Question

In a pollution control experiment, minute solid particles (typical mass 1×10-13 slug) are dropped in air. The terminal speed of the particles is measured to be 0.2 ft/s. The drag force of these particles is given by FD=kV, where V is the instantaneous particle speed. Find the value of the constant k. Find the time required to reach 99 percent of terminal speed.

Answer 1

1) $161 \frac{lbf \cdot s}{ft}$

The particle experiences two forces during its fall:

- The force of gravity, $mg$, pointing downward, where m is the mass of the particle and g is the acceleration of gravity

- The drag force, $kv$, pointing upward, where k is a constant and v is the instantaneous speed of the particle

When the particle reaches its terminal speed, the acceleration becomes zero: this means that the net force is also zero, so the two forces are balanced. Therefore we can write

$mg = kv_t$

where $v_t$ is the terminal velocity. We have the following data:

$m = 1\cdot 10^{-13} slug= 1\cdot 10^{-13} \frac{lbf \cdot s^2}{ft}$ is the mass

$g = 32.2 ft/s^2$ is the acceleration

$v_t = 0.2 ft/s$ is the terminal speed

Solving for k, we find

$k=\frac{mg}{v_t}=\frac{(1 \frac{lbf \cdot s^2}{ft})(32.2 \frac{ft}{s^2})}{0.2 \frac{ft}{s}}=161 \frac{lbf \cdot s}{ft}$

2) 28.6 ms

We now want to find the time at which the particle reaches 99 % of the terminal speed, so a speed of

$v' = 0.99 v_t = (0.99)(0.2)=0.198 ft/s$

The net force acting on the particle is

$F=mg-kv$

So the acceleration is

$a=\frac{dv}{dt}=g-\frac{k}{m}v$

This differential equation can be rewritten as

$\frac{dv}{g-\frac{k}{m}v}=dt$

And by integrating on both sides and performing the calculation, we can get the final expression for the velocity as

$v=\frac{mg}{k}(1-e^{-\frac{k}{m}t})$

We want to find the time t at which v = 0.198 ft/s, so we now need to re-arrange the formula for t. This gives:

$t = -\frac{m}{k} ln (\frac{g-\frac{k}{m}v}{g})$

And substituting all the values, we find:

$t = -\frac{1}{161} ln (\frac{32.2-\frac{161}{1}(0.198)}{32.2})=0.0286 s = 28.6 ms$