Chapter 6, Problem 69CP

(a)

To determine

The terminal speed for water droplets with a radius of $10.0\text{μm}$.

Answer to Problem 69CP

The terminal speed for water droplets with a radius of $10.0\text{μm}$ is $0.0132\text{m/s}$.

Explanation of Solution

Write the expression for the magnitude of resistive force exerted on a sphere.

    $F=arv+b{r}^2{v}^2$                                                                           (I)

Here, $F$ is the magnitude of resistive force exerted on a sphere, $a$ is a constant, $b$ is a constant, $r$ is the radius of the water droplet and $v$ is speed of the water droplet.

Write the expression for the force.

    $F=mg$                                                                                       (II)

Here, $m$ is the mass of the water droplet and $g$ is the acceleration due to gravity.

Equating equations (I) and (II),

    $mg=arv+b{r}^2{v}^2$                                                                        (III)

Write the expression for the density of the water droplet.

    $\rho =\frac{m}{V}$                                                                                     (IV)

Here, $\rho$ is the density of the water droplet and $V$ is the volume of the water droplet.

Write the expression for the volume of the water droplet.

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

Rewrite the expression (IV) for the mass of the water droplet by substituting equation (V).

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

Conclusion:

Substitute $1000{\text{kg/m}}^{\text{3}}$ for $\rho$ and $10.0\text{μm}$ for $r$ in equation (VI) to find $m$.

    $m=\frac{4}{3}\left(1000{\text{kg/m}}^{\text{3}}\right)\pi {\left(10.0\text{μm}\right)}^3$

Substitute $\frac{4}{3}\left(1000{\text{kg/m}}^{\text{3}}\right)\pi {\left(10.0\text{μm}\right)}^3$ for $m$ and $9.8{\text{m/s}}^2$ for $g$ in equation (II) to find $mg$.

    $\begin{array}{c}mg=\left(\frac{4}{3}\left(1000{\text{kg/m}}^{\text{3}}\right)\pi {\left(10.0\text{μm}\right)}^3\right)\left(9.8{\text{m/s}}^2\right)\\ =4.11\times {10}^{-11}\text{kg}\cdot {\text{m/s}}^{\text{2}}\end{array}$

Substitute $3.10\times {10}^{-4}$ for $a$ , $0.870$ for $b$, $4.11\times {10}^{-11}\text{kg}{\text{.m/s}}^{\text{2}}$ for $mg$ and $10.0\text{μm}$ for $r$ in equation (III) to find $v$.

    $\begin{array}{c}4.11\times {10}^{-11}\text{kg}{\text{.m/s}}^{\text{2}}=\left(3.10\times {10}^{-4}\right)\left(10.0\text{μm}\right)v+\left(0.870\right){\left(10.0\text{μm}\right)}^2{v}^2\\ =\left(3.10\times {10}^{-9}\right)v+\left(0.870\times {10}^{-8}\right)v^2\end{array}$

Here $v$ is small, so we can ignore the $v^2$ term in the above equation.

    $v=0.0132\text{m/s}$

Therefore, the terminal speed for water droplets with a radius of $10.0\text{μm}$ is $0.0132\text{m/s}$.

(b)

To determine

The terminal speed for water droplets with a radius of $100\text{μm}$.

Answer to Problem 69CP

The terminal speed for water droplets with a radius of $100\text{μm}$ is $1.03\text{m/s}$.

Explanation of Solution

Write the expression for the magnitude of resistive force exerted on a sphere.

    $F=arv+b{r}^2{v}^2$                                                                            (VII)

Here, $F$ is the magnitude of resistive force exerted on a sphere, $a$ is a constant, $b$ is a constant, $r$ is the radius of the water droplet and $v$ is speed of the water droplet.

Write the expression for the force.

    $F=mg$                                                                                         (VIII)

Here, $m$ is the mass of the water droplet and $g$ is the acceleration due to gravity.

Equating the equation (VII) and (VIII),

    $mg=arv+b{r}^2{v}^2$                                                                           (IX)

Write the expression for the density of the water droplet.

    $\rho =\frac{m}{V}$                                                                                             (X)

Here, $\rho$ is the density of the water droplet and $V$ is the volume of the water droplet.

Write the expression for the volume of the water droplet.

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

Rewrite the expression (X) for the mass of the water droplet and substitute equation (XI).

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

Conclusion:

Substitute $1000{\text{kg/m}}^{\text{3}}$ for $\rho$ and $100\text{μm}$ for $r$ in equation (XII) to find $m$.

