A metal sphere of radius 1 mm and mass 50 mg falls vertically in glycerine. Find (a) the viscous force exerted by the glycerine on the sphere when the speed of the sphere is 1 cm s−1, (b) the hydrostatic force exerted by the glycerine on the sphere and (c) the terminal velocity with which the sphere will move down without acceleration. Density of glycerine = 1260 kg m−3 and its coefficient of viscosity at room temperature = 8.0 poise.
Solution
Given:
Radius of metallic sphere r = 1 mm = 10−3 m
Speed of the sphere v = 10−2 m/s
Coefficient of viscosity η = 8 poise = 0.8 decapoise
Mass m = 50 mg = 50 × 10−3 kg
Density of glycerin σ = 1260 kg/m3
(a) Viscous force exerted by glycerine on the sphere F = 6πηrv
⇒ F= 6 × (3.14) × (0.8) × 10−3 × (10−2)
= 1.50 × 10−4 N
(b) Let V be the volume of the sphere.
Hydrostatic force exerted by glycerin on the sphere
\[F' = V\sigma g\]
\[\Rightarrow F' = \frac{4}{3}\pi r^2 \sigma g\]
\[= \left( \frac{4}{3} \right) \times \left( 3 . 14 \right) \times \left( {10}^{- 6} \right) \times 1260 \times 10\]
\[ = 5 . 275 \times {10}^{- 5} \text{N}\]
(c) Let the terminal velocity of the sphere be v'.
The forces acting on the drops are
(i) The weight mg acting downwards
(ii) The force of buoyance, i.e., \[\frac{4}{3}\pi r^3 \sigma g\] acting upwards
(iii) The force of viscosity, i.e., 6πηrv' acting upwards
From the free body diagram:
\[6\pi\eta r v' + \frac{4}{3}\pi r^3 \sigma g = \text{ mg}\]
\[ \Rightarrow v = \frac{\text{ mg }- \frac{4}{3}\pi r^2 \sigma g}{6\pi\eta r}\]
\[ = \frac{50 \times {10}^{- 3} - \frac{4}{3} \times 3 . 14 \times {10}^{- 6} \times 1260 \times 10}{6 \times 3 . 14 \times 0 . 8 \times {10}^{- 3}}\]
\[ = \frac{500 - \frac{4}{3} \times 3 . 14 \times {10}^{- 3} \times 1260 \times 10}{6 \times 3 . 14 \times 0 . 8}\]
\[ = 2 . 3 \text{ cm/s }\]
