Physics Book - User contributions [en]

Real Systems

2022-04-09T22:37:02Z

Rcooper47 1: /* The Main Idea */

This page describes real systems and how they can be used to model certain aspects of a system's motion.

==The Main Idea==

A real system is a system in which every part of the system is modeled separately, allowing for the internal behaviors of the system to be analyzed in addition to the system's motion through its environment.

In [[Point Particle Systems]], forces are analyzed as though they act directly on the system's center of mass, and only the translational kinetic energy of a system can change. Through week 9, we have modelled most bodies as point particles, as we have been primarily interested in their translational motion. However, we may now encounter problems where we want to analyze the internal behaviors of systems as well. For example, we may want to calculate a change in a system's [[Translational, Rotational and Vibrational Energy]], [[Potential Energy]], or [[Thermal Energy]]. In a real system, the point of application of each force must be considered. Furthermore, when calculating the work done by a force, the distance over which the force is applied is not the distance traveled by the system's center of mass, but rather the distance traveled by the force's point of contact. These two key differences lead to a mathematical model that can be more complicated than that of a point particle system, but yields insights about the internal behavior of the system.

===A Mathematical Model===

The mathematical concepts used to analyze point particle systems depend on the system. Often, the [[Work/Energy|work-energy theorem]] is used. This section explains how to use the work-energy theorem for a point particle system because this is the concept that varies the most significantly from its application to point particle systems.

The work done on a real system should be calculated separately for each force acting on it because each force might be exerted over a different distance due to the geometry of the system. The work done on a real system by each force is defined as

<math>W = \int \vec{f} \cdot d \vec{r}</math>,

where <math>\vec{r}</math> is the position of the point on the system on which the force acts.

Let us assume that the force acting on the system is constant, so that we can get rid of the integral, and that the force acts in the direction of the system's motion, so that we can replace the dot product with regular multiplication. This will be the case for most real system problems involving work. With these assumptions, the work done on a part of a real system by a force is given by

<math> W = f * d </math>

where <math>F_{net}</math> is the magnitude of the net force acting on the particle and <math>d</math> is the distance over which the force is exerted. For real systems, that is the distance traveled by the point of contact of the force, which may not be the same as the distance traveled by the system's center of mass, depending on the movement of the systems' different parts. This contrasts with point particle systems, where the distance over which the force is exerted is simply the displacement of the system's center of mass.

The work-energy theorem states that work done on a system increases that system's energy. For real systems, this is the system's total energy. Since each force's work must be found separately in a real system, remember to sum the values of the work done by the different forces acting on the system.

In other words, for a real system,

<math> \Delta E = \sum_i f_i * d_i </math>.

The equation above is the basis for answering work/energy questions using real systems. It is important to remember that when modeling a system as a real system, calculating the work done on the system as described above finds all of the work done on the system- both translational kinetic energy and internal types of energy.

===A Computational Model===

Most computer simulations treat systems as real systems, modeling each part of a system separately. This is because [[Iterative Prediction]] makes keeping track of each part of the system separately easy even when it might be difficult to do so analytically. The last example on the [[Point Particle Systems]] page describes a system in which two masses are connected by a spring and one of them is pulled by a force. Consider the following simulation of the system:

[https://www.glowscript.org/#/user/YorickAndeweg/folder/PhysicsBookFolder/program/TwoMassSystem Two mass system simulation]

If you click "view source" on the top left corner, you will see that the process of iterative prediction is performed separately for each of the two masses- they each have their own position vectors, and the forces acting on each are calculated. There is no need to treat them as a single particle and thereby lose information about their internal behavior- in this case, their oscillations.

==Examples==

Some of the examples on the [[Point Particle Systems]] page can also be analyzed as real systems. Modeling each system as a real system makes the math more complicated but allows us to calculate the internal energies of the systems. Let us look at each example from the point particle systems page. (If you haven't already looked at them on the point particle systems page, do that first.)

