By: Annie Huang, Heesue Kim, Sophie Gilliam, and Sylvia Towey

Hi guys!

Welcome to the Girls Talk Math blog today! This blog is to show you guys what we have learned and accomplished with fluid dynamics. At first, we (Annie, Heesue, Sophie, Sylvia) thought this was a very difficult topic but after some explanation and experiment, we learned how easy it is to work with the different topics thanks to the Girls Talk Math Camp held on the UNC Chapel Hill campus. Today we will be giving you a brief intro to mathematical modeling, Bernoulli’s principle, Dimensional Analysis, and Projectile motion.

**So… what is a mathematical model?**

A mathematical model is a description of a system using mathematical concepts and language. It usually helps explain the characteristics and properties of a system and is used to make predictions about its behavior. Mathematical models are used in the sciences (physics, biology, earth science, meteorology) and engineering (economics, psychology, sociology, political science.)

**How accurate are they?**

Models can either be very simplified or very accurate, but not at the same time. There can’t be too much of one characteristic, but must have an equal balance. The ideal model is simple enough to understand, but also complex enough to correctly represent the situation.

*Ockham’s Razor *is the idea that among models with equal predictive power, the simplest one is the most desirable. Adding complexity makes the model more difficult to understand and analyze.

Fit to experimental data-It is essential to check if a model fits experimental measurements.

Parameters-A parameter is any characteristic that can help in defining/classifying a particular system. It’s an element of a system that is useful when evaluating the identity, performance, or condition of a system.

In mathematical models, the quantities in equations are sorted into parameters or variables. Quantities that can be measured independently are the variables. “Constants” stand for inherent properties of nature, or of the materials and equipment used in a given experiment. These are the parameters.

For example, in the equation h(t)=20t, 20 is a fixed value so it is a parameter.

Scope of the model- determining what situations the model can be used for is not straightforward. Interpolation is whether the model well describes the properties of the system between data points and extrapolation is the same question but for events outside of the observed data.

Philosophical Considerations- The validity of a model rests on the ability to extend to situations beyond those described on the model.

**What is fluid dynamics? **

Fluid dynamics is the field of study that deals with fluid flow.

What counts as a fluid? A fluid is a substance that continually flows under an applied shear stress. Fluids include liquids, gases, and plasmas.

Shear stress-Shear stress arises when a force is applied to an object in a parallel direction to its cross section.

Strain vs. strain rate-The strain of an object is the measure of deformation representing the displacement between particles in the body relative to a reference length. The strain rate is the change in strain of a material with respect to time. It includes the rate at which the material is expanding, shrinking and being deformed without changing its volume.

**Bernoulli’s Principle**

For the first part of our math problems, we looked at something called Bernoulli’s Principle. None of us knew what it was or what mathematical phenomena it explained so we were introduced to this topic by an experiment we did called the balloon experiment. To do the balloon experiment, we held two blown up balloons at face level, six inches apart and blew air between the balloons. The full balloon experiment is shown below.

Balloon Experiment

Materials:

- Two inflated balloons

Experimental Procedure:

- Hold the two balloons at face level, approximately 6 inches apart from each other.
- Blow air between the balloons. It is important that you blow air between them rather than on them.

When we blew air between the balloons, we observed something funny. The balloons actually drew closer to each other rather than getting blown farther away from each other. We wondered, why does this happen? Well, the reason why the balloons actually moved closer together is because there was a difference in pressure, or force applied per unit area, when air was blown between the balloons. The pressure was actually lower between the balloons since when we blew air between the two balloon causing the air to move at a higher velocity, the pressure decreased and it made the two balloons actually move closer together. This is explained by Bernoulli’s Principle, which basically states that velocity and pressure are inversely proportional for a fluid; the pressure decreases if the velocity increase.

Bernoulli’s Principle can be illustrated using an equation: gz+p/ρ+v2/2=constant where *g *is the acceleration of gravity (9.81m/s2), *z* is the elevation from an arbitrary reference point, *p* is the pressure, ρ is the density, and *v *is the velocity of the fluid. To show you how this equation is used, we will be doing an example.

Example: Now let’s use Bernoulli’s Principle to analyze the balloon experiment. Assuming the balloons were held 1.5 meters off the ground and that you blew air at a velocity of 7 meters per second between the balloons, what was the pressure in between them? To compute the constant in Bernoulli’s Principle, use the fact that atmospheric pressure is 101,325 Pascals at zero air velocity. Also, the density of air is 1.225 kg/m3. How much lower was the pressure between the balloons than the pressure on the other sides?

