Question

Derive an expression for terminal velocity of a spherical object falling under gravity through a viscous medium.

Hint:

Terminal velocity balances gravity, buoyancy, and Stokes’ drag; valid for laminar flow conditions.

Answer 1

When a spherical object falls through a viscous medium, three forces act:
1. Gravitational force: \( mg = \frac{4}{3} \pi r^3 \rho g \), where \( \rho \) is the density of the object, \( r \) is radius, \( g \) is gravitational acceleration.
2. Buoyant force: \( \frac{4}{3} \pi r^3 \sigma g \), where \( \sigma \) is the density of the medium.
3. Viscous drag (Stokes’ law): \( F_d = 6 \pi \eta r v \), where \( \eta \) is the viscosity, \( v \) is velocity.
At terminal velocity \( v_t \), net force is zero:
\[ \frac{4}{3} \pi r^3 \rho g = \frac{4}{3} \pi r^3 \sigma g + 6 \pi \eta r v_t. \] Simplify:
\[ \frac{4}{3} \pi r^3 g (\rho - \sigma) = 6 \pi \eta r v_t. \] \[ v_t = \frac{2 r^2 g (\rho - \sigma)}{9 \eta}. \] Answer: \( v_t = \frac{2 r^2 g (\rho - \sigma)}{9 \eta} \).