===1.===

A 60kg person jumps straight up in the air from a crouching position. From the time the person begins to push off of the ground to the time their feet leave the ground, their center of mass moves up 2m, and the normal force between the ground and the person's feet has a constant magnitude of twice the person's weight. Find the velocity of the jumper at the moment their feet leave the ground. Use 10m/s2 for g.

In this problem, the internal energy of the person does not change, so modeling the jumper as a real system offers few advantages. In fact, it makes it difficult to take an energy approach to answer the question, since when the person is modeled as a real system, the point of contact of the normal force does not move and so does not do work. Instead, forces in the person's legs do work, but the amount of work they do is difficult to calculate since different parts of the legs move different amounts. This problem is best analyzed as a point particle system.

===2.===

A giant 20kg yo-yo floats at rest in space. Its string (whose mass is negligible compared to the mass of the yo-yo) is pulled with a constant force of 8N. What is the speed of the yo-yo when it has travelled 5m?

[[File:Pointparticlesystemsyoyo.png]]

This problem can be analyzed as a real system, although it requires knowledge of rotational physics and is more difficult than treating it like a point particle.

The work done on the yo-yo is no longer simply 40Nm because the point of the force's application travels faster than the yo-yo's center of mass due to the yo-yo's rotation. It makes sense that the work done on the real yo-yo is greater than the work we found when treating it as a point particle because this work includes the increase in its rotational kinetic energy as well as the increase in its translational kinetic energy Let us find the distance over which the force is exerted in the real system.

Because the force applies a constant torque, the yo-yo undergoes constant angular acceleration <math>\alpha</math>. Let us say it takes <math>t_1</math> seconds for the yo-yo to travel the 5m. The average angular velocity of the yo-yo over that time period is

<math>\omega_{avg} = \frac{1}{2} \alpha t_1</math>.

This means that the distance traveled by the point of contact between the yo-yo and the string due to the yo-yo's rotation during the <math>t_1</math>-second interval is given by

<math>d_{rot} = \frac{1}{2} \alpha {t_1}^2 r</math>,

where r is the radius of the yo-yo (the radius of the string wrapping around it, not the radius of its outer rim).

The total work done on the yo-yo is given by

<math>W = f * (d_{trans} + d_{rot})</math> (<math>d_{trans}</math> is the 5m traveled by the yo-yo's center of mass and <math>d_{rot}</math>) is the distance traveled by the edge of the yo-yo due to its rotation, found above.)

<math>W = f * (d_{trans} + \frac{1}{2} \alpha {t_1}^2 r)</math>.

Let us manipulate this expression. Let us start by distributing out the force at the beginning:

<math>W = f * d_{trans} + f * \frac{1}{2} \alpha {t_1}^2 r</math>.

The second term has both f and r as factors. Since the product of these two is the torque exerted by the force, we can replace <math>f*r</math> with <math>T</math> for torque:

<math>W = f * d_{trans} + \frac{1}{2} \alpha {t_1}^2 T</math>.

Next, since <math>T = I\alpha</math> for fixed-plane rotations, we can replace <math>T</math> with <math>I\alpha</math>, where <math>I</math> is the moment of inertia of the yo-yo:

<math>W = f * d_{trans} + \frac{1}{2} \alpha^2 {t_1}^2 I</math>.

Finally, we can replace the product <math>\alpha * t_1</math> with <math>\omega_1</math>, the angular velocity of the yo-yo when it reaches the 5m mark:

<math>W = f * d_{trans} + \frac{1}{2} I {\omega_1}^2</math>.

This expression is the amount of work done on the yo-yo, and since its initial energy of the yo-yo is 0, this expression also gives the final total energy of the yo-yo. Some of it is translational kinetic energy, and some of it is rotational kinetic energy. We recognize the expression <math>\frac{1}{2} I {\omega_1}^2</math> as the final rotational kinetic energy of the yo-yo, so that leaves <math>f * d_{trans}</math> as the final translational kinetic energy of the yo-yo. This is 8 * 5 = 40 Joules, which is the same value for the final translational kinetic energy of the yo-yo that we got by analyzing the yo-yo as a point particle on the point particle systems page. The yo-yo's translational speed can be easily calculated from this value.

