Question

The terminal velocity of sphere of radius R ,falling in a viscous fluid , is proportional to ????????????????????????????????????????????????????????????????????

Answer 1

QUESTION:

The terminal velocity of sphere of radius $R$, falling in a viscous fluid, is proportional to

SOLUTION:

According to the Stokes' law of viscosity the frictional force exerted on spherical objects in a viscous fluid is given by

$$F_{\mathrm{fr}} = 6 \pi \mu R \upsilon$$

Here $\mu$ is the fluid viscosity, $\upsilon$ is the particle's velocity and $R$ is the particle's radius. There are three forces acting on a sphere falling in a fluid:

1. $F_{\mathrm{fr}}$ (acting upward)

2. the gravity force $mg$ (acting downward)

3. buoyancy force $F_{\mathrm{b}} = \rho_{\mathrm{fluid}} \cdot g \cdot V$ (acting upward)

According to the Newton's second law of motion and taking in account that the velocity of the particle remains constant (terminal velocity), we can write:

$$F_{\mathrm{fr}} + F_{\mathrm{b}} - mg = 0$$ $$m = \rho_{\text{particle}} \cdot V$$ $$V = \frac{4}{3} \pi R^3$$ $$6 \pi \mu R \upsilon + \rho_{\text{fluid}} \cdot g \cdot \frac{4}{3} \pi R^3 - \rho_{\text{particle}} \cdot g \cdot \frac{4}{3} \pi R^3 = 0$$ $$\upsilon = \frac{4}{3} R^2 g \left( \rho_{\text{particle}} - \rho_{\text{fluid}} \right) / 6\mu = \frac{2}{9} R^2 \frac{\rho_{\text{particle}} - \rho_{\text{fluid}}}{\mu}$$

Hence the terminal viscosity is proportional to the particle's radius squared.

Answer

The terminal viscosity is proportional to the particle's radius squared.