· Web viewWord count: 3656 Abstract This essay investigates the how friction force is dependent from velocity of a falling object in water and whether the air resistance theory applies

Extended essay in physicsInvestigating air resistance phenomenon simulated in water and

determining how force of resistance is dependent from objects velocity.

Word count: 3656

Tomas Bilkauskas Extended essay in Physics 002390-0006

Abstract

This essay investigates the how friction force is dependent from velocity of a falling object in water

and whether the air resistance theory applies to falling objects in other mediums such as air. The

experiment conducted was dropping a ball in a glass vessel full of water and using camera capture

tools to measure the displacement versus time data that was later used graphically to get the velocity

and acceleration of the falling body. The cameras frame rate was not small enough to make

calculations accurate therefore I used a method of interpolation of the graphs to make slope

measuring accurate. The balls were dropped overall six times but only two drops were calculated

later on in the essay since they were the only ones really straight and since the end graphs of both

are very similar I assume that it is not needed to conduct the experiment more times as the end

results would have very small differences.

The friction force versus time graphs got were close to theory and are explained by the formula for

drag/friction force of water. The most interesting part of my results is that the terminal velocity got

is similar to what would be expected to get using the formula of terminal velocity so overall I can

say that the experiment proves that the air resistance theory is the same in water it only differs by

the conditions of water which is that it is more dense and more viscous forces are acting on the ball

during the fall than in air. This is probably the same for other mediums as well but should be

investigate furthermore.

Word count: 274

2Siauliai Didzdvaris gymnasium School Code: 002390


Table of contents..........................................................................................31. Introduction..........................................................................................41.1. Equipment description...............................................................................................................4

1.2. Procedure...................................................................................................................................4

2. Data collection, processing, presentation and analysis........................62.1. Raw data table...........................................................................................................................6

2.2. Displacement versus time..........................................................................................................7

2.3. Interpolated data tables and graphs...........................................................................................9

2.4. Velocity versus time................................................................................................................12

2.5. Acceleration versus time.........................................................................................................14

2.6. Friction force versus velocity..................................................................................................15

2.7. Analysis...................................................................................................................................18

3. Conclusion and Evaluation................................................................213.1. Conclusion...............................................................................................................................21

3.2. Evaluation................................................................................................................................21

4. Bibliography:.....................................................................................22



1. Introduction

At first I wanted to do an experiment involving air resistance like throwing different objects off the

balcony and measuring the difference in falling. But to get reasonably distinguishable

measurements I would need to drop these objects from a lot higher than a balcony which leads to

my inability to do the experiment because I cannot get the heights needed. So I decided to do it in a

medium in which the change in force and acceleration would be noticeable more quickly and in

lower heights. This gave me an interesting idea to determine if the theory of air resistance applies in

other mediums than air like water by simulating it and see if the simulations are accurate, according

to the theory, or not and what is the reason for inaccuracies. So my research question is how force

of resistance is dependent from objects velocity in water and whether the mechanics of air

resistance are similar in water like in air.

1.1.Equipment description

The equipment I used was:

1) Glass vessel (long and thin so that dropped object has enough distance to fall vertically and

its acceleration decrease noticeably but it does not have space to move horizontally)

2) Metal ball (for an object to drop, the shape of a ball was picked because I want it to move

only vertically and that it does not get too strong resistance from the shape that it instantly is

at the terminal velocity)

3) Video camera (In order to determine the velocity changes I decided to use a video camera

and then work with the footage using program called VideoPoint because balls fell down the

vessel very quickly and using something that can stop frames helps a lot)

4) Water

5) Pincers (Pincers were used to drop balls because the neck of the vessel was very thin to fit

the ball with fingers)

1.2.Procedure

The experiment itself is not very complicated but it has a lot of uncertainties that involve the

experiment itself and the calculations and graphical conversions of data. I filled the vessel with

water and started dropping the balls while recording them with a stable camera. First ball used was

very big and this lead to a lot of horizontal movement which was not good for accurate

measurement. Then I decided to use smaller ones. All of the small balls were the same size for

similar results and dropping them resulted in a straight fall. This might be compromised because the 4

Siauliai Didzdvaris gymnasium School Code: 002390


camera sees the ball in a 2D plane really. Ball is going from top to bottom and can move to left and

to right according to camera but it can also move towards the camera and away from it which the

camera would not capture it as the balls position in that plane is not visible. This movement can

lead to small inaccuracies between the drops if one drop is completely straight and the other is

moving in the plane that camera can see which I assume looking at the footage as straight.

