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Velocity of a falling object after 1 meters

  1. Oct 7, 2009 #1
    Hello,

    I have got an question. How can I calculate the velocity of an object which I'll drop from a height of 1 meters with a starting velocity of 0 m/s?

    I did some research myself and found out that I can calculate this with using the following formula:

    St = 0.5 * 9.81 * t^2
    St = distance
    t = time in seconds

    So: t^2 = 1 / 0.5 * 9.81 = 0.20 -> t = 0.45 sec
    Velocity after 1 meter: v = 9.81 * t -> 9.81 * 0,45 = 4.42 m/s

    However, I understand that this is in a vacuum. But I want to know the real velocity in air. Therefore I need to know more about the shape and area of the object and the density of air. Which are:

    Shape: half-sphere
    Diameter: 2 cm
    Mass of the object: 0.2 kg
    Area: 0.0006 m^2
    Density of air: 1.293 kg*m-3

    However, I also found a drag coefficient at: http://en.wikipedia.org/wiki/Drag_coefficient" [Broken] which is 0.42

    I would like to know if I'm right with what I wrote above. How do I calculate the velocity of the object after 1 meters? Can I use the drag coefficient above and how? Or does the coefficient depend too much on the area of the object? Can someone give me an example how I can calculate this?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Oct 7, 2009 #2
    I dealt with this question rather long time ago. You may have to setup a PDE equation and with that, it is easy
    Edit: sorry, not PDE but DE (differential equation)
     
    Last edited: Oct 7, 2009
  4. Oct 7, 2009 #3
    your first work is correct. You could also use first principles of energy convservation, ie. potential energy before dropping is equal to kinetic energy at impact. mgh=1/2mv². The masses cancel and you're left with a velocity of 4.43m/s.

    The drag coeffiecient is a dependant on the geometry (and they are usual determined experimentally for different shapes) and the flow velocity/viscosity/etc. (Reynold's number) and will be different in laminar or turbulent flow. If you assumed that the Cd was 0.42 in the situation you're calculating, you can use this value together with you other information in the drag equation: F=1/2*rho*v²*Cd*A.
    F - Force due to drag (N)
    rho - density of air (kg/m³)
    v - velocity of free stream (around your object) (m/s)
    Cd - drag coefficient (dimensionless)
    A - frontal area (the area that sees the flow, in your case the area of the circle with diameter 2cm (m²)

    Now there are many ways to go form there.
    In your method you use accelerations. Before drag, the only acceleration is the 9.81m/s² of gravity. The force of the drag will also generate an acceleration in the opposite direction to your gravity. F=ma (newton's 3rd law). a=F/m. So, your new accleration would be (9.81 - F/m). You know m, but the tricky part is that your F increases as the velocity increases. So, before you drop it, the acceleration is (9.81 - 0/m), because F= 0, there is no velocity and therefore no drag. The most accurate way, would be to solve the velocitiy and the Force simulataneously so that you know the drag force (and hence the deceleration) for every time as the object drops. This can be set up by using a differential equation and the then integrating over the dropped distance (or time) to calculate the total velocity.
     
  5. Oct 7, 2009 #4
    I see. Thanks for your comment.
    Unfortunately, I don't know much about integrating so I find it very difficult. So it would be really great if one could help me out with that, or at least give an example how I can solve my question.

    Thanks
     
  6. Oct 8, 2009 #5
    what do you need to know this for? What kind of accuracy do you need? Can you perform some experiments?
     
  7. Oct 8, 2009 #6
    It's for a project at my study where we are doing experiments for a company that makes a new material which we have to test on different kind of properties. One of these experiments is an impact test (to see a visual property).
    As the experiment has been developed by other students a long time ago, and there are no theoretical and practical information available with this test, we are trying to find out what this experiment can tell to us.
    We can drop weights (in tubes with airholes of 1 and 3 meters long), and then see what happens to the surface and measure the dent it makes. But we would like to add some more information to it like the velocity at impact and after that we can also calculate the force by impact etc.
    The weights have a weight of 200 grams and 500 grams, both will be used for a fall from 1 and 3 meters high.
    First we wanted to measure the time of the fall, but that's not possible because it goes too fast to register so that's not accurate.
     
  8. Oct 8, 2009 #7
    If you have access to spreadsheet software (excel), I believe this can be done quite easily. You take the 1 meter drop and break it up into tiny pieces maybe each piece as 1 milimeter. For each piece you assume a constant acceleration, which allows you to calculate change in velocity for each piece. Each cell in the spreadsheet will use information from the previous cell (final velocity at the end of the 1mm segment).

