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

To investigate the relationship between the velocity of a parachute and the drag force Essay 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 = (2grà ¯Ã‚ ¿Ã‚ ½)(d1-d2)/9à ¯Ã‚ ¿Ã‚ ½whereV = velocity of fall (cm sec-à ¯Ã‚ ¿Ã‚ ½),g = acceleration of gravity (cm sec-à ¯Ã‚ ¿Ã‚ ½),r = equivalent radius of particle (cm),dl = density of particle (g cm -à ¯Ã‚ ¿Ã‚ ½),d2 = density of medium (g cm-à ¯Ã‚ ¿Ã‚ ½), andà ¯Ã‚ ¿Ã‚ ½ = viscosity of medium (dyne sec cm-à ¯Ã‚ ¿Ã‚ ½).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. The acceleration becomes zero because the resultant force on the object becomes zero. The speed therefore becomes constant; this value is referred to as the Terminal Velocity.TaskTo investigate the effect of a change in mass on the time taken for a parachute to fall a set distance.Other variables that could be investigated are:à ¯Ã‚ ¿Ã‚ ½ Surface area of the parachuteà ¯Ã‚ ¿Ã‚ ½ Length of string (between the parachute and mass), which might control the volume of air under the parachute.à ¯Ã‚ ¿Ã‚ ½ Distribution of mass, i.e. perhaps on the parachute itself as opposed to on string attached to the parachute (this of course would not be a continuous variable so it would not be of great value).ApparatusA square of bin liner, thread, sticky tape, plasticene, and weighing scales.MethodOne parachute w as assembled using a square of bin liner, thread and sticky tape. The thread was tied in such a way that plasticene masses could be attached. For each mass, the experiment was performed three times and after completion, the entire investigation was repeated. The actual experiments consisted of timing how long the parachute took to travel from the ceiling to the floor, a distance of 2.85 metres. The measurements were taken in grams and then converted into Newtons for more accurate results.In order to make this a fair test I am going to keep a number of things constant, e.g., the bin liner parachute, the length of the string, the distance for it to fall, the surface area of the parachute, and the distribution of mass.DiagramPredictionsà ¯Ã‚ ¿Ã‚ ½ The larger the mass, the shorter the time because when the mass is larger the parachute accelerates to a higher speed due to the terminal velocity being higher.TheoryVelocity = DistanceTimeAcceleration = Increase in VelocityTimeResultsExperi ment 1Mass (N) Time 1 (s) Time 2 (s) Time 3 (s) Average Time (s) Average Velocity (m/s)* Average Acceleration (m/s2)0.02 3.35 3.29 3.31 3.32 0.86 0.260.04 2.17 2.35 2.18 2.23 1.28 0.570.06 1.72 1.88 1.64 1.75 1.63 0.930.08 1.58 1.65 1.62 1.62 1.76 1.090.10 1.46 1.41 1.23 1.37 2.08 1.520.12 1.26 1.29 1.31 1.29 2.21 1.710.14 1.11 1.27 1.08 1.15 2.48 2.160.16 1.15 1.13 1.04 1.11 2.57 2.320.18 1.04 1.18 1.05 1.09 2.61 2.390.20 1.03 0.97 1.10 1.04 2.74 2.63Experiment 2Mass (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 0.370.04 2.18 2.30 1.67 2.05 1.39 0.680.06 1.57 1.40 1.50 1.49 1.91 1.280.08 1.09 1.14 1.25 1.16 2.46 2.120.10 1.19 1.31 1.29 1.26 2.26 1.790.12 1.13 1.20 1.14 1.16 2.46 2.120.14 1.09 1.07 1.13 1.10 2.59 2.350.16 0.91 1.08 1.10 1.03 2.77 2.690.18 0.88 1.01 1.06 0.98 2.91 2.970.20 0.93 0.97 1.00 0.97 2.94 3.03Averages Over Experiments 1 and 2Mass (N) Average Time (s) Average Velocity (m/ s)* Average Acceleration (m/s2)0.02 3.06 0.93 0.300.04 2.14 1.33 0.620.06 1.62 1.76 1.090.08 1.39 2.05 1.470.10 1.32 2.16 1.640.12 1.23 2.32 1.890.14 1.13 2.52 2.230.16 1.07 2.66 2.490.18 1.04 2.74 2.630.20 0.97 2.94 3.03Notes* This was calculated using the formula above (in the Theory section) using the Average Time. Unfortunately, in this case, it is not possible (without further study into complex formulae) to calculate the actual change in velocity due to the fact that the finishing velocity, or in this case the terminal velocity, remains unknown. Therefore, in order to give a very rough idea of the average acceleration, the average velocity was used as the finishing velocity and, obviously, 0 m/s used as the starting velocity (which in this case is correct).Analysisà ¯Ã‚ ¿Ã‚ ½ The first prediction, albeit rather basic, was correct and, although it was not tested, it is safe to presume that this is due to the fact that when the mass is larger, so is the terminal velocity. This means that the parachute can accelerate to a higher velocity resulting in a shorter time.à ¯Ã‚ ¿Ã‚ ½ As can be seen from the graph above it can be seen that the drop in time is rather large to begin with but gets smaller as the mass increases. This cervical result leads one to believe that there is a limit to the terminal velocity. This would imply that once a larger mass is added, a terminal terminal velocity is achieved beyond which a parachute cannot accelerate. This is presumably due to the lesser effect of air resistance at higher masses.à ¯Ã‚ ¿Ã‚ ½ The same pattern can be seen average velocities, but obviously going up rather than down, but to a lesser extent.à ¯Ã‚ ¿Ã‚ ½ The mass is directly proportional to the velocity (as the mass increases the speed increases) , and the velocity and mass is indirectly proportional to the time ( as the speed and mass increases the time decreases).Evaluationà ¯Ã‚ ¿Ã‚ ½ As was said in the Notes section above, it would be highly preferable to be able to calculate the final velocity, and even better the terminal velocity. The final velocity could be calculated with the use of computer sensors to measure the velocity in the last, say, 10cm. In order to calculate the terminal velocity it would be sensible to increase the distance travelled in order to ensure that the parachute does indeed reach terminal velocity before the velocity at the end is measured.à ¯Ã‚ ¿Ã‚ ½ As far as inaccuracies are concerned, it is obvious to see, from the Average Times graph, that the most problematic results are those measured for a mass of 0.08 N. Fortunately, they even out to provide a good average curve.à ¯Ã‚ ¿Ã‚ ½ Another problem could be the results for a mass of 0.20 N where you can see that the results seem to converge as opposed to following the otherwise reasonably error-free curve.à ¯Ã‚ ¿Ã‚ ½ Lastly, it must be further re-iterated that the Average Accelerations, and to a lesser extent the Average Velocities, use very inaccurate results due to the fact that the final velocity, and therefore the acceleration, is unknown. Therefore, the graphs of those results show very little of value other than to highlight the aforementioned inaccuracies, because they show up much more on those graphs.Conclusionsà ¯Ã‚ ¿Ã‚ ½ This theory could be proved, as well as the terminal terminal velocity calculated by using the usual mechanics formulae:i) s = ut + 1/2at2ii) v2 = u2 + 2asiii) s = (u + v)2Unfortunately, without knowledge of the terminal velocity, or the real acceleration, this cant 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 = 0.465Quite obviously, this distance is nowhere near the actual distance o f 2.85m but, of course, it shouldnt 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 = 1.16Again, 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 = 1.47It 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.