Question

A gold sphere of radius R=100 μm and density 19g/cm^3 falls through water. Given the viscosity of water is about 10^-3​ Pa s and the density of water is 1.0g/cm^3, calculate the terminal velocity of the sphere. The viscous drag coefficient for a sphere is α=6π.
Don't forget to consider the buoyant force on the sphere when calculating net force!

Answer 1

The terminal velocity of gold sphere is 39.2 cm/s

What is terminal velocity?

Terminal velocity is the maximum velocity attainable for an object as it falls through a fluid.

How to calculate the terminal velocity of the gold sphere?

The terminal velocity of the gold sphere is given by v = 2gr²(ρ - σ)/9η where

• g = acceleration due to gravity = 9.8 m/s²,
• r = radius of sphere = 100 μm = 100 × 10⁻⁶ m = 10⁻⁴ m = 10⁻² cm,
• ρ = density of sphere = 19 g/cm³,
• σ = density of water = 1.0 g/cm³ and
• η = viscosity of water = 10⁻³ Pa-s

So, susbtituting the values of the variables into the equation, we have that

v = 2gr²(ρ - σ)/9η

v = 2 × 9.8m/s²× (10⁻² cm)²(19 g/cm³ - 1.0 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 9.8 m/s² × 10⁻⁴ cm² × (18 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 980 cm/s² × 10⁻⁴ cm² × 2 g/cm³/(1 × 10⁻³ Pa-s)

v = 3920 g/s² × 10⁻⁴/(1 × 10⁻³ Pa-s)

v = 392 cm/s × 10³ × 10⁻⁴

v = 392 × 10⁻¹ cm/s

v = 39.2 cm/s

So, the terminal velocity is 39.2 cm/s

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