Viscosity & Terminal Velocity
Drop a sphere into a viscous fluid and watch it settle to a steady terminal speed.
About this simulation
Release a ball through a fluid and drag grows with speed until it balances gravity minus buoyancy, so acceleration stops. Varying radius, viscosity and densities shows terminal velocity scaling with r² — double the radius and the ball falls four times as fast, while a thicker fluid gives a gentler approach.
How to use it
- Watch the sphere fall through the fluid column: the arrows labelled W, B and F_d show its weight, the buoyant force and the viscous drag, with F_d growing as the sphere speeds up.
- Follow the v–t graph as the curve rises and flattens towards the dashed v_t line; the τ mark shows one time constant, where it reaches about 63% of v_t.
- Change Sphere radius, or press 'Double the radius', and check that the v_t value on the dashed line becomes four times as large.
- Raise Fluid viscosity, or press 'Thick as honey', and see the terminal velocity drop: doubling η halves it.
- Set Sphere density below Fluid density, or press 'Lighter than the fluid', and watch the sphere rise instead of sink.
What to notice
- Drag grows with speed, so the net downward force shrinks as the sphere speeds up until drag plus buoyancy equal the weight and the acceleration becomes zero.
- Terminal velocity is proportional to r², so a small change in radius makes a large change in speed.
- Terminal velocity is inversely proportional to viscosity: the thicker the fluid, the slower the steady fall.
- If the sphere is less dense than the fluid, ρ − σ is negative and the terminal velocity points upward.
The formulas
- F = 6πηrv
- Stokes' law: viscous drag F on a sphere of radius r moving at speed v through a fluid of viscosity η.
- v_t = 2r²(ρ − σ)g ⁄ 9η
- Terminal velocity of a sphere of density ρ in a fluid of density σ and viscosity η; g is 9.8 m/s².
Questions
What is terminal velocity?
Terminal velocity is the constant speed a body reaches when falling through a fluid, once the upward viscous drag and buoyancy together balance its weight. The net force is then zero, so it stops accelerating. In the simulation, the v–t graph flattens onto the dashed v_t line when the forces balance.
What is Stokes' law?
Stokes' law gives the viscous drag on a small sphere moving slowly through a fluid: F = 6πηrv. The drag is proportional to viscosity, radius and speed. It holds only for slow, streamlined flow around the sphere; for large or fast objects in thin fluids such as water and air, the drag grows faster than v and the formula no longer applies.
How do you derive the expression for terminal velocity?
At terminal velocity, weight = buoyancy + drag. Weight is (4/3)πr³ρg, buoyancy is (4/3)πr³σg and drag is 6πηrv_t. Setting (4/3)πr³(ρ − σ)g = 6πηrv_t and solving gives v_t = 2r²(ρ − σ)g/9η. The v_t value on the simulation's dashed line uses exactly this expression.
Why do bigger raindrops fall faster than small ones?
Weight grows with r³ but Stokes drag only with r, so a larger drop needs a higher speed before drag can balance its weight. Under Stokes' law terminal velocity goes as r². Doubling the radius in the simulation makes terminal velocity four times larger. Real raindrops that are large fall fast enough that Stokes' law stops being exact, but the trend is the same.
What is the coefficient of viscosity?
The coefficient of viscosity η measures a fluid's internal friction, its resistance to layers sliding past one another. Its SI unit is the pascal-second (Pa·s), also written N·s/m². Water at room temperature is about 0.001 Pa·s, while glycerine and honey are around 1 Pa·s or more. Use the Fluid viscosity slider to compare them. Slide it down towards water and the Reynolds number turns red: a steel ball in water falls too fast for Stokes' law to hold.
