to gravity is $g$.

Write the expression for velocity.

$v=\Omega R$ ……. (II)

Here, angular velocity is $\Omega$ and the radius is $R$.

Substitute $\Omega R$ for $v$ in Equation (I).

$\begin{array}{c}{h}_{total}=\frac{{\left(\Omega R\right)}^2}{2g}\\ =\frac{{\Omega }^2{R}^2}{2g}\end{array}$ ……. (III)

Write the expression for change in height.

$\Delta h_{outer}=-\frac{\Delta h_{centre}}{2}$ …… (IV)

Here, the change in height of outer half is $\Delta h_{outer}$ and the change in height of the center tube is $\Delta h_{centre}$.

Write the expression for rise of water in outer leg.

$\Delta h_{outer}=\frac{1}{3}{h}_{total}$ ……. (V)

Substitute $\frac{{\Omega }^2{R}^2}{2g}$ for $h_{total}$ in Equation (V).

$\begin{array}{c}\Delta h_{outer}=\frac{1}{3}\frac{{\Omega }^2{R}^2}{2g}\\ =\frac{1}{6}\frac{{\Omega }^2{R}^2}{g}\end{array}$

Substitute $\frac{1}{6}\frac{{\Omega }^2{R}^2}{g}$ for $\Delta h_{outer}$ in Equation (IV).

$\begin{array}{l}\frac{1}{6}\frac{{\Omega }^2{R}^2}{g}=-\frac{\Delta h_{centre}}{2}\\ \Delta h_{centre}=-2\times \frac{1}{6}\frac{{\Omega }^2{R}^2}{g}\\ \Delta h_{centre}=-\frac{1}{3}\frac{{\Omega }^2{R}^2}{g}\end{array}$

Conclusion:

The formula for change in height of the tube is $\Delta h_{centre}=-\frac{1}{3}\frac{{\Omega }^2{R}^2}{g}$.

To determine

(b)

The height in each tube.

Answer to Problem 2.159P

The height of water in outer tube is $0.0268\text{m}$.

The height of water in center tube is $-0.0536\text{m}$.

Explanation of Solution

Given information:

The depth of water is $20\text{cm}$, the angular velocity is $120\text{r}/\text{min}$ and the radius is $10\text{cm}$.

Write the expression for displacement in outer tube.

$\Delta h_{outer}=\frac{1}{6}\frac{{\Omega }^2{R}^2}{g}$ ……. (VI)

Here, displacement in outer tube is $\Delta h_{outer}$, acceleration due to gravity is $g$, angular velocity is $\Omega$ and the radius is $R$.

Write the expression for displacement of water in central tube.

$\Delta h_{centre}=-\frac{1}{3}\frac{{\Omega }^2{R}^2}{g}$ ……. (VII)

Here, displacement of water height in center tube is $\Delta h_{centre}$.

Calculation:

Substitute $120\text{r}/\text{min}$ for $\Omega$, $10\text{cm}$ for $R$ and $9.8{\text{m}/\text{s}}^2$ for $g$ in Equation (VI).

$\begin{array}{c}\Delta h_{outer}=\frac{1}{6}\frac{{\left(\left(120\text{r}/\text{min}\right)\right)}^2{\left(\left(10\text{cm}\right)\right)}^2}{9.8{\text{m}/\text{s}}^2}\\ =\frac{1}{6}\frac{{\left(\left(120\text{r}/\text{min}\right)\left(\frac{2\pi }{60}\text{radians}\right)\right)}^2{\left(\left(10\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}{9.8{\text{m}/\text{s}}^2}\\ =0.0268\text{m}\end{array}$

Substitute $120\text{r}/\text{min}$ for $\Omega$, $10\text{cm}$ for $R$ and $9.8{\text{m}/\text{s}}^2$ for $g$ in Equation (VII).

$\begin{array}{c}\Delta h_{center}=-\frac{1}{3}\frac{{\left(\left(120\text{r}/\text{min}\right)\right)}^2{\left(\left(10\text{cm}\right)\right)}^2}{9.8{\text{m}/\text{s}}^2}\\ =-\frac{1}{3}\frac{{\left(\left(120\text{r}/\text{min}\right)\left(\frac{2\pi }{60}\text{radians}\right)\right)}^2{\left(\left(10\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}{9.8{\text{m}/\text{s}}^2}\\ =-0.0536\text{m}\end{array}$

Conclusion:

The height of water in outer tube is $0.0268\text{m}$.

The height of water in center tube is $-0.0536\text{m}$.

The negative sign shows that the height in center tube is decreased.