Step-by-Step Terminal Velocity Formula Using Stokes' Law With Examples
Terminal velocity is a fundamental concept in the study of fluid mechanics. It describes the constant speed that a body attains while falling through a fluid, such as air or a viscous liquid, when the forces acting on it become balanced. This concept plays a key role in understanding motion, resistance, and equilibrium in fluids, and it has practical implications in topics ranging from rain drops to parachutes.
What is Terminal Velocity?
When a sphere or any object falls through a viscous fluid, it is subjected to several forces. At first, it accelerates due to gravity. As its speed increases, the resistive forces also increase. Eventually, the object reaches a stage where its acceleration drops to zero and it continues to move with a constant speed. This steady speed is known as terminal velocity.
Forces Acting on a Sphere in a Fluid
Three forces act on a sphere falling through a viscous liquid:
- Gravitational Force (Weight): Downward force, given by \( F_g = mg \)
- Buoyant Force: Upward force due to displaced fluid, \( F_b = \rho V g \)
- Viscous (Drag) Force: Upward resistive force according to Stokes' Law, \( F_v = 6 \pi \eta r v \)
– \( m \) is the mass of the sphere
– \( g \) is acceleration due to gravity
– \( \rho \) is the density of the fluid
– \( V \) is the volume of the sphere
– \( \eta \) is the viscosity of the fluid
– \( r \) is the sphere's radius
– \( v \) is the velocity of the sphere
Derivation: Expression for Terminal Velocity
- At terminal velocity, the sphere moves at a constant speed, so the net force on it is zero.
- Balance of forces:
\( mg = F_b + F_v \)
- Substitute for buoyant and viscous forces:
\( mg = \rho V g + 6\pi \eta r v_t \)
- The volume of a sphere is \( V = \frac{4}{3} \pi r^3 \), so:
\( mg = \rho \left(\frac{4}{3}\pi r^3\right)g + 6\pi \eta r v_t \)
- Express mass in terms of the sphere's density (\( \sigma \)):
\( m = \sigma V = \sigma \left(\frac{4}{3}\pi r^3\right) \)
- Combine and simplify the equation:
\( \left[\sigma \left(\frac{4}{3}\pi r^3\right)g - \rho \left(\frac{4}{3}\pi r^3\right)g \right] = 6\pi \eta r v_t \)
- Extract common factors:
\( \frac{4}{3}\pi r^3 g (\sigma - \rho) = 6\pi \eta r v_t \)
- Solve for terminal velocity (\( v_t \)):
\( v_t = \frac{2}{9} \frac{(\sigma - \rho)gr^2}{\eta} \)
This formula shows that terminal velocity depends on the radius of the sphere, the difference in densities, gravitational acceleration, and the viscosity of the fluid.
Key Terminal Velocity Formula
| Quantity | Formula | Variables |
|---|---|---|
| Terminal Velocity | \( v_t = \frac{2}{9} \frac{(\sigma - \rho)gr^2}{\eta} \) | \( r \): radius of sphere \( \sigma \): density of sphere \( \rho \): density of fluid \( g \): gravity \( \eta \): viscosity |
| Stokes' Drag Force | \( F_v = 6\pi \eta r v \) | \( \eta \): viscosity \( r \): radius \( v \): velocity |
| Buoyant Force | \( F_b = \rho V g \) | \( \rho \): density (fluid) \( V \): volume \( g \): gravity |
Physical Interpretation and Practical Example
Suppose a steel ball of radius 0.5 cm is dropped in a viscous liquid. The terminal velocity can be quickly calculated using the formula above by substituting the relevant values for densities, gravity, radius, and viscosity.
This principle explains why small particles settle slowly in water and why raindrops do not accelerate indefinitely but fall at a constant speed.
Application and Variation of Velocity with Time
Initially, when an object is released, its velocity increases rapidly as gravity dominates. As resistance from fluid builds up, the rate of increase slows down.
The velocity vs. time graph for the falling sphere starts at zero, rises steeply, and then levels off, approaching a straight horizontal line—this constant value corresponds to terminal velocity.
| Stage | Velocity Behaviour |
|---|---|
| Initial Drop | Velocity increases quickly |
| Approaching Terminal Velocity | Increase slows due to drag |
| After Terminal Velocity | Remains constant |
Factors Affecting Terminal Velocity
| Factor | How It Affects Terminal Velocity |
|---|---|
| Radius of sphere (\( r \)) | Terminal velocity increases with square of radius |
| Difference in densities (\( \sigma - \rho \)) | Higher difference gives higher terminal velocity |
| Viscosity (\( \eta \)) | Higher viscosity reduces terminal velocity |
| Gravity (\( g \)) | Directly proportional to terminal velocity |
Related Concepts and More Practice
For a deeper understanding of velocity, fluids, and motion, explore these related resources:
- Velocity – Concept & Differences
- Velocity-Time Graph
- Drag Force and Its Effects
- Fluid Properties
- Fluid Friction
- Viscosity and Applications
To strengthen problem-solving, practice finding terminal velocity for different spheres and fluids. Pay attention to units, substitute values accurately, and verify results using the formula provided. Reviewing solved examples and testing your knowledge with numerical questions will boost your understanding and readiness for exams.
