Topic 6.2.115 min read
Terminal Velocity
Terminal velocity is the maximum constant velocity acquired by a freely falling object in a viscous medium when upward drag balances its weight. In the textbook's Stokes model, v_t = 2ρgr²/(9η).
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Beginner Level
A falling raindrop first speeds up because its weight is greater than air resistance. As its speed increases, drag increases. Eventually drag becomes equal to weight, the net force becomes zero and the drop continues downward at a constant terminal velocity.
- F_d
- upward drag (N)
- W
- weight (N)
When to use: Terminal-speed force balance.
Recommended for Class 11
6.2.1 Terminal Velocity
Definition
Learning Outcomes
- Define terminal velocity.
- Explain the changing force balance during a fall.
- Derive terminal velocity using Stokes' law.
- State how radius, density and viscosity affect v_t.
- Solve terminal-velocity calculations.
- Distinguish zero acceleration from zero velocity.
How Terminal Velocity Develops
| Stage | Weight | Drag | Motion |
|---|---|---|---|
| Just released | Downward and unchanged | Nearly zero | Large downward acceleration |
| Speed increasing | Unchanged | Increasing upward | Acceleration decreases |
| Terminal velocity | Equal to drag | Equal to weight | Constant downward velocity |
At terminal velocity, acceleration is zero:
Intermediate Level
Derivation Using Stokes' Law
The textbook model balances Stokes drag with weight and neglects buoyancy of air.
For a uniform sphere of density ρ and radius r:
Substitute m into the first terminal-velocity expression:
- ρ
- sphere density in the textbook model (kg m⁻³)
- r
- sphere radius (m)
- η
- fluid viscosity (Pa s)
When to use: Uniform small sphere, Stokes regime, textbook model neglecting buoyancy.
Dependence of Terminal Velocity
| Change | Effect on v_t |
|---|---|
| Radius doubles | v_t becomes 4 times |
| Radius triples | v_t becomes 9 times |
| Sphere density doubles | v_t doubles |
| Viscosity doubles | v_t becomes half |
| Viscosity increases | Terminal speed decreases |
Buoyancy-Corrected Form
For a sphere falling in a fluid whose density is not negligible, upthrust also acts. Replacing sphere density ρ_s by the density difference gives the more complete relation:
- ρ_s
- sphere density (kg m⁻³)
- ρ_f
- fluid density (kg m⁻³)
When to use: Use when fluid buoyancy must be included.
Solved Numerical Examples
Textbook Assignment 6.2: Raindrop
Find the terminal velocity of a water raindrop of radius 0.3 mm falling through air of viscosity 1.8 × 10⁻⁵ kg m⁻¹ s⁻¹. Use ρ_water = 1000 kg m⁻³ and neglect air buoyancy.
- r = 0.3 mm = 3.0 × 10⁻⁴ m
- η = 1.8 × 10⁻⁵ Pa s
- ρ = 1000 kg m⁻³
- g = 9.8 m s⁻²
vt = 2ρgr²/(9η)
vt = 2(1000)(9.8)(3.0 × 10⁻⁴)²/(9(1.8 × 10⁻⁵))
vt = 2(1000)(9.8)9.0 × 10⁻⁸1.62 × 10⁻⁴
vt = 1.764 × 10⁻³/(1.62 × 10⁻⁴)
Final answer: vt ≈ 10.9 m s⁻¹
Example 2: Radius Comparison
Two droplets of the same liquid fall in the same fluid. Their radii are r and 2r. Compare their terminal velocities.
- r₂ = 2r₁
- Same ρ, g and η
v_t ∝ r²
v₂v₁ = (r₂r₁)²
v₂v₁ = 2²
Final answer: v₂ = 4v₁
Example 3: Viscosity Comparison
A sphere has terminal speed 6.0 cm s⁻¹ in a liquid. What is its terminal speed in a liquid with three times the viscosity, all else unchanged?
- v₁ = 6.0 cm s⁻¹
- η₂ = 3η₁
vt ∝ 1η
v₂v₁ = η₁η₂
v₂ = 6.03
Final answer: v₂ = 2.0 cm s⁻¹
Real-Life Applications
- Small raindrops reach safe terminal speeds instead of accelerating indefinitely.