To begin this problem, we need to find the pressure of the air between the balloons and to do that, we know that we are going to have use Bernoulli’s equation, gz+p/ρ+v2/2=constant. First, we can list out what we know: *g*=9.81 m/s2, *z*=1.5 m, *v*=7 m/s, ρ=1.225 kg/m3. We must also find the constant so we can set the equation equal to a number in order for us to solve for *p*. Because we need to solve for the constant before we can solve for *p*, we will “use the fact that atmospheric pressure is 101,325 Pascals at zero air velocity” to solve for the constant. So now we have: (9.81m/s2)x(1.5m)+(101,325kg/ms2)/(1.225kg/m3)+(0/2)=constant. When we simplify that, we get: 14.715 m2/s2+82,714.28571m2/s2=constant. When we add these numbers together, we have: constant=82,729.001m2/s2. Now that we have our constant, we can set up our next equation using Bernoulli’s equation to find the pressure; so now we set it up as: (9.81m/s2)(1.5m)+(*p*)/(1.225kg/m3)+(7m/s)2/2=82,729m2/s2. We simplify that equation so that now it is: (14.715m2/s2+(*p*)/(1.225kg/m3)+24.5m2/s2=82,729m2/s2. We then simplify it even more to: (*p*)/(1.225kg/m3)=82,689.785m2/s2. Now we can find the pressure, we get that *p* is equal to 101,295kg/ms2 which is equal to 101,295 Pa. The pressure between the two balloons is 101,295 Pa and the pressure between the balloons is 30 Pa lower than the pressure surrounding the balloons. This is an example of how you can apply Bernoulli’s experiment to real life problems, as long as you can find the constant and know all of the variables except the one you are looking for, you will be able to find what you need to find!

Two other experiments we did were the ping-pong ball experiment and the inflating plastic bag experiment. To do the ping-pong ball experiment, you will need two plastic cups and a ping-pong ball. You then want to set the ping-pong ball in a cup that is about 6 inches away from the other cup and try to move the ping-pong ball from one cup to the other cup without touching the ball or the cup! To conduct the plastic bag experiment, you will only need a plastic bag with no tears or hole. You will then close up the opening of the bag until it’s just small enough to put around your mouth and blow into it to try to fill the whole bag up with air. This will be difficult so can you try to find another way to fill up the bag? Both of these experiments can be completed using Bernoulli’s principle. For the ping-pong ball experiment, try blowing across the top of the cup containing the ping-pong ball; this will move the ball since the pressure is lower when you blow across the cup causing the ping-pong ball to “jump” from one cup to the other. For the plastic bag experiment, try holding the plastic bag 6 inches away from your face and blow. It should be a lot easier to fill up the plastic bag with air! This happens because of the lower air pressure that causes air to want to fill up the bag more quickly.

**Dimensional Analysis**

Dimensional analysis is the idea of simplifying units into what they measure. These are the fundamental dimension(s) of the unit.

For example, 8 seconds is a measurement of time so the fundamental dimension is time. 30 meters per second is a measurement of speed but more simply it describes how far something travels in an amount of time so the fundamental dimension is distance/time.

Dimensional analysis can be used to simplify deriving formulas.

Let’s look at drag force as an example. Drag force’s dimension is (kilograms*meters)/seconds^{2}. Kilograms measure mass, meters measure distance, and seconds measure time. So drag force’s fundamental dimension is (mass*distance)/time^{2}.

The area of the object affects how quickly it falls so it is an important thing to include in the equation. Area is length*width which are both measurements of distance so it becomes just distance^{2}.

Velocity of the object affects how much it is pushing against the air so it should be included. Velocity measures speed which is how far you travel in an amount of time so it becomes distance/time.

Density of air affects how much the object’s motion is resisted to it should be included. Density is mass/volume. Volume is length*width*height these are all measurements of distance so volume can be simplified to distance^{3}. So density becomes mass/distance^{3}.

What we want to do is combine the fundamental dimensions of area, velocity, and density in a way so that they equal (mass*distance)/time^{2}.

If we multiply density and volume’s fundamental dimensions we have (mass*distance)/time*distance^{3} which has the right top but not the right bottom. If we multiply (mass*distance)/time*distance^{3} and area’s fundamental dimensions then length cancels out and we get mass/time which is not right but if we multiply by volume’s fundamental dimension we get mass*distance/time^{2}.

