Air Drag on Falling Cones – Experiment copy PDF

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Investigating the Air Drag on Falling Cones

Abstract: This experiment aimed to find the drag coefficient of four paper cones of different masses and solid angles while exploring the effects of drag coefficients for objects in air. The drag coefficient found in this experiment was found for solid angle of 0.5 steradians, which has a corresponding half angle of the cone of 23 . The drag coefficient CD was found to be equal to 0.6 0.6 , and as this is a dimensionless quantity it has no units. This drag coefficient was found by using the knowledge that when an object has reached terminal velocity, its weight force is equal to its gravity force. Equating these two equations we obtain a relationship between the terminal velocities of the masses and their mass, which, when plotted linearly, has a gradient that can be rearranged to find CD. This experiment was conducted in the stairwell of the West building at the University of Canterbury by dropping four paper cones of four different masses down from the eighth floor, and taking appropriate measurements in order to find the terminal velocity of each cone. Using this data, a graph showing the relationship between the square of the terminal velocity and the cones masses was used to find the drag coefficient for the cones of the solid angle 0.5 steradians. Comparing this value to other drag coefficients obtained by other groups in the class, we see that there is a positive linear relationship between the solid angle of a cone and its drag coefficient which shows that as the cross-sectional area of the cone exposed to the air increases, so too does the cones drag coefficient.

Intro & Theory: When an object moves through air with a velocity v relative to the air, that object is subject to a drag force that acts in the opposite direction to v . This air drag force is given by 1 F D = C D Aρ v2 2 [1] and its size is affected by several factors. CD is the drag coefficient of the object - a dimensionless value that depends on the shape of the object, the viscosity of the gas that it is moving through, and also on the objects velocity v . A in equation [1] refers to the cross sectional area [in m2] presented to the air flow,  is the air density [in kg/m3] , and v is the velocity of the object [in m/s2]. Altogether, these factors combine to produce an air drag force opposing any object in motion through the air – whether that object is a missile1, a tropical cyclone2, or just a simple paper cone. For a falling paper cone, this drag force can be used to express the cone’s equation of motion using Newton’s Laws:  F =m a [2]

 F = Fg+  Fd

[3] 1 2 ma=mg − C D Aρ v 2 [4] This equation can be complicated to solve initially, however when the drag force equals the gravitational force it simplifies down considerably and can be re-written as 1 2 mg= C D Aρ v 2 [5] At this point the net force on the falling cone is zero and it will no longer accelerate as it has reached terminal velocity, which can be expressed from equation [5] as; v 2=

2 gm C D Aρ

[6] Since

g ,

C D , and

are all constants in this equation, we can see that if we m want to measure C D , we could find a range of values for v 2 and and plot A them linearly on a graph, which would have a gradient equal to ρ

m= [7]

2g CD ρ

In this experiment, the cross-sectional areas of each cone were kept constant so a range of values only had to be obtained for v 2 and m , the masses of four different 2 cones. This meant that a graph of v against m had a gradient that was equal to m=

2g C D ρA

[8] The drag coefficient, CD, and its associated uncertainty of the four paper cones can then be found by rearranging this gradient equation. In addition to investigating the equation of motion for a falling paper cone, this experiment also included some analysis of solid angles while constructing the paper cones used in the experiment. The solid angle of a cone, , is the three-dimensional analogue of an angle and is measured in steradians. It is related to the angle cut out of a section of a circle, , shown in figure (a) below through 4π =ϕ −1 ❑ [9] Figure (b) below shows how a cone can be constructed from a flat circular piece of paper.

√

A (B) C

B O h

A

m r

h

O (a)

(b)

C

Figure 1. Construction of a cone from flat piece of paper The arc-length of a section of a circle like that in figure (a) is given by s = h. The radius of the cross-sectional area in figure (b) can be found using trigonometry to be r = h sinm The circumference of the circle of the cross-sectional area in figure 1(b) is the arclength shown in figure 1(a), and can be expressed as s=2 πhsin θm [10] And therefore, using s = h, we see that ∅=2 πsin θm [11] The cross-sectional area of the cone in figure 1(b) is given by A=π r 2 And since r = h sinm this cross-sectional area can be written as 2 A=π (hsin θm ) [12]

A=π h2 si n2 θm [13]

Experimental Procedure: The aim of this experiment was to find the drag coefficient of four paper cones of the same solid angle. The first step in this experiment was to construct the four paper cones that each had a solid angle of  = 0.5 steradians. This was done by first converting the solid angle into the angle cut out from the section of a circular piece of paper using equation [9] and then constructing a cone by cellotaping the ends of this section of paper together. 2 gm , it was decided that our group would C D Aρ use four cones that each had the same surface area A but had four different masses. This simplified calculations as it meant that A in this equation was a constant, and the only variables were the terminal velocities of the cones, v, and their masses, m.

