1. Stokes Law OBJECTIVES: Must be able to: (Grades D/E)
Describe the motion of a moving object under the effect of its weight, upthrust and viscous drag.
State the conditions for body to be moving with terminal velocity,
SHOULD be able to: (Grades B/C)
Apply Stokes Law in conjunction with the equations for weight and upthrust to objects falling through a fluid.
COULD be able to: (Grades A/B)
Extend the application to new and unfamiliar situations.

2. Terminal velocity Forces acting on a skydiver. Fv U W Weight, W = mg
Upthrust, U
Viscous drag, Fv
Applying vector addition and, say, down as positive
Resultant force,
ΣF = W + U + Fv
a = ΣF/m Fv U W

3. Terminal velocity Viscous drag force, F = 6πηrv
Consider a sphere falling through a fluid.
What determines the size of the viscous drag force on the sphere?
The coefficient of viscosity, η of the fluid.
The velocity, v
The cross sectional area, A of the sphere.
A = πr2
So viscous drag also depends on r.
STOKES LAW:
Viscous drag force, F = 6πηrv

4. Terminal velocity Weight and Upthrust are both constant.
Viscous drag, Fv increases with speed.
The faster the skydiver falls the greater the combined upward force, Fv + U.
If, eventually, the resultant force ΣF = 0, then,
acceleration, a = 0
velocity is constant (terminal velocity).
Now since ΣF = 0
W + Fv + U = 0
Rearranging: Fv = - (W +U)
We can use this as an alternative way of finding η.

5. Finding the coefficient of viscosity using Stokes Law
Measure the mass, m and radius, r of a ball bearing,
Measure the density, ρ of the fluid to be tested.
Measure the terminal velocity,v of the ball bearing as it falls through the fluid.
Repeat to obtain an average. ρ

6. Calculation m ρ V Fv = - (W +U) W = weight of the ball bearing. W = mg
U = weight of fluid displaced.
Volume of fluid displaced = volume of the ball bearing
= πr3
Density of the fluid = ρ mf = V ρ
So mass of fluid displaced, mf = πr3ρ
Weight of fluid displaced = mf g = πr3ρg
Therefore,
Upthrust, U = πr3ρg 4 3 m ρ V 4 3 4 3 4 3

7. Calculation continued
From Stokes Law
Fv = 6πηrv
So now,
W + Fv + U = 0
can be written as,
mg + (-6πηrv) + (- πr3ρg) = 0
The brackets are superfluous, so:
mg - 6πηrv - πr3ρg = 0
Try rearranging this to make η the subject,
Or making the terminal velocity, v the subject.
- signs due to direction 4 3 4 3

8. Calculation continued4 3
mg - 6πηrv - πr3ρg = 0
To make η the subject,
Add 6πηrv to both sides.
mg - πr3ρg = 6πηrv
You could factorise the left hand side
(m - πr3ρ) g = 6πηrv
Divide both sides by 6πrv
(m - πr3ρ) g = η
6πrv 4 3 4 3 4 3

9. Calculation continued4 3
mg - 6πηrv - πr3ρg = 0
To make v the subject,
Add 6πηrv to both sides.
mg - πr3ρg = 6πηrv
You could factorise the left hand side
(m - πr3ρ) g = 6πηrv
Divide both sides by 6πηr
(m - πr3ρ) g = v
6πηr 4 3 4 3 4 3

10. Stokes Law OBJECTIVES: Must be able to: (Grades D/E)
Describe the motion of a moving object under the effect of its weight, upthrust and viscous drag.
State the conditions for body to be moving with terminal velocity,
SHOULD be able to: (Grades B/C)
Apply Stokes Law in conjunction with the equations for weight and upthrust to objects falling through a fluid.
COULD be able to: (Grades A/B)
Extend the application to new and unfamiliar situations.