Chapter 3, Problem 135P

If the rate of rotational speed of the 3-tube system shown in Fig. P3-135 is $\omega =10$ rad/s, determine the water heights in each tube leg. At what rotational speed will the middle tube be completely empty?

To determine

The water height in each tube.

The speed at which the middle limb is completely empty.

Answer to Problem 135P

The height of water in left tube is $23.15\text{cm}$.

The height of water in right tube is $19.08\text{cm}$

The height of water in middle tube is $2.77\text{cm}$.

The angular speed at which the middle limb is completely empty is $15.07\text{rad}/\text{s}$.

Explanation of Solution

Given information:

The rotational speed of the system is $10\text{rad}/\text{s}$, the height of water in the right tube before rotation is $15\text{cm}$, the bottom length between middle and right tube is $10\text{cm}$ and the bottom length between left and middle tube is $20\text{cm}$.

Write the expression for the water level rises in the left limb.

  $\left(\frac{L_1+L_2}{2}+\frac{2x}{3}\right)=\left(\frac{L_1+L_2}{2}-x\right)+\frac{1}{2g}{V}_L^2$   ...... (I)

Here, the distance of the left limb from the axis is $L_1$, the distance of the right limb from the axis is $L_2$, the axis that is located at $\frac{2}{3}$ distance from the midpoint of the system is $x$, the acceleration due to gravity is $g$ and the velocity in the left limb is $V_L$.

Write the expression for the velocity in the left limb.

  $V_L=\omega L_1$   ...... (II)

Here, the angular speed is $\omega$.

Write the expression for the rise of liquid in the left limb.

  $h_L=\left(\frac{L_1+L_2}{2}+\frac{2x}{3}\right)$   ...... (III)

Here, the rise in the left limb is $h_L$.

Write the expression for the water level rises in the right limb.

  $h_R=\left(\frac{L_1+L_2}{2}+\frac{x}{3}\right)$   ...... (IV)

Here, the rise of water in the right limb is $h_R$.

Write the expression for the rise of water in middle limb.

  $h_m=\frac{L_1+L_2}{2}-x$   ...... (V)

Here, the rise of water in the middle limb is $h_m$.

Write the expression for the angular velocity at which the water height in the middle limb is zero.

  $\omega =\sqrt{\frac{4g\left(h_L-h_m\right)}{R^2}}$   ...... (VI)

Here, the distance between the center and the left limb is $R$.

Calculation:

Substitute $20\text{cm}$ for $L_1$ and $10\text{rad}/\text{s}$ for $\omega$ in Equation (II).

  $\begin{array}{c}{V}_L=\left(10\text{rad}/\text{s}\right)\times \left(20\text{cm}\right)\\ =\left(10\text{rad}/\text{s}\right)\times \left(20\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =2\text{m}/\text{s}\end{array}$

Substitute $20\text{cm}$ for $L_1$, $10\text{cm}$ for $L_2$, $9.81\text{m}/{\text{s}}^2$ for $g$ and $2\text{m}/\text{s}$ for $V_L$ in Equation (I).

  $\begin{array}{l}\left(\frac{20\text{cm}+10\text{cm}}{2}+\frac{2x}{3}\right)=\left(\frac{20\text{cm}+10\text{cm}}{2}-x\right)+\frac{1}{\left(2\times 9.81\text{m}/{\text{s}}^2\right)}{\left(2\text{m}/\text{s}\right)}^2\\ \left[\left(15\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)+\frac{2x}{3}\right]=\left[\left(15\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)-x\right]+\left(0.2038\text{m}\right)\\ \left(0.15\text{m}+\frac{2x}{3}\right)=\left(0.15\text{m}-x\right)+0.2038\text{m}\\ x=0.1223\text{m}\end{array}$

Substitute $20\text{cm}$ for $L_1$, $10\text{cm}$ for $L_2$ and $0.1223\text{m}$ for $x$ in Equation (III).

  $\begin{array}{c}{h}_L=\left(\frac{20\text{cm}+10\text{cm}}{2}+\frac{2\times 0.1223\text{m}}{3}\right)\\ =\left(15\text{cm}+\left(0.08152\text{m}\times \frac{100\text{cm}}{1\text{m}}\right)\right)\\ =23.15\text{cm}\end{array}$

Substitute $20\text{cm}$ for $L_1$, $10\text{cm}$ for $L_2$ and $0.1223\text{m}$ for $x$ in Equation (IV).

  $\begin{array}{c}{h}_R=\left(\frac{20\text{cm}+10\text{cm}}{2}+\frac{0.1223\text{m}}{3}\right)\\ =\left(15\text{cm}+\left(0.04077\text{m}\times \frac{100\text{cm}}{1\text{m}}\right)\right)\\ =19.08\text{cm}\end{array}$

Substitute $20\text{cm}$ for $L_1$, $10\text{cm}$ for $L_2$ and $0.1223\text{m}$ for $x$ in Equation (V).

  $\begin{array}{c}{h}_m=\left(\frac{20\text{cm}+10\text{cm}}{2}-0.1223\text{m}\right)\\ =\left(15\text{cm}-\left(0.1223\text{m}\times \frac{100\text{cm}}{1\text{m}}\right)\right)\\ =2.77\text{cm}\end{array}$

Substitute $9.81\text{m}/{\text{s}}^2$ for $g$, $23.15\text{cm}$ for $h_L$, $0\text{cm}$ for $h_m$ and $20\text{cm}$ for $R$ in Equation (VI).

  $\begin{array}{c}\omega =\sqrt{\frac{4\times \left(9.81\text{m}/{\text{s}}^2\right)\times \left(23.15\text{cm}-0\text{cm}\right)}{{\left(20\text{cm}\right)}^2}}\\ =\sqrt{\frac{\left(39.24\text{m}/{\text{s}}^2\right)\times \left(\left(23.15\text{cm}\right)\times \left(\frac{1\text{m}}{100\text{cm}}\right)\right)}{{\left(20\text{cm}\times \left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}}\\ =\sqrt{227.10}\text{rad}/\text{s}\\ =15.07\text{rad}/\text{s}\end{array}$

Conclusion:

The height of water in left tube is $23.15\text{cm}$.

The height of water in right tube is $19.08\text{cm}$

The height of water in middle tube is $2.77\text{cm}$.

The angular speed at which the middle limb is completely empty is $15.07\text{rad}/\text{s}$.