As you can see, we were able to save a lot of work by treating the yo-yo as a point particle system.

===3.===

A 50kg metal sphere is suspended in the inside of a large cubic box of negligible mass by six rubber bands- one attaching the sphere to each face of the box. The box and the sphere are initially at rest. A 200N rightward force is applied to the side of the box, causing it to accelerate. At time <math>t_1</math>, the box has been displaced by 10m to the right. At this time, the sphere is no longer exactly in the center of the box; it is 1m to the left of the center of the box due to its inertia. What is the speed of the sphere at time <math>t_1</math>?

[[File:Pointparticlesystemspherebox.png]]

Analyzing this system as a real system allows us to see that 2000Nm of work was exerted on the system rather than the 1800nM found using a point particle system. This is because the point of the force's application is on the box, which traveled all 10m. The 1800J found using the point particle system is the translational kinetic energy of the sphere, and the remaining 200J was an increase in the potential energy of the rubber bands, which became stretched when the sphere was displaced by 1m. This energy may become vibrational as the light box jiggles around the sphere.

===4.===

A uniform disk of mass <math>M</math> starts at rest and rolls without slipping down a ramp. The plane of the ramp makes an angle of <Math>\Theta</Math> with the horizontal. The force of [[Static Friction]] acting on the bottom of such a disk, causing it to rotate, always has a magnitude of 1/3 the force acting to move the disk down the ramp. What is the translational speed of the disk after it has traveled a distance of <math>d</math> diagonally down the ramp?

[[File:Pointparticlesystemsramp.png]]

Using a point particle system, we calculated the work done by the net force acting on the particle. The net force had a magnitude of

<math>f_{net} = \frac{2}{3} Mg \sin \Theta</math>.

However, when treating the disk as a real system, the work done by the friction force and by the gravitational force need to be calculated separately because the forces act on different parts of the system and are exerted over different distances. The work done by the gravitational force is

<math>W = Mg \sin \Theta * d</math>,

since the gravitational force force acts on the center of the disk's mass, which travels a distance of d.

The work done by the friction force, however, is actually 0. This is because the friction force is applied to the point on the edge of the disk: the point that is in contact with the ramp. Since the disk rolls instead of slides and is affected by [[Static Friction]] with the ramp, this point is not moving relative to the ramp, and so the friction force is exerted over a distance of 0.

When modeling the disk as a real system, therefore, the final energy of the disk is

<math>E = Mg \sin \Theta * d</math>.

This is higher than the energy found when we treated the disk like a point particle system because this energy includes both translational kinetic energy and the rotational kinetic energy of the disk.

Let us find (without looking at our answer from the point particle analysis) how much of this energy is rotational and how much is translational. Let us say that the final translational speed of the disk is <math>v</math>.

In that case, the translational kinetic energy of the disk is given by

<math>KE_{trans} = \frac{1}{2}Mv^2</math>

and the rotational kinetic energy of the disk is given by

<math>KE_{rot} = \frac{1}{2}I \omega^2</math>

<math>KE_{rot} = \frac{1}{2} \frac{1}{2}MR^2 (\frac{v}{R})^2</math> where R is the radius of the disk.

<math>KE_{rot} = \frac{1}{4} M v^2</math>

<math>KE_{rot} = \frac{1}{2}KE_{trans}</math>, so 2/3 of the disk's kinetic energy is translational and 1/3 is rotational.

This means the translational kinetic energy of the disk is

<math>E = \frac{2}{3} Mg \sin \Theta * d</math>.

This is the same value for the final translational kinetic energy of the disk that we got by analyzing the disk as a point particle on the point particle systems page. The disk's translational speed can be easily calculated from this value.

===5.===

A pair of 10kg masses are connected by a massless spring of unknown spring constant. They begin at rest at the spring's equilibrium length. A 30N rightward force is applied to the mass on the right for 4s. At the end of the 4 second period, the force ceases to act and the mass on the right has traveled 20m. The system continues to travel to the right, and the distance between the two masses oscillates sinusoidally. How much vibrational kinetic energy does the system have? (Vibrational kinetic energy is the energy of the mass' oscillatory movement, as opposed to their translational kinetic energy, which is the energy of the rightward movement of their center of mass.)

