Applications of Terminal Velocity

The greatest velocity obtained by a body falling through a fluid is known as terminal velocity. When the total drag and buoyancy equal the downward gravitational force exerted on the item, terminal velocity is noticed. The object’s acceleration is zero since the net force exerted on it is zero.

Expression for terminal velocity

Let us consider a water droplet that is dropped from a height. In that condition, three forces are acting on the droplet, viscous force (upward), upthrust force(upward), and weight force(downwards). Viscous force is directly proportional to the velocity. Initially, the speed is zero due to which the viscous force is also zero.

The drop keeps falling, hence at a certain height, the drop will have a certain amount of velocity, because of that the viscous force is acting upward along with the upthrust force (upward), and weight force (downwards).

W ＞U+Fv

The weight force on the droplet that is so high causes the net force to act downwards but it’s less than that in the initial condition.

As the drop keeps falling, a condition occurs when the weight is equal to the upthrust force and the viscous force;

W =U+Fv

Where,

W is the weight force

U is the upthrust force

Fv is the viscous force

From the above equation, it is observed that the net force acting on the object is zero while the acceleration is zero and the droplet falls with constant velocity, which is known as terminal velocity – let’s say y.

Let’s consider a spherical droplet having a radius of r, the coefficient of viscosity of the fluid, the density of the solid, the density of the liquid, the volume of the fluid displaced. The mass of the droplet is equal to the product of the density of and volume. The expression for the terminal velocity will be found as;

W =U+Fv

mg=Vρg+6πηrvT

σVg=Vρg+6πηrvT

vT=σVg-νρg / 6πη

vT=Vg(σ-ρ)/6πηr              (V=4/3πr3.)

vT=4/3πr3g(σ-ρ)/6πηr 

vT=2r3g(σ-ρ)/9η

In comparison with other forces, the inertia of the fluid is insignificant for very slow fluid motion. Such flows are known as creeping or Stokes flows.

The equation of the creeping flow is given as;

∇p=μ∇2v

Where,

v is the vector field of the velocity

p is the pressure field of the fluid

μ is the viscosity of fluid

Example of terminal velocity

(Gravity and drag)

When an object is freely falling under the influence of darg with the weight, w, the equation is found as:

F=D-W

D=Cd ρV2A2

When the drag is equal to the weight, the net force will be zero.

D=Cd ρ V2A/ 2

W=Cd ρ V2A/ 2

V=√2W/Cd ρ A

Problems on terminal velocity

1. A guy stands 2000 metres above the ground. What would be his maximum speed?

Sol. The given data in the problem are:

V is the terminal velocity=?

g is the acceleration due to gravity=9.81 m/s2

h is the height above the ground =2000 m

V=√2gh

V=2 9.812000

V=197.98 m/sec

Hence the terminal velocity will be 197.98 m/sec

2 . If the body’s terminal velocity is 100 m/s, calculate its height.

Sol. The given data in the problem are:

V is the terminal velocity=100 m/sec

g is the acceleration due to gravity=9.81 m/s2

H is the height above the ground =?

The height will be found as;

h=v2/2g

h=(100)2/2×9.81

h=510.204 m

Hence, the height will be 510.24 m.

3. Consider a spherical entity moving through water. At any given moment, the body’s velocity is 2ms–1. What will be the drag force of the fluid on the body? Assume Stokes’ law is correct.

Sol: The given data in the problem are:

v is the velocity of the body=20 m/sec

F is the drag force =?

From the Stoke’s law the drag force is foundas;

F=6πηrv

F=63.140.001210–3 ×2

F=75.36 10–6 N

Hence, the drag force on the body will be 75.36 10–6 N.

Conclusion

Terminal velocity is the highest velocity attained by a body falling through a fluid. It equals the downward gravitational force exerted on the object when the combined drag and buoyancy match the downward gravitational force exerted on the item.