and 1kg respectively they will have gravitational forces of 100N and 10N,
they will begin to accelerate at the same speed when air resistance is 0N,
but if both encounter the same air resistance the object with the smaller
mass will slow more quickly whereas the larger object will slow more
slowly as you can see from these two tables below which shows the
resultant downward force which is the gravitational force minus air
resistance and acceleration.
Object with mass of 10Kg
Object with mass of 1Kg
Resultant Force in N
Acceleration in m/s/s
Resultant force in N
Acceleration in m/s/s
100
10
10
10
99
9.9
9
9
98
9.8
8
8
97
9.7
7
7
96
9.6
6
6
95
9.5
5
5
94
9.4
4
4
93
9.3
3
3
92
9.2
2
2
91
9.1
1
1
90
9
0
0
This shows that more massive objects accelerate at a larger rate and for
longer than less massive objects meaning that they will have a higher
speed. The force diagram below also shows this.
Air Resistance force is the same
it = 1N
Mass = 10Kg Mass = 1Kg
Gravitational force = 100N Gravitational force = 10N
Resultant force = 99N Resultant force = 9N
So the acceleration can be worked out by:
99/10 = 9.9/s2 9/1=9m/s2
so it is clear to see that the more massive object has a greater acceleration.
I have also found how to work out the acceleration that an object will fall
at by using certain equations.
Weight is constant W= M x G
Resistance (Drag) depends D = cd (r x V2) A
on square of velocity 2
cd = drag coefficient (worked out experimentally) by using the formula:
cd = D
r A V2/2
r = air density (which is 1.29kg/m3 at room temperature.
V = Velocity
A = area which coefficient is based
Finally we can put the results that we have obtained from the previous
steps into the following equation to find the acceleration:
a = F a = (W - D)
M M
Now that I have researched the general points that are relevant to my
experiment I will state the fact or that will affect my experiment in more
detail. The first factor is the bottom surface area of the object that I will be
dropping which will be in contact with the air. If the area is bigger then it
will come into contact with more air particles and will use up more energy
moving them thus affecting the rate that it will fall at. In my experiment this
is the factor that I will be changing. All other factors will remain constant.
The height that I will drop the object from would affect the time that the
object would take to reach the ground, as this is what I will be measuring
this will remain constant as each one will be dropped from an equal height.
The mass of the object will remain the same as more massive objects fall
faster than less massive objects as I explained earlier, so the mass of the
object will remain constant. I will do this by using the same piece of paper
for all experiments and simply folding it to adjust the surface area. The
temperature will have to remain constant so that the air density will also
remain constant. This will have to remain constant as if the density were to
change then the amount of air particles hitting the falling object could
increase or decrease depending on how the density changes.
To decide which object I will drop I have done some preliminary work. I
decided to choose a range of objects and see how easy it would be to
time them falling and change there surface area without changing there
weight. Firstly I thought of using some kind of different sized balls such
as football squash ball etc. but found it impossible to keep there weight
the same also they tended to fall very quickly making it difficult to time. I
then thought of rather than changing surface area changing weight I tried
filling a tennis ball with various amounts of sand to change it weight it was
easy to do but again very difficult to time accurately over a small height. I
then decided to try paper I first found it difficult to keep weight constant
as I added glue to stick the write weight of paper together to make the
write surface areas but then found I could fold the same piece of paper
into different areas meaning the weight stayed exactly the same. I also
found this easy to time as it fell slowly. So I have decided to use paper for
the object I will drop.
In this experiment all the factors that I have mentioned will be controlled.
The only factor that will change is the surface area. I have chosen to
change this because I know that surface area affects air resistance and
therefore I will be able to change the surface area and if there is a change
in my results I will know that air resistance does have a significant affect
on the time it will take for objects of the same mass to fall. The data that I
will record will be the time that it takes the objects to fall. This diagram
shows the equipment that I will use and how I will set it out.
Piece of Paper
Meter
Rule
Table
1.60M
00:00:00 Stopwatch
Floor
Using this equipment I will drop 6 different sizes of paper, all of the same
mass, from the same height which will be 1.6m every time and record the
time it takes the paper to fall to the ground. For each size of paper I will
drop the paper 10 times to account for any anomalous results. I will start
the stop watch when I release the paper and stop it when the paper hits the
floor. I will use the same piece of paper for all the tests to ensure that the
mass remains constant, I will change the surface area by folding the paper.
Also to ensure that the paper falls in a straight line downwards I will fold
the paper so that the middle is higher than the edges so that it does not
drift, like in this diagram.
The shape of the paper on the left will work better as it will catch the air
and this will prevent the paper from drifting. However the one on the right
will simply deflect the air so that it will drift and it will also reduce air
resistance because it will travel faster.
The surface areas that I will be using will be to start with 21cm x 21cm, I
will then use ½ this area then ¼ followed by 1/8th, 1/16th and finally 1/32
of the area. I will then take an average of these times for each piece of
paper and record this average on a graph against surface area.