Sometimes it got a bit of a curve to left or right as well which might be because my hand shook

while letting go. I dropped overall 6 balls then used the footage in VideoPoint program to measure

the time it took for the ball to fall a certain distance at different points in falling. This allows me to

graphically calculate the velocity and acceleration of the ball at different points in the fall. Other

inaccuracies can be for example the bending of light. If the ball drops not completely in the middle

the bending of the glass vessel the glass will bend the light making the camera capture inaccurate

movement of the ball. The inaccuracies in calculating come in as I am making a lot of conversions

to graphs. Taking one data table’s information putting it into a graph then getting its slope making

another graph from it which can result in greater inaccuracies if there was some errors in

interpolation of functions. So the experiment has to be conducted very carefully because a small

mistake at the start can make the end results vary a lot.



2. Data collection, processing, presentation and analysis

Using the footage from the experiment I noticed that some drops were very curved while some were

straight. So I only used three that were the most straight and the rest left as inaccurate.

2.1.Raw data table

Table 1.

First drop Second drop Third droptime,s ∆t=0.001

displacement,m ∆s=0.005

time,s ∆t=0.001


time,s ∆t=0.001


0.04 0.006

0.04 0.005

0.04 0.081

0.08 0.030

0.08 0.026

0.08 0.157

0.12 0.069

0.12 0.058

0.12 0.222

0.16 0.104

0.16 0.099

0.16 0.288

0.20 0.153

0.20 0.148

0.20 0.350

0.24 0.211

0.24 0.201

0.24 0.416

0.28 0.272

0.28 0.260

0.28 0.483

0.32 0.335

0.32 0.319

0.36 0.396

0.36 0.383

0.40 0.458

0.40 0.448

0.44 0.506

0.44 0.508

Digital SONY camera was used. I looked up the specifications of the model of the camera, which

was used in the experiment, from SONY page1 and found out that shutter speed varies from 1/50s to

1/4000s so the time uncertainty during filming could be around 0.001s. From the video the

uncertainty of displacement is somewhere 0.005m because I had to measure from the same point of

the ball which is hard to do and can vary. I process data using graphical methods. And in order to

get the change in velocity or acceleration I try to get the slope of a function of displacement versus

1 http://www.sony.co.uk/support/en/content/cnt-specs/DCR-TRV245E/list (Last accessed 4th February, 2015)6


http://www.sony.co.uk/support/en/content/cnt-specs/DCR-TRV245E/list


time. Slope=∆s∆ t

=v (this is instantanious velocity which menas I need very small differences in

time) My data is dependent on the frame speed of the camera which means my measurement is

every 0.04 seconds which is not small enough. In order to get more data values at smaller intervals I

interpolated the graph using curve fitting which lead to calculate values every 0.01 interval using

this function: f ( t )=1.94 t2−0.115 t+0.00262



2.2.Displacement versus time

Now using the data from the raw data table first of all I need to construct graphs which have the

function of displacement versus time. Using these graphs I can evaluate whether the data got from

the video point are useful and similar to theory. Also before I can use Graphical analysis to

accurately measure the slope therefore measuring the instantaneous velocity I will need these

graphs to get the function which I then will use for interpolation. The graphs constructed are shown

in Graph 1. , Graph 2. and Graph 3. below.

Graph 1.



Graph 2.

Graph 3.



The third drop graph is a straight line which is not useful for me because I need the change of force

against the change of velocity and in this graph its constant. So I decided not to interpolate the third

drop as it gives no results. The first and second graph look similar and seem to be accurate to the

theory. At first the balls have zero velocity as they are at rest. After they are dropped the the slope

of the function is increasing because displacement depend from the speed and since earths gravity

force is acting on them they have acceleration so the the velocity is inreasing. The rate at which

slope is changing is decreasing and at around 0.3 seconds reaching terminal velocity. This means

that acceleration of the balls is decreasing until it reaches zero which leads to an assumption that

some other force (Friction force) is acting on it that is increasing as the balls velocity increases.

When that force, because of speed, is equal to the gravity force the balls no longer have acceleration

leaving them at terminal velocity.

