What Is Terminal Velocity?
Terminal velocity is the maximum constant velocity attained by a body falling through a fluid (such as air or water) when the resultant force acting on it becomes zero.
When an object falls through a fluid, its velocity initially increases due to gravity. As its speed increases, the resistive force exerted by the fluid also increases. Eventually, the resistive forces balance the effective weight of the object. The object then continues to fall with constant velocity, called terminal velocity.
Definition: Terminal velocity is the constant maximum velocity attained by a falling body when the downward gravitational force is balanced by the upward resistive forces acting on it.
How Does a Falling Object Reach Terminal Velocity?
When an object is released from rest in a fluid, it experiences different forces during its downward motion.
Motion Before Reaching Terminal Velocity
- Initially, the object is released from rest, and its velocity is zero.
- The gravitational force causes the object to accelerate downwards.
- As its velocity increases, the fluid resistance also increases.
- The net downward force gradually decreases, causing the acceleration to decrease.
- Finally, the resistive forces balance the effective weight of the object.
- The resultant force becomes zero, and the object moves with constant terminal velocity.
Forces Acting at Terminal Velocity
For an object falling through a fluid, the forces are:
- Weight (W): Acts vertically downwards.
- Buoyant force (FB): Acts vertically upwards.
- Drag or viscous force (F): Acts vertically upwards.
At terminal velocity:
Weight = Buoyant force + Resistive force
The resultant force is zero, so acceleration is also zero.
Important: Zero acceleration does not mean zero velocity. The object continues to move with constant velocity.
Terminal Velocity of a Falling Object in Air
When an object falls through air, it experiences gravitational force downwards and air resistance upwards. As the speed increases, air resistance increases.
At terminal velocity, the downward weight is balanced by the upward drag force and, more precisely, any buoyant force due to air.
Terminal Velocity Formula for Quadratic Air Resistance
For an object moving through air at speeds where drag is approximately proportional to the square of velocity, the drag force is:
Drag force = ½ × ρ × Cd × A × v²
At terminal velocity:
mg = ½ × ρ × Cd × A × vt²
Therefore:
Terminal velocity, vt = √[2mg / (ρ × Cd × A)]
Where:
- m = Mass of the falling object in kg
- g = Acceleration due to gravity in m/s²
- ρ = Density of air in kg/m³
- Cd = Coefficient of drag (dimensionless)
- A = Cross-sectional area of the object in m²
- vt = Terminal velocity in m/s
This expression neglects the buoyant force of air. It is useful for understanding the terminal velocity of objects such as skydivers and parachutists.
Terminal Velocity of a Sphere in a Viscous Liquid
When a small spherical body falls through a viscous liquid, it experiences three forces:
- Weight acting vertically downwards.
- Buoyant force acting vertically upwards.
- Viscous force acting vertically upwards, opposing the motion.
As the sphere falls, its velocity increases and the viscous resistance also increases. Eventually, the downward weight becomes equal to the sum of the upward buoyant and viscous forces.
At this stage, the sphere attains terminal velocity.
Stokes’ Law
According to Stokes’ law, the viscous force acting on a small sphere moving slowly through a viscous fluid is directly proportional to its radius and velocity.
Viscous force = 6 × π × η × r × v
Where:
- η = Coefficient of viscosity of the liquid in Pa s
- r = Radius of the sphere in metres
- v = Velocity of the sphere in m/s
Derivation of Terminal Velocity Using Stokes’ Law
Consider a small spherical body of radius r and density ρs falling through a liquid of density ρf.
The weight of the spherical body acts downwards.
Weight = Volume × Density of sphere × g
Weight = (4/3) × π × r³ × ρs × g
The buoyant force acting upwards is equal to the weight of the displaced liquid.
Buoyant force = (4/3) × π × r³ × ρf × g
According to Stokes’ law:
Viscous force = 6 × π × η × r × vt
At terminal velocity, the resultant force is zero.
Therefore:
Weight = Buoyant force + Viscous force
Substituting the expressions:
(4/3) × π × r³ × ρs × g = (4/3) × π × r³ × ρf × g + 6 × π × η × r × vt
Rearranging:
(4/3) × π × r³ × (ρs − ρf) × g = 6 × π × η × r × vt
Simplifying:
Terminal velocity, vt = 2r²(ρs − ρf )g / 9η
This is the expression for the terminal velocity of a small sphere falling through a viscous liquid according to Stokes’ law.
Conditions for Stokes’ Law
The above expression is valid under the following conditions:
- The body should be spherical.
- The sphere should be small compared with the dimensions of the container.
- The motion should be slow and laminar.
- The Reynolds number should be sufficiently small.
- The liquid should behave as a Newtonian fluid.
- The influence of the container walls should be negligible.
Stokes’ law should not be applied indiscriminately to large raindrops, rapidly moving bodies or flows with significant turbulence.
Factors Affecting Terminal Velocity
The terminal velocity of a falling sphere through a viscous liquid depends on several factors.
1. Radius of the Sphere
Terminal velocity is directly proportional to the square of the radius.
vt ∝ r²
If the radius is doubled, the terminal velocity becomes four times its original value, provided all other conditions remain unchanged.