[[File:Pointparticlesystemtwomass.png]]

[https://www.glowscript.org/#/user/YorickAndeweg/folder/PhysicsBookFolder/program/twomasssystem Here is a simulation of the system to help you visualize it.]

By modeling the system as both a real system and a point particle system (which was done on the point particle systems page), we are able to answer the question. Finding more detailed information about the oscillations, such as their frequency, amplitude, and phase, is much more difficult and requires an entirely real model. A difficult-to-solve system of differential equations would have to be devised, describing the accelerations of the masses in terms of their positions.

==See also==

*[[Point Particle Systems]]
*[[Work/Energy]]
*[[Translational, Rotational and Vibrational Energy]]

===External links===

A helpful page for additional info: http://p3server.pa.msu.edu/coursewiki/doku.php?id=183_notes:pp_vs_real

A helpful video lecture: https://www.youtube.com/watch?v=T780lL5FlLg&index=41&list=PL9HgJKLOnKxedh-yIp7FDzUTwZeTeoR-Y

==References==

Chabay, Ruth W., and Bruce A. Sherwood. "9." Matter & Interactions. N.p.: n.p., n.d. N. pag. Print.

[[Category:Energy]]

Air Resistance

2022-04-09T21:42:06Z

Rcooper47 1: /* Main Idea */

==Sidarth Rajan Fall 2020==
==Main Idea==
[[File:Air resistance.jpg|300px|right]]
Based on what we've learned about Newton’s Second Law and its affiliation to gravity and acceleration, we assume that an object in freefall will fall at a constant acceleration without any other forces acting upon it; however, this assumption is wrong. As an object moves through a medium(whether it be gas or liquid), forces that oppose the motion of the object come into play such as viscosity, drag, and air resistance; moreover, these principles form the basis of the field of physics centered around fluid dynamics, which examines this topic in great detail. Air Resistance is the force we see when we throw an object in the air and it is falling down, if we were to measure the acceleration at which an object is falling, we can see that the magnitude of the acceleration is decreasing due to a force acting in the opposite direction, known as air resistance. Air resistance is particularly important when examining objects in flight as to infer data on the motion of an object, we must properly be able to calculate the magnitude of the drag force in the form of air resistance.

===Formulation===
Air Resistance is built upon a relationship of a few variables all pertinent towards the motion of an object falling. These include
::::<math>\rho =</math> a measurement of the density of the medium
::::<math>v =</math> the velocity of the object
::::<math>A =</math> the cross-sectional area(looking for the area coming into contact with the air)
::::<math>C_{D} =</math> which is a non-dimensional constant that determines a relative drag depending on the shape of the object.

===Mathematical Model===
A typical drag force can be described by:

:::<math>F_{D} = \frac{1}{2}C_{D} A \rho v^2 </math>

====Cross Sectional Area====
We have encountered the term of 'cross-sectional area' in many cases: for example, when you need to calculate the flux of a liquid flowing through a pipe, you need the cross-sectional area to determine the volume of the liquid flowing through. However, the 'cross-sectional area' here is different. We'd better see it as the 'frontal area': the area facing the direction of the object's movement.

It is easy to understand: we can relate this idea to the choice of the system and moving objects. For example, when a plane is flying forward with a speed <math>v</math> in the <math>+x</math> direction and the air is static, only the surface of the plane pointing in the <math>+x</math> direction is subject to the air resistance since most air particles are colliding with that area with relatively high speed in the opposite direction.

Essentially, what we are describing buy the cross-sectional area is the portion of the surface area that is facing the direction of the object's motion, as this will be the area that has the most in contact with external mediums such as air. When we examine air resistance, we look at the area of the object that is facing the air. Conceptually speaking, we can see from the mathematical model that as the cross-sectional area increases the magnitude of the drag force also increases, this is due to the fact that there is more area of the object coming into contact with the specific medium.

=====Skydiver/Parachute Example=====
When we view skydivers, we can see that when they move downwards headfirst they tend to travel much faster than when they move stomach first. Why is this?