2.3.Interpolated data tables and graphs

Now in order to evaluate how force of friction is dependant from the velocity I need to get the

instantanious velocity at specific time intervals. This I decided to do graphically because the slope

of the function of displacement versus time gives velocity v=∆s∆ t . I use Graphical Analysis 3.1

program which allows me to manipulate graphics made from x and y cordinates points that I write

in. This tool helps me to find the slope but before it can be used accurately I need to have more

points. Instantanious velocity is gathered by calculating slope at very small time intervals and my

raw data table has too big of the intervals. But the graph that I got from raw data table lets me

determine the function of the grapth that I can interpolate. Using interpolation I can get as many

points as I want at any time intervals. The interpolated data tables are shown in Table 2. and Table

3. and drawn graphs from that data are shown in are shown in Graph 4. and Graph 5. below. The

uncertainties are the same as for raw data as the data varies by the same amount because this was

used to get more points and the data was got from function.



Table 2.

First graph interpolationt,s ∆t=0.001

s,m ∆s=0.005

t,s ∆t=0.001

s,m ∆s=0.005

t,s ∆t=0.001

s,m ∆s=0.005

t,s ∆t=0.001

s,m ∆s=0.005

0.00

0.000

0.11

0.053

0.22

0.184

0.33

0.350

0.01

0.001

0.12

0.063

0.23

0.198

0.34

0.365

0.02

0.002

0.13

0.073

0.24

0.213

0.35

0.381

0.03

0.005

0.14

0.083

0.25

0.228

0.36

0.396

0.04

0.008

0.15

0.094

0.26

0.243

0.37

0.411

0.05

0.012

0.16

0.106

0.27

0.258

0.38

0.426

0.06

0.017

0.17

0.118

0.28

0.273

0.39

0.440

0.07

0.023

0.18

0.130

0.29

0.288

0.40

0.455

0.08

0.029

0.19

0.143

0.30

0.304

0.41

0.469

0.09

0.037

0.20

0.156

0.31

0.319

0.42

0.483

0.10

0.045

0.21

0.170

0.32

0.335

0.43

0.496

0.4

40.509

Graph 4.





Table 3.

Second graph interpolationt,s ∆t=0.001

s,m ∆s=0.005

t,s ∆t=0.001

s,m ∆s=0.005

t,s ∆t=0.001

s,m ∆s=0.005

t,s ∆t=0.001

s,m ∆s=0.005

0.02

0.001

0.13

0.068

0.24

0.201

0.35

0.368

0.03

0.003

0.14

0.077

0.25

0.215

0.36

0.384

0.04

0.007

0.15

0.088

0.26

0.230

0.37

0.400

0.05

0.011

0.16

0.099

0.27

0.245

0.38

0.416

0.06

0.015

0.17

0.110

0.28

0.260

0.39

0.432

0.07

0.021

0.18

0.122

0.29

0.275

0.40

0.447

0.08

0.027

0.19

0.134

0.30

0.290

0.41

0.463

0.09

0.034

0.20

0.147

0.31

0.306

0.42

0.478

0.10

0.041

0.21

0.160

0.32

0.321

0.43

0.494

0.11

0.050

0.22

0.173

0.33

0.337

0.44

0.509

0.12

0.058

0.23

0.187

0.34

0.353

Graph 5.





2.4.Velocity versus time

Now that I have many points at very small intervals I use the Graphic analysers tool to get the slope

at specific time intervals which are the instantaneous velocity at those intervals the uncertainties

now for those velocities are equal to sum of proportional uncertainties of time and displacement

multiplied by the magnitude of velocity (calculated values displayed in Table 4. and Table 5.) and

time uncertainty is the same. The data tables of instantaneous velocity of first and second drops are

shown in Table 4. and Table 5. and the graphs of the instantaneous velocity of the balls created

from those tables are shown in Graph 6. and Graph 7. below. Now the graphs that I expect to get

from the instantaneous velocity should have the slope that is equal to instantaneous acceleration at

specific time intervals as a=∆v∆ t . The function should at first be increasing quickly and then

decrease because acceleration at first should be close to 9.8 m/s2 and as the balls are reaching

bottom that is when the balls speed is increasing the acceleration should reach zero. Interpolation

for these graphs are no longer necessary since the graph is already created from the interpolated

graph of displacement versus time which means graph will already have many points with small

enough time intervals.

Table 4.

