Sunday, May 18, 2014

Using Math to Describe Gravity (from Playing With Math)

We are nearing completion of the book, Playing With Math: Stories from Math Circles, Homeschoolers, and Passionate Teachers. Our copy edit process was really a deeper editing process, and took over a year, with us working through a few chapters a week at first, getting everything just right. We finished it on May 9, Mother's Day. (My son was born on Mother's Day. I didn't realize until this moment how much I like it that our copy edit finished on Mother's Day too!)

One of the last pieces to go through copy edit was Sean's catapult activity that goes with Kate Nowak's chapter, Better Teaching Through Blogging. I had to build a catapult to test out the instructions Sean gave. I am not a crafts sort of person, so that had to wait until I had plenty of time to deal with it. Last fall, after I finally made my catapult, made my small adjustments to Sean's piece, and sent it to our copy editor, she  asked a lot of questions about the math. I began to realize that we needed an explanation of the math behind the catapult project.

So I wrote one last piece for the book, Using Math to Describe Gravity. I had fun writing this one. Most of what I wrote for the book took me a long time (and lots of agonizing) to write. This one was easy and quick. I finally realized that I really enjoy writing explanations. (I think I know my next book project...) I thought this might be useful to have online, so I'm including a modified version here. Enjoy!

Using Math to Describe Gravity

In the picture above, we see water shooting upward. Whenever you squirt water from a hose pointed up at an angle, it follows a similar path. Have you ever wondered why? Math is helpful whenever we want to think about how something is changing.

For example, the idea of velocity tells us how position is changing with respect to time. (Unlike speed, which is just a positive number, velocity has direction, and can be negative to indicate a downward direction.) If you are driving at 70 miles per hour, in one hour you will have driven 70 miles. In two hours, 140 miles. Mathematicians generalize this idea by writing distance = rate * time.

The path of the water in the fountain makes the shape of a parabola. The physics of gravity explains why the water follows that path. As the force of gravity pulls us toward the center of the Earth, it creates an acceleration – a change in velocity. When we are near the surface of the Earth, that acceleration is always 32 feet per second squared (downward). That’s a weird unit, isn’t it? It means that if you are headed straight down, your velocity will increase by 32 feet per second each second. So when you drop something, one second later it has already gone from a speed of 0 feet per second to a speed of 32 feet per second. Over 20 miles per hour! In metric units, that would be 9.8 meters per second squared. (We’ll mostly stick with metric from here on out.)

This acceleration affects the relationship between distance and time because the speed, or rate, is changing. In the driving example, if your speed goes from 55 miles per hour to 60 miles per hour to 70 miles per hour, it makes it harder to calculate how far you have driven. (Calculus is great for understanding situations where your rate of change is changing, but I've written this for people who aren't familiar with the concepts of calculus, sticking to algebraic ideas.)

When considering the physics of situations like the fountain, we can analyze the vertical and horizontal motion separately. Gravity isn’t affecting the horizontal motion, so that stays constant. (If you were moving very fast, air resistance would slow you down. But at these speeds, we can ignore the effect of air resistance.) When you throw something upward at an angle, gravity pulls straight down, changing the vertical component of the velocity. Since the horizontal part of the motion is constant, this gradually changes the direction the object is headed, making the parabolic path you see above.

But how do we know that it’s in exactly the shape of a parabola? To see why it is, we’ll start with a simpler experiment, throwing a ball directly up. Now the path is no longer a parabola, because the horizontal position is not changing. But, amazingly, if we were to draw a graph of height versus time, that graph would still be a parabola. To describe parabolas algebraically, we use equations like y = at2+bt+c. In this case, y is the height and t is the time.

[Note: Depending on the flavor we want, we say the same thing in lots of different ways. rate * time = distance, R*T=D, velocity * time = height, h=v*.]

We can figure out a lot about a, b, and c by using what we know about the physical situation. When t=0, y = a*02+b*0+c = c, so c can be filled in by knowing your initial height. If we had no gravity (and no air resistance), what we threw upward would keep going up with a constant speed. So its height would be given by rate (velocity) times time plus initial height. Do you see why b is the initial velocity the ball has as it leaves your hand?

That leaves a. The value of a will always be half the value of the gravitational constant - hmm, why half? In one second, the acceleration of gravity would increase velocity from 0 to 9.8 meters per second. So the average velocity during that second is 4.9 meters per second. In t seconds, we would increase from 0 to 9.8t meters per second, with an average of 4.9t meters per second times t seconds, for a height change of 4.9t2 meters.

Now we can see the effects of initial height, initial velocity, and gravity combining to make an equation of the form y = at2+bt+c for height versus time. Gravity will actually affect an object in this way no matter which direction it’s pointed. And since the horizontal motion is constant, this same sort of relationship holds when we look at height versus horizontal position, although the values for a, b, and c will change.

When an object is launched at an angle, the value for a is determined by both gravity and the launch angle, b is determined by both the initial speed of launch and the launch angle, and c is still the initial height. To find the values for a and b, we can use the symmetry of parabolas across their vertex. If we can find the coordinates for the position of the vertex, we can use that to help use find the values for a and b.

All this thinking can help us understand the catapult activity better:
  • When launching from the floor, where beginning and ending heights are the same, the x-coordinate of the vertex is just half the distance. So the vertex will be reached halfway through the time in the air.
  • From the time the projectile reaches the vertex until the time it hits the floor, its height is decreasing at the same rate as an object that has been dropped, so you just use 4.9t2 to find how far it dropped, which tells you how high it was.
  • The vertex form for the equation of a parabola is y = a(x - h)2 + k, where (h,k) represents the vertex, which we just found. If we assume that we launched from the origin, plugging zero in for x and y allows us to find the value for a.
  • With the values for a, h, and k filled in, we have an equation in x and y. We can simplify it (change it to the form y = ax2+bx+c), and then modify it for raised launches by simply changing the value of c from 0 to the height of the launch surface. 
  • We can use the new equation to figure out where to put a target we want to hit! The target will be on the ground, where the height is 0, so we can plug in y=0, and find x. In real life, the numbers that show up are almost never simple enough for factoring to work, so we’d need the quadratic formula. Measuring the horizontal distance from the floor right below the catapult, to the positive x-value from the quadratic formula, and centering the target there, we should be able to hit it. Candy bombing, here we come! 

My thanks to John Golden for his help improving this.

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