2. Difference in Density
Terminal velocity is directly proportional to the difference between the density of the sphere and the density of the liquid.
vt ∝ (ρs − ρf)
A greater density difference produces a greater terminal velocity under the same conditions.
3. Coefficient of Viscosity
Terminal velocity is inversely proportional to the coefficient of viscosity.
vt ∝ 1/η
If the viscosity is doubled, the terminal velocity becomes half its original value.
4. Acceleration Due to Gravity
Terminal velocity is directly proportional to acceleration due to gravity.
vt ∝ g
For the same sphere and liquid, terminal velocity decreases when gravitational acceleration decreases.
Difference Between Terminal Velocity in Air and in a Viscous Liquid
| Terminal velocity in air | Terminal velocity in a viscous liquid |
| Air acts as the surrounding fluid. | A liquid acts as the surrounding fluid. |
| Air resistance or drag opposes motion. | Viscous resistance opposes motion. |
| Drag often depends approximately on the square of speed at higher Reynolds numbers. | Stokes’ law applies at sufficiently low Reynolds numbers. |
| Depends on mass, cross-sectional area, drag coefficient and air density in the quadratic-drag model. | Depends on sphere radius, density difference, viscosity and gravity under Stokes’ law. |
| Examples include skydivers and falling objects. | Examples include small spheres falling through glycerine or oil. |
Solved Numerical Problems on Terminal Velocity
Example 1: Calculating Terminal Velocity Using Stokes’ Law
A small spherical ball of radius 0.001 m and density 8000 kg/m³ falls through a liquid of density 1000 kg/m³. If the coefficient of viscosity of the liquid is 0.8 Pa s, calculate its terminal velocity. Take g = 9.8 m/s².
Given:
Radius, r = 0.001 m
Density of sphere, ρs = 8000 kg/m³
Density of liquid, ρf = 1000 kg/m³
Coefficient of viscosity, η = 0.8 Pa s
Acceleration due to gravity, g = 9.8 m/s²
Formula:
Terminal velocity = 2r²(ρs − ρf)g / 9η
Substitution:
vt = [2 × (0.001)² × (8000 − 1000) × 9.8] / (9 × 0.8)
vt = 0.1372 / 7.2
Answer: Terminal velocity = 0.0191 m/s (approximately).
Example 2: Effect of Radius on Terminal Velocity
The radius of a small sphere falling through a viscous liquid is doubled. What happens to its terminal velocity?
Principle:
Terminal velocity is directly proportional to the square of the radius.
vt ∝ r²
If the radius becomes 2r:
New terminal velocity ∝ (2r)²
= 4r²
Answer: The terminal velocity becomes four times the original value.
Example 3: Effect of Viscosity on Terminal Velocity
A sphere falls through a liquid with terminal velocity 0.04 m/s. If the viscosity of the liquid is doubled, calculate the new terminal velocity, assuming all other factors remain unchanged.
Principle:
Terminal velocity is inversely proportional to viscosity.
vt ∝ 1/η
New terminal velocity = Original terminal velocity / 2
= 0.04 / 2
Answer: 0.02 m/s.
Frequently Asked Questions
1. What is terminal velocity?
Terminal velocity is the constant maximum velocity attained by a falling object when the resultant force acting on it becomes zero.
2. Why does a falling object attain terminal velocity?
As the object speeds up, the resistive force increases. Eventually, the resistive forces balance the effective weight, making the resultant force zero.
3. Is acceleration zero at terminal velocity?
Yes. Since the resultant force is zero, acceleration is zero. However, the object continues moving with constant velocity.
4. What is the formula for terminal velocity in a viscous liquid?
For a small sphere falling under Stokes’ law:
vt = 2r²(ρs − ρf)g / 9η
5. What happens to terminal velocity if the radius of a sphere is doubled?
Under Stokes’ law, terminal velocity becomes four times the original value because it is proportional to the square of the radius.
6. How does viscosity affect terminal velocity?
Terminal velocity is inversely proportional to the coefficient of viscosity under Stokes’ law. Greater viscosity results in lower terminal velocity.
7. What is the difference between terminal velocity and uniform velocity?
Uniform velocity means constant velocity in a specified direction. Terminal velocity is a particular constant velocity reached by an object falling through a fluid when the resultant force becomes zero.
8. Does a parachute increase or decrease terminal velocity?
A parachute increases air resistance and decreases the terminal velocity of the falling person.
9. Is terminal velocity independent of mass?
No. In the quadratic-drag model for a falling object in air, terminal velocity depends on mass. For a sphere falling through a viscous liquid under Stokes’ law, it depends on the difference in densities and the sphere’s radius.
10. What is Stokes’ law?
Stokes’ law states that the viscous drag force on a small sphere moving slowly through a viscous fluid is proportional to the coefficient of viscosity, the radius of the sphere and its velocity.
Conclusion
Terminal velocity is an important concept in fluid mechanics. It explains why falling objects eventually move with constant velocity when the resistive forces balance their effective weight.
The study of terminal velocity using Stokes’ law helps us understand the motion of small spherical bodies through viscous liquids and provides a method for determining the viscosity of fluids. It also explains practical phenomena such as parachute motion, sedimentation and the settling of particles.
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