    $m=\frac{4}{3}\left(1000{\text{kg/m}}^{\text{3}}\right)\pi {\left(100\text{μm}\right)}^3$

Substitute $\frac{4}{3}\left(1000{\text{kg/m}}^{\text{3}}\right)\pi {\left(100\text{μm}\right)}^3$ for $m$ and $9.8{\text{m/s}}^2$ for $g$ in equation (VIII) to find $mg$.

    $\begin{array}{c}mg=\left(\frac{4}{3}\left(1000{\text{kg/m}}^{\text{3}}\right)\pi {\left(100\text{μm}\right)}^3\right)\left(9.8{\text{m/s}}^2\right)\\ =4.11\times {10}^{-8}\text{kg}\cdot {\text{m/s}}^{\text{2}}\end{array}$

Substitute $3.10\times {10}^{-4}$ for $a$, $0.870$ for $b$, $4.11\times {10}^{-8}\text{kg}\cdot {\text{m/s}}^{\text{2}}$ for $mg$ and $100\text{μm}$ for $r$ in (IX) to find $v$.

    $\begin{array}{c}4.11\times {10}^{-11}\text{kg}\cdot {\text{m/s}}^{\text{2}}=\left(3.10\times {10}^{-4}\right)\left(10.0\text{μm}\right)v+\left(0.870\right){\left(10.0\text{μm}\right)}^2{v}^2\\ =\left(3.10\times {10}^{-9}\right)v+\left(0.870\times {10}^{-8}\right)v^2\end{array}$

Here taking positive root values.

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

Therefore, the terminal speed for water droplets with a radius of $10.0\mu m$ is $1.03\text{m/s}$.

(c)

To determine

The terminal speed for water droplets with a radius of $1.00\text{mm}$.

Answer to Problem 69CP

The terminal speed for water droplets with a radius of $1.00\text{mm}$ is $6.87\text{m/s}$.

Explanation of Solution

Write the expression for the magnitude of resistive force exerted on a sphere.

    $F=arv+b{r}^2{v}^2$                                                                          (XIII)

Here, $F$ is the magnitude of resistive force exerted on a sphere, $a$ is a constant, $b$ is a constant, $r$ is the radius of the water droplet and $v$ is speed of the water droplet.

Write the expression for the force.

    $F=mg$                                                                                         (XIV)

Here, $m$ is the mass of the water droplet and $g$ is the acceleration due to gravity.

Equating equations (XIII) and (XIV),

    $mg=arv+b{r}^2{v}^2$                                                                           (XV)

From the part b the expression for the mass of the water droplet,

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

Conclusion:

Substitute $1000{\text{kg/m}}^{\text{3}}$ for $\rho$ and $1.00\text{mm}$ for $r$ in equation (XVI) to find $m$.

    $m=\frac{4}{3}\left(1000{\text{kg/m}}^{\text{3}}\right)\pi {\left(1.00\text{mm}\right)}^3$

Substitute $\frac{4}{3}\left(1000{\text{kg/m}}^{\text{3}}\right)\pi {\left(1.00\text{mm}\right)}^3$ for $m$ and $9.8{\text{m/s}}^2$ for $g$ in equation (XIV) to find $mg$

    $\begin{array}{c}mg=\left(\frac{4}{3}\left(1000{\text{kg/m}}^{\text{3}}\right)\pi {\left(1.00\text{mm}\right)}^3\right)\left(9.8{\text{m/s}}^2\right)\\ =4.11\times {10}^{-5}\text{kg}\cdot {\text{m/s}}^{\text{2}}\end{array}$

Substitute $3.10\times {10}^{-4}$ for $a$, $0.870$ for $b$, $4.11\times {10}^{-5}\text{kg}{\text{.m/s}}^{\text{2}}$ for $mg$ and $1.00\text{mm}$ for $r$ in equation (XV) to find $v$.

    $\begin{array}{c}4.11\times {10}^{-5}\text{kg}{\text{.m/s}}^{\text{2}}=\left(3.10\times {10}^{-4}\right)\left(1.00\text{mm}\right)v+\left(0.870\right){\left(1.00\text{mm}\right)}^2{v}^2\\ =\left(3.10\times {10}^{-9}\right)v+\left(0.870\times {10}^{-8}\right)v^2\end{array}$

Here assuming $v>1$ and taking second value only.

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

Therefore, the terminal speed for water droplets with a radius of $1.00\text{mm}$ is $6.87\text{m/s}$.