Stoke's Law vl6j

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). At

this

Viscous

drag

speed: =

6πηrv

=

Weight

=

mg

The following formula can be proved (see dimensional proof)

Frictional force (F) = 6πηrv

(Stokes' law)

If the density of the material of the sphere is  and that of the liquid σ, then effective gravitational force = weight - upthrust = 4/3πr3 (ρ – σ)] Therefore we have for the viscosity (η):

Viscosity (η) = 2gr2(ρ - σ)/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! Figure 3 shows how the velocity of an object will increase with time as it falls through a viscous

fluid. 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.

schoolphysics Stokes' Law animation To see an animation of Stokes' Law click on the animation link.

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 = 6rv = Weight = mg 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( – ) Therefore we have for the viscosity (h):

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

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Stokes' Law When small spherical bodies move through a viscous medium, the bodies drag the layers of the medium that are in with them. This dragging results in relative motion between different layers, which are away from the body. Therefore, a viscous drag comes into play, opposing the motion of the body. It is found that this backward force or viscous drag, increases with increase in velocity of the body. According to Stoke, the viscous drag 'f'', depends on the coefficient of viscosity 'h' of the medium, the velocity (v) of the body and radius (r) of the spherical body.

By methods of dimensions, the values of a, b and c are 1,1,1 respectively. Therefore,

Sub Topics

1. Applications of Stoke's law 2. Expression for terminal velocity

Applications of Stoke's law Back to Top

Stokes' law explains why the speed of a raindrop is less than a freely falling body with constant velocity, from the height of clouds.

The same law helps a man coming down with the help of a parachute, to slow down.

Terminal velocity refers to the constant velocity, acquired by a freely falling body in a viscous medium. Consider a small spherical body, falling freely due to gravity in a viscous medium.

The various forces acting on the body are: 

Weight of the body, acting downwards



Viscous drag FV, acting upwards (opposing motion of the body)



Upthrust or Buoyant force (FT) of liquids, equal to weight of the displaced liquid.

As the body falls, its velocity and the viscous drag increase due to gravity. There comes a stage, when all the three forces balance each other i.e., the net forces acting on the sphere is zero. When these

conditions are achieved, the body starts moving with a constant velocity. This constant velocity is called as the terminal velocity.

Expression for terminal velocity Back to Top

Weight of the body =mg

where r is the radius of the body, r is density, g is the gravity due to upward viscous drag F v = 6phvr (Stokes' law). where h is coefficient of viscosity, v is the velocity of body, r is radius of the body. Upthrust or Buoyant force FT = weight of displaced liquid = Volume of body x density of liquid x acceleration due to gravity FT

When the body moves with terminal velocity

Note that if r < s , the body moves up with constant velocity. For example, gas bubbles rise up through soda water bottle.