Experiment to Investigate Air Resistance

[Jimmy87]

Homework Statement

I collected data for an experiment involving air resistance. We made paper disks and timed how long they took to fall over a fixed distance. We changed the radius of the disk. We found that as you increase the radius they take longer to fall. I am trying to explain my results but with some difficulty.

Homework Equations

The Attempt at a Solution

We have been taught about forces expeienced by falling object e.g. weight and drag. I was thinking that if you double the radius of the circle then the area will increase by a factor of 4 which means the weight force will increase by a factor of 4. Since we found that it took longer for a bigger radius, the air resistance must not be proportional to the area. We have not been given any equations for air resistance but I did some research and found this one (https://en.wikipedia.org/wiki/Drag_equation). You can see that the drag force is in fact proportional to the surface area. However, if you double the radius the area should increase by a factor of 4 which means the weight will increase by a factor of 4 but according to this equation so should the drag force which means a paper circle of a bigger radius should take the same time to fall. What am I missing here? I know I am missing somehting because if this were true then for a given material e.g. a parachute it wouldn't matter how big you make the surface area and we all know that it does.

 

[mfb]

Usually paper circles doesn't fall in a smooth and repeatable way. How did you perform your experiment?

 

[PeroK]

What about the coefficient of drag? Which is part of the drag equation.

 

[Jimmy87]

Usually paper circles doesn't fall in a smooth and repeatable way. How did you perform your experiment?

Hi. We actucally used a pair of scissor and made a single cut to the centre of the circle and gently twisted it into a cone. We then dropped the cone with the tip pointing downwards so that it fell nicely. