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B Investigating Drag with Paper Cones

  1. Nov 14, 2017 at 11:16 AM #1
    Hi,

    I have been doing an experiment at school looking into how the angle of a cone affects the time it takes for it to fall 2m. The data consistently shows an inverse relationship but it is unclear whether or not they are inversely proportional i.e. if the angle doubles, the time taken to fall halves. I wanted to see if I could relate the angle of the cone to the force of drag in some way and wanted to know if someone could help me?

    The drag equation we are given is:

    FD = CDApv2 (FD = force of drag, CD = drag coefficient, A = cross-sectional area, p = density of fluid, v = velocity)

    From looking at the equation I am thinking that the angle of the cone will affect the cross sectional area. So I need some way of expressing the cross-sectional area in terms of the angle of the cone so I see what the relationship would be between the angle of cone and the force of drag. Any ideas?

    Many thanks!
     
  2. jcsd
  3. Nov 14, 2017 at 11:36 AM #2

    jbriggs444

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    What are you holding constant as you change the angle of the cone?
     
  4. Nov 14, 2017 at 1:00 PM #3
    Yeh good point, I didn't really consider that. I'm holding velocity constant as I am timing it for the last 2m of a 3m fall so should have reached terminal velocity. The density of air is constant. I know that changing the angle is like an indirect measure of cross sectional area. So I'm actually unsure if changing the angle changed the drag force or the drag coefficient or both?
     
  5. Nov 14, 2017 at 1:42 PM #4

    jbriggs444

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    What else are you holding constant as you change the angle of the cone? Is it the same cone canted at an angle? Or is it the same paper rolled up into a pointier cone? Or what?
     
  6. Nov 14, 2017 at 3:09 PM #5

    CWatters

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    Is the area of the base of the cone constant?
     
  7. Nov 14, 2017 at 4:36 PM #6
    No it isn't and the cone is actually dropped upside down (i.e. the pointed tip facing down) otherwise it tumbles and the data is awful. The radius of the flat circle I originally cut out is the same. I basically cut 6 circles each with a radius of 6cm. I then made a straight cut along the radius. I then twisted the paper to form a cone. I measured the height of the cone and new radius once twisted to find the angle. I have attached a diagram to show what I mean.
     

    Attached Files:

  8. Nov 14, 2017 at 4:40 PM #7
    The last one - the same bit of paper rolled into a pointier cone as shown by the attached diagram on post #6.
     
  9. Nov 14, 2017 at 4:50 PM #8
    Also just realised that if the velocity is constant after falling 1m that drag will equal the weight and therefore be constant! So the only variables changing are the angle (which affect cross-sectional area) and the drag coefficient I guess?
     
  10. Nov 14, 2017 at 5:08 PM #9

    CWatters

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    Ok so in addition to the angle, the frontal area is also changing. I would try and make cones that have the same frontal area (eg same base diameter) and same weight. May need to add ballast to them.
     
  11. Nov 14, 2017 at 5:21 PM #10
    Thanks. But then I don't see how the angle could be changed. The angle changes as a result of changing the height and base diameter. How can you make cones which have the same base diameter but a different angle?
     
  12. Nov 14, 2017 at 6:55 PM #11

    jbriggs444

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    Modify the height.

    If you are constructing a cone by taking a circular piece of paper, carving out a wedge and taping the cut edges then the size of the wedge affects the height and circumference of the cone. Increase the size of the wedge and you increase the height while reducing the circumference. The diameter of the circle from which you cut affects the height and the circumference. Increase the diamter and you increase both height and circumference. For any desired height and circumference, there is a way to construct a cone to match.
     
  13. Nov 15, 2017 at 5:48 AM #12
    Thanks. So should I change my method or not? Seeming as though all variables are constant except for angle (which surely is a function of cross-sectional area) then the only thing to change would be drag coefficient. I didn't actually cut out a wedge, I simply made a cut along the radius (so no paper was taken away). I then moved one side over the other to twist into a cone. This will decrease the original radius and the angle as shown in the diagram. Does it matter if they both change? Because I was thinking I could get an equation for the cross-sectional area - which is just one variable - or is this hard to do? So are you saying you would keep the base area the same but just change the height by cutting out different sized wedges? Because I can't see how you can get the base area and angle the same by cutting out the same wedge? If you did cut out different sized wedges would this not change the mass though?
     
