Q.A small steel ball bearing of radius $1.0\ \text{mm}$ (density $7800\ \text{kg/m}^3$) is released from rest at the top of a tall column of glycerine (density $1260\ \text{kg/m}^3$, coefficient of viscosity $1.5\ \text{Pa·s}$) and falls vertically through it. Using Stokes' law, calculate the terminal velocity attained by the ball bearing as it falls through the glycerine. (Take $g = 9.8\ \text{m/s}^2$.)

Stokes' law gives the viscous drag force on a small sphere falling through a viscous fluid: $F_v = 6\pi\eta r v$, where r is the sphere's radius, v its speed, and η the fluid's coefficient of viscosity — the constant $6\pi$ being an empirically (and dimensionally) determined proportionality factor.

A sphere falling through a viscous medium experiences three forces: its weight $F_g$ (downward, constant), a buoyant upthrust $F_u$ (upward, constant), and the viscous drag $F_v$ (upward, growing as speed grows). Starting from rest, the sphere accelerates as long as $F_g$ exceeds $F_v + F_u$, but because $F_v$ keeps rising with speed, the net force shrinks and eventually vanishes once $F_g = F_v + F_u$ — from that point on the sphere falls at a constant speed, its terminal velocity. For a sphere of radius r and density ρ falling through a medium of density σ and viscosity η, this force balance gives:

which can equally be rearranged to give η itself from a measured terminal velocity — a practical method for measuring a fluid's viscosity in the laboratory. Terminal velocity is always the ceiling on a falling object's speed in a given viscous fluid: an object's actual speed rises toward it but never overshoots it.