Lab 8: Motion in One Dimension with Air Drag

Purpose: To analyze how changing force affects motion in one dimension.

Introduction: 
So far we have assumed objects move with constant acceleration. This assumes that a constant force is applied to the object. What if a changing force (which causes a changing acceleration) is applied to an object?

An object under the influence of air drag is one example of this. Calculus provides a way to solve this problem where variables are changing with respect time. A spreadsheet can be used to perform this calculus. Two basic relationships can be used to analyze the object:  

                                                           v_new = v_old + a_avg△t     (1)


Where v_old  is the previous value of velocity and position, and a_avg  is the average value of the acceleration and velocity during the time interval △t.


1. By unit analysis, show that the above equation (1) is valid:

            Left side unit = m/s
            Right side unit = (m/s) + (m/s^2) × s = (m/s) + (m/s) = m/s
          
    So the units of two sides are identical.
 

2. Why do we use a_avg in equation (1).
    
            Because the can be changing all the time, so the average data is more accurate.


3. Come up with an analogous equation relating y_new and y_old:

            y_old = y_o
             
                y_new = y_o + v_old△t + (1/2)a_avg(△t)^2
            or   y_new = y_o + v_avg△t


4. What is the benefit of changing a small △t:

            Because the acceleration is changing all the time, so in a small △t, the a_avg is closer to real value.     So, the v_new will also be more accurate.



Since the acceleration depends on the forces acting on the object, we must specify precisely what these forces are. In our problem the drag force will at first be assumed to depend on the first power of the velocity and can be written as:
                                                                     F_D= -kv
Where k is a proportionality constant.


1. Draw a detailed motion diagram of the object falling down:

a. Now draw a force diagram for the object falling down. Include vectors for all forces, and write a  statement of Newton's Second. Solve for acceleration.

    a= F / m = D - G = (-kv-mg) / m = - (g + kv / m )


b. Give the condition for the object at terminal velocity (a=0). Using this condition solve for k.

    a= 0, So net force is 0, mg = kv_t , k= mg / v_t


c. Substitute k into your expression for the acceleration from part a.
 
     a = - (g + kv / m )
     k = mg / v_t
     So, a = -g (1+ v / v_t )


2. Open the spreadsheet in the Physics Apps file called air drag. Describe what the spreadsheet is  calculating.

a. What are the assumption? (i.e. initial values? )

     t_o = 0s , g = -9.8m/s^2 , initial velocity = 20m/s , terminal velocity = -40m/s , △t = 0.1s

b. What is v-halfstep?

     V-halfstep is the average velocity in 0.1s.

c. What is the a-halfstep?
 
    A-halfstep is the average acceleration in 0.1s.


3. Using the graph paper provided, draw scales graphs of position vs. time, velocity vs. time and acceleration vs. time for the object no air drag.

position vs. time graph when there is no drag

                                                             p= -4.9t^2 + 20t +1000

velocity vs. time graph when there is no drag

                                                                 v= -9.8t + 20

Make predictions on another sheet of graph paper about how position and velocity graphs would change if you include air drag (D= -kv).

position vs. time graph when drag is equal to -kv

velocity vs. time graph when drag is equal to -kv

Now look at a drag force that is dependent on the square of the velocity. Assuming a drag force,  
F_D= | kv^2 |, find the new formula for the acceleration.

a= F / m = D - G = (kv^2 - mg) / m
mg = k v_t^2 , k = mg / v_t ^2
So, a= [(v^2/ v_t^2) -1] g

position vs. time graph when drag is equal to |kv^2|

velocity vs. time graph when drag is equal to |kv^2|

Conclusion:

     In this lab, we analyzed how changing force affects motion in one dimension. We plotted v-t and p-t graphs in  three cases: with no drag; drag is equal to -kv and drag is equal to | kv^2 |. 

     When there is no drag, the net force on the ball is gravity force, so the acceleration is equal to -9.8m/s^2.  The motion trajectory is parabola. The velocity is changing with constant slope, decreasing to 0 first then increasing in negative position. 

     When drag is equal to -kv, a= -g (1+ v / v_t ). Because the velocity is changing all the time, so the acceleration is also changing. When the ball goes up, velocity is decreasing, so the acceleration will increase in negative position. The maximum height of the ball will be lower than the ball with no drag. Then the velocity will increase in negative position. The velocity will be constant when drag is equal to the gravity force.

     When drag is equal to |kv^2|, according to the v-t graph, we find its acceleration is lower than D= -kv before it reaches the highest point. So, the its maximum height is higher than the ball when D= -kv and D= 0.For balls with  D= -kv and D= |kv^2|, their terminal velocities are the same, and at the highest point their velocities are all equal to 0. According to the v-t graph, when balls are going down the one with D= |kv^2| has a smaller average acceleration. Because v_t^2- v_o^2= 2ax, so the ball with lower acceleration will go deeper. That's why in this case the ball goes deeper than the ball with D= -kv.

      Causes of error:

      We use average velocities in 0.1s to plot our graphs, so the velocities are approximate values, which are not accurate.