When a skydiver is traveling headfirst, the cross-sectional area coming into contact with the air is simply the area on the skull, which is substantially smaller than the surface area of the anterior side of your body. Therefore since less area is coming into contact with the air, the drag force is much less, which causes a greater acceleration as opposed to one with a greater surface area.

This also explains why parachutes open up with a wide surface area on the inside, as the entirety of the inside is coming into contact with the air, allowing for a large drag force, which causes the acceleration to fall greatly until one reaches terminal speed.

====Drag Coefficient====
The drag coefficient is a dimensionless number that engineers use to model all of the complex dependencies of shape and flow conditions on motion resistance (drag). It is usually determined by experiment, since the factors are very complex. The drag coefficient is a function of several parameters like shape of the body, Reynolds Number for the flow, Froude number, Mach Number and Roughness of the Surface. In short, different materials, substrates, and solvents have different drag coefficients.

===Computational Model===
:[[https://trinket.io/glowscript/f95f56408b]]

==Examples==

===Simple===
You are standing at the top of a <math>20</math> meter high building, and throw a ball with an initial speed of <math>v_{0} = 10 \frac{m}{s}</math> in the <math>+x</math> direction.

:'''a) If you neglect air resistance, where would you expect the ball to hit on the flat surface below?'''

::Without air resistance, the acceleration of the ball will be about a constant <math>g = 9.81 \ \frac{m}{s^2}</math>

::We can use the kinematic equations under constant acceleration:

:::<math>v_{f} = v_{0} + at \ \ \mathbf{\text{(1)}} \quad \And \quad \Delta x = v_{0}t + \frac{1}{2}at^2 \ \ \mathbf{\text{(2)}} \quad \And \quad {v_f}^2 = {v_0}^2 + 2a \Delta x \ \ \mathbf{\text{(3)}}</math>

::We will break this into two parts:

:::#''The y-direction (vertical)''
:::#''The x-direction (horizontal)''

::''The y-direction:''

:::From the problem statement, we are told the ball is thrown horizontally, so there is no initial velocity along the y-direction:

::::<math>v_{y_{0}} = 0</math>

:::Also, the only acceleration affecting the system, the acceleration due to gravity is along the negative y-direction:

::::<math>a_{y} = -g = -9.81</math>

:::Also, we are told the initial height:

::::<math>y_{0} = 20</math>

:::Using equation '''3''', we can find the final velocity along the y-direction of the ball:

::::<math>v_{y_{f}} = - \sqrt{{v_{y_{0}}}^2 + 2a_{y} \Delta y} = -\sqrt{0 + 2 \times (-9.81) \times (0 - 20)} = -\sqrt{392.4} = -19.81 \ \frac{m}{s}</math>

:::Now, we can plug this final velocity into '''1''' to find the time of flight for the ball:

::::<math>v_{y_{f}} = v_{y_{0}} + a_{y}t</math>

::::<math>t = \frac{\Delta v_{y}}{a_{y}} = \frac{-19.81}{-9.81} = 2.02 \ s</math>

:::Knowing the time of flight will allow us to solve for the distance traveled horizontally.

::''The x-direction:''

:::We know the initial velocity in this direction:

::::<math>v_{x_{0}} = 10</math>

:::Also, there is no acceleration along the x-direction:

::::<math>a_{x} = 0</math>

:::Using equation '''2''', we can find how far the ball goes horizontally:

::::<math>\Delta x = v_{x_{0}}t + \frac{1}{2}a_{x}t^2 = v_{x_{0}} t = 10 \times 2.02 = 20.2 \ m</math>

::Therefore, the ball will land a distance of 20.2 meters away horizontally and 20 meters below the roof of the building.

:'''b) Do you think your prediction without air resistance is too large or too small? Explain.'''

::If air resistance was taken into consideration, it would impede the motion of the ball, slowing it down. Thus, the previously predicted distance would be further than the ball would really go.

===Middling===
John is going sky diving for the first time. His mass is 70 kg and his terminal speed is 38 m/s.