Because equation [6] shows that

v 2=

The masses of the four cones were changed by inserting blue-tack into the bottom of the cone. The four values of the masses for the four cones is given in table 1 of the appendix. We then had to devise a way to measure the velocities of the four paper cones as they d fell down the stairwell from the 8th floor of the West building. Because v = we t could calculate the distance between each floor, as well as the time of travel for each cone between each floor, and from this find the velocity of each cone as it travelled between each floor. It was decided that there would be a person on each floor of the stairwell, who would each start their stopwatch when the cone was dropped from the 8th floor and then stop their stopwatch when the cone passed them at the bannister level of the floor that they were on. This raw data is shown in tables 1.1 to 1.4 in the appendix. From this data, the time taken for each cone to travel between each floor can be calculated by subtracting the time it takes to reach one floor from the time it takes for the cone to reach the floor below. The distance between each floor was also measured by lowering a measuring tape down from the 8th floor. This raw data is given in part 2 of the appendix. From this raw data, the terminal velocities of each cone could be found, and using the fact that at the terminal velocity, equations [5] to [8] hold, the drag coefficient of paper cones with a solid angle  = 0.5 steradians can be found. This experimental set-up has been depicted in figure 2 below. Stairwell in the West building: 8 7 6 5 4 3 2

Figure 2. Experimental set-up of stairwell where the cones were dropped (not to scale).

Results & Analysis 2 gm C D Aρ As this experiment was conducted close to earth g was taken to me 9.81ms-2. The values that had to be gathered before CD could be computed using equation [8] were therefore the surface area of the cones, A, air density, ρ , mass of the cones, m, and also v, the terminal velocities of the cones. From equation [6] we see that

v 2=

Cross-Sectional Surface Area Measurements: The solid angle assigned to our group was  = 0.5 steradians. Using equation [9] this was converted into the arc angle of the curve

√

4π −1=140.7 ° 0.5 From equation [11] we can convert this into the half angle of the cone m ϕ=0.5

❑m=sin−1

( 140.7 360 )

so ❑m=23 ° The slant height of the flat circular piece of paper, h, (as shown in figure 1(a)) was found to be h = (0.12  0.00005) m 2 2 Using equation [13] A=π h si n θm We get 2 2 A=π ( 0.12 ) si n (23) A = 6.9066 x 10-3 m2

Uncertainty calculations: As A is a function of two variables r and δA=

√

❑m ;

∂A ∂A (δ❑m )2 (δr)2 + ∂r ∂❑m ∂A =2 si n2 ( ❑m ) r ∂r

∂A =r 2 sin ( 2m ) ∂❑m

0.05 ¿ ¿ 2 2 2 2 ( 2 si n ( 23) ×0.12 ) ( 0.00005 ) +( 0.122 sin ( 46 ) ) ¿ δA=√ ¿ −3 δA=1.627× 10 m

So the cross-sectional area of the cone is given by A=( 0.007 ±0.001 ) m 2 Air density measurements: As this experiment was conducted on a rainy day, it was decided that the density of air in the West building would also need to be calculated from the ideal gas equation in order to accurately reflect the experimental conditions. Using the following air density equation3 p M +p M ρhumid air= d d v v RT With pd = (100952  1) Pa Md = 0.028964 pv = 129.18Pa Mv = 0.018 R = 8.314 T = (296.15  274.15) K The density of air in the stairwell was found to be ρhumid air=

(100952 × 0.028964 +129.18 × 0.018) 8.314 ×296.15 ρhumid air=1.18849 kg/m3

Uncertatinty Calculations:

Uncertainties were found using

( )

δP d ¿ ¿

δp 2 δp 2 2 ) ¿ (δT ) +( δT δ Pd δp= √¿

Md δp = δT R T K

(

δp −P d M d + Pv M v = 2 δT RTk

0.01 ¿ ¿ 2 100952 × 0.028 +129 × 0.018 2 0.0289 2 ) ¿ (1) +( 2 8.314 ×296.15 8.314 × 296.15 δp=√ ¿

)

So the density of air in the stairwell is given by

1.1885 ±0.0004 ) kg/m3 ρ=¿

Terminal Velocity Measurements: In order to measure the terminal velocity of four different cones of different masses, it was decided that there would be a person on each floor of the stairwell, who would each start their stopwatch when the cone was dropped from the 8th floor and then stop their stopwatch when

the cone passed them at the bannister level of the floor that they were on. This raw data is shown in tables 1.1 to 1.4 in the appendix. From this data, the time taken for each cone to travel between each floor can be calculated by subtracting the time it takes to reach one floor from the time it takes for the cone to reach the floor below. This data, showing the time it took for the cones to travel between each floor, is given in the following table.

