Terminal Velocity: Formula, Definition and Examples

Terminal Velocity Formula: To understand terminal velocity, let us imagine a situation. A heavy iron ball is thrown into a deep sea (assume infinite depth). The downward speed of the ball will start increasing. As predicted by Sir Isaac Newton, when an object falls, the force of gravity initially causes it to continuously speed up. As it gets faster and faster, the air drag force increases as per Stokes’s law until eventually, the air drag force is exactly equal to the force of gravity, and there is no net force acting on the object. If these two forces are exactly balanced, the object will no longer speed up or slow down but will continue falling at a constant velocity.

What is Terminal Velocity?

When the ball is thrown into the sea, It accelerates initially due to gravity. As the velocity increases, the retarding force also increases (by Stokes’s Law). Finally, the net force becomes zero when the viscous force plus buoyant force becomes equal to the force due to gravity, and so does the acceleration. Then, the sphere descends with a constant velocity. Thus in equilibrium, this constant velocity is called terminal velocity. Terminal velocity is the maximum velocity of a body moving through a viscous fluid.

Stokes’s Law

This law gives an expression for the viscous force experienced by a body (a spherical) moving through a fluid. This expression was given by Sir George G. Stokes. When a body falls through a fluid, as shown in Fig. (a), it drags the layer of the fluid in contact with it, and the body experiences a retarding force when there is a relative motion between the different layers of the fluid is set. The force experienced by falling raindrops and swinging pendulum bob are some common examples of such motion. We observe that the viscous force is proportional to the velocity of the object, and it is opposite to the direction of motion. When a spherical object is dropped in a fluid, it is seen that the viscous force \(\left( {{F_v}{\rm{ or }}\;{F_d}} \right)\) experienced by the object is proportional to the

1. Velocity \(\left( v \right)\) of the object  passing through the fluid
2. Viscosity \(\left( \eta \right)\) of the fluid and
3. Radius \(\left( r \right)\) of the sphere.

Here \(6\pi \) is the constant of proportionality. Thus we can write 

\(F = 6\pi \rho rv\,\, \ldots {\rm{(1) }}\)

This is known as Stokes’s law. It is an interesting example of retarding force which is proportional to velocity.

Terminal Velocity Formula (v_t)

Consider a small sphere shown in Fig. (b) is falling from rest through a large column of viscous fluid. The forces acting on the sphere are:

(i) Weight \(\left( w \right)\) of the sphere which is acting vertically downwards.
(ii) Upthrust or buoyancy force \(\left( {{F_t}} \right)\) acting vertically upwards.
(iii) Viscous force \(\left( {{F_v}} \right)\)  acting vertically upwards, i.e. in a direction opposite to the velocity of the sphere.

Initially, \({F_v} = 0\) (as \(v = 0\))                    

And \(w > {F_t}\)

So, initially, ball starts accelerating, and the speed starts increasing. This also causes an increase in drag force. After some time, equilibrium is attended.

Thus in equilibrium, when the viscous force plus buoyant force becomes equal to force due to gravity, the net force becomes zero, so we can write:

\(w = {F_t} + {F_v}\,\,…….(i)\)

After this net force on the sphere is zero and it moves downwards with a constant velocity called terminal velocity \(\left( {{v_t}} \right)\).

Substituting proper values in Eq. \((i)\) we have,

\(\frac{4}{3}\pi {r^3}\rho g = \frac{4}{3}\pi {r^3}\sigma g + 6\pi \eta r{v_t} \ldots \ldots (ii)\)

Where,

 \(r = \) density of sphere,

\(\rho = \) density of falling body,

\(\sigma = \) density of fluid and,

\(\eta = \) coefficient of viscosity of fluid.

Now, from equation \((ii)\), we get

\({v_t} = \frac{2}{9}\frac{{{r^2}(\rho – \sigma )g}}{\eta }\, \ldots \ldots (iii)\)

So the terminal velocity \(\left( {{v_t}} \right)\) is directly proportional to the square of the radius of the sphere. That is why bigger raindrops fall with a greater velocity compared to the smaller raindrops and hurt more. It is inversely proportional to the viscosity of the medium that means the more viscous the fluid, the smaller the terminal velocity attained by a body.

The graph in Fig. (c), shows how the velocity of a falling object in a viscous fluid first increases and then becomes constant. The constant velocity is called terminal velocity.

Terminal Velocity Formula– Sample Problems

Q.1. Assume that a spherical body is flowing through the water. The velocity of the body at a particular instant is \(2\;{\rm{m}}\;{{\rm{s}}^{ – 1}}\). What will be the drag force on the body due to the fluid? Assume that Stokes’s law is valid. (Given: viscosity of water \( = 0.001\;{\rm{kg}}\;{{\rm{m}}^{ – 1}}\;{{\rm{s}}^{ – 1}}\), the radius of spherical object \( = 2\;{\rm{mm}}\))
Ans: According to Stokes’s law, we know that the drag force on the body will be 
\(F = 6\pi \eta rv\)

Now, substituting all the values in this expression we have,

\( \Rightarrow F = 6 \times 3.14 \times 0.001 \times 2 \times {10^{ – 3}} \times 2\)

\( \Rightarrow F = 75.36 \times {10^{ – 6}}{\rm{N}}\).

