# Hydroynamics: Stoke's Law, Find work done by drag force

1. Nov 12, 2012

### Blandongstein

1. The problem statement, all variables and given/known data
A water droplet starts falling from rest from a height $h$ and a quires the terminal speed just before reaching the ground. If $g$(acceleration due to gravity) is assumed to be constant and the radius of the drop is $r$, find the work done by air drag. The density of air is $\rho_{\text{air}}$ and the density of water is $\rho_{\text{W}}$

2. Relevant equations
Stoke's Law: $F_{\text{drag}}=6\pi \eta r v$

3. The attempt at a solution
Let $a$ be the acceleration of the droplet. Then,

\begin{aligned}mg-F_{\text{upthrust}}-F_{\text{drag}} &= ma \\ ma &= mg-mg(\frac{\rho_{\text{air}}}{\rho_{\text{W}}}) -6\pi \eta rv \\ a &= g\left(1-\frac{\rho_{\text{air}}}{\rho_{\text{W}}}\right)-\frac{6\pi \eta r }{m}v \\ v\frac{dv}{ds} &=g\left(1-\frac{\rho_{\text{air}}}{\rho_{\text{W}}}\right)-\frac{6\pi \eta r }{m}v \\ ds &=\frac{v}{g\left(1-\frac{\rho_{\text{air}}}{\rho_{\text{W}}}\right)-\frac{6\pi \eta r }{m}v } dv \end{aligned}

The work done by the drag force is

\begin{aligned} dW &= -F_{\text{drag}} ds \\ &= -(6\pi \eta r v)\left[ \frac{v}{g\left(1-\frac{\rho_{\text{air}}}{\rho_{\text{W}}}\right)-\frac{6\pi \eta r }{m}v } dv\right] \\ \end{aligned}

Am I doing correct? The answer given in my book is

$$\frac{4}{3}\pi r^3\rho_{\text{W}}g\left[ \frac{2r^4 g}{81\eta^2}(\rho_{\text{W}}-\rho_{\text{air}})^2-h\left( 1- \frac{\rho_{\text{air}}}{\rho_{\text{W}}}\right) \right]$$