For this investigation using the scientific research I have done I can
predict what the results will be. From the information I have gathered I can
say that because of the inertia in air particles when an object passes
through the air it must expand energy upon these particles to move them
out of its way. When we change the surface area we can also change the
amount of particles that the object will hit as it falls and thus the amount of
energy it will have to expend on these particles to get them to move out of
its way. So I can predict that the larger the surface area the more particles
it will have to move to pass throughout the air thus it will incur a greater air
resistance and accelerate through the air more slowly and take longer to
fall to the ground. So the larger the surface area the slower an object will
fall through the air and in my experiment this means that the larger the
surface area of the paper the longer it will take to fall.
Obtaining
In my experiment I did the following. I set up the equipment first in the
way in which I showed in my method. Then I cut the paper that I was
using. I did this by folding a piece of A4 paper from the corner to create a
square and cutting of the excess. I then folded it in the way shown earlier
so that it would not drift. I then found the point on the ruler which I knew
was 1.29m and held the paper there. I checked that there was no draft. I
took the stopwatch in my hand and made sure it was reset to 0. I then held
the paper by the tip I had made in the middle in one hand held the
stopwatch in the other hand. I then let go of the paper and started the
stopwatch simultaneously. I then watched the paper fall down to the
ground and stopped the stopwatch when I saw it hit the ground. I then
repeated this 10 times for the first area. Then I folded the piece of paper
so that the surface area on the bottom side was half as big but the mass
was still the same. I then repeated the steps I mentioned earlier 10 times
for this area. I then folded it in half again to make it ¼ of the size and
tested it 10 times then folded it in half again to make 1/8th and tested it 10
times. I then folded it in half again to make 1/16th and tested it 10 times
and finally folded it in half again to make 1/32nd and tested it 10 times.
Results for experiment
Surface Area in cm2
Time to fall in seconds
Average Time of Fall in Seconds
441
1.65
1.488
1.44
1.41
1.44
1.41
1.50
1.47
1.56
1.53
1.47
220.50
1.15
1.046
0.97
1.03
1.10
1.09
1.09
0.97
1.03
0.97
1.06
110.25
0.88
0.853
0.81
0.88
0.90
0.82
0.87
0.85
0.87
0.84
0.81
55.13
0.69
0.673
0.69
0.65
0.62
0.72
0.63
0.75
0.66
0.63
0.69
27.56
0.59
0.573
0.53
0.60
0.56
0.59
0.50
0.60
0.59
0.57
0.60
13.78
0.44
0.469
0.50
0.50
0.46
0.47
0.41
0.44
0.47
0.50
0.50
Analysis
The results from my experiment as shown on my graph, show that the
greater the surface area the greater the time taken to fall. You can see this
as there is a curve which connects all but one of the results which is
slightly off the curve. The curve starts at the lowest surface area, 13.78cm2
which took an average of 0.469 seconds to fall up to the largest surface
area of 441cm2 which took an average of 1.488 seconds to fall. This
shows that the surface area is proportional to the time taken for the paper
to fall. These results agree with my prediction.
The scientific reasons for why this happened are that because when the
pieces of paper with the bigger surface areas fall they incur more air
resistance so the resultant downward force they have that they have is less
than the pieces of paper with smaller surface area and less air resistance.
There is a larger air resistance on a larger surface area because there are
more air particles hitting the piece of paper. This means that because each
of these particles has its own inertia more energy will be used to accelerate
more of these particles away from the falling piece of paper. However the
piece of paper with the smaller surface area will have to move a smaller
amount of these particles so it will not loose as much energy and therefor
the resultant downward force will be greater. This is shown in this
diagram:
air resistance
air resistance
paper with paper with large surface area
small area
Particles of air
Particles of air
Resultant
force
Resultant force
Gravity Gravity
We can also use a formula to work out the resistance so that we can
compare the amount of resistance in each of the surface areas. The
formula is;
D= cd ( r x V2 ) A
2
cd = drag coefficient, worked out experimentally by using the formula
cd = D
r A V2/2
r = air density ( which is 1.29kg/m3 at room temperature)
V = velocity
A = area which coefficient is based
D = drag force in newtons
I can work out the drag coefficient first by using my results from my
experiment for the largest surface area, 44l cm2. I can work out its actual
acceleration and then take this away from the acceleration it would have
had from gravity and find the force of air resistance. How ever because of
the nature of my experiment I can only use the average velocity.
Velocity = displacement acceleration = change in velocity
time time
Velocity = 1.60m = 1.075268817 m/s
1.488 sec
Acceleration = 1.075268817 m/s = 0.722626893 m/s2
1.488 sec
I also need to work out the mass of the piece of paper. The area and mass
of an A4 piece of paper is 623.7cm2 and 0.005kg so I can find the mass
of my 441cm2 piece of paper as follows;
0.005kg x 441cm2 = 0.003535353kg
623.7cm2
Now with this information I can work out the force of drag so that I can
work out the drag coefficient.