If we look at what we did to get (mass*distance)/time^{2} from the fundamental dimensions of area, velocity, and density it was velocity^{2}*area*density*constant = (mass*distance)/time^{2} or velocity^{2}*area*density*constant = drag force. We needed to include a constant as well. The constant doesn’t have a dimension so it was not in the fundamental dimension equations. The actual equation is drag force = v^{2}*a*d*1/2*c_{d}. The only difference between our equation and the real one is c_{d} which is a drag coefficient that tells you what kind of drag it is.

By using dimensional analysis to derive this formula we were able to derive it using only basic algebra. If we had derived the formula a different way it would have been harder to get. This is the benefit of using dimensional analysis instead of something else.

**Projectile Motion**

To explain what vectors are, we will begin with a scenario. Here’s the situation: You are walking past a middle school and you see that the students are at recess. Some of them are seeing who can throw a rock the farthest. You observe that each rock creates an arch formation as it is moving through the air, and you observe that the arch changes shape based on the angle the rock is thrown at and its velocity (which is changed based on how hard the rock is thrown).

Vectors are often used to give the position of a projectile at a given time, *t*. A vector is a quantity that is plugged into an equation and determines the location of one point in relation to another. The vector equation that tells you where, at time *t*, the horizontal position or the *x*-position, the projectile can be found, *x*(*t*)*=v**x**t+x**0*. The *y*-position is given by *y*(*t*)*=*(-*g/2*)(*t*2)+*v**y**t*+*y**0*. From these two equations, we know that *g* is the acceleration of gravity (9.81m/s2), *v**x ** *and *v**y* are the initials velocities of the projectile given by the equations *v**x**=*rcosθ and *v**y*=rsinθ, and *x**0* and *y**0* are the initial positions of the projectile. We will be doing an example below to help you gain a better understanding of how to use vectors in a math problem that simulates a real life situation.

Example: If a projectile is launched at an angle of 25° with speed 15m/s, how far will it travel horizontally before it hits the ground?

- First find
*v**x*and*v**y*.

We know the two equations that will give us the values of *v**x* and *v**y*.

*v**x*=rcosθ=(15m/s)cos(25°)=13.6m/s

*v**y*=rsinθ=(15m/s)sin(25°)=6.3m/s

2. What does it mean for the projectile to have hit the ground? Can you write out an equation that can be solved for the time at which the projectile hits the ground?

When the projectile hits the ground, it means that *y=*0 so the equation is 0=(-*g/2*)(*t*2)+*v**y**t*+*y**0*.

3. Which of our equations tells us how far our projectile travels horizontally?

The *x*(*t*)*=v**x**t+x**0 *equation tells us how far the projectile has traveled horizontally.

4. Use your work from above to figure out how far the projectile travels horizontally by the time it hits the ground.

0=(1/2)(-9.81)*t*2+6.3*t*+0, so we get that *t*=1.2844037 seconds.

We then plug this value in for t in the *x*(*t*) equation. So we can now solve: *x*(*1.2844037*)*=*13.6(1.2844037)*+*0=17.5 meters.

5. What factors are we ignoring in this model?

We are ignoring the drag force, the various characteristics of the object, and wind.

We did this experiment to help us learn about the factors, such as drag force, that affect the motion of a projectile.

Here are some materials that you will need. We used a scale to measure each item we dropped, and a stopwatch to time how long it took to reach the ground. We used a tape measure to ensure that each item was dropped from the same height. The items we dropped were a wadded up piece of paper, a paper lantern, a tennis ball, a stress ball, 4 varying sizes of bouncy balls, and a ping-pong ball. To complete this experiment, the first thing we did was weigh each item on the kitchen scale. We also took each item’s circumference, which we used to calculate the area of the largest cross-section, the volume, and the radius of each item we dropped. Next we measured the height of the doorway, and dropped each item from that height to be consistent. Three timers were used to time the objects travel, then the three times were averaged. The third and final step was graphing the different possible factors in relation to the objects travel time, then looking for a trend in one of the graphs.

After concluding this experiment we observed that there were trends in two graphs: the circumference and area graphs. Since there is a trend, we know they are related to the air drag of an object (essentially the time it takes for an object to hit the ground). For spherical objects, the formula for area involves the circumference, which is why there is a trend in both graphs. Really, area is the factor that affects the time of an object’s travel. The reason area affects how long it takes for an object to reach the ground is simple: the more area there is, the more the air molecules, or air drag, can push against it.

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