Air Resistance

What is Air Resistance?

Thus far, in most physics problems involving motion, it's explicitly stated that air resistance is negligible. In real life that is not the case as all objects experience some level of resistance as they pass through the air.

Drag and air resistance are also types of friction but the word is usually used to refer to how an object is slowed down when it moves against a rough surface or how rough surfaces moving against each other will slow down. These drag forces cause the object to move more slowly by acting in the direction of the incoming flow and are proportional to the velocity. It is a type of non-conservative force since it makes the energy dissipate.

Frictional forces between surfaces occur because they are not perfectly smooth. If you were to look at them on a microscopic scale you would see lots of little bumps and an uneven surface. When surfaces slide across each other, they become stuck a little bit due to not being completely flat and a force is required to push them past one another. As the surfaces are forced to move, they may become damaged a little.

This line of reasoning also applies too when objects move through fluids (gases and liquids). As mentioned above, the type of friction that acts when an object moves through a fluid is called drag. For example, to swim through water, you have to push the water out of the way and as you move forwards, it will move against your body causing a drag force, which results in you slowing down.

Air resistance is the name given to the drag acting on something when it is moving through the air. It has a much weaker effect than the drag experienced in water as air is a lot less dense than water so it contains much fewer particles per unit volume and is, therefore, easier to push aside. Planes experience air resistance when flying but this can be used to their advantage as they can be shaped so that the air around them is distorted in a way that lifts them up, as shown in the diagram above.

Let's say we have a ball with mass \(m\). We drop it and as it falls, it's going to experience a resistive force. The resistive force mathematically is equal to

$$ \vec{F}_{\mathrm{r}} = - k \vec{v} $$

where \(k\) is a positive constant, and \(v\) is the velocity of the object relative to the medium. The negative sign indicates that the resistive force is in the opposite direction to the velocity.

Air Resistance Coefficient

As discussed earlier, \(k\) is a constant of proportionality. Its value is determined by the properties of the medium and the unique characteristics of the object. The main contributing factors are the density of the medium, the surface area of the object, and a dimensionless quantity known as the drag coefficient. In a real-life example involving a skydiver, the medium would be the air and the surface area would refer to either the skydiver or the parachute.

Now we can explain the effectiveness of a parachute when it comes to slowing down a skydiver. As the surface area \(A\) of the object falling increases,

$$ A_{\mathrm{skydiver}} \ll A_{\mathrm{parachute}},$$

\(k\) increases, so the magnitude of the resistive force increases as well, therefore slowing the object down.

Let's look at a free-body diagram to understand its motion better.

Air Resistance Free Body Diagram

What happens to an object as it gets dropped and is falling down? It experiences a downward force in the form of weight and a resistive force in the opposite direction of the motion due to air resistance, both of which are visualized in the free-body diagram visible below.

Fig. 1 - As the object falls, the resistive force acts upward on it, meanwhile the weight pulls it downward.

According to Newton's second law, the net force acting on an object \(\vec{F}_{\mathrm{net}}\) is equal to the mass \(m\) of the object times its acceleration \(\vec{a}\). So knowing all that, we can obtain the following expression

$$ m\vec{g} - k\vec{v} = m\vec{a}.$$

When we start the motion at \(t=0\), its initial velocity is \(\vec{v}_0=0\), therefore, the initial air resistance force is also zero. As time passes and the object starts moving, eventually it will reach a constant velocity, which is called terminal velocity \(\vec{v}_\mathrm{T}\). Because the velocity is constant, the acceleration will be zero. The right-hand side of the expression becomes zero, and we can rearrange the remaining terms

$$ m\vec{g} = k\vec{v}_\mathrm{T} $$

to find the equation for terminal velocity

$$ \vec{v}_\mathrm{T}= \frac{m\vec{g}}{k}. $$

Terminal velocity is reached when there is no net force applied to the object, meaning that the acceleration is zero. Let's look at an example problem involving terminal velocity.

