Terminal Speed

claimed by cshimkus3

Main Idea

Mathematical Model

In Projectile Motion, we considered the motion of an object under the influence of a constant gravitational force [math]\displaystyle{ g = 9.8 m/s^2 }[/math], an approximation which is essentially accurate on the Earth's surface (the more accurate understanding, provided by Newton, may be found in Gravitational Force). However, we did not take into account a very important factor, because in everyday life, objects in motion do not stay in motion: this is because they are feeling the effects of air resistance (and, if they are in contact with other objects, friction). If, on the other hand, and object falls in a vacuum, it can be observed to have the same acceleration regardless of its mass or size. In the following video, a feather and a ball bearing are dropped inside a vacuum, and one may observe that they fall at exactly the same rate. (Click on the picture)

The friction due to air is a non-constant force on a falling object, and is determined by a number of factors. The force of air on a falling object is defined as follows:

[math]\displaystyle{ \vec{F}_{drag} = -\frac{1}{2} C \rho A v^2 \hat{v} }[/math]

Where [math]\displaystyle{ C }[/math] is the numerical drag coefficient which is dependent on the shape of the object, [math]\displaystyle{ \rho }[/math] is the density of air, [math]\displaystyle{ A }[/math] is the cross sectional area of the object, and [math]\displaystyle{ v }[/math] is its velocity. The vectors and sign enforce the correct direction of the force. Naturally, these coefficients vary greatly between a feather and a ball, and this is why in normal experience they fall at a substantially different rate:

One of the important features of this equation is the dependence upon the velocity of the object: since the force of air depends on the square of the velocity, while the force of gravity remains constant, the force of air resistance will increase until the two forces cancel each other out and the net force is zero. At this point, the acceleration of an object is zero and the object has reached terminal velocity. An object's terminal velocity is defined by:

[math]\displaystyle{ v_t = \sqrt{\frac{2mg}{C\rho A}} }[/math]

Computational Modelling

Using the same computational techniques discussed in Fundamentals of Iterative Prediction with Varying Force and utilized in Simple Harmonic Motion and Iterative Prediction of Spring-Mass System, it is simply necessary to define and input the newly defined force of air resistance. This is done so here using numpy, and may also be done with VPython.

Examples

Simple

Compute the drag due to air resistance (assume quadratic drag) for a spherical object (at low speeds and normal scales, a sphere has a coefficient of drag of approximately [math]\displaystyle{ C_D = 0.47 }[/math]), with a cross sectional area of [math]\displaystyle{ 1 \; m^2 }[/math], in air at STP ([math]\displaystyle{ \rho = 1.225 \; kg/m^3 }[/math]), travelling at [math]\displaystyle{ 10 \; m/s }[/math].

Solution

We plug this into the standard equation for quadratic drag:

[math]\displaystyle{ \vec{F}_D = -\frac{1}{2} C_D A \rho v^2 \hat{v} }[/math]

When we substitute the values, we find that

[math]\displaystyle{ \vec{F}_D = -\frac{1}{2} \cdot(0.47)\cdot(1 \; m^2) \cdot (1.225 \; kg/m^3) \cdot (10 \; m/s)^2 \hat{v} = 28.7875 \; N }[/math]

From the vectors, this force is directed opposite the direction of motion.

Middling

Assume the object described above has a mass of [math]\displaystyle{ 4 \; kg }[/math]: what is the terminal velocity? If it instead had a mass of [math]\displaystyle{ 16 \; kg }[/math], what would the new terminal velocity be?

Solution

For clarity's sake, let's rederive the equation for terminal velocity by setting the force of gravity equal to the force of drag:

[math]\displaystyle{ |F_D| = |F_G| \Rightarrow 1/2 C_D A \rho v^2 = m g }[/math]

Solving this for [math]\displaystyle{ v }[/math] gives

[math]\displaystyle{ v = \sqrt{\frac{2 m g}{C_D A \rho}} }[/math]

Here we have discarded the plus or minus sign, and simply choose the appropriate sign at the end based on the vectors. Now, let's plug in our values:

[math]\displaystyle{ v = \sqrt{\frac{(2)(4\;kg)(9.8\;m/s^2)}{(0.47)(1)(1.225)}} = 11.67 \; m/s }[/math]

This then is the terminal velocity. We can repeat this computation for the new mass, but there is actually an easier way: the mass has increased by a factor of 4, and the equation for terminal velocity depends on the square root of the mass, so the terminal velocity will double, going to [math]\displaystyle{ 23.34 \; m/s }[/math]

