Anthony Stanich
Department of Physics and Earth Space Sciences
University of Indianapolis
May 4, 2011

Abstract
The results of the experiment show that the average drag coefficient is 0.001436236 with a standard deviation of 0.001039045. The results also show that the average terminal velocity of a ping pong ball in free fall is
0.331375 m/s with a standard deviation of 0.013159312 m/s.
Introduction
This experiment was performed in order to gain an understanding of the properties of a fluid dynamics, especially drag coefficients. The purpose was also to determine the dependency of mass on the drag coefficient. In fluid dynamics, the drag coefficient is a dimensionless quantity used to quantify the resistance of an object in a fluid environment such as air or water.
Description of Experiment and Calculations
A ping pong ball in free fall experiences fluid friction caused by the molecules in air. The force is called the drag force. The drag coefficient depends on the type of surface and the surface area. To test this force a ping pong ball was used. The ball was released from a height of 2 meters and recorded using a high speed video camera. The videos were then analyzed using LoggerPro video analysis.
The force on a moving object due to a fluid is described by equation 1.
Equation 1
Where
F
D
is the drag force
ρ is the mass density of the fluid for air approximately 1kg/m 3 v is the velocity of the object relative to the fluid in m/s
A is the reference area in m
2
C
D
is the drag coefficient dimensionless
Typically, the reference area A for simple shapes, such as a sphere, is exactly the same as a cross sectional area. For other objects (for instance, a rolling tube or the body of a cyclist), A may be significantly larger than the area of any cross section along any plane perpendicular to the direction of motion.
In order to formulate a differential equation describing motion of an object falling in the atmosphere near sea level there are two variables to look at time, t and velocity, v.
Newton’s 2 nd
Law
states that the sum of the forces is equal to the mass times the acceleration, which can be written as the time derivative of the velocity function and is represented by equation 2.
F = ma = m (d v /d t )
 net force
Equation 2
The forces acting on an object in free fall are gravity and air resistance. Air resistance always opposes the motion and is depicted in figure 1.
2

Figure 1
The force of gravity on an object is related to the mass of the object as explained in Newton’s Universal
Law of Gravity and shown in equation 3.
F = mg
 downward force
Equation 3
The force of air resistance can expressed as equation 4, where ϒ = (1/2)ρvC
D
A .
F =
 v
 upward force
Equation 4
Substituting equations 3 and 4 into Newton’s Second Law (equation 2), equation 5 can be derived.
Equation 5
When the object reaches terminal velocity or the point of maximum velocity, the acceleration of the object is 0m/s
2
. This allows equation 5 to be simplified to equation 6 resulting in an equation that relates the velocity of the object to the drag coefficient. mg= ϒ v
Equation 6
By substituting equation 1 in for the drag force in equation 6, equation 7 is derived. mg=(1/2)ρv 2
C
D
A
Equation 7
Solving equation 7 for C
D
, equation 8 is derived.
C
D
=ρv 2 A/2mg
Equation 8
The first step in the experiment was the construction of the release apparatus. I used a piece of 2in PVC pipe cut to a 2ft length and placed a hole at 1 in from the bottom and affixed it to the ceiling using zip ties.
In the hole I used a zip tie attached to a string that I could pull from the position of the camera. I started the camera and then pulled the string waited for the ball to drop out of view of the camera and stopped recording. After that I used the camera to remove frames before and after the drop where the ball was not in view. I then uploaded the videos into LoggerPro and used the video analysis function to obtain the time and position of the ball. I was able to change the settings in the analysis to match the frame speed of the camera, 420 frames/s. I exported the data to excel and created a graph of the terminal velocity. To do this,
I used the trend line function and removed data at the beginning and end until the R 2 value was above
3

0.9900. The slope of the position versus time graph is equal to the terminal velocity at this point. I then used those terminal velocities to find the drag coefficients. Finally, I graphed the drag coefficients as a function of mass. In order to change the mass of the ping pong ball, I inject water into the ball in increments of 1mL to reach a final volume of 10mL. For the last trial I completely filled the ping pong ball with water.
Analysis of Results
Figures 2 through 13 are the graphs of position versus time for a specific amount of water injected into the ping pong ball and the data set where the R
2
value was greater than 0.9900.
1,2
1
0,8
0,6
0,4
0,2
0
1 1,2 1,4 1,6 1,8
Time (s)
2 x = -0.3201t + 1.3336
R² = 0.9931
2,2 2,4 2,6
Figure 2 Graph of terminal velocity for no water added,
0,9
0,8
0,7
0,6
0,5
0,4
0,3
0,2
0,1
0
1 1,2 1,4 1,6 1,8
Time (s)
2 x = -0.3291t + 1.1002
R² = 0.9915
2,2 2,4 2,6
Figure 3 Graph of terminal velocity for 1mL of water added
4

1
0,8
0,6
0,4
0,2
0
1
x= -0.3199t + 1.0972
R² = 0.9911
1,2 1,4 1,6 1,8
Time (s)
2 2,2 2,4 2,6
Figure 4 Graph of terminal velocity for 2mL of water added
1
0,8
0,6
0,4
0,2
0
1
x = -0.3254t + 1.1539
R² = 0.9909
1,2 1,4 1,6 1,8
Time (s)
2 2,2 2,4 2,6
Figure 5 Graph of terminal velocity for 3mL of water added
0,9
0,8
0,7
0,6
0,5
0,4
0,3
0,2
0,1
0
x = -0.3405t + 1.1254
R² = 0.9932
1 1,2 1,4 1,6 1,8
Time (s)
2 2,2 2,4 2,6
Figure 6 Graph of terminal velocity for 4mL of water added
5