Instantaneous velocity of the first drop

t,sv,m/s ∆v, m/s t,s

v,m/s ∆v, m/s t,s

v,m/s ∆v, m/s t,s

v,m/s ∆v, m/s

0.03 0.3 0.3

0.14 1.08 0.07

0.25 1.49 0.04

0.36 1.51 0.02

0.04 0.4 0.2

0.15 1.13 0.07

0.26 1.50 0.04

0.37 1.49 0.02

0.05 0.5 0.2

0.16 1.18 0.06

0.27 1.52 0.04

0.38 1.47 0.02

0.06 0.5 0.2

0.17 1.23 0.06

0.28 1.53 0.03

0.39 1.45 0.02

0.07 0.6 0.1

0.18 1.27 0.06

0.29 1.54 0.03

0.40 1.43 0.02

0.08 0.7 0.1

0.19 1.31 0.05

0.30 1.54 0.03

0.41 1.40 0.02

0.09 0.8 0.1

0.20 1.35 0.05

0.31 1.55 0.03

0.42 1.36 0.02

0.10 0.8 0.1

0.21 1.38 0.05

0.32 1.55 0.03

0.43 1.33 0.02

0.11 0.90 0.09

0.22 1.41 0.04

0.33 1.54 0.03

0.44 1.31 0.02

0.12 0.96 0.08

0.23 1.44 0.04

0.34 1.54 0.03



0.13 1.02 0.08

0.24 1.46 0.04

0.35 1.52 0.02



Graph 6.

Table 5.

Instantaneous velocity of the second dropt,s v,m/s ∆v, m/s t,s v,m/s ∆v, m/s t,s v,m/s ∆v, m/s

0.05 0.4 0.2 0.15 1.06 0.07 0.26 1.47 0.040.06 0.5 0.2 0.16 1.11 0.06 0.27 1.49 0.040.07 0.6 0.1 0.17 1.16 0.06 0.28 1.51 0.030.08 0.7 0.1 0.18 1.20 0.06 0.29 1.53 0.030.09 0.7 0.1 0.19 1.24 0.05 0.30 1.54 0.030.10 0.8 0.1 0.20 1.28 0.05 0.31 1.56 0.030.11 0.84 0.09 0.21 1.32 0.05 0.32 1.57 0.030.12 0.90 0.08 0.22 1.35 0.05 0.33 1.57 0.030.13 0.96 0.08 0.23 1.39 0.04 0.34 1.58 0.030.14 1.01 0.07 0.24 1.42 0.04 0.35 1.58 0.03

0.25 1.44 0.04 0.36 1.58 0.02



Graph 7.

I got very similar graphs to what I was expecting above. At first the balls’ speed was increasing

quickly and as they got closer to the ground acceleration slowly decreased and finally reach zero at

around 0.30 seconds as expected from the displacement versus time graphs because at that time the

displacement was changing at a constant speed.

2.5.Acceleration versus time

As said above these graphs now allow me to get the instantaneous acceleration values of each drop

in order to see how the acceleration is decreasing. This is also done using Graphical analyser and

measuring the slope at every time interval. The data values of the instantaneous acceleration are

shown in Table 6. below (uncertainties are calculated in the same way the velocity uncertainties

were calculated. It is the sum of proportional uncertainties of time and velocity, which were

calculated before, times the magnitude of acceleration at that time interval).



Table 6.

Instantaneous acceleration of the first dropt,s a,m/s2 ∆a m/s2 t,s a,m/s2 ∆a m/s2 t,s a,m/s2 ∆a m/s2

0.04 9 6 0.13 5.8 0.5 0.22 2.9 0.10.05 8 4 0.14 5.5 0.4 0.23 2.61 0.090.06 8 3 0.15 5.1 0.3 0.24 2.29 0.070.07 8 2 0.16 4.8 0.3 0.25 1.98 0.060.08 7 1 0.17 4.5 0.2 0.26 1.66 0.050.09 7 1 0.18 4.2 0.2 0.27 1.34 0.040.10 6.7 0.9 0.19 3.9 0.2 0.28 1.03 0.030.11 6.4 0.7 0.20 3.6 0.1 0.29 0.71 0.020.12 6.1 0.6 0.21 3.2 0.1 0.30 0.39 0.01

Instantaneous acceleration of the second dropt,s a,m/s2 ∆a m/s2 t,s a,m/s2 ∆a m/s2 t,s a,m/s2 ∆a m/s2