Next Steps and Further Learning
To master this concept further, explore topic collections on Fluid Mechanics and review detailed derivations, physical explanations, and advanced applications. Build a strong foundation in Physics by connecting related topics and practicing across question types.
For additional help and exam-focused preparation, access more resources and live sessions on terminal velocity and fluid dynamics on Vedantu.
FAQs on Terminal Velocity Derivation Made Easy (Class 11 & 12 Physics)
1. Define terminal velocity.
Terminal velocity is the constant speed attained by a falling object when the downward force of gravity is exactly balanced by the upward forces of viscous drag and buoyancy, resulting in zero net acceleration. At terminal velocity, the object continues to move at a uniform speed through the fluid.
2. What are the forces acting on a sphere falling through a viscous liquid?
Three main forces act on a sphere falling through a viscous liquid:
- Gravitational force (Weight): Acts downward, given by Fg = mg or (4/3)πr³σg.
- Buoyant force: Acts upward, given by Fb = ρVg, where ρ is the density of fluid.
- Viscous force (Stokes' drag): Acts opposite to motion, Fv = 6πηrv.
3. State Stokes’ law and give its mathematical expression.
Stokes’ law states that the viscous force experienced by a small spherical object moving slowly through a fluid is directly proportional to its radius, the viscosity of the fluid, and the velocity of the object. The formula is:
F = 6πηrv
- F = viscous force
- η = viscosity of fluid
- r = radius of sphere
- v = velocity
4. Derive the expression for the terminal velocity of a sphere falling through a viscous liquid.
Terminal velocity (vt) is derived by setting the net force acting on the sphere to zero:
- Gravitational force = Buoyant force + Viscous force
- (4/3)πr³σg = (4/3)πr³ρg + 6πηrvt
- Solving for vt:
vt = [2r²(σ − ρ)g] / (9η)
- r = radius of sphere, σ = density of sphere, ρ = density of fluid, g = gravity, η = viscosity
5. Which factors affect the terminal velocity of a sphere in a fluid?
The terminal velocity depends on the following factors:
- Radius of the sphere (r): vt increases with r²
- Density difference (σ − ρ): Greater difference increases vt
- Viscosity (η) of the fluid: Higher viscosity decreases vt
- Acceleration due to gravity (g): Directly proportional to vt
6. What is the difference between terminal velocity and escape velocity?
Terminal velocity is the constant speed reached by an object falling through a fluid when forces balance. Escape velocity is the minimum speed required for an object to escape the gravitational field of a planet without further propulsion. They occur in different contexts and have different formulae.
7. How does velocity change with time when a sphere is dropped into a viscous liquid?
When a sphere is dropped:
- Initially, velocity increases as the sphere accelerates under gravity.
- As time passes, viscous and buoyant forces increase, reducing acceleration.
- Eventually, the sphere reaches terminal velocity and moves at constant speed.
- Graphically: The velocity-time curve rises steeply at first and flattens as it approaches terminal velocity.
8. Give a practical example where the concept of terminal velocity is applied.
A common example is rain drops falling from clouds. Due to air resistance, raindrops quickly reach their terminal velocity, which limits their falling speed and prevents them from causing injury on impact. Parachutists also rely on terminal velocity to descend safely.
9. How can you calculate the terminal velocity of a steel ball of radius 0.01 m in glycerin (density 1200 kg/m³, viscosity 1.2 Pa·s), if density of steel is 7800 kg/m³?
Using the formula: vt = (2r²(σ − ρ)g) / (9η)
- r = 0.01 m, σ = 7800 kg/m³, ρ = 1200 kg/m³, η = 1.2 Pa·s, g = 9.8 m/s²
- vt = [2 × (0.01)² × (7800 − 1200) × 9.8] / (9 × 1.2) = 12.936 / 10.8 ≈ 1.2 m/s
10. What is the mathematical form of Stokes' drag force?
The Stokes' drag force on a sphere of radius r moving at velocity v in a fluid of viscosity η is:
Fv = 6πηrv
11. What are the units of the coefficient of viscosity?
The SI unit of the coefficient of viscosity (η) is Pascal second (Pa·s) or equivalently kg·m⁻¹·s⁻¹.
12. Explain why the acceleration of a body becomes zero after attaining terminal velocity in a fluid.
After reaching terminal velocity, the sum of the upward (drag and buoyant) forces equals the downward gravitational force, making the net force and hence acceleration zero. The object then falls at a constant speed.
