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Terminal Velocity & Parachutes

  1. Oct 5, 2008 #1
    I cant really follow the template with this question.

    Im dropping parachutes out of a window with weights attatched. I'm assuming that they travel at terminal velocity from the moment i drop them, and i know this becomes less and less true the heavier weights i use.

    I plotted the results i got as weight (newtons) against terminal velocity (m/s).

    [​IMG]

    The lightest weight i dropped was 0.048 newtons, and the heavest was 10. This makes this graph imossible to see any results from and i know that nothing above 30ish newtons reached terminal velocity.

    I worked out the resistive force, and when it was equal to the weight, i knew they had reached terminal velocity. Below is the graph that focuses on the drops that reached terminal velocity.

    [​IMG]

    Now from that graph, it is not clear wether the graph should be linear or curved. If it is linear it will cross the X axis at around 1 m/s, is that pheasable? Would it travel at 1 m/s?? with extremely small mass? I think my result / trend line falls appart at lower weights but i am still not sure of the line of best fit.

    You can ignore the last point on the second graph as that is questionable as to wether it reaches terminal velocity or not. I think that the graph might be linear where the results reach terminal velocity, and when the weights don't reach terminal velocity the graph is curved, but i also don't know if this is right. Can anyone add any insight to my investigation on what you think the results show? Just a little bit stuck thats all.

    Maybe it should look like this:

    [​IMG]


    Thanks alot for your time!!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 7, 2008 #2
    Any clue?? I'm still not too sure!
     
  4. Oct 7, 2008 #3

    PhanthomJay

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    Hi, doc. I'm not sure how you are determining or measuring terminal velocity. Terminal velocity is aproached rather quickly with the lighter masses, and not so quickly with the heavier ones. The terminal velocity using the lighter masses is apt to increase linearly with increasing mass (directly proportional to the weight), but perhaps increasing non linearly with the heavier masses (directly proprtional to the square root of the weight). Thus, if you were truly measuring terminal velocity, you might get a linear graph of weight vs. terminal velocity at the lighter masses, but a convex (not concave, as you have shown) curve at the heavier masses. You thus may not be measuring terminal velocity in your experiment. Actually, terminal velocity is never reached, only approached. The equation for the objects velocity at a certain time or distance is a rather complex differential equation involving the natural log/hyperbolic functions.
     
  5. Oct 8, 2008 #4
    Hi, I did something like that last year in highschool. At low velocity, frictional force increases linearly with speed, quadratically at higher speeds (eg. 50km/hr). I'm not positive from my memory but the F friction = crosssectional area*constant depending on the shape / roughness of object*velocity^(1 or 2)
     
  6. Oct 8, 2008 #5
    I know i'm not actualy measuring terminal velocity properaly as that is quite difficult, with the lighter weights i know that terminal velocity is reached very early on, and with my heavier weights it doesnt reach it at all. So i worked out the GPE and the KE on impact, and the difference of that must have been given off in a frictional force. If the frictional force is equal to weight, then they did reach terminal velocity, as they are balanced and acceleration stopped. I know all this becomes less accurate the heavier my weights get but this isn't really important its just the method and my ideas that matter.

    Would you expect this lienar graph of the lighter weights to go through the origin?

    Also i did another experiment with glycerol and steel balls dropping through a tube.. I plotted the graph of that too and it is curved too, and i suspect that these do reach terminal velocity by the time i start measuring their fall as i dont measure their fall for quite some distance. After some point they no no matter how much i increase the weight, the terminal velocity doesn't increase at all.. so im assuming that they all accelerate through the same point at the same acceleration, so thus all reach the same speed, that is not their terminal velocity.

    For some reason in glycerol i can't seem to prove their terminal velocity by finding GPE and KE and finding the difference as i did before.. But i use the value for gravity as 9.8 still... is this value 9.8 only for the acceleration through air? or all gasses? is it different in glycerol?

    Sorry if my information isn't very structured, i'm not an english student!! :)

    Thanks alot for your help so far guys!!
     
  7. Oct 8, 2008 #6
    Also how would i explain this concex graph? I can pretty much ignore the results that arn't terminal velocity just by saying they arn't terminal velocity thus they don't concern me, and think this is after 34g that they don't reach terminal velocity. But its the lower end of the graph that confuses me, it cant intercect at 1. something with nearly 0 mass will not travel 1 m/s. So i dont know how to explain what my graph shows me really. :|
     
  8. Oct 8, 2008 #7

    PhanthomJay

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    yes
    The acceleration of gravity at earth's surface is 9.8m/sec^2.The medium through which it propagates is not relevant. When you drop an object through water or a fluid, there is the buoyancy force to consider as well as the drag force .
    For linear drag, terminal velocity is equal to cM, where c is a constant dependent on the density of the medium thru which the object falls, gravity, and other factors noted by ice ace above. Thus , for this case, term vel increases linearly with mass: double the mass, you double the speed. When the mass is very small, terminal velocity is very small (theoretically it passes thru the origin at m=0).
    For quadratic drag, term vel = c(sq root of M); Double the mass, and the term velocity goes up by root 2; this yields a convex curve.
     
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