0.06 7 3 0.16 4.8 0.3 0.26 2.38 0.070.07 7 2 0.17 4.5 0.3 0.27 2.12 0.060.08 7 1 0.18 4.3 0.2 0.28 1.87 0.050.09 6 1 0.19 4.1 0.2 0.29 1.63 0.040.10 6.2 0.9 0.20 3.8 0.2 0.30 1.39 0.030.11 6.0 0.7 0.21 3.6 0.1 0.31 1.15 0.030.12 5.8 0.6 0.22 3.3 0.1 0.32 0.90 0.020.13 5.5 0.5 0.23 3.1 0.1 0.33 0.66 0.010.14 5.3 0.4 0.24 2.84 0.09 0.34 0.418 0.0080.15 5.0 0.4 0.25 2.60 0.08 0.35 0.176 0.003

As the tables show at start balls’ acceleration is around 7-9 m/s2 which is close to free falling

acceleration. As the ball was closer to the bottom of the glass vessel the acceleration got smaller

until it reached zero and the balls were in terminal velocity.

2.6.Friction force versus velocity

From these tables I use the formula to get the Friction force and then construct a graph with

function of Friction force versus velocity which show how Force of friction is dependent from

velocity of each drop. The formula I use is: F f = m(g-a) where (Ff) is Friction force, (m) is mass of

the metal balls, (g) is free falling acceleration and (a) is the instantaneous acceleration of the metal

ball. So using the table of instantaneous acceleration and inserting the acceleration into the formula

and taking that balls mass was 0.075kg and both balls have the same mass I get the table of velocity

and force which is shown in Table 7. below.



Table 7.

Instantanious velocity and Friction force for the first dropv,m/s ∆v, m/s Ff,N ∆F, N v,m/s ∆v, m/s Ff,N ∆F, N v,m/s ∆v, m/s Ff,N ∆F, N

0.4 0.2 0.09 0.06 1.02 0.08 0.30 0.03 1.41 0.04 0.52 0.020.5 0.2 0.11 0.05 1.08 0.07 0.33 0.02 1.44 0.04 0.54 0.020.5 0.2 0.14 0.04 1.13 0.07 0.35 0.02 1.46 0.04 0.56 0.020.6 0.1 0.16 0.04 1.18 0.06 0.37 0.02 1.49 0.04 0.59 0.020.7 0.1 0.18 0.04 1.23 0.06 0.40 0.02 1.50 0.04 0.61 0.020.8 0.1 0.21 0.03 1.27 0.06 0.42 0.02 1.52 0.04 0.63 0.020.8 0.1 0.23 0.03 1.31 0.05 0.44 0.02 1.53 0.03 0.66 0.02

0.90 0.09 0.25 0.03 1.35 0.05 0.47 0.02 1.54 0.03 0.68 0.020.96 0.08 0.28 0.03 1.38 0.05 0.49 0.02 1.54 0.03 0.71 0.02

Instantanious velocity and Friction force for the second dropv,m/s ∆v, m/s Ff,N ∆F, N v,m/s ∆v, m/s Ff,N ∆F, N v,m/s ∆v, m/s Ff,N ∆F, N

0.5 0.2 0.19 0.07 1.11 0.06 0.38 0.02 1.47 0.04 0.56 0.020.6 0.1 0.21 0.06 1.16 0.06 0.39 0.02 1.49 0.04 0.58 0.020.7 0.1 0.23 0.05 1.20 0.06 0.41 0.02 1.51 0.03 0.59 0.020.7 0.1 0.25 0.04 1.24 0.05 0.43 0.02 1.53 0.03 0.61 0.020.8 0.1 0.27 0.04 1.28 0.05 0.45 0.02 1.54 0.03 0.63 0.02

0.84 0.09 0.29 0.03 1.32 0.05 0.47 0.02 1.56 0.03 0.65 0.010.90 0.08 0.30 0.03 1.35 0.05 0.49 0.02 1.57 0.03 0.67 0.010.96 0.08 0.32 0.03 1.39 0.04 0.50 0.02 1.57 0.03 0.69 0.011.01 0.07 0.34 0.03 1.42 0.04 0.52 0.02 1.58 0.03 0.70 0.011.06 0.07 0.36 0.03 1.44 0.04 0.54 0.02 1.58 0.03 0.72 0.01

Using this table I can construct the graph with function Friction force versus velocity which shows

how Friction force is dependent from velocity. The problem with this is that I could not create a

function that would allow me to get the Friction force graphically. This means my calculations for

Friction force can have a big inaccuracy if any data error were entered in graphical

calculations/measurements before or if there was any inaccuracies with the conducted experiment.

