Very small objects, such as dust particles, experience a linear drag force,${\overrightarrow{F}}_{\text{duag}}=$(bv, direction opposite the motion), where $b$is a constant. That is, the quadratic model of drag of Equation $6.15$fails for very small particles. For a sphere of radius $R$, the drag constant can be shown to be $b=6\pi \eta R$, where $\eta$is the viscosity of the gas.

a. Find an expression for the terminal speed$v_{\text{term}}$of a spherical particle of radius $R$and mass $m$falling through a gas of viscosity $\eta$.

b. Suppose a gust of wind has carried$50-\mu \mathrm{m}$a -diameter dust particle to a height of $300\mathrm{m}$. If the wind suddenly stops, how long will it take the dust particle to settle back to the ground? Dust has a density ofrole="math" localid="1647776411884" $2700\mathrm{kg}/{\mathrm{m}}^3$the viscosity of${25}^{\circ }\mathrm{C}$air is $2.0\times {10}^{-5}\mathrm{Ns}/{\mathrm{m}}^2$, and you can assume that the falling dust particle reaches terminal speed almost instantly.

Drag force is given as,$F_{\text{drag}}=bv$.

Drag constant,$b=6\pi \eta R.$

Newton's second law $F=mg.$

The force on the spherical particle due to gravity is given by Newton's second law$F=mg$.....(1)

Here,

$m$is the mass of the spherical particle.

$g$is the acceleration due to gravity.

The drag force on the particle is given as

$F_{\text{drag}}=b{v}_{\text{terminal}}\dots \dots \left(2\right)$

$=6\pi \eta R{v}_{\text{terminal}}$.

In the condition of equilibrium, the drag force on the particle must balance the force of gravity

$\begin{array}{c}mg=6\pi \eta R{v}_{\text{tcrminal}}\\ v_{\text{tcrminial}}=\frac{mg}{6\text{ise}R}\end{array}$.

Diameter of the particle, $D=50\mu \mathrm{m}.$

Height,$h=300\mathrm{m}.$

Density of the dust, $\rho =2700\mathrm{kg}/{\mathrm{m}}^3.$

Viscosity,$\eta =2\times {10}^{-5}\mathrm{Ns}/{\mathrm{m}}^2.$

The terminal speed of the spherical particle is given by:

$v_{\text{lerminal}}=\frac{mg}{6mpR}$.

Here,

$m$is the mass.

$g$is the acceleration due to gravity.

$\eta$is the viscosity.

$R$is the radius of the spherical particle.

The mass of the particle is calculated as

$\begin{array}{c}m=\rho \times \frac{4}{3}\times \pi \times {\left(\frac{D}{2}\right)}^3\\ =2700\times \frac{4}{3}\times \pi \times {\left(\frac{s1\times {10}^{-6}}{2}\right)}^3\\ =1.76\times {10}^{-10}\mathrm{kg}.\end{array}$

The terminal velocity of the particle is calculated as:

$v_{\text{lerminal}}=\frac{mg}{6mpR}.$

Plugging the values in the above equation:

role="math" localid="1647775897792" $\begin{array}{rl}{v}_{\text{terminal}}& =\frac{1.76\times {10}^{-10}\times 9.81}{6\times \mathrm{m}\times 2\times {10}^{-3}\times 25\times {10}^{-6}}\\ & =0.183\mathrm{m}/\mathrm{s}.\end{array}$

The time required by the particle to settle down is calculated as

$\begin{array}{c}t=\frac{h}{v_{\text{lerminal}}}\\ =\frac{300}{0.183}\\ =1640\sec \\ =27.32\min \\ \approx 27\min .\end{array}$

Part (a): Expression of the terminal speed of the spherical particle is $V_{terminal}=\frac{mg}{6\mathrm{\pi mR}}.$

Part (b): The time required by the particle to settle back to the ground is$27\min .$