Difficult

Using an iterative prediction program, find the distance traveled by the [math]\displaystyle{ 4 \; kg }[/math] object described above after [math]\displaystyle{ 30 \; s }[/math]. Calculate how far it would have gone if it traveled the entire distance at terminal velocity. What does this number tell us about how quickly it reached terminal velocity? Check a graph of velocity versus time to estimate what that time was (it's also pretty useful for your intuition to look at it generally). Next, think about whether the discrepancy between the constant speed assumption and the actual case would have been larger or smaller for the [math]\displaystyle{ 16 \; kg }[/math] object, and explain your reasoning. Check the answer with an iterative prediction program.

Solution

First, using a program such as this, we may find that the object will travel about [math]\displaystyle{ 340 \;m }[/math] (your result may vary based on your time step width, and if you are using a different program which converges at a different rate; this computation was done with a time step width of [math]\displaystyle{ 0.001 \; s }[/math]). If we assume that the object traveled the entire distance at terminal velocity, then we would have, using the result from above, that it traveled [math]\displaystyle{ (11.67 \; m/s)(30 \; s) = 350 \; m }[/math]. This discrepancy is fairly small, so it indicates that the object reached terminal velocity fairly quickly. When we check the graph, we see that it had reached terminal velocity within the first 4 seconds, so this logic checks out.

Now, let's think about what happens when the mass increases. We know that the terminal velocity will increase by a factor of two, so the object will have to have more acceleration to reach it. Since drag does not depend directly on mass, it will not be varied directly. However, the change in the velocity will affect the drag over time, so we don't expect the duration of the transient to fully double. Checking the simulation, we find that the object falls [math]\displaystyle{ 662 \; m }[/math] in [math]\displaystyle{ 30 \; s }[/math], whereas in the constant approximation we would expect it to fall [math]\displaystyle{ 700 \; m }[/math]. Thus the net discrepancy has quadrupled, and the relative discrepancy (percent error) has slightly less than doubled. Thinking about this physically, we are basically asking what happened if we increased the density of the object - we have changed its mass without changing surface area after all - and it makes sense physically that a feather reaches terminal velocity much faster - and therefore is much closer to the constant speed approximation - than a bowling ball.

Connectedness

How is this topic connected to something that you are interested in?

I have always been interested in skydiving and this topic is a vital concept for the spot. Just as in the example above, the way a skydiver positions their body determines how fast they will fall. Although it may seem easy to reposition your body mid-air, at such high velocities it takes very slow calculated movements so that you do not start plummeting out of control.

How is it connected to your major?

As a chemical engineer, I may encounter having to develop materials that will be used in free fall. If I am helping develop the metal coating for a spacecraft or vessel that will free fall to Earth's surface at high velocities, I will need to ensure that this material has an appropriate heat capacity to withstand the friction from the air.

Is there an interesting industrial application?

Although this is not quite an industrial application, the concept of terminal velocity is very important for the military when they perform air drops. In a war zone, supplies may need to be delivered to civilians and military forces behind enemy lines. To do so, very careful calculations must be made based on the speed of the aircraft making the delivery, the sturdiness of the material the containers are made of, and when the parachutes should be deployed. If anything with the calculations is off, the packages may end up miles away from the target zone, or may be shot down if the parachute is deployed too early.

History

The study of free fall began many centuries ago by the experiments of Aristotle and Galileo. They both monitored the motion of falling objects and Aristotle noted that if a feather and a hammer are dropped from the same height, the hammer will reach the ground first. Galileo, however, noted that if the hammer and feather had similar shapes, they would reach the ground at approximately the same time. He hypothesized that in the absence of air resistance, all objects falling from the same height will reach the ground at the same time. Many centuries later, when it became possible for his theory to be tested in a vacuum, he was proved to be correct. At this point, physicists knew that it was air resistance and not mass that was the factor affecting the speed of a free fall. Further experiments were performed to determined the relationship between an objects shape and size and its terminal velocity. It was found that objects with a larger surface area perpendicular to free fall will be exposed to more air force than an object with a smaller surface areas. From these experiments and relationships, the equation for force due to air and terminal velocity were developed.

Terminal Speed

Contents

Main Idea

Mathematical Model

Computational Modelling

Examples

Simple

Middling

Difficult

Connectedness

History

See also

Further reading

External links

References

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