Floors 87 76 65 54 43 3 2 21

Cone 1 (s) ( 0.05s)

Cone 2 (s) ( 0.05s)

1.46 1.40 1.49 1.14 1.46 0.85 1.61

1.28 1.14 1.21 0.77 1.16 1.09 1.23

Cone 3 (s) ( 0.05s) 1.08 1.11 0.70 1.14 0.93 0.75 1.30

Cone 4 (s) ( 0.05s) 0.82 1.07 0.65 0.43 1.31 0.29 1.85

Table 1. Time taken to travel between each floor for each cone As the distances between each floor, given by table 2.0 in the appendix, were also been measured, the velocity of each cone between each floor could be found from the velocity equation

v=

d t

This produced the following table, showing the velocity between each floor for each cone

Cone 1 (ms- Cone 2 (ms1 ) ) 8  7 2.7392 3.1250

Floors

1

76 65 54 43 3 2 21

2.7428 2.5704 3.3421 2.6095 4.4941 2.4534

Cone 3 (ms1 ) 3.7037 3.4594 5.4714 3.3421 4.0967 5.0933 3.0384

Cone 4 (ms1 ) 4.878 3.5887 6.0793 8.8604 3.7722 13.1724 2.1351

3.3684 3.1652 4.9480 2.3664 3.5045 3.2113 Table 2. Velocity between each floor for each cone

As these velocity values varied, we took an average of the velocity values of the cones flight between the last four floors, that is, from floors 5  4 through to floors 2  1 in order to find an average terminal velocity of the cones that could be plotted. It was also noticed that the values of the velocities of cone 4 varied so much within these last four floors, from 3.7733ms -1 between floors 4  3 and then 13.1724ms-1 from floors 3  2 that the data we had gathered for this cones flight was unreliable and could not be used as a terminal velocity point on the graph because it was not clear where, or whether the cone had even reached terminal velocity. Also, since this velocity v is a function of both distance measurements and time measurements, we could calculate an uncertainty using the appropriate uncertainty measurement

√

δv=

∂v 2 2 ∂v ( δt ) (δd ) + ∂t ∂d

And as

v=

d t

,

∂v ∂d

is given by

1 t

and

∂v ∂t

by

d . t2

For each cone, an average time between floors, t, for the bottom four floors was found, and an average distance, d, was also found. These values could then be used in this uncertainty

equation given above to produce uncertainties on the terminal velocities for the three cones. This is summarized in the table below taverage (s)  0.05 daverage (m)  0.021 Terminal Velocity av. (ms-1)

Cone 1 (2.32 kg) 1.27 3.847

Cone 2 (3.31 kg) 1.063 3.847

Cone 3 (4.15kg) 1.031 3.847

3.2247  0.2989

3.5075  0.5432

3.8926  0.3723

Table 3. Collated data for finding the terminal velocities of each cone. With this information, we could plot a graph using MatLab of the cones mass (in kg), given in table 1 of the appendix, against the square of their terminal velocity ((m/s)2) and the gradient of this graph would help us find the drag coefficient of each cone, as we knew

v 2=

2 gm C D Aρ

from equation [6].

This graph is shown below.

Graph 1. Determining the drag coefficient from the terminal velocity squared and masses of four paper cones.

The relevant MatLab code for this graph is given in the appendix. The gradient of this graph is equal to 3952  50 m.