  14. Nov 15, 2017 at 6:07 AM #13

    jbriggs444

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    If you want to keep the area of the base of the cone fixed but vary the angle then you will have to vary both the diameter of the circle you use to construct the cone and the overlap when you roll it up.

    If you want to then keep the mass of the cone fixed then you will need to add a enough ballast (sand? chewing gum?) so that each cone masses the same.
     
  15. Nov 15, 2017 at 6:16 AM #14
    I'm going to first try to tackle your biggest problem which is the controlling of the variables. I must say at this point that I think you are following the wrong variable, since changing the angle of the cone will probably not effect the velocities, since this is not what the drag force depends on. It is the area of the base of the cone you should be trying to change instead of the angle. You could reason in your school experiment that it is not required to control the angle, while you change the base area. In simple words, change the base area, forget the angle. The angle has little to no effect on the drag force and this should have been obvious.

    Secondly, your issue with relating the drag force and velocity should be easier to solve if you simply use the base area. It should be easy to calculate the drag force knowing the area, and therefore the terminal velocity.

    I have studied in the IB where school experiments make up the majority of the final grades you receive. I have three years worth of creating experiments and labs, try to keep it simple.
     
  16. Nov 15, 2017 at 6:43 AM #15
    I don't quite understand your reasoning. None of the variables affect the drag force as the drag force is constant isn't it? This experiment where t=0s is terminal velocity so drag force is constant? If I keep the base area the same wouldn't the only variable that can change be the drag coefficient?
     
  17. Nov 15, 2017 at 6:54 AM #16
    I don't understand what you mean. If you change the surface area of the cone (the base area effects this most), the drag increases, therefore the terminal velocity decreases. If you simply change base area, the terminal velocities will change. You will have a working experiment and be able to identify a trend. If you change the cone angle, the change in terminal velocity will be close to 0. Try and work with the base area instead. Even your calculations for drag will be easier since you have your cross-sectional area right in front of you. Angles will complicate things.
     
  18. Nov 15, 2017 at 7:36 AM #17

    CWatters

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    If you want to know how the angle affects the time it takes to fall 2m then you should try and keep all other variables constant. Otherwise you can't tell if the angle or something else is responsible for the change/different performance.

    So I would make a set of cones that only differ by the cone angle and have the same base area and the same mass. That way you know they aren't effecting the results.

    Edit: Likewise if you wanted to know how the frontal area affects the time you should vary that and keep everything else (mass and cone angle) constant.
     
  19. Nov 15, 2017 at 7:53 AM #18
    Ah yes I agree the velocity will change so this will be the dependent variable but the drag force won't change at terminal velocity. If the cones have the same mass then they have the same weight and terminal velocity means drag = weight so drag will be a constant in my experiment won't it?
     
  20. Nov 15, 2017 at 8:08 AM #19
    The drag is exactly what you hope to see changing. Only if the drag changes will you see a result. Separate your independent and dependent and controlled. Your independent will be the base area. The dependent will be the velocity which is determined by drag ( drag is determined by base area). The controls will be mass, height, etc. These are controlled and should not be affected by base area (ideally).

    I think I understand what you mean by "drag=weight". Even though drag could be increasing, the weight definitely is not. You are confusing their relationship. You can change the weight which could affect drag, but it doesn't work the other way around. Drag could increase since you are increasing cross-sectional area, but that doesn't mean you are effecting the weight. You can still keep that constant. Drag is not controlled, weight is.
     
  21. Nov 15, 2017 at 8:09 AM #20

    jbriggs444

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    If you are investigating how the point angle affects things, the independent will be the point angle. That's pretty basic.
     
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