Project 24: Falling with Air Resistance: Data and a Linear Model
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Galileo's law written as a second-order differential equation can be solved to give an explicit formula for the distance an object has fallen.
For example, if we start from rest and position zero, measuring down, then
with g=9.8 in meter-kilogram-second units. (The unit of force is called a "newton" in mks units, so 1 kilogram exerts a gravitational force of 9.8 newtons at sea level on earth.) You may have learned this formula or extensions of it in high school physics.
It is interesting because it comes from such a simple law of speeding up, namely that acceleration is constant.
It clears up the false intuitive notion that heavy objects fall faster; they don't - the constant g is the same for all objects, at least in vacuum.
Figure 24.1: Free Fall with Air Friction
Light and heavy objects fall at the same speeds if we can neglect air friction.
Often we can't neglect that.
The computer program AirResistance contains data for a wooden ball thrown off the same cliff as the lead ball from the Gravity program of Chapter 10 of the main text.
We know that a parachute radically affects the speed with which an object falls - air friction often does matter.
The graph of the data from AirResistance on our website is shown in Figure 24.1. Notice that it "flattens out" as time increases.
- 1.
Compute the velocities v[t] of the object in the data from the AirResistance program from t=0 to
, from
to t=1, etc.
What happens to speed as time increases?
- 2.
Use your velocity list to compute the accelerations a[t] from
to t=1, from t=1 to
, etc.
What happens to acceleration as time increases?
After Galileo, Newton discovered calculus and derived a more general acceleration law commonly known by
"Force equals mass times acceleration." Newton also related mass to force due to gravity by
so massive objects weigh more but still fall by the same acceleration because m cancels on both sides of
The differential equation
is Galileo's law expressed in the language of calculus (which, hopefully, you observed in text Chapter 10). Newton's law in the language of calculus is
We want to use Newton's law to formulate a more realistic model of a falling body that IS affected by air friction.
The new model does not have a simple explicit solution like
. It does have a messy one that you will be able to compute later this semester, but the important thing is the use of calculus as a language describing the changes caused by gravity and air.
More complicated models will not have any explicit solutions but still can be described by simple differential equations.
What can we say about air friction? The faster we go, the more the air resists.
If you walk and hold your arm out, you hardly notice the resistance of air.
If you bike and hold your arm out, you notice, but the force is mild.
If you drive on the interstate and stick your arm out the car window at 65 mph, you notice a substantial force.
About the simplest law we can formulate with this property is that the air friction force is proportional to the speed, but we need to keep track of directions.
The force from air is opposite to the direction of motion, so
Air friction force is opposite to the direction of the velocity and proportional to its magnitude.
An object falling under gravity has two forces acting on it, gravity pulls down and wind resistance pushes up - if we move,
where the force due to air is negative (because of its direction). We write Newton's "F=ma" to get
divide by the mass m and let the unknown constant
,
where
, the velocity.
Find a constant b so that the model
fits the data in the program AirResistance. You computed velocity and acceleration in the previous exercise using the computer.
You can do list computations to estimate b.
24.1 Terminal Velocity
There is an interesting way to measure b. If you look carefully at the data for a longer fall, you will see that the wooden ball approaches a limiting velocity.
This is just another way to say that the acceleration "flattens out." This limiting velocity is called the terminal velocity, vT, of the falling object.
Unfortunately, we would have to drop the ball out of an airplane to get enough data. (The data we have do not flatten out all the way, the final accelerations are not quite zero.)
The differential equations and initial conditions describing the motion of the ball are
Problem 24.1
Terminal Computations
- 1.
If v[t] tends to a limit, then its rate of change becomes less and less as it does so.
Why?
- 2.
What is the calculus expression for the time rate of change of v? According to part (1), this expression tends to zero as the velocity tends to its limiting value.
- 3.
At terminal velocity, vT,
. Use the differential equation for v to find vT in terms of g and b.
- 4.
What is the terminal velocity of the wooden ball in air? In other words, how fast would you expect it to be going when it hit the ground if dropped from an airplane? Hint: Use your computation of b from the data.
24.2 Comparison with the Symbolic Solution
A trick called "separation of variables" that we will learn later shows that the solution to the air resistance initial value problem is
Matching Theory and Data
1) Show that y[0]=0.
2) Use symbolic rules of calculus to show that
.
3) Show that v[0]=0.
4) Show that
and that g-bv[t]=ge-bt also, so that
.
Now wrap up your whole project.
Explain why
is the symbolic solution to the differential equations describing the acceleration due to gravity and air resistance.
Then use your measured value of b and plot y[t] together with the data from the AirResistance program.
: Comparison of Symbolic Solution and Data
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