From equation [6]; v 2=

2g ×m C D Aρ

2g =395250 C D Aρ If we call the gradient N, this equation can therefore be arranged for CD and written as We know that

CD=

2g A ρN

Substituting in the correct values for A,

ρ , g and N we get

CD=

2 × 9.81 =0.5966 (0.007 ) × ( 1.18849) × 3592

Since CD is a function of A, ρ and also N (the gradient value that has its own uncertainty value), then the appropriate uncertainty for C D can be calculated from

√

2

2

2

2

∂ CD ∂C D ∂ CD 2 ) (δρ) +( ) (δA)2 +( ) (δN ) δ CD= ( ∂ρ ∂A ∂N Where: ∂C D −2 g = ∂ ρ NA ρ2

∂C D −2 g = ∂ A N ρA2

∂C D −2 g = ∂ N A ρN 2

and with A = (0.007  0.001) m  = (1.1884  0.0004) kg/m3 N = (3952  50) (m/s/kg) The uncertainty on CD is therefore ¿

√(

)

2

2

(

)

2

2

(

)

2

2× 9.81 2× 9.81 2× 9.81 (0.0004 ) + ( 0.001 ) + (50)2 2 2 2 3952× 0.007 ×1.188 4 3952 × 0.00 7 × 1.1884 395 2 × 0.007 ×1.1884

¿ 0.61

Therefore C D =0.6 0.6

Conclusions The drag coefficient for paper cones in this experiment that had a solid angle of 0.5 steradians was found to be C D =0.6 0.6 . This has an uncertainty of 100%, which was not the desired uncertainty, however given the context of this experiment this large uncertainty value is not entirely unexpected. Not being able to use the data gathered from the motion of the fourth cone as it was unreliable and seemingly did not reach its terminal velocity would have affected this value of CD, and in addition to this we know that the way in which the experiment was conducted – in the stairwell of the West building on a windy day where the bottom floor was often exposed to a draft as doors opened and closed would have affected the results we obtained as the motion of the paper cones would also have been affected by this wind in the stairwell. If this experiment was to be conducted again, it would be beneficial to take more readings of the terminal velocity measurements and attempt to properly find where the cones reached terminal velocity for more masses, as this terminal velocity data was used to find CD along with its associated uncertainty, and with more accurate terminal velocity data, a more accurate CD value would be found. Another factor contributing to the large uncertainty on CD was just the natural error in the measurements we took along the stairwell due to the fact that 8 different people all have different reaction times when using a stopwatch. In addition to this is the fact that calibrating the experiment so that every person on each floor of the stairwell

started their stopwatch at the same time proved to be difficult, as the method this experiment used was to have one person on the middle floor yell loudly to signal when the cone had first been dropped. This method could have been improved by using a different method of communication, as making sure everybody started their stopwatch at the same time was important because it determined the time difference of the cones flight between the floors, and therefore the terminal velocity measurements that were gathered. We can see that plotting the CD value gathered in this experiment alongside the other values obtained by the class, there is a linear relationship between the solid angle value of a cone and its corresponding drag coefficient. A graph depicting the class values gathered is shown below. However, as there were disagreements between the group on how to report this large uncertainty, the value of CD that we obtained in our experiment has not been shown on this graph of class measurements.

Graph 2. Class drag coefficient values against their solid angle We can see from this graph that as the omega value of the solid angle increases, so does the drag coefficient of the cone. We know from equation [9] that there is a relationship between this solid angle and the arc angle of the circular section of paper, ∅ , and that this angle can then be converted into the half-angle of the crosssectional area of the cone using equation [11]. This cross-sectional area of the cone exposed to the air-flow is therefore dependent on the solid angle, and equation [13] can be re-written as

2

4π −1) ( ❑ 2 A=π h 2π Which, when used in the equation for finding the drag coefficient, 2g CD= A ρN Shows the relationship between the drag coefficient CD and the solid angle of the cone.

√

From this experiment, the drag coefficient of paper cones with a solid angle of 0.5 steradians was found to be equal to 0.6 0.6 , a unitless quantity that reflects the extent to which the cones motion down the stairwell was affected by a drag force opposing the downwards force of gravity.

Appendix: Masses of the Four Cones: Cone

Mass (kg)

1 2 3 4

2.32 3.31 4.15 5.14

x x x x

10-3 10-3 10-3 10-3

Raw data time measurements: The tables below show the times measured by each person from when the cone was dropped from the 8th floor to when it passed them on their floor. Cone 1 Trial #2 1.47 2.78

Trial #3 1.49 2.78

Avera ge 1.46 2.86

4.20 5.34 6.86 7.70 9.30

4.32 5.47 6.84 7.99 9.33

4.35 5.49 6.95 7.80 9.41

Trial #1 1.37 2.40 3.53 4.37 5.47 6.55 7.95

Trial #2 1.21 2.28 3.69 4.41 5.68 6.83 7.7...