Example

Question:

The terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at 20°C is \(6.5\,\mathrm{cm\,s^{-1}}\). Compute the viscosity of the oil at 20°C. Density of oil is \(1.5 \times 10^3\,\mathrm{kg\,m^{-3}}\), density of copper is \(8.9 \times 10^3\,\mathrm{kg\,m^{-3}}\).

Solution:

We have \[ v_t = 6.5 \times 10^{-2}\,\mathrm{m\,s^{-1}},\quad a = 2 \times 10^{-3}\,\mathrm{m},\quad g = 9.8\,\mathrm{m\,s^{-2}}, \] \[ \rho = 8.9 \times 10^3\,\mathrm{kg\,m^{-3}},\quad \sigma = 1.5 \times 10^3\,\mathrm{kg\,m^{-3}} \]
From Eq. (9.18): \[ \eta = \frac{2}{9} \frac{a^2 g (\rho - \sigma)}{v_t} \] \[ = \frac{2}{9} \frac{(2 \times 10^{-3})^2 \times 9.8 \times 7.4 \times 10^3}{6.5 \times 10^{-2}} \] \[ = 9.9 \times 10^{-1}\,\mathrm{kg\,m^{-1}\,s^{-1}} \]