The constructed graphs of Friction force versus velocity are shown in Graph 8. and Graph 9. below.



Graph 8.

Graph 9.



The two graphs above are very similar to each other. At first velocity is increasing quickly as is the

friction force but as balls fall the speed at which velocity changes (acceleration) is decreasing. The

speed of change of friction force at the end is decreasing slower than velocity therefore it bends

upwards. At the end of the graph the graph cuts out because terminal velocity is reached. At this

point velocity does not change any more just like friction force so the graph cannot go further

beyond that point. Since the two graphs are very similar it seems that more drops were not

necessary as the result graph would not give any particular difference that could be analysed. So the

two drops are sufficient for determining the dependency of friction force versus time.

2.7.Analysis

The Table 7. is now the most important table as it holds my research question’s answer of how is

friction force dependant from velocity. At the start it is expected that there is no friction force

exerted on the ball from Table 7. it is shown that the friction force is a little bigger than zero that is

0.08886N for the first drop and 0.19458N for the second. This is because the calculations are not

exactly at the start. The starts of the graphs are thrown away as it needs two additional points to

calculate the slope and for the first point overall only two points are available and to make it even

more accurate I left out two points from the start. This was also done for the ends of the graphs.

And since the slope was calculated not only for one but for two different graphs the start and end

cut off two times leaving our last graph starting from around 0.04 seconds of the fall instead of the

start. Having that in mind the start friction forces are small enough to assume that at start they were

zero. Now the maximum friction force that can be exerted on the balls cannot exceed the weight of

the ball that is around 0.75 N (The balls mass is 0.075kg. 0.075 * 9.8m/s2). The friction force at the

end of the falls are 0.70545N and 0.7218N which are close to the expected weight of the balls.

Friction forces at start and end of the falls in air should be the same so trying to compare these to

some data gathered from research is not leading to better understanding of how the friction force is

different in water. Now terminal velocity on other hand could give a lot of information. At the end

of the fall where the balls reach terminal velocity speeds are 1.544593 m/s and 1.582196 m/s

considering that they are not exactly reached terminal velocity 1.6 m/s is a close approximation of

the true terminal velocity in my experiment. Terminal velocity is dependant from four variables the

weight of the body, drag coefficient, density of the gas, and area of the object (V=√ 2∗WCd∗ρ∗A

, V –

Terminal velocity, W – weight of the body, Cd – drag coefficient, ρ – density of the gas, A –

Frontal area)2 („The drag coefficient is a number that aerodynamicists use to model all of the

2 Source http://www.grc.nasa.gov/WWW/k-12/airplane/termv.html (Last accessed 4th February, 2015)22


http://www.grc.nasa.gov/WWW/k-12/airplane/termv.html


complex dependencies of shape, inclination, and flow conditions on aircraft drag“3). These

determinants of the terminal velocity were given for the gasses. If in our case we take water, what

determinants of terminal velocity change: the weight is the same, frontal area is the same. The only

changes are the density of the medium, which is a lot bigger than air, and Coefficient of the drag.

The drag coefficient is changed as it is dependent not only from the shape of the object but from the

flow of the object in the medium in other words it depends from the viscosity of the medium. In

water viscous forces should be stronger than in air because the surface of the object is being acted

by a bigger amount atoms of water therefore I assume that drag coefficient should be bigger as well.

A small golf ball weighs around 0.046 g which is a little bit less than weight of my ball and in air its

terminal velocity is around 32 m/s.4 Now assuming that my ball‘s terminal velocity in air is of the

golf ball‘s that is 32 m/s the ratio is VaVw

= 32m /s1.6m / s

=20. That is in water ball has 20 times slower

terminal velocity. Now assuming if it was only dependant from the density

VaVw

=√ ( 2∗WCd∗ρa∗A

)

√( 2∗WCd∗ρw∗A

)=√ ρw√ ρa

= √ 998√1.293

=28 (ρw – water density, ρa – air density).5 If the only

difference was density terminal velocity should be 28 time slower which knowing that because of

the drag coefficient difference would be even bigger makes our 20 time slower terminal velocity

seem inaccurate. This is of course because I took terminal velocity of a golf ball. My ball was

heavier and is smoother and has smaller frontal area than the golf ball therefore meaning that

terminal velocity of metal ball in air would be greater than 32 m/s leaving with the assumption that

the 28 time slower or more is plausible as VaVw of my ball in reality is bigger than 20. Also any more

points that I could have gotten from the experiment would have been put at the same point in the

graph when terminal velocity was reached as the velocity and friction force would not change past

that point. Thus in the graphs near that point where terminal velocity is reached that bending of

function complicates my finding of a function that could define the relation of velocity and friction

force, so I decided to only take the beginning of the graph, that is v ≈ 0.5m/s, till near the end, v ≈

1.41m/s. At this part the relation seems to be close to quadratic function which seems viable as drag

function in fluids is depending on the square of velocity (As seen in Graph 10. and Graph 11.