    For the first 1 milimeter I would assume that no force due to air resistance exists, which means a = 9.8 m/s^2. Then I would calculate the change in velocity for the first 1 milimeter. The final velocity over the first milimeter would be used to calculate the force due to air resistance to be used for the second milimeter. From that you can calculate a new acceleration that will be slightly less than 9.8 m/s^2. Using that new acceleration you calculate the final velocity over the second milimeter which is used to calculate the force due to air resistance for the third milimeter and so on. This process can be easily automated in excel.

    I am not sure how accurate/precise this method is, but it's an idea. I know there are different methods that can be used in order to reduce the % error, but I can't remember them.
     
  9. Oct 8, 2009 #8
    I agree with Bishopuser in using excel. My way is divide the time to delta t (say 0.01 sec) in the first column. In 2nd column, you put in accelaration which equal g in the first line. From line 2 and after that, accelaration equals to g - v*k, with k = dragging coefficient. In 3rd column, you put in velocity which calculated by accelaration * time adding the previous velocity. In the 4th column, you put in the distance corresponding to that delta t. And finally, the total distance is put in column 5th.

    From the sheet you can check velocity at 1 meter or whatever you want. You can also draw a graph of velocity / accelaration etc vs time.
     
  10. Oct 9, 2009 #9
    Excellent advice from both BishopUser and Pixel01. I would probably also use a spreadsheet for the graphing abilities. The only problem is that the calculations you make will be based on assumed Cd values and the times you measure won't be accurate. A falling object in a tube with a diameter around about the diameter of the object will also produce very different drag coefficients than from an object in a free stream.

    You might have to do some experiments to calculate the drag coefficient of your objects in the tube that they will be used in. You could do this using water instead of air, so that the drop can be timed more accurately, but then you'll have to read up a little on dimensional analysis and how changing the density and viscosity of your fluid can be related back to an experiment in air.
     
  11. Oct 12, 2009 #10
    Hi,

    I did my best and tried something in excel. I would like to ask you guys to have a look at it to see if I did the correct thing. If so, my problem has been solved :). However, the drag due to the tube has not been calculated, indeed this should matter though.

    Anyway, just to assume there is no tube:
    http://bannier.homelinux.net/files/weights.xls" [Broken]

    If this is right, then the final velocity would be 4,4280555 m/s (time: 0,452117) right?
    So in that case, there's not much difference between the values in vacuum and the values with drag.

    Please confirm.

    Thanks
     
    Last edited by a moderator: May 4, 2017
  12. Oct 12, 2009 #11
    ummm, somethings not right.

    your velocity is now higher with the drag than when you calculated it without the drag ??

    but anyway, you clearly show that even at it's maximum, the drag force is about 0.0032N compared to your gravitational force of 1.962N, which is about 0.16%! Is it really necessary to include the drag in your equation?

    Again, you are also using a Cd value that was gathered using experiment probably in a freestream, not in a tube, so the calcualtion of drag is pretty useless unless you find it experimentally for your specific situation (It's still going to be a lot smaller than your gravitational force). It really depends on what you are testing and what you want to get out of the results. Saying it's a drop test to "see a visual property" I would assume that the accuracy is not that significant.
     
  13. Oct 12, 2009 #12
    Velocity in vacuum should be: 4.429447 m/s
    Sorry for this fault in my first post. So, with the drag, the velocity is slower after all. But yes, the different is not much for just 1 meters. But at least I might have some idea for how the velocity at impact should be around 1 or 3 meters.
     
  14. Oct 12, 2009 #13
    Hi Dutchli,
    it seems there are two mistakes:
    the area should be 0.0003 m^2 , and that density corresponds to 0 Celsius degrees.
    If the ambient temperature = 25 degrees, the density should be 1.184 kg*m-3.
    (see http://en.wikipedia.org/wiki/Density_of_air" [Broken] )
     
    Last edited by a moderator: May 4, 2017
  15. Oct 12, 2009 #14
    Are you sure the area is 0.0003 m^2?

    Diameter = 2 cm -> r = 0,01 m
    4*π*r^2 = 0,00126 m^2
    0,00126 / 2 = 0,0006 m^2
     
  16. Oct 12, 2009 #15
    Yes, I'm sure.
    The frontal area of a sphere (or your half-sphere) is π*r^2.
    (see http://en.wikipedia.org/wiki/Drag_coefficient" [Broken])
     
    Last edited by a moderator: May 4, 2017
  17. Oct 13, 2009 #16
    Rogerio is right, as I said in post #3, you need the frontal Area of the sphere (the area that "sees" the flow) which is the area of the circle with diameter 2cm.
     
    Last edited by a moderator: May 4, 2017
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