- Sedimentation separates particles according to size and density.
- Falling-sphere viscometers infer viscosity from terminal speed.
- Parachutes increase drag and reduce terminal speed.
- Dust and mist remain suspended longer because very small particles have small terminal speeds.
Common Mistakes
- Saying the object stops at terminal velocity.
- Setting drag equal to mass instead of weight mg.
- Forgetting the square on radius.
- Using diameter for r.
- Not converting millimetres to metres.
- Using the textbook formula outside the Stokes regime.
- Mixing object density with fluid viscosity units.
Short-Question Preparation
Q. Define terminal velocity.
A. It is the maximum constant velocity acquired by a freely falling body in a viscous medium.
Q. Why does acceleration decrease during the fall?
A. Drag increases with speed while weight remains constant, so the downward net force decreases.
Q. What is the force condition at terminal velocity?
A. In the textbook model, upward drag equals downward weight, so net force and acceleration are zero.
Q. How do radius and viscosity affect terminal velocity?
A. v_t is proportional to r² and inversely proportional to η.
Long-Question Preparation
Q. Derive terminal velocity for a spherical body.
A. 1. At terminal velocity, Fd = W. 2. Use Stokes' law Fd = 6πηrvt. 3. Use W = mg. 4. Therefore 6πηrvt = mg. 5. Rearrange vt = mg/(6πηr). 6. For a sphere, m = ρV. 7. Use V = 4πr³/3. 8. Substitute: vt = ρ(4πr³/3)g6πηr. 9. Cancel π and r. 10. Hence v_t = 2ρgr²/(9η). 11. State assumptions and proportionalities.
Sticky Notes
- At release: drag ≈ 0.
- As speed rises: drag rises.
- At vt: Fd = W.
- At vt: a = 0 but v ≠ 0.
- vt = 2ρgr²/(9η).
- vt ∝ r² and vt ∝ 1η.
60-Second Revision
Summary
A falling body accelerates until speed-dependent drag balances its weight. It then moves at constant terminal velocity. Combining this balance with Stokes' law gives v_t = 2ρgr²/(9η) for the textbook's uniform-sphere model.
Ask-the-Tutor Modes
- Explain terminal velocity simply.
- Walk through the force stages.
- Show the derivation one line at a time.
- Compare two droplets.
- Check my unit conversion.
- Test me with an MCQ.
Exam Focus
Exam Focus
Important points
- Definition
- Force stages
- Terminal balance
- Derivation
- Dependencies
Important formulas
- Fnet = W - Fd
- Fd = 6πηrv
- vt = mg/(6πηr)
- vt = 2ρgr²/(9η)
Common exam traps
- Zero velocity claim
- Missing r²
- Wrong units
- Ignoring assumptions
Short question tips
- State Fd = W and a = 0.
Long question structure
Definition → force stages → balance → Stokes law → sphere mass → simplification → dependencies.
Common Mistakes
Common Mistakes
- Mistake
Terminal velocity is not zero.
- Mistake
Use weight mg.
- Mistake
Square radius.
- Mistake
Convert radius to metres.
- Mistake
State Stokes-model assumptions.
Revision Zone
60-Second Revision
Quick cues from this topic — use them for last-minute revision.
Fd = W.
a = 0.
v_t ∝ r².
vt ∝ 1η.
Terminal velocity is constant falling velocity when drag balances weight.
vt = 2ρgr²/(9η)
- Definition
- Terminal velocity is constant falling velocity when drag balances weight.
- Formula
- vt = 2ρgr²/(9η)
- Example
- A 0.3 mm water drop in the stated air model has v_t ≈ 10.9 m s⁻¹.
- Critical warning
- Zero acceleration does not mean zero velocity.
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Exam Practice
Important Short Question
Q1. Define terminal velocity.
Conceptual Question
Q2. Why does acceleration become zero?
Long Question Preparation
Q3. Derive terminal velocity for a sphere.
Important Short Question
Q4. State the dependencies of terminal velocity.