3 Source http://www.grc.nasa.gov/WWW/k-12/airplane/dragco.html (Last accessed 4th February, 2015)4 Source http://hyperphysics.phy-astr.gsu.edu/hbase/airfri2.html (Last accessed 4th February, 2015)5 Source of the densities http://hyperphysics.phy-astr.gsu.edu/hbase/tables/density.html (Last accessed 4th February, 2015)


http://hyperphysics.phy-astr.gsu.edu/hbase/tables/density.html

http://hyperphysics.phy-astr.gsu.edu/hbase/airfri2.html

http://www.grc.nasa.gov/WWW/k-12/airplane/dragco.html


below.)(Formula for drag force in fluids - FD=12ρu2CD A; FD- the drag force, which is by

definition the force component in the direction of the flow velocity, ρ - the mass density of the

fluid, u - is the flow velocity relative to the object, CD- is the drag coefficient – a dimensionless

coefficient related to the object's geometry and taking into account both skin friction and form drag,

A- is the reference area)6.

Graph 10.

Graph 11.

6 Formula taken from http://en.wikipedia.org/wiki/Drag_equation (Last accessed 4th February, 2015)24


http://en.wikipedia.org/wiki/Drag_equation


So I would assume my Graph 8. and Graph 9. are accurately showing the dependence of friction

force versus velocity.

3. Conclusion and Evaluation

3.1.Conclusion

Calculations made in analysis and theory seem to prove accuracy of my graphs showing the

dependency of friction force versus velocity. As ball is falling its acceleration is decreasing leaving

velocity increasing in a slower rate but the friction force is dependant from the velocity squared

which means that as velocities rate of change slows down friction force is still increasing more by

square (F f=Cd∗ρ∗V 2∗A

2).7 This leads to that strange tail curve upwards at the end of the graph

also the middle of the graph show a relation of a quadratic function expected from the formula of

drag in fluids. The terminal velocity reached and relationship of friction force versus time is

plausible which lets me assume that the experiment conducted was successful.

3.2.Evaluation

Although at first it I expected great inaccuracies because experiment had a lot dependencies from

my dropping and many transformations of graph could have led to errors in calculations the

conclusion is plausible. Now it would be interesting to do the same experiment with other variables

like shape or size or material of the ball. The most interesting change would be try it in more than

one medium. Something like food oils, or honey also maybe try to increase density of water with

trying to dissolve some salts in it or other kinds of mean changing the density of the medium and

getting the graph from those and then comparing them could lead to interesting results.

7 Formula from http://www.grc.nasa.gov/WWW/k-12/airplane/termv.html (Last accessed 4th February, 2015)25




4. Bibliography:

http://www.grc.nasa.gov/WWW/k-12/airplane/termv.html (Last accessed 4th February,

2015)

http://www.grc.nasa.gov/WWW/k-12/airplane/dragco.html (Last accessed 4th

February, 2015)

http://hyperphysics.phy-astr.gsu.edu/hbase/airfri2.html (Last accessed 4th February,

2015)

http://hyperphysics.phy-astr.gsu.edu/hbase/tables/density.html (Last accessed 4th

February, 2015)

http://www.sony.co.uk/support/en/content/cnt-specs/DCR-TRV245E/list (Last

accessed 4th February, 2015)

http://en.wikipedia.org/wiki/Drag_equation (Last accessed 4th February, 2015)


http://en.wikipedia.org/wiki/Drag_equation

http://www.sony.co.uk/support/en/content/cnt-specs/DCR-TRV245E/list

http://hyperphysics.phy-astr.gsu.edu/hbase/tables/density.html

http://hyperphysics.phy-astr.gsu.edu/hbase/airfri2.html

http://www.grc.nasa.gov/WWW/k-12/airplane/dragco.html


Documents

· Web viewWord count: 3656 Abstract This essay investigates the how friction force is dependent from velocity of a falling object in water and whether the air resistance theory applies