Dynamic Fluids.

Question

A small ball of radius 8.7 mm and mass 3.15 g falls at terminal velocity through a fluid with a density of 869 kg/m³ and viscosity of 0.075 Pa·s. What is the ball's speed?

Unsure as to what equations to use.

Answer 1

George Stokes found (in 1845) the resistive force F_r on a tiny sphere sinking in a viscous

fluid as F_r = 6πηrv with (η, r) denoting (viscosity coefficient, sphere radius).

While falling through the viscous medium, the sphere encounters the frictional force F_r

of the surrounding fluid, the buoyant force B of that fluid and the weight w of the sphere

itself. The magnitude of the sphere's weight is w = ρgV equal to ρg(4/3)πr³.

By Archimedes' Principle, the magnitude of the buoyant force is the weight of the fluid

that the sphere displaces: B = ρ_fgV equal to ρ_fg(4/3)πr³ with ρ_f the fluid density.

When the fall begins, frictional resistance is zero since sphere

velocity is zero. The instantaneous rate of change of velocity with respect to time

(or acceleration) increases both sphere velocity and the resistive force F_r.

Upon reaching its maximum (or terminal) velocity v_t, the resultant force

goes to zero with B and F_r balancing w.

Then write F_r + B = w as 6πηrv_t + ρ_fg(4/3)πr³ = ρg(4/3)πr³.

Develop this last to obtain v_t = g(4/3)πr³(ρ − ρ_f) ÷ 6πηr or 2gr²(ρ − ρ_f)/9η.

The density of the small ball is gained via 0.00315 kg ÷ [(4/3)π(0.0087m)³]

or 1141.994958 kg/m³.

Now determine 2gr²(ρ − ρ_f)/9η as 2(9.80665 m/s²)(0.0087m)²(1141.994958 − 869) kg/m³

divided by 9(0.075 N•s/m²) which crushes into 0.600399096 meter per second

(roughly 60 centimeters per second).