Chapter 6, Problem 6.69CP

(a)

To determine

The terminal speed for water droplets falling under their own weight in air for the drop radii equal to $10.0\text{μm}$.

Answer to Problem 6.69CP

The terminal speed for water droplets falling under their own weight in air for the drop radii equal to $10.0\text{μm}$ is $0.0132\text{m}/\text{sec}$.

Explanation of Solution

Given info: The expression of magnitude of resistive force exerted on a sphere is $F=\left(arv+b{r}^2{v}^2\right)\text{newtons}$, the radius of the sphere is $r\text{meters}$, the speed of the stream of air is $v\text{m}/\text{sec}$, the numerical value of constant $a$ is $3.10\times {10}^{-4}$, the numerical value of constant $b$ is $0.870$ and the radii of the water droplets is $10.0\text{μm}$.

The expression of the resistive force is,

$F=\left(arv+b{r}^2{v}^2\right)\text{N}$

Here,

$a$ is a constant.

$b$ is a constant.

$r$ is the radii of the droplet.

$v$ is the terminal speed of the water droplet.

Substitute $3.10\times {10}^{-4}$ for $a$ and $0.870$ for $b$ in the above equation.

$F=\left(3.10\times {10}^{-4}rv+0.870r^2{v}^2\right)\text{N}$ (1)

The expression of mass of the water droplet is,

$m=\rho V$ (2)

Here,

$\rho$ is the density of the water.

$V$ is the volume of the water droplet.

The expression of volume of the water droplet is,

$V=\frac{4}{3}\pi r^3$

Substitute $10.0\text{μm}$ for $r$ in the above equation.

$\begin{array}{c}V=\frac{4}{3}\pi {\left[10.0\text{μm}\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)\right]}^3\\ =\frac{4}{3}\pi {\left(10.0\times {10}^{-6}\text{m}\right)}^3\\ =4.188\times {10}^{-15}{\text{m}}^3\\ V\approx 4.2\times {10}^{-15}{\text{m}}^3\end{array}$

Substitute $1.0\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $4.2\times {10}^{-15}{\text{m}}^3$ for $V$ in equation (2).

$\begin{array}{c}m=\left(1.0\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(4.2\times {10}^{-15}{\text{m}}^3\right)\\ m=4.2\times {10}^{-12}\text{kg}\end{array}$

Thus, the mass of the water droplet is $4.2\times {10}^{-12}\text{kg}$.

The expression of force that acts on the water droplet is,

$F=mg$

Here,

$g$ is the acceleration due to gravity.

Compare and equate the equation (1) and the above equation.

$mg=\left(3.10\times {10}^{-4}rv+0.870r^2{v}^2\right)\text{N}$ (3)

Substitute $10.0\text{μm}$ for $r$, $4.2\times {10}^{-12}\text{kg}$ for $m$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above equation.

$\begin{array}{c}\left(4.2\times {10}^{-12}\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)=\left(3.10\times {10}^{-4}\left(10.0\text{μm}\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)\right)v+0.870{\left(10.0\text{μm}\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)\right)}^2{v}^2\right)\text{N}\\ \text{ 41}\text{.16}\times {10}^{-12}=\left(3.10\times {10}^{-4}\left(10.0\times {10}^{-6}\text{m}\right)v+0.870{\left(10.0\times {10}^{-6}\text{m}\right)}^2{v}^2\right)\\ \text{41}\text{.16}\times {10}^{-12}=\left(3.10\times {10}^{-9}v+0.870\times {10}^{-10}{v}^2\right)\end{array}$

The contribution of the second term of $v$ to the air resistance is very small and it is ignored. The new equation is,

$\begin{array}{c}\text{41}\text{.16}\times {10}^{-12}=\left(3.10\times {10}^{-9}v\right)\\ v=\frac{\text{41}\text{.16}\times {10}^{-12}}{3.10\times {10}^{-9}}\text{m}/\text{sec}\\ =13.277\times {10}^{-3}\text{m}/\text{sec}\\ v\approx 0.0132\text{m}/\text{sec}\end{array}$

Conclusion:

Therefore, the terminal speed for water droplets falling under their own weight in air for the drop radii equal to $10.0\text{μm}$ is $0.0132\text{m}/\text{sec}$.

(b)

To determine

The terminal speed for water droplets falling under their own weight in air for the drop radii equal to $100\text{μm}$.

Answer to Problem 6.69CP

The terminal speed for water droplets falling under their own weight in air for the drop radii equal to $100\text{μm}$ is $1.03\text{m}/\text{sec}$.

Explanation of Solution

Given info: The expression of magnitude of resistive force exerted on a sphere is $F=\left(arv+b{r}^2{v}^2\right)\text{newtons}$, the radius of the sphere is $r\text{meters}$, the speed of the stream of air is $v\text{m}/\text{sec}$, the numerical value of constant $a$ is $3.10\times {10}^{-4}$, the numerical value of constant $b$ is $0.870$ and the radii of the water droplets is $100\text{μm}$.

From equation (1) the expression of resistive force is,

$F=\left(3.10\times {10}^{-4}rv+0.870r^2{v}^2\right)\text{N}$

The expression of volume of the water droplet is,

$V=\frac{4}{3}\pi r^3$

Substitute $100\text{μm}$ for $r$ in the above equation.

$\begin{array}{c}V=\frac{4}{3}\pi {\left[100\text{μm}\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)\right]}^3\\ =\frac{4}{3}\pi {\left(100\times {10}^{-6}\text{m}\right)}^3\\ =4.188\times {10}^{-12}{\text{m}}^3\\ V\approx 4.2\times {10}^{-12}{\text{m}}^3\end{array}$

From equation (2), the expression of mass of the water droplet is,

$m=\rho V$

Substitute $1.0\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $4.2\times {10}^{-12}{\text{m}}^3$ for $V$ in above equation.