Q 2. What will be the terminal velocity of an air bubble \(0.8\,{\rm{mm}}\) in diameter rise in a liquid of viscosity \(0.15\;{\rm{N}} – {\rm{s}}\,{{\rm{m}}^{ – 2}}\) and specific gravity \(0.9\)? If the density of air is given as \(1.293\,{\rm{kg}}\;{{\rm{m}}^{ – 3}}\)  and density of the fluid is \(0.9 \times {10^3}\;{\rm{kg}}\;{{\rm{m}}^{ – 3}}\) and \({\rm{g}} = 9.8\;{\rm{m}}\;{{\rm{s}}^{ – 2}}\).
Ans: The terminal velocity of the bubble is given by Stokes’s law as:

\({v_t} = \frac{{2{r^2}(\rho – \sigma )g}}{9}\)

Here,   

\(r = 0.4 \times {10^{ – 3}}\;{\rm{m}},\,\sigma = 0.9 \times {10^3}\;{\rm{kg}}\;{{\rm{m}}^{ – 3}},\,\rho = 1.293\;{\rm{kg}}\;{{\rm{m}}^{ – 3}},\,\eta = 0.15\;{\rm{N}} – {\rm{s}}\,{{\rm{m}}^{ – 2}}\) and \(g = 9.8\;{\rm{m}}\;{{\rm{s}}^{ – 2}}\)

Substituting the values, we have

\( \Rightarrow {v_t} = \frac{2}{9} \times \frac{{{{\left( {0.4 \times {{10}^{ – 3}}} \right)}^2}\left( {1.293 – 0.9 \times {{10}^3}} \right) \times 9.8}}{{0.15}}\)

\( \Rightarrow {v_t} = – 0.0021\;{\rm{m}}\;{{\rm{s}}^{ – 1}}\)

\( \Rightarrow {v_t} = – 0.21\;{\rm{cm}}\;{{\rm{s}}^{ – 1}}\)

Here, the negative sign implies that the bubble will rise.

Summary

The property of a fluid by which an internal frictional force acts between its different layers which oppose their relative motion is called viscosity. The ratio of the shear stress to the time rate of shearing strain is known as the coefficient of viscosity \(\left( \eta \right)\). The coefficient of viscosity \(\left( \eta \right)\) falls when the temperature rises and vice versa. In a gas, the temperature rise increases the random motion of atoms and viscosity increases. When an object falls under the influence of gravity or is subjected to some other constant driving force is subject to a resistance or drag force that increases with velocity.

The viscous force plus buoyant force becomes equal to the force due to gravity, and so does the acceleration. And at the equilibrium, the object moves with a constant velocity, and this constant velocity is called terminal velocity, and when the viscous force plus buoyant force becomes equal to the force due to gravity, and so does the acceleration. Then, the sphere descends with a constant velocity. Thus in equilibrium, this constant velocity is called terminal velocity.

FAQs About Terminal Velocity

Let’s look at some of the commonly asked questions about Terminal Velocity:

Q.1. What is terminal velocity?
Ans: Terminal velocity is defined as the highest velocity which can be attained by an object during its falling through the air. It happens when the sum of the dragged force (\(\left( {{F_d}} \right)\) and the buoyancy is equal to the downward force of gravity \(\left( {{F_g}} \right)\) acting on the body and the net force acting on the object is zero.

Q.2. Does terminal velocity exist in a vacuum?
Ans: In a vacuum, since there is no drag force, the terminal velocity does not exist.

Q.3. What does Terminal Velocity depend on?
Ans: Terminal velocity is the point at which the drag force equals the force of gravity. Hence, terminal velocity will depend on the mass, cross-sectional area, and drag coefficient of the object, as well as the density of the fluid through which the object is falling and gravitational acceleration.

Q.4. Why a flat piece of paper will fall more slowly as compared to the same paper after it has been crumpled into a ball?
Ans: The paper weighs the same, but the air drag forces on the ball have decreased because its surface area has decreased. Due to this, the crumpled paper will have a higher terminal velocity than the flat paper.

Q.4. Give an application of the terminal velocity in our life?
Ans: Parachutes and hang gliders. As it is possible to increase or decrease the terminal velocity by making some changes in your weight or shape, or altitude. Thus if the shape of the object changes (e.g. By opening a parachute), then by opening a parachute, you have a lot of surface area for your mass, and thus terminal velocity is low enough that you can’t get seriously injured hitting a flat surface no matter how far you fall.

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