:'''a) What is the magnitude of the drag force on John?'''

::Since John has reached terminal speed, he is not accelerating. Equivalently, this means his acceleration is <math>0 \ \frac{m}{s^2}</math>, his change in momentum is <math>0 \ \frac{kg \cdot m}{s}</math>, and the net force on him is <math>0 \ \text{N}</math>. Thus, we know the two forces acting on him, the gravitational force and the drag force, are balanced:

:::<math>F_{net} = F_{D} - F_{g} = 0</math>

:::<math>F_{D} = F_{g}</math>

::Since John is close to the surface of Earth, we will use <math>mg</math> as the gravitational force:

:::<math>F_{D} = mg = 70 \times 9.81 = 686.7 \ \text{N}</math>

::Therefore, the drag force on John is about <math>686.7</math> Newtons pointing away from the Earth.

:'''b) What is an expression for the value of the cross-sectional area of John?'''

::Using our drag force equation and setting it equal to the gravitational force, we see:

:::<math>F_{D} = \frac{1}{2}\rho v^2 A C_{D}</math>

:::<math>\frac{1}{2}\rho v^2 A C_{D} = mg</math>

::Solving this for <math>A</math> gives:

:::<math>A = \frac{2mg}{\rho v^2 C_{D}}</math>

::If we knew the density of the air and the drag coefficient, we could give a value for the cross-sectional are of John.

===Difficult===
Sarah is doing an air resistance experiment in class. The experiment requires Sarah to drop a coffee filter from a height of 2 meters. Let's say that the mass of the coffee filter was 2.0 grams, the radius of its base was 6 cm, and it reached the ground with a speed of 1.0 m/s.

:'''a) How much energy did the air gain as the coffee filter floated down?

::This is a conservation of energy problem. We can describe the coffee filter as the system and the air as the surroundings to a good approximation. Thus, by the Energy Principle:

:::<math>\Delta E_{c} + \Delta E_{a} = 0</math>

:::<math>\Delta E_{c} = - \Delta E_{a}</math>

::The initial energy of the coffee filter can be described as a combination of the initial Thermal, Gravitational Potential, and Kinetic energies:

:::<math>E_{c_{0}} = U_{c_{0}} + K_{c_{0}} + Q_{c_{0}}</math>

::The initial kinetic energy will be <math>0</math> since the coffee is dropped from rest:

:::<math>E_{c_{0}} = U_{c_{0}} + Q_{c_{0}}</math>

::The final energy of the coffee filter can be described as a combination of the final Thermal, Gravitational Potential, and Kinetic energies:

:::<math>E_{c_{f}} = U_{c_{f}} + K_{c_{f}} + Q_{c_{f}}</math>

::Subtracting the final energy from the initial energy will give us the change in energy of the coffee filter:

:::<math>\Delta E_{c} = E_{c_{f}} - E_{c_{0}} = (U_{c_{f}} + K_{c_{f}} + Q_{c_{f}}) - (U_{c_{0}} + Q_{c_{0}}) = (U_{c_{f}} - U_{c_{0}}) + K_{c_{f}} + (Q_{c_{f}} - Q_{c_{0}}) = \Delta U_{c} + K_{c_{f}} + \Delta Q_{c}</math>

::Using common expressions for Gravitational Potential energy and Kinetic energy, we see:

:::<math>\Delta E_{c} = mg(h_{f} - h_{0}) + \left(\frac{1}{2}\right)m{v_f}^2 + \Delta Q_{c}</math>

::We will assume the change in Thermal energy is negligible:

:::<math>\Delta E_{c} = mg(h_{f} - h_{0}) + \left(\frac{1}{2}\right)m{v_f}^2</math>

::We have all the needed values for this:

:::<math>\Delta E_{c} = 0.002 \times 9.81 \times (0 - 2) + \left(\frac{1}{2}\right) \times 0.002 \times (1.0)^2 = -0.03924 + 0.001 = -0.03824 \ J</math>

::Using:

:::<math>\Delta E_{c} = - \Delta E_{a}</math>

::We see:

:::<math>- \Delta E_{c} = \Delta E_{a}</math>

:::<math>\Delta E_{a} = -(-0.03824) = 0.03824 \ J</math>

::We see the coffee filter lost <math>0.03824 \ J</math> of energy, and the air in contact with the coffee filter gained it.

