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Stokes' law and terminal velocity
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. The
mathematics of the viscous drag on irregular shapes is
difficult; 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.
Consider a sphere falling through a viscous fluid. 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 (shown
by Figure 2).
Figure 1
At this speed:
Viscous drag = 6rv = Weight = mg
Viscous drag (6rv)
The following formula can be proved (see dimensional proof)
Frictional force (F) = 6rv
(Stokes' law)
If the density of the material of the sphere is  and that of the
liquid , then
effective gravitational force = weight – upthrust = 4/3r3 ( – )
m
Velocity v
Gravitational pull (weight) mg
Figure 2
Therefore we have for the viscosity ():
Viscosity = 2gr2( - )
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.
Stokes' law is important in Millikan's experiment for the measurement of the charge on an
electron, and it also explains why large raindrops hurt much more than small ones when they
fall on you - it's not just that they are heavier, they are actually falling faster.
People falling through the atmosphere will also eventually reach their terminal velocity. For lowlevel air (below about 3000 m) this is around 200 km/hour flat out and just over 320 km/hour
head down. However at high altitudes around 30 000m this can reach almost 1000 km/hour!
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Figure 3 shows how the velocity of an object will increase with time as it falls through a viscous
fluid.
Velocity of
falling sphere
Terminal velocity
Figure 2
Example problem
Calculate the terminal velocity for an air bubble of radius 0.5 mm rising through a glass of fizzy water.
Density of air = 1.2 kgm-3, density of water = 1000 kgm -3, viscosity of water = 1x10-3 Pas, g = 9.8 ms-2).
Using the equation: = [2gr2( - )]/9v
Terminal velocity (v)
= [2gr2( - )]/9= [2x9.8x(0.5x10-3)2x(1000 – 1.2)]/(9x1x10-3)
= 4.89x10-3/9x10-3
= 0.5 ms-1
It is interesting to consider the effect of various shapes of objects falling through a fluid. These
can be made from plasticene.
See the section on subsonic, supersonic and hypersonic vehicles and the shape of the bulbous
bow on a nuclear powered submarine.
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