# B Investigating Drag with Paper Cones

#### Jimmy87

Ohhh, I think I get it now. All of the accelerations and velocity are completely useless, since in the end we are looking at terminal velocity regardless of time and which reaches the ground first. Am i right?

I think you should carry ahead with any independent variable, execute the experiment and hope that your independent variable had an effect. If not, then try and look at your equations and see why. Go ahead with the experiment, if anything doesn't go perfectly, you can just find the problem and explain this in whatever you might submit to your teacher. There's no harm in being wrong or not getting the results you expected.
Ok, well thanks for all peoples help. I think I will assume terminal velocity has been reached so drag is constant. However, the terminal velocity will change and this will simply be the last 2m of the fall divided by the time it takes to fall these last 2m. I will now change my independent variable to the radius of the CONE but will keep the original radius of the flat CIRCLE I cut out the same. Going back to the OP now. I have two questions:

1) The drag equation shows that cross-sectional area is inversely proportional to the square of the velocity (if everything else is held constant). Should I get a graph of y = 1/x2 if I plot terminal velocity against base area of cone. I only ask because the base area is not exactly the same as the cross-sectinoal area in contact with the air since the cone is dropping with its point facing down. Is there a simple mathematical way to relate base area of the cone to velocity (if it is not inversely proportional to the square of the velocity)?
2) At terminal velocity, will the drag coefficient Cd be a constant (as it needs to be if my independent variable is base area of cone and dependent variable is terminal velocity)

Thanks again!

#### lekh2003

Gold Member
Ok, well thanks for all peoples help. I think I will assume terminal velocity has been reached so drag is constant. However, the terminal velocity will change and this will simply be the last 2m of the fall divided by the time it takes to fall these last 2m. I will now change my independent variable to the radius of the CONE but will keep the original radius of the flat CIRCLE I cut out the same. Going back to the OP now. I have two questions:

1) The drag equation shows that cross-sectional area is inversely proportional to the square of the velocity (if everything else is held constant). Should I get a graph of y = 1/x2 if I plot terminal velocity against base area of cone. I only ask because the base area is not exactly the same as the cross-sectinoal area in contact with the air since the cone is dropping with its point facing down. Is there a simple mathematical way to relate base area of the cone to velocity (if it is not inversely proportional to the square of the velocity)?
2) At terminal velocity, will the drag coefficient Cd be a constant (as it needs to be if my independent variable is base area of cone and dependent variable is terminal velocity)

Thanks again!
You're welcome.

#### CWatters

Science Advisor
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1) The drag equation shows that cross-sectional area is inversely proportional to the square of the velocity (if everything else is held constant). Should I get a graph of y = 1/x2 if I plot terminal velocity against base area of cone. I only ask because the base area is not exactly the same as the cross-sectinoal area in contact with the air since the cone is dropping with its point facing down.
You are on the right lines except the "A" in the drag equation is the maximum cross sectional area (sometimes referred to as the frontal area) which is the area of the base of a cone when it's pointing downwards.

Is there a simple mathematical way to relate base area of the cone to velocity (if it is not inversely proportional to the square of the velocity)?
The drag equation is..

FD = ½ρ CDAV2

At terminal velocity FD = mg

so
mg = ½ρCDAV2

Rearrange to give

1/V2 = ½mgρCDA

The drag coefficient typically depends on the shape of the object so I would expect it to vary with the cone angle. If you keep the mass and cone angle constant then I would simplify the equation to..

1/V2 = KA

where K is a constant = ½mgρCD

If you plot y = 1/V2 and x = A you should get a straight line with slope = K.

You could also consider square rooting both sides and ploting 1/V against R (the radius of the base of the cone). That works because the area is proportional to R2 .

2) At terminal velocity, will the drag coefficient Cd be a constant (as it needs to be if my independent variable is base area of cone and dependent variable is terminal velocity)
Over the narrow speed and size range of this experiment I would hope the drag coefficient Cd is reasonably constant. Over a wider speed range or a large range of sizes Cd isn't constant. eg sub sonic vs super sonic, or model vs full size aircraft.

• lekh2003 and Jimmy87

#### Jimmy87

You are on the right lines except the "A" in the drag equation is the maximum cross sectional area (sometimes referred to as the frontal area) which is the area of the base of a cone when it's pointing downwards.
Thank you - this is extremely useful! Just to make sure I understand what 'A' represents - let's say I have two cones, each with an equal base diameter (and therefore area) but one is simply a flat piece of paper (a circle) and the other is a cone. Would I enter the same 'A' into the drag equation? Would I also enter the same 'A' even though the cone is falling with its tip facing downward? In this example, the paper circle will fall very slowly compared to the cone so what would explain the greater drag - would it be the drag coefficient?

#### CWatters

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Thank you - this is extremely useful! Just to make sure I understand what 'A' represents - let's say I have two cones, each with an equal base diameter (and therefore area) but one is simply a flat piece of paper (a circle) and the other is a cone. Would I enter the same 'A' into the drag equation? Would I also enter the same 'A' even though the cone is falling with its tip facing downward?
Yes (provided you can get the flat piece of paper to fall in the horizontal position and not swoop about like a falling leaf.).

In this example, the paper circle will fall very slowly compared to the cone so what would explain the greater drag - would it be the drag coefficient?
Yes I would expect a cone with a wider angle to fall more slowly than a cone with a narrow angle because Cd is higher.

• Jimmy87

#### CWatters

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There are some drag coefficients here..

https://en.wikipedia.org/wiki/Drag_coefficient
The Cd of a cone point first is given in the top right hand table as 0.5.
The Cd of a flat plate is given in the lower table as = 1.28

Note these are only a rough guide. The angle of the cone isn't specified.

• Jimmy87

#### Jimmy87

Yes I would expect a cone with a wider angle to fall more slowly than a cone with a narrow angle because Cd is higher.
Would a reasonable scientific explanation of this be to discuss how the rate in change of momentum of air particles changes between the flat circle and cone. Looking at the flat circle, air particles will undergo a change in momentum of 2mv when they strike the surface. The force would be the rate of change of momentum which would be dictated by the velocity of the object. If we look at the cone then the air particles strike the surface of the cone which is now at some angle compared to the normal of the cone. Therefore the momentum will go down by the sine of the angle between the air particle striking the surface and the normal. Is this ok to explain why a cone falls faster through the air?

#### CWatters

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Perhaps but its not that simple. The shape behind the point of maximum cross sectional area also matters, perhaps not in this case (the back of a disk and the back of a cone are similar) but certainly for wing sections.

#### NTL2009

You might also consider adding a short string ( ~ 10 CM?) with a little weight on one end, and the other pulled through the tip of the cone, and attached on the inside. This would help keep the cone pointed down, and the weight would be more consistent cone-to-cone, as small variations in cone weight would not affect total weight very much (%-wise).

#### f todd baker

There seems to be a lot more floundering around than necessary in this discussion. You want to know how drag depends on the angle of your cone. It sounds like all your cones are constructed of the same amount of paper so their masses are equal. The area A is the orthographic projection which, simply put, is the area presented to the onrushing wind and that is the area of the base. The area of the base is A=πR2 where R=rsinθ and r is the radius of the paper circle (6" for you) and θ is the angle of the cone. Therefore A=π(rsinθ )2. You should therefore expect the measured drag to be proportional to the square of the sine of θ, FD∝sin2θ.

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