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To investigate the relationship between the velocity of a parachute and the drag force.

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Introduction

James De Souza        Science 1        B.E.Tyler

To investigate the relationship between the velocity of a parachute and the drag force.

Viscosity :- The viscosity of a fluid is a measure of its resistance to flow. Viscous forces acting on bodies moving through a fluid and in fluids moving through pipes and channels. The pressure in a fluid decreases where the speed increases.

Stokes Law :-  An equation relating the terminal settling velocity of a smooth, rigid sphere in a viscous fluid of known density and viscosity to the diameter of the sphere when subjected to a known force field. It is used in the particle-size analysis of soils by the pipette, hydrometer, or centrifuge methods. The equation is:

V = (2gr²)(d1-d2)/9µ

where

V = velocity of fall (cm sec-¹),

g = acceleration of gravity (cm sec-²),

r = "equivalent" radius of particle (cm),

dl = density of particle (g cm -³),

d2 = density of medium (g cm-³), and

µ = viscosity of medium (dyne sec cm-²).

A falling object has an acceleration equal to g, provided air resistance is negligible. If air resistance is significant, the force due to air resistance drags on the object. This drag force increases as the object speeds up, until the force becomes equal and opposite to its weight.

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Middle


0.20         1.03         0.97         1.10                 1.04                 2.74                         2.63


Experiment 2
Mass (N) Time 1 (s) Time 2 (s) Time 3 (s)   Average Time (s)    Average Velocity (m/s)*  Average Acceleration (m/s2)…
0.02         2.78         2.32         3.28               2.79                        1.02                         0.37
0.04         2.18         2.30         1.67               2.05                 1.39                         0.68
0.06         1.57         1.40         1.50               1.49                 1.91                         1.28
0.08         1.09         1.14         1.25               1.16                 2.46                         2.12
0.10         1.19         1.31         1.29               1.26                 2.26                         1.79
0.12         1.13         1.20         1.14               1.16                        2.46                         2.12
0.14         1.09         1.07         1.13               1.10                        2.59                         2.35
0.16         0.91         1.08         1.10               1.03                        2.77                         2.69
0.18         0.88         1.01         1.06               0.98                        2.91                         2.97
0.20         0.93         0.97         1.00               0.97                        2.94                         3.03


Averages Over Experiments 1 and 2
Mass (N)         Average Time (s)

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Conclusion


ii)         v2 = u2 + 2as
iii)         s = (u + v)
                       2

Unfortunately, without knowledge of the terminal velocity, or the real acceleration, this can’t be done properly. However, to give a rough idea of how it could be used, the test is detailed below:

In an attempt to acquire the most accurate results possible, albeit a futile attempt, the third formula will be used and the average velocity used in place of the terminal velocity.

1) To begin with, try the first set of results, i.e. a mass of 0.02 N:

s =
(0 + 0.93)
           2
\ s = 0.465

Quite obviously, this distance is nowhere near the actual distance of 2.85m but, of course, it shouldn’t be because with such a small mass, air resistance is still playing a major part.

2) Next, the results for a mass of 0.12 N will be tried:

s =
(0 + 2.32)
            2
\ s = 1.16

Again, this is nowhere near the actual distance but it is getting closer.

3) Lastly, the results for the last mass, 0.20 N, will be tried:

s =
(0 + 2.94)
             2
\ s = 1.47

It would appear then, in conclusion, that this test was a failure. The question is, though, is this because of the fact that the final velocity is obviously false, or because this is not the way to go about finding the “terminal” terminal velocity, which of course may not exist. In all likelihood, however, looking at the results, it does exist but without the actual values for the final, or terminal, velocity, it is difficult to prove its existence.

image02.pngimage00.png

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