Air Resistance Formula

Let's now find the velocity as a function of time. To achieve that, we need to convert Newton's second law into a differential equation. Acceleration is the first derivative of velocity, so \(\vec{a}=\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}\). Then we can write

$$ m\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}=m\vec{g}-k\vec{v}. $$

Let's separate our variables:

$$ \frac{\mathrm{d}v}{mg- kv}=\frac{\mathrm{d}t}{m}.$$

Here, it is important to set the integration limits. The time goes from zero to time \(t_{\mathrm{f}}\). When time is equal to zero, our initial velocity is zero as well, and as time goes to \(t_{\mathrm{f}}\), our velocity becomes velocity \(v_{\mathrm{f}}\).

$$\int_{0}^{v_\mathrm{f}} \frac{\mathrm{d}v}{mg-kv} = \int_{0}^{t_{\mathrm{f}}} \frac{\mathrm{d}t}{m}$$

If we take the antiderivative, we will obtain a natural logarithm

$$\left.\frac{\ln(mg-kv)}{-k}\right|_0^{v_\mathrm{f}} = \left.\frac{t}{m}\right|_0^{t_\mathrm{f}}$$

Now let's apply the limits

$$ \begin{align} \frac{\ln(mg-kv_{\mathrm{f}})}{-k} - \frac{\ln(mg)}{-k} & = \frac{t_{\mathrm{f}}}{m}, \\ \ln \left ( \frac{mg-kv_{\mathrm{f}}}{mg} \right ) & = \frac{-kt_{\mathrm{f}}}{m}. \end{align} $$

Finally, get rid of the natural logarithm:

$$ \begin{align} \mathrm{e}^{\ln \left ( \frac{mg- kv_{\mathrm{f}}}{mg} \right )} &= \mathrm{e}^{\frac{-kt_{\mathrm{f}}}{m}} \\ \frac{mg-kv_{\mathrm{f}}}{mg} &= \mathrm{e}^{\frac{-kt_{\mathrm{f}}}{m}} \\ 1- \frac{kv_{\mathrm{f}}}{mg}&= \mathrm{e}^{\frac{-kt_{\mathrm{f}}}{m}} \\ \frac{kv_{\mathrm{f}}}{mg} & = 1- \mathrm{e}^{\frac{-kt_{\mathrm{f}}}{m}} \\ v_{\mathrm{f}} &= \frac{mg}{k} \left ( 1-\mathrm{e}^{\frac{-kt_{\mathrm{f}}}{m}} \right ). \end{align} $$

The final version of the equation including all the vector values is as follows

$$ \vec{v_{\mathrm{f}}}=\vec{v}_\mathrm{T} \, (1-\mathrm{e}^{-\frac{t_{\mathrm{f}}}{T}}) $$

where \(T\) is the time constant and equal to \(\frac{m}{k}\).

And that's how we derive the velocity expression as a time function! The final equation confirms our previous conclusions about the terminal velocity. If the value of \(t_{\mathrm{f}}\) is set to zero, \(\vec{v_{\mathrm{f}}}\) also will be zero, meanwhile if \(t_{\mathrm{f}}\) is set to something huge, let's say infinity, we will be left with \(\vec{v_{\mathrm{f}}} = \vec{v_\mathrm{T}}\).

What would happen though if the initial velocity was not zero?

Let's say we have a car with an initial velocity \(\vec{v}_0\) against some resistive force \(\vec{F}_\mathrm{r}\) that is again equal to \(-k\vec{v}\). When we draw a free-body diagram of the car, the weight is downward, the normal force is upward, and the air resistance force is in the opposite direction of the motion.

