Chapter 11, Problem 42P

To determine

The terminal velocity of the sphere in the water.

Answer to Problem 42P

The terminal velocity of the sphere in the water $0.1551\text{m}/\text{s}$.

Explanation of Solution

Given information:

The diameter of sphere is $6\text{mm}$, the density of plastic sphere is $1150\text{kg}/{\text{m}}^{\text{3}}$, temperature of water is $20°\text{C}$

Write the expression for frontal area of sphere.

   $A=\frac{\pi D^2}{4}$     ...... (I)

Here, diameter of sphere is $D$.

Write the expression for drag force acting on the sphere.

   $F_D=\frac{1}{2}{C}_{D}A{\rho }_{\text{w}}{V}^2$     ...... (II)

Here, coefficient of drag is $C_D$, density of water is ${\rho }_{\text{w}}$, terminal velocity of sphere is $V$.

Write the expression for weight of sphere.

   $W={\rho }_{\text{s}}{V}^{\prime }g$     ...... (III)

Here, volume of sphere is $V^{\prime }$, density of sphere is ${\rho }_{\text{s}}$, acceleration due to gravity is $g$.

Write the expression for buoyant force.

   $F_B={\rho }_{\text{w}}{V}^{\prime }g$     ...... (IV)

Write the expression for volume of sphere.

   $V^{\prime }=\frac{\pi D^3}{6}$     ...... (V)

Write the relation of weight, buoyant force and drag force for terminal velocity.

   $W=F_D+F_B$     ...... (VI)

Calculation:

Substitute $6\text{mm}$ for the $D$ in Equation (I).

   $\begin{array}{c}A=\frac{\pi {\left(6\text{mm}\right)}^2}{4}\\ =\frac{\pi {\left(6\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}^2}{4}\\ =\frac{\pi {\left(0.006\text{m}\right)}^2}{4}\\ =2.83\times {10}^{-5}{\text{m}}^{\text{2}}\end{array}$

Refer to Table-11-2 “Drag coefficient of various three-dimensional bodies” to obtain drag coefficient of sphere as $0.5$.

Refer to Table-A-3 “Properties of saturated water” at $20°\text{C}$ to obtain density of water as $998\text{kg}/{\text{m}}^{\text{3}}$.

Substitute $0.5$ for $C_D$, $2.83\times {10}^{-5}{\text{m}}^{\text{2}}$ for $A$, $998\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{w}}$ in Equation (II).

   $\begin{array}{c}{F}_D=\frac{1}{2}\left(0.5\right)\left(2.83\times {10}^{-5}{\text{m}}^{\text{2}}\right)\left(998\text{kg}/{\text{m}}^{\text{3}}\right)V^2\\ =\left(0.025\right)\left(2.83\times {10}^{-5}{\text{m}}^{\text{2}}\right)\left(998\text{kg}/{\text{m}}^{\text{3}}\right)V^2\\ =7.06\times {10}^{-3}\text{kg}/\text{m}\times V^2\end{array}$

Substitute $6\text{mm}$ for the $D$ in Equation (V).

   $\begin{array}{c}{V}^{\prime }=\frac{\pi {\left(6\text{mm}\right)}^3}{6}\\ =\frac{\pi {\left(6\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}^3}{6}\\ =\frac{\pi {\left(0.006\text{m}\right)}^3}{6}\\ =1.13097\times {10}^{-7}{\text{m}}^{\text{3}}\end{array}$

Substitute $1.13097\times {10}^{-7}{\text{m}}^{\text{3}}$ for $V^{\prime }$, $1150\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{s}}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (III).

   $\begin{array}{c}W=\left(1150\text{kg}/{\text{m}}^{\text{3}}\right)\left(1.13097\times {10}^{-7}{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =0.00127\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\times \frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\\ =0.00127\text{N}\end{array}$

Substitute $1.13097\times {10}^{-7}{\text{m}}^{\text{3}}$ for $V^{\prime }$, $998\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{w}}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in Equation (IV).

   $\begin{array}{c}{F}_B=\left(998\text{kg}/{\text{m}}^{\text{3}}\right)\left(1.13097\times {10}^{-7}{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =0.0011\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}\times \frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\\ =0.0011\text{N}\end{array}$

Substitute $7.06\times {10}^{-3}\text{kg}/\text{m}\times V^2$ for $F_D$, $0.00127\text{N}$ for $W$ and $0.0011\text{N}$ for $F_B$ in Equation (VI).

   $\begin{array}{l}0.00127\text{N}=7.06\times {10}^{-3}\text{kg}/\text{m}\times V^2+0.0011\text{N}\\ 7.06\times {10}^{-3}\text{kg}/\text{m}\times V^2=0.00127\text{N}-0.0011\text{N}\\ 7.06\times {10}^{-3}\text{kg}/\text{m}\times V^2=0.00017\text{N}\times \frac{1\text{N}}{1\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}\\ V^2=\frac{0.00017\text{kg}\cdot \text{m}/{\text{s}}^{\text{2}}}{7.06\times {10}^{-3}\text{kg}/\text{m}}\end{array}$

   $\begin{array}{l}{V}^2=\frac{0.00017{\text{m}}^2/{\text{s}}^{\text{2}}}{7.06\times {10}^{-3}}\\ V=\sqrt{0.02407932{\text{m}}^{\text{2}}/{\text{s}}^{\text{2}}}\\ V=0.1551\text{m}/\text{s}\end{array}$

Conclusion:

The terminal velocity of the sphere in the water $0.1551\text{m}/\text{s}$.