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.

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

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

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.

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

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

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

Calculation continued 4 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

Calculation continued 4 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

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.