$\begin{array}{c}m=\left(1.0\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(4.2\times {10}^{-12}{\text{m}}^3\right)\\ m=4.2\times {10}^{-9}\text{kg}\end{array}$

From equation (3) the final equation is,

$mg=\left(3.10\times {10}^{-4}rv+0.870r^2{v}^2\right)\text{N}$

Substitute $100\text{μm}$ for $r$, $4.2\times {10}^{-9}\text{kg}$ for $m$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above equation.

$\begin{array}{c}\left(4.2\times {10}^{-9}\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)=\left(3.10\times {10}^{-4}\left(100\text{μm}\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)\right)v+0.870{\left(100\text{μm}\left(\frac{{10}^{-6}\text{m}}{1\text{μm}}\right)\right)}^2{v}^2\right)\text{N}\\ \text{ 41}\text{.1}\times {10}^{-9}=\left(3.10\times {10}^{-4}\left(100\times {10}^{-6}\text{m}\right)v+0.870{\left(100\times {10}^{-6}\text{m}\right)}^2{v}^2\right)\\ \text{4}\text{.11}\times {10}^{-8}=\left(3.10\times {10}^{-8}v+0.870\times {10}^{-8}{v}^2\right)\end{array}$

Further solve the above equation.

$\begin{array}{c}0.870\times {10}^{-8}{v}^2+3.10\times {10}^{-8}v-\text{4}\text{.11}\times {10}^{-8}=0\\ 0.870v^2+3.10v-\text{4}\text{.11}=0\end{array}$

Apply quadratic formula to solve the above equation.

$\begin{array}{c}v=\frac{-3.10\pm \sqrt{{\left(3.10\right)}^2-4\left(0.870\right)\left(-\text{4}\text{.11}\right)}}{2\left(0.870\right)}\\ v\approx 1.03\text{m}/\text{sec}\end{array}$

Conclusion:

Therefore, the terminal speed for water droplets falling under their own weight in air for the drop radii equal to $100\text{μm}$ is $1.03\text{m}/\text{sec}$.

(c)

To determine

The terminal speed for water droplets falling under their own weight in air for the drop radii equal to $1.00\text{mm}$.

Answer to Problem 6.69CP

The terminal speed for water droplets falling under their own weight in air for the drop radii equal to $1.00\text{mm}$ is $6.87\text{m}/\text{sec}$.

Explanation of Solution

Given info: The expression of magnitude of resistive force exerted on a sphere is $F=\left(arv+b{r}^2{v}^2\right)\text{newtons}$, the radius of the sphere is $r\text{meters}$, the speed of the stream of air is $v\text{m}/\text{sec}$, the numerical value of constant $a$ is $3.10\times {10}^{-4}$, the numerical value of constant $b$ is $0.870$ and the radii of the water droplets is $1.00\text{mm}$.

From equation (1) the expression of resistive force is,

$F=\left(3.10\times {10}^{-4}rv+0.870r^2{v}^2\right)\text{N}$

The expression of volume of the water droplet is,

$V=\frac{4}{3}\pi r^3$

Substitute $1.00\text{mm}$ for $r$ in the above equation.

$\begin{array}{c}V=\frac{4}{3}\pi {\left[1.00\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right]}^3\\ =\frac{4}{3}\pi {\left(1.00\times {10}^{-3}\text{m}\right)}^3\\ =4.188\times {10}^{-9}{\text{m}}^3\\ V\approx 4.2\times {10}^{-9}{\text{m}}^3\end{array}$

From equation (2), the expression of mass of the water droplet is,

$m=\rho V$

Substitute $1.0\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $4.2\times {10}^{-9}{\text{m}}^3$ for $V$ in above equation.

$\begin{array}{c}m=\left(1.0\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(4.2\times {10}^{-9}{\text{m}}^3\right)\\ m=4.2\times {10}^{-6}\text{kg}\end{array}$

From equation (3) the final equation is,

$mg=\left(3.10\times {10}^{-4}rv+0.870r^2{v}^2\right)\text{N}$

Substitute $1.00\text{mm}$ for $r$, $4.2\times {10}^{-6}\text{kg}$ for $m$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above equation.

$\begin{array}{c}\left(4.2\times {10}^{-6}\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)=\left(3.10\times {10}^{-4}\left(1.00\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)v+0.870{\left(1.00\text{mm}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\right)}^2{v}^2\right)\text{N}\\ \text{ 41}\text{.16}\times {10}^{-6}=\left(3.10\times {10}^{-4}\left(1.00\times {10}^{-3}\text{m}\right)v+0.870{\left(1.00\times {10}^{-3}\text{m}\right)}^2{v}^2\right)\\ \text{41}\text{.16}\times {10}^{-6}=\left(3.10\times {10}^{-7}v+0.870\times {10}^{-6}{v}^2\right)\end{array}$

The contribution of the first term of $v$ to the air resistance is very small and it is ignored. The new equation is,

$\begin{array}{c}\text{41}\text{.16}\times {10}^{-6}=\left(0.870\times {10}^{-6}{v}^2\right)\\ v^2=\frac{\text{41}\text{.16}\times {10}^{-6}}{0.870\times {10}^{-6}}\text{m}/\text{sec}\\ v=\sqrt{\frac{\text{41}\text{.16}\times {10}^{-6}}{0.870\times {10}^{-6}}}\text{m}/\text{sec}\\ v\approx 6.87\text{m}/\text{sec}\end{array}$

Conclusion:

Therefore, the terminal speed for water droplets falling under their own weight in air for the drop radii equal to $1.00\text{mm}$ is $6.87\text{m}/\text{sec}$.