1
0,8
0,6
0,4
0,2
0
1
x = -0.3212t + 1.091
R² = 0.9933
1,5 2 2,5
Time (s)
Figure 7 Graph of terminal velocity for 5mL of water added
1
0,8
0,6
0,4
0,2
0
1
x= -0.3377t + 1.1024
R² = 0.9928
1,2 1,4 1,6 1,8
Time (s)
2 2,2 2,4 2,6
Figure 8 Graph of terminal velocity for 6mL of water added
1
0,8
0,6
0,4
0,2
0
1 x = -0.3203t + 1.1138
R² = 0.9918
1,2 1,4 1,6 1,8
Time (s)
2 2,2
Figure 9 Graph of terminal velocity for 7mL of water added
2,4 2,6
6

1
0,8
0,6
0,4
0,2
0
1
yx= -0.3179t + 1.1051
R² = 0.9912
1,2 1,4 1,6 1,8
Time (s)
2 2,2 2,4 2,6
Figure 10 Graph of terminal velocity for 8mL of water added
1
0,8
0,6
0,4
0,2
0
1
x= -0.3409t + 1.1279
R² = 0.9925
1,2 1,4 1,6 1,8
Time (s)
2 2,2 2,4 2,6
Figure 11 Graph of terminal velocity for 9mL of water added
1,00
0,80
0,60
0,40
0,20
0,00
x = -0.343t + 1.1369
R² = 0.9978
1 1,5 2 2,5
Time (s)
Figure 12 Graph of terminal velocity for 10mL of water added
7

1
0,8
0,6
0,4
0,2
0
1
1,2 1,4 1,6 1,8
Time (s)
2 x= -0.3605t + 1.1546
R² = 0.991
2,2 2,4 2,6
Figure 13 Graph of terminal velocity for full of water added
Figure 14 is a table that lists the volume of water added in mL, the mass of the ball in kg, and the terminal velocity in m/s from the graphs.
Volume of H2O
(mL)
0
Mass of Ball
(kg)
0.00181
Terminal Velocity
(m/s)
0.3201
0.96
1.9
2.96
0.00277
0.00371
0.00477
0.3291
0.3199
0.3254
3.94 0.00575 0.3405
4.93
5.98
0.00674
0.00779
0.3212
0.3377
6.91
8.16
8.89
9.99
0.00872
0.00997
0.0107
0.0118
0.3203
0.3179
0.3409
0.343
21.87
Average
Standard
Deviation
0.02368 0.3605
0.331375
0.013159312
Figure 14 Values of volume, mass, and terminal velocity for each trial.
The Grubbs Test determines if a data point is an outlier, equation 9. If the calculated Grubbs value is greater than the Grubbs table value, the data point is an outlier.
Equation 9 The Grubbs test formula
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The Grubbs calculated value for the highest and lowest velocities are 2.213 and 1.024, respectively. The table value for N=12 is 2.412. Therefore there are no outliers in the data set. Figure 15 shows the relationship between terminal velocity and mass. The data shows that there is a slight positive correlation between mass and terminal velocity, meaning that as the mass goes up so does the terminal velocity.
0,37
0,36
0,35
0,34
0,33
0,32
0,31
0
5 y = 0,0017x + 0,3175
R² = 0,5647
10
Mass (g)
15 20 25
Figure 15 Graph of the relationship between mass and terminal velocity.
Using equation 8, the drag coefficients were found. The results for the drag of each mass are shown in figure 15. The radius of the ball is 0.021m and the cross sectional area is 0.00139m
2
.
Mass
Drag
Coefficient
0.00181 0.004001522
0.00277 0.002763811
0.00371 0.001949786
0.00477 0.001569095
0.00575 0.001425275
0.00674 0.001081991
0.00779 0.001034802
0.00872 0.000831629
0.00997 0.000716503
0.0107 0.00076772
0.0118 0.000704756
0.02368 0.000387937
Average 0.001436236
Standard
Deviation 0.001039045
Figure 16 Results for the drag coefficients for each mass
Figure 17 shows the relationship between mass and the drag coefficient. The graph shows that there is a definite correlation between mass and the drag coefficient. The coefficient appears to decrease exponentially as mass increases.
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0,005
0,004
0,003
0,002
0,001
0
0
0,005 0,01
Mass (kg)
0,015 0,02 0,025
Figure 17 Relationship between mass and drag coefficients
Conclusions
The results of the experiment show that the average drag coefficient is 0.001436236 with a standard deviation of 0.001039045. The results also show that the average terminal velocity of a ping pong ball in free fall is
0.331375 m/s with a standard deviation of 0.013159312 m/s. The results show that the terminal velocity of the ping pong ball has a slight correlation with increasing mass. The results also show that the drag coefficient has a definite exponential decay relationship with mass.
References
Edward & Penny. Differential Equations. 4 th
Ed.
Fowles & Cassiday. Analytical Mechanics. 7 th
Ed.
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