Question

A steel ball of diameter $d=3.0mm$ starts sinking with zero initial velocity in olive oil whose viscosity is $\eta =0.90P$. How soon after the beginning of motion will the velocity of the ball differ from the steady-state velocity by $\eta =1.0\%$?

Answer 1

$m\frac{dv}{dt}=mg-6\pi \eta rv$
or $\frac{dv}{dt}+\frac{6\pi \eta r}{m}v=g$
or $\frac{dv}{dt}+kv=g$, $k=\frac{6\pi \eta r}{m}$
or $e^{kt}\frac{dv}{dt}+k{e}^{kt}v=g{e}^{kt}$ or $\frac{d}{dt}{e}^{kt}v=g{e}^{kt}$
or $v{e}^{kt}=\frac{g}{k}{e}^{kt}+C$ or $v=\frac{g}{k}+C{e}^{-kt}$ (where $C$ is constant)
Since $v=0$ for $t=0$, $0=\frac{g}{k}+C$
So $C=-\frac{g}{k}$
Thus $v=\frac{g}{k}(1-e^{-kt})$
The steady state velocity is $\frac{g}{k}$.
$v$ differs from $\frac{g}{k}$ by $n$ where $e^{-kt}=n$
or $t=\frac{1}{k}\ln n$
Thus $\frac{1}{k}=\frac{\frac{4\pi }{3}{r}^{3}P}{6\pi \eta r}=-\frac{4r^2\rho }{18\eta }=-\frac{d^2\rho }{18\eta }$
We have neglected buoyancy in olive oil.