# How do you derive eta=(2r^2g(rho_s-rho_l))/(9v) for a sphere obeying stokes law in a liquid, falling at terming velocity so that F+U=W?

Dec 3, 2017

Use the fact that three forces are in equilibrium - weight, buoyant force and the Stoke's drag force.

#### Explanation:

A perfectly smooth sphere of radius $R$ is dropped in a liquid of dynamic viscosity $\setminus \eta$. Three forces act on it - its weight, the Stokes drag force and buoyant force.

The constant weight, ${\vec{F}}_{g}$, acts down while the velocity dependent drag force, ${\vec{F}}_{d} \left(v\right)$, and the constant buoyant force, ${\vec{F}}_{b}$, act up.

Initially the drag force is small. So the ball starts accelerating down. But the drag force increases in strength with speed. At some point the sum of the buoyant force and viscous drag equilibrate with the weight and the ball achieves a constant terminal velocity ${v}_{t}$.

Dynamic Equilibrium Condition: vecF_g + vec F_b + vec F_d = vec 0; ...... (1)

Weight: Writing the mass of the ball in terms of its volume and the density of the material ($\setminus {\rho}_{s}$) of which it is made of,

vec F_g = -mg = -4/3\piR^3\rho_sg;

Buoyant Force : By the Archimedes Principle the buoyant force acting on an object is equal to the weight of the liquid it displaces,

${\vec{F}}_{b} = + \frac{4}{3} \setminus \pi {R}^{3} \setminus {\rho}_{l} g$

Stoke's Drag Force: The upward viscous drag force on a smooth sphere of radius $R$ travelling with a speed ${v}_{t}$ downward in a fluid of dynamic viscosity $\setminus \eta$ is -

${\vec{F}}_{d} = + 6 \setminus \pi \setminus \eta R {v}_{t}$

Expanding (1),
\qquad -4/3\piR^3\rho_sg + 4/3\piR^3\rho_lg + 6\pi\etaRv_t = 0;

$6 \cancel{\setminus \pi} \setminus \eta R {v}_{t} = \frac{4}{3} \cancel{\setminus \pi} {R}^{3} \left(\setminus {\rho}_{s} - \setminus {\rho}_{l}\right) g$
$\setminus \eta = \frac{2}{9} \frac{{R}^{2} g \left(\setminus {\rho}_{s} - \setminus {\rho}_{l}\right)}{v} _ t$