Chapter 5, Problem 61P

(a)

To determine

Terminal velocity of water droplets at radii $10.0\mu \text{m}$.

Answer to Problem 61P

Terminal velocity of water droplets at radii $10.0\mu \text{m}$ is $\underset{\_}{0.0132\text{m}/\text{s}}$.

Explanation of Solution

At the terminal velocity, the drag force is balanced by the force due to gravity.

    $mg=arv+b{r}^2{v}^2$        (I)

Here, $m$ is the mass, $g$ is the acceleration due to gravity, $r$ is the radius, $v$ is then speed, and $a,b$ are constants.

Write the expression for mass water drop.

    $m=\rho \frac{4}{3}\pi r^3$        (II)

Here, $\rho$ is the density.

Conclusion:

Substitute, $1000\text{kg}/{\text{m}}^3$, ${10}^{-5}\text{m}$ for $r$ in the equation (II), to find $m$.

    $\begin{array}{c}m=\left(1000\text{kg}/{\text{m}}^3\right)\frac{4}{3}\pi {\left({10}^{-5}\text{m}\right)}^3\\ =4.186\times {10}^{-12}\text{kg}\end{array}$

Substitute, $4.186\times {10}^{-12}\text{kg}$ for $m$, $10.0\mu \text{m}$ for $r$, $3.10\times {10}^{-4}$ for $a$, $0.870$ for $b$, and $9.80\text{m}/{\text{s}}^2$ for $g$ in the equation (I), and solve for $v$.

    $\begin{array}{c}\left(4.186\times {10}^{-12}\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)=\left(3.10\times {10}^{-4}\right)\left(10.0\mu \text{m}\right)v+\left(0.870\right){\left(10.0\mu \text{m}\right)}^2{v}^2\\ 4.11\times {10}^{-11}=\left(3.10\times {10}^{-9}\right)v+\left(0.870\times {10}^{-10}\right)v^2\end{array}$

Assume that $v$ is small, so neglect the second term in the right hand side.

    $\begin{array}{c}4.11\times {10}^{-11}=\left(3.10\times {10}^{-9}\right)v\\ v=\frac{4.11\times {10}^{-11}}{3.10\times {10}^{-9}}\\ =0.0132\text{m}/\text{s}\end{array}$

Therefore, terminal velocity of water droplets at radii $10.0\mu \text{m}$ is $\underset{\_}{0.0132\text{m}/\text{s}}$.

(b)

To determine

Terminal velocity of water droplets at radii $100\mu \text{m}$.

Answer to Problem 61P

Terminal velocity of water droplets at radii $100\mu \text{m}$ is $\underset{\_}{1.03\text{m}/\text{s}}$.

Explanation of Solution

Conclusion:

Substitute, $1000\text{kg}/{\text{m}}^3$, $100\times {10}^{-6}\text{m}$ for $r$ in the equation (II), to find $m$.

    $\begin{array}{c}m=\left(1000\text{kg}/{\text{m}}^3\right)\frac{4}{3}\pi {\left(100\times {10}^{-6}\text{m}\right)}^3\\ =4.186\times {10}^{-9}\text{kg}\end{array}$

Substitute, $4.186\times {10}^{-9}\text{kg}$ for $m$, $100\times {10}^{-6}\text{m}$ for $r$, $3.10\times {10}^{-4}$ for $a$, $0.870$ for $b$, and $9.80\text{m}/{\text{s}}^2$ for $g$ in the equation (I), and solve for $v$.

    $\begin{array}{c}\left(4.186\times {10}^{-9}\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)=\left(3.10\times {10}^{-4}\right)\left(100\times {10}^{-6}\text{m}\right)v+\left(0.870\right){\left(100\mu \text{m}\right)}^2{v}^2\\ 4.11\times {10}^{-8}=\left(3.10\times {10}^{-8}\right)v+\left(0.870\times {10}^{-8}\right)v^2\\ \left(0.870\times {10}^{-8}\right)v^2+\left(3.10\times {10}^{-8}\right)v-\left(3.10\times {10}^{-8}\right)v=0\end{array}$

Solve the above quadratic equation.

    $\begin{array}{c}v=\frac{-3.10\pm \sqrt{{\left(3.10\right)}^2+4\left(0.870\right)\left(4.11\right)}}{2\left(0.870\right)}\\ =1.03\text{m}/\text{s}\end{array}$

Therefore, terminal velocity of water droplets at radii $100\mu \text{m}$ is $\underset{\_}{1.03\text{m}/\text{s}}$.

(c)

To determine

Terminal velocity of water droplets at radii $1.00\text{mm}$.

Answer to Problem 61P

Terminal velocity of water droplets at radii $1.00\text{mm}$ is $\underset{\_}{6.87\text{m}/\text{s}}$.

Explanation of Solution

Conclusion:

Substitute, $1000\text{kg}/{\text{m}}^3$, $1.00\times {10}^{-3}\text{m}$$1.00\text{mm}$ for $r$ in the equation (II), to find $m$.

    $\begin{array}{c}m=\left(1000\text{kg}/{\text{m}}^3\right)\frac{4}{3}\pi {\left(1.00\times {10}^{-3}\text{m}\right)}^3\\ =4.186\times {10}^{-6}\text{kg}\end{array}$

Substitute, $4.186\times {10}^{-6}\text{kg}$ for $m$, $1.00\times {10}^{-3}\text{m}$$1.00\text{mm}$ for $r$, $3.10\times {10}^{-4}$ for $a$, $0.870$ for $b$, and $9.80\text{m}/{\text{s}}^2$ for $g$ in the equation (I), and solve for $v$.

    $\begin{array}{c}\left(4.186\times {10}^{-6}\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)=\left(3.10\times {10}^{-4}\right)\left(1.00\times {10}^{-3}\text{m}\right)v+\left(0.870\right){\left(1.00\times {10}^{-3}\text{m}\right)}^2{v}^2\\ 4.11\times {10}^{-5}=\left(3.10\times {10}^{-7}\right)v+\left(0.870\times {10}^{-6}\right)v^2\end{array}$

Assume that, $v>1\text{m}/\text{s}$, and neglect the first term in the above equation.

    $\begin{array}{l}4.11\times {10}^{-5}=\left(0.870\times {10}^{-6}\right)v^2\\ v=6.87\text{m}/\text{s}\end{array}$

Therefore, terminal velocity of water droplets at radii $1.00\text{mm}$ is $\underset{\_}{6.87\text{m}/\text{s}}$.