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Viscosity and Stokes' Law

When any object rises or falls through a fluid it will experience a viscous drag, whether it is a parachutist or spacecraft falling through air, a stone falling through water or a bubble rising through fizzy lemonade. We will consider here only the case of a falling sphere. The formula was first suggested by Stokes and is therefore known as Stokes' law.

As the sphere falls so its velocity increases until it reaches a velocity known as the terminal velocity. At this velocity the frictional drag due to viscous forces is just balanced by the gravitational force and the velocity is constant. At this speed:

Viscous drag = 6phrv = Weight = mg

Frictional force (F) = 6phrv (Stokes' law)

If the density of the material of the sphere is r and that of the liquid s, then effective gravitational force = weight upthrust = 4/3pr3(r s) Therefore we have for the viscosity (h):

Viscosity (h) = 2gr2(r s)/9v

where v is the terminal velocity of the sphere.

From the formula it can be seen that the frictional drag is smaller for large spheres than for small ones, and therefore the terminal velocity of a large sphere is greater than that for a small sphere of the same material

© Keith Gibbs/John Bourne 2008