:'''b) What is the drag force at the end of the object's motion? Assume the drag constant is 2.3 and air has a density of 1.22 <math>\frac{kg}{m^3}</math>

::::<math>\rho =</math> 1.22
::::<math>v =</math> 1.0
::::<math>A = (0.06)^2 \pi </math>
::::<math>C_{D} =</math> 2.3

:::We now have everything we need to calculate the drag force

::::<math>F_{D} = \frac{1}{2}C_{D} A \rho v^2</math>
::::<math>F_{D} = \frac{1}{2}(2.3)((0.06)^2 \pi)(1.22)(1)^2</math>
::::<math>F_{D} = \frac{1}{2}(2.3)(0.01131)(1.22)(1)^2</math>

:::After calculating everything we can see that

::::<math>F_{D} = 0.01587</math> N

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

*Air resistance plays a big factor in sports, and physical activity. We can see that during sports such as football, rugby, and soccer when the ball is kicked upwards, it travels downwards with an amount of drag, which is important to take into consideration when playing these sports. Furthermore, the sport of ultimate frisbee is entirely based upon the principle of air resistance, as really good players know how to manipulate the air to cause the best throws and motion of the frisbee.

*We can also see that the concept of air resistance in specific its relation to the overall concept of drag, is present throughout the universe in many different forms, one of them being the drag you feel against water. We know that it is much harder to move underwater than it is in air, and that is due to the fluid mechanics behind the motion between both mediums, and the increased drag in water.

'''How is it connected to your major?'''

*I am a computer science major who is interested in speciation in artificial intelligence and computer vision. Dealing with forces involved in a drone's or vehicle's motion is exactly what I will be having to deal with in the future. Learning how drag force is dependent on the speed proves that it is constantly changing, and must constantly be updated and factored into calculations regardless of whether the motion is vertical or lateral.

'''Is there an interesting industrial application?'''

*Air resistance has a lot of applications throughout academia, entertainment, security, and many other things. Through academia, we can examine how air resistance impacts the motion of objects in freefall and we can see the magnitude that it has on an object's net speed. Moreoever, through the sports world, we can see how knowing how to control the amount of drag that will be on a ball is imperative towards reaching higher athletic goals.

==History==

Aristotle was the first to write about air resistance in the 4th century BC. In the 15th century, Leonardo da Vinci published the Codex Leicester, in which he rejected Aristotle's theory and attempted to prove that the only effect of air on a thrown object was to resist its motion. The first equation for air resistance was:

:::<math>F_{D} = \rho S V^2 \text{sin}^2(\theta)</math>

This equation overestimates drag in most cases, and was often used in the 19th century to argue the impossibility of human flight.

Louis Charles Breguet's paper in 1922 began efforts to reduce drag by streamlining. A further major call for streamlining was made by Sir Melvill Jones who provided the theoretical concepts to demonstrate emphatically the importance of streamlining in aircraft design. The aspect of Jones’s paper that most shocked the designers of the time was his plot of the horse power required versus velocity, for an actual and an ideal plane.

==See also==

===Further reading===
*http://www.physicsclassroom.com/class/newtlaws/Lesson-3/Free-Fall-and-Air-Resistance 
*http://www.forbes.com/sites/chadorzel/2015/09/29/the-annoying-physics-of-air-resistance/[http://www.forbes.com/sites/chadorzel/2015/09/29/the-annoying-physics-of-air-resistance/] 
*https://www.youtube.com/watch?v=mP1OTzdB0oI

===External links===

==References==
*Chabay, Ruth, and Bruce Sherwood. "Internal Energy." Matters and Interactions. 4th ed. Vol. 1. Wiley, 2015. Print. 
*More information can be found on drag[https://en.wikipedia.org/wiki/Drag_(physics)] and aerodynamics[https://en.wikipedia.org/wiki/History_of_aerodynamics]