Question

A small spherical steel ball of radius
a of a material of density à   falls from rest in still air of viscosity à Ž ·. Now (i) Write up
the equation of motion of the ball, (ii) Estimate its terminal velocity, and (iii) Find out
the time and distance traversed when the ball has acquired 90% of the terminal
velocity.

Answer 1

(i) For a small ball and low speed

$F=ma\to (mg-\rho_agV)-(6\pi\eta r)v=m\frac{dv}{dt}\to a-bv=m\frac{dv}{dt}$

$v(t)=\frac{a}{b}(1-e^{-bt/m})=\frac{mg-\rho_agV}{6\pi\eta r}(1-e^{-6\pi\eta rt/m})=v_0(1-e^{-6\pi\eta rt/m})$

$m=(4/3)\pi r^3\rho$

$v(t)=v_0(1-e^{-9\eta t/2\rho r^2})$

(ii) $v_0=\frac{mg-\rho_agV}{6\pi\eta r}=\frac{2gr^2(\rho-\rho_a)}{9\eta}$

(iii) $0.9v_0=v_0(1-e^{-9\eta t/2\rho r^2})\to t=\frac{2\rho r^2\ln10}{9\eta}$

$l=\int_0^tv(t)dt=\int_0^tv_0(1-e^{-9\eta t/2\rho r^2})dt$