Acceleration = force so force = mass x acceleration
mass
gravitational force = 0.003535353535kg x 9.8m/s2
= 0.0346464646N
resultant downward force = 0.003535353535kg x 0.722626893 m/s2
for 441 area
= 0.002554741N
so the air resistance = gravitational force - resultant downward force
or drag force
= 0.0346464646N - 0.002554741 N
=0.032091723N
Now I can put this information into the drag coefficient equation;
cd = D
r A V2/2
cd = 0.032091723N
1.29kg/m3 x 0.0441m2 x ( 1.075268817m/s)2/2
cd = 0.487899791
Using this I can now work out the drag whatever velocity, area or air
density. I can also work out the terminal velocity. However for the
equation to work the mass of the object must not change. Below are my
results for all the different areas. These results are based on the fact that all
pieces of paper are travelling at the same velocity of 1.075268817m/s.
Surface area in m2
Air resistance in N
0.0441
0.032091723
0.02205
0.016045861
0.011025
0.008022931
0.005513
0.004011465
0.002756
0.002005733
0.001378
0.001002866
These results show that air resistance is directly proportional to the
surface area. These results show that when the surface area halves the air
resistance also halves exactly. This is true for every result. The results for
air resistance are also proportional to the amount of time it took for each
surface area to fall. The 0.0441m2 area had the largest air resistance and
also took the longest amount of time to fall. This shows that the greater
the surface area the greater the air resistance and this is the reason why the
larger surface areas took longer to fall than the smaller ones.
I have also found a formula that can work out the time it takes an abject to
fall in a vacuum without air resistance. This will show the affect of air
resistance on the time it takes for an object to fall. The formula is;
s = ut + ½at2
s = height of fall
u = initial velocity
t = time to fall
a = acceleration of fall
To find the time for the fall I must re-arrange the formula. Because the
value of u is 0 the value of ut must also be 0 so this can be taken out of
the equation so we are left with;
s = ½at2
This re- arranges to:
t = s
½ x a
I can now put the variables from my experiment into this equation to find
the time it takes for an object to fall without air resistance over the same
height that my object fell. Because there is know air resistance this will
apply to all objects of different surface areas and masses.
t = 1.60m
0.5 x 9.8N
t = 0.258145115 seconds
This shows that with no air resistance an abject would take 0.258 seconds
to fall 1.6m . This is about 0.2 seconds less than the smallest of my
surface areas. When I saw the formula used I decided that it may be
related to my results as well I found that the graph that I had was similar to
the graph of:
y = x
Below are the graphs of my results and y = x
This tells me that there is a relationship between the surface area of the
paper and the time it takes for paper to fall. This relationship would
include tacking the square route of the area and then a constant would
have to be found which would translate the number into an accurate
estimate of the time it would take the paper to fall in real life.
In conciliation my results have agreed with my prediction, the larger the
surface area the larger the air resistance and the smaller the time taken for
the paper to fall. I have also seen that the surface area and the time taken
to fall are related to the formula y= x . I also found that this could be
used to estimate the time when the area is known if a constant was found.
Evaluation
On the whole my experiment was successful, I was able to draw relevant
conclusions from my results that allowed me to explain why I got the
results that I did. I felt that the results were accurate enough as I repeated
each test 10 times and made an average this would have accounted for any
anomalous results. However when I began to put the data into the relevant
equations that I used to show that the results from other experiments
could be accurately predicted I found that if the results had been more
accurate I would have been able to use the equations to a fuller extent and
been able to predict the results with greater success. The problems which
I encountered during this experiment were that the paper did not always
drop straight down because of any slight movement in the air that
surrounded the test area. I also found it difficult to drop the paper and
time it by myself.
The way in which I conducted my experiment could have been better in
the following ways, when dropping the paper I used my eyes to judge
whether the height was the same and then dropped it with my hand, I
could have used a mechanism that dropped the paper in exactly the same
way from exactly the same height for each of my experiments. Also the
way in which I measure the time for the paper to drop could have been
better. The way in which I measured it was to judge with my eyes when
the paper was released and then when it hit the floor, I would start and
stop the stop-watch accordingly. I could have mad this more accurate by
using pressure sensitive sensors on the releasing mechanism and on the
floor that would start a stop watch.
Other variables such as the room temperature air density and any wind
could also be controlled better in my experiment. They could be measured
and kept constant for each experiment. My experiment could be extended
by controlling more data from the same variables that I used. I could use
another measuring device such as a ticker timer to measure acceleration
and velocity of the falling object at any point. This would mean that I
could use the equations that I have used more accurately and fully. I could
also conduct further experiments where I would try and accurately predict
the results using the formula and test it to see how accurate it is. I would
do this with other variables and test them.
Overall my experiment allowed me to answer the question that was put to
me and my results were accurate enough to allow me to make conclusions
that answered this question.