In this case, the final velocity will be zero, and the car will stop. The only force acting on the object in the direction of the motion is the resistive force, so it will be our net force. Then we can write

$$ m\vec{a} = -k\vec{v}.$$

We are going to repeat the same procedure as previously since this becomes a differential equation when we write acceleration as \(\vec{a}=\frac{\mathrm{d}\vec{v}}{\mathrm{d}t}\) and obtain

$$ \begin{align} m \frac{\mathrm{d}\vec{v}}{\mathrm{d}t} & = - k\vec{v} \\ \frac{\mathrm{d}v}{v} & =\frac{-k}{m} \mathrm{d}t. \end{align}$$

Once again, for the calculations, we'll consider the scalar version of the equation. Here we have to take integrals of both sides, but first, we need to decide on the limits. Time once again goes from zero to \(t\). However, now we have an initial velocity, so our velocity limit is from \(v_0\) to \(v\)

$$\int_{v_0}^{v_{\mathrm{f}}} \frac{\mathrm{d}v}{v} = \int_{0}^{t_{\mathrm{f}}} \frac{-k}{m} \mathrm{d}t. $$

Again, take the derivative to have a natural logarithm, apply the limits and obtain the following expression

$$ \ln \left ( \frac{v_{\mathrm{f}}}{v_0} \right ) = \frac {-kt_{\mathrm{f}}}{m}.$$

We can rewrite this as:

$$ \begin{align} \mathrm{e}^{\ln \left (\frac{v_{\mathrm{f}}}{v_0} \right )} & = \mathrm{e}^{\frac{-kt_{\mathrm{f}}}{m}} \\ \frac{v_{\mathrm{f}}}{v_0} & =\mathrm{e}^{\frac{-kt_{\mathrm{f}}}{m}} \end{align}$$

where the final expression including all the vector quantities becomes

$$ \vec{v_{\mathrm{f}}} = \vec{v}_0 \mathrm{e}^{\frac{-kt_{\mathrm{f}}}{m}}.$$

Air Resistance Example

Let's look at an example problem involving the same skydiver mentioned earlier, to check our knowledge!

Some other examples of air resistance in our everyday lives would be

1. Walking in a storm makes walking challenging quite frequently. A significant amount of resistance is experienced by the individual walking against the wind, making it difficult to walk forward. The same reason makes it challenging to hold an umbrella in your hand when there is a strong wind present.

2. A feather falling to the ground has a tendency to float and move slowly, rather than fall within seconds like other objects, of slightly bigger mass. The gravitational force pulls the feather towards the earth; however, the air resistance force prevents the feather from falling or moving while in motion.

3. Paper planes, if built correctly, fly effortlessly in the air. To accomplish this, the front surface of the paper plane is sharpened. As a result, the paper plane cuts through the air and escapes the air resistance force just enough to keep it in the air for longer.

4. A real airplane's engine, wings, and propellers are all built to provide enough thrust to help the plane overcome the force of air resistance. Turbulence is also caused by the friction that the air creates. Spacecrafts, however, only have to worry about air resistance during launching and landing, as there is no air in space.

Friction and Air Resistance

Remember that air resistance is a type of friction that happens in air, and drag is a type of friction that happens in liquids.

Friction and Air Resistance Similarities

Although friction between solid surfaces and air resistance seem very different, they are very similar and can be related to each other in many ways:

• Friction between solid surfaces and air resistance both oppose the motion.
• They both cause objects to lose energy - hence slowing them down.
• They both cause heat to be produced - the objects lose energy when they release thermal energy.
• Both air resistance and friction act all of the time. There are some situations where their effects are so small that they can be neglected but there is always at least some resistive force acting on moving objects.

Friction and Air Resistance Differences

• Air resistance acts when an object moves through air (drag is the more general term for the resistive force acting on an object moving through a fluid) and the process usually referred to as 'friction' occurs between solids (although air resistance is also a type of friction).

• Air resistance often depends on the speed of the object, the relationship between the force and the velocity can change in different situations depending on other factors. Friction between solid surfaces does not depend on the relative speed of the surfaces.
• Air resistance increases as the cross-sectional area perpendicular to the direction of motion increases. The area does not affect friction between solids.
• Friction between an object and a surface depends on the weight of the object.