dimensional analysis of stokes law
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Dropping the Ball (Slowly)
Michael Fowler, UVa
Stokes’ Law
We’ve seen how viscosity acts as a frictional brake on the rate at which water flows through a pipe, let us now examine its frictional effect on an object falling through a viscous medium. To make it simple, we take a sphere. If we use a very viscous liquid, such as glycerin, and a small sphere, for example a ball bearing of radius a millimeter or so, it turns out experimentally that the liquid flows smoothly around the ball as it falls, with a flow pattern like:

(The arrows show the fluid flow as seen by the ball. This smooth flow only takes place for fairly slow motion, as we shall see.)
If we knew mathematically precisely how the velocity in this flow pattern varied near the ball, we could find the total viscous force on the ball by finding the velocity gradient near each little area of the ball’s surface, and doing an integral. But actually this is quite difficult. It was done in the 1840’s by Sir George Gabriel Stokes. He found what has become known as Stokes’ Law: the drag force F on a sphere of radius a moving through a fluid of viscosity η at speed v is given by:
F=6πaηv.
Note that this drag force is directly proportional to the radius. That’s not obvious—one might have thought it would be proportional to the cross-section area, which would go as the square of the radius. The drag force is also directly proportional to the speed, not, for example to v2.
Understanding Stokes’ Law with Dimensional Analysis
Is there some way we could see the drag force must be proportional to the radius, and to the speed, without wading through all of Sir George’s mathematics? The answer is yes—by using dimensions.
First we must ask: what can this drag force depend on?
Obviously, it depends on the size of the ball: let’s say the radius is a, having dimension L.
It must depend on the speed v, which has dimension LT−1.
Finally, it depends on the coefficient of viscosityη which has dimensions ML−1T−1.
The drag force F has dimensions [F]=MLT−2:: what combination of [a]=L, [v]=LT−1 and [η]=ML−1T−1 will give [F]=MLT−2?
It’s easy to see immediately that F must depend linearly on η,, that’s the only way to balance the M
Michael Fowler, UVa
Stokes’ Law
We’ve seen how viscosity acts as a frictional brake on the rate at which water flows through a pipe, let us now examine its frictional effect on an object falling through a viscous medium. To make it simple, we take a sphere. If we use a very viscous liquid, such as glycerin, and a small sphere, for example a ball bearing of radius a millimeter or so, it turns out experimentally that the liquid flows smoothly around the ball as it falls, with a flow pattern like:

(The arrows show the fluid flow as seen by the ball. This smooth flow only takes place for fairly slow motion, as we shall see.)
If we knew mathematically precisely how the velocity in this flow pattern varied near the ball, we could find the total viscous force on the ball by finding the velocity gradient near each little area of the ball’s surface, and doing an integral. But actually this is quite difficult. It was done in the 1840’s by Sir George Gabriel Stokes. He found what has become known as Stokes’ Law: the drag force F on a sphere of radius a moving through a fluid of viscosity η at speed v is given by:
F=6πaηv.
Note that this drag force is directly proportional to the radius. That’s not obvious—one might have thought it would be proportional to the cross-section area, which would go as the square of the radius. The drag force is also directly proportional to the speed, not, for example to v2.
Understanding Stokes’ Law with Dimensional Analysis
Is there some way we could see the drag force must be proportional to the radius, and to the speed, without wading through all of Sir George’s mathematics? The answer is yes—by using dimensions.
First we must ask: what can this drag force depend on?
Obviously, it depends on the size of the ball: let’s say the radius is a, having dimension L.
It must depend on the speed v, which has dimension LT−1.
Finally, it depends on the coefficient of viscosityη which has dimensions ML−1T−1.
The drag force F has dimensions [F]=MLT−2:: what combination of [a]=L, [v]=LT−1 and [η]=ML−1T−1 will give [F]=MLT−2?
It’s easy to see immediately that F must depend linearly on η,, that’s the only way to balance the M
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