• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

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

Extracts from this document...

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.

...read more.

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)

...read more.

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

...read more.

This student written piece of work is one of many that can be found in our GCSE Forces and Motion section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related GCSE Forces and Motion essays

  1. Investigating the effect of mass on a parachute

    The average of each of the six masses shows this on the graph I drew. However, some results are inaccurate as shown on the graph and in the table of results. This is likely to be because I may have released the cone from a lower/higher distance than the required

  2. Pressure distribution over a symmetrical airfoil.

    Where p? is the static pressure in the free stream and U is the velocity. The coefficients may be determined directly from the monometer reading as follows: Cp = (h-hs) / (ha-hs) Where h is the reading for the tapping, hs is the static pressure in the tunnel and ha is the atmospheric pressure.

  1. The effect of the temperature on the viscosity of the syrup.

    To measure the time that the sphere travels from it's starting point to the end Tweezers To remove the sphere from the syrup easily Thin marker A thinner line would enable me to make a more accurate measurement of how far the sphere reaches, as a thicker line would cloud the final reading.

  2. Find out how a variable affects the terminal velocity of a parachute.

    Cut 4, 30cm strings 4. Put one string through each hole in the square and tie a knot so that the string will not come off the plastic sheet. 5. Take plasticine and use a weighing scale and weigh out 5g, 6g, 7g, and 8g of plasticine, correct to 1mg.

  1. Investigation into the effect of temperature on viscosity

    Prediction: From the derivation I can predict that the higher the viscosity the slower the ball, due to the derived inverse relationship: ?? 1 v ? = 2 r2 (? steel - ?fluid) g 9 (s/t) I predict that the lower the temperature the higher the viscosity.

  2. Practical Investigation Into Viscosity

    Caster oil can change its viscosity about 8% per degree c! The equations covered so far work as long as the motion is slow enough to keep the flows in the laminar domain. Once the speeds increase past a limit the drags grow at enormous rates.

  1. Practical investigation into Viscosity in liquids (Stokes Law).

    0.20 0.20 0.200 50-70 0.20 0.19 0.2 0.197 Very Large Distance Timed (cm) Time taken(s) for ball bearing to pass through distance measured 1 2 3 Average 0-20 0.33 0.32 0.30 0.317 10-30. 0.28 0.29 0.27 0.280 20-40 0.27 0.26 0.26 0.263 30-50 0.23 0.24 0.25 0.240 40-60 0.22 0.23

  2. To determine the relationship between mass and acceleration when force is kept constant.

    Like all frictional forces, the force of air resistance always opposes the motion of the object. And the three main factors of air resistance are: 1. Aerodynamic [streamline] properties of the object 2. Density of the object 3. Velocity of the object The more aerodynamic, dense, and fast the object

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work