§ 24. Motion of a Sphere in a Viscous Fluid (Stokes' Law) (Grade 10 · Uni)
Sphere falling through viscous oil, Stokes drag force & approach to terminal velocity. When buoyancy plus Stokes viscous drag balances gravity, net acceleration vanishes and the sphere descends at constant terminal velocity.
Theory Summary & Scientific Mechanism
When a rigid sphere of radius r and density ρₛ settles through a viscous fluid of density ρ_f at low Reynolds number (Re = (2ρ_f v r)/(η) ≪ 1), the flow remains smooth and laminar (creeping flow). Internal fluid friction arises from momentum diffusion across adjacent shearing fluid layers.
By solving the linearized Navier–Stokes equations around a sphere, George Gabriel Stokes (1851) derived Stokes’ Law for viscous drag: F_d = -6π η r v, where η is the dynamic viscosity (in Pa·s). Vertical motion is governed by m(dv)/(dt) = (ρₛ - ρ_f)(4)/(3)π r³ g - 6π η r v.
As the sphere accelerates from rest, viscous drag F_d ∝ v grows until it exactly balances net gravity minus buoyancy (F_g = F_A + F_d). Acceleration then vanishes and the sphere descends at constant terminal velocity vₜerm = (2 r² g (ρₛ - ρ_f))/(9η), scaling as r². This relationship supports falling-sphere viscometry and Millikan’s oil-drop experiment.
Everyday Analogy
Drop a steel marble into a jar of thick honey: it accelerates only briefly and then glides down at a steady constant speed! Skydivers and raindrops reach this same terminal velocity when fluid drag balances gravity minus buoyancy.
Interactive Question & Derivation
A steel sphere of radius r settles through a tall cylinder of viscous glycerin (η). How does its steady terminal velocity vₜerm change if we use a sphere of the same material with twice the radius 2r?
- A: Terminal velocity remains unchanged because all bodies fall at the same rate
- B: Because gravity and buoyancy scale with volume r³ while Stokes drag F_c = 6πη r v scales with r, terminal velocity increases 4-fold (vₜerm ∝ r² / η)
- C: The larger sphere falls slower due to greater fluid resistance
Study Hint: Once Stokes’ viscous drag F_S = 6πη r v balances net gravity minus buoyancy, what is the sphere’s acceleration a, and how does its velocity behave afterward?
Proof & Explanation: Three vertical forces act on the sphere: gravity F_g = (4)/(3)π r³ ρₛ g downward, buoyancy F_A = (4)/(3)π r³ ρ_f g upward, and Stokes viscous drag F_c = 6πη r v upward. At terminal velocity F_g - F_A = F_c, giving Stokes' formula: vₜerm = (2 r² g (ρₛ - ρ_f))/(9η). Thus vₜerm scales as r²!
Key Equations & 3D Simulation Parameters
F_c = 6πη r v, vₜerm = (2 r² g (ρₛ - ρ_f))/(9η), v(t) = vₜerm≤ft(1 - e⁻t/τ )
- Sphere radius r: 0.25 … 0.65 cm (default: 0.42 cm)
- Fluid dynamic viscosity η: 0.4 … 2.8 Pa·s (default: 1.2 Pa·s)
- Sphere density ρₛphere: 1500 … 4000 kg/m³ (default: 2600 kg/m³)
- Viscous fluid density ρ_fluid: 700 … 1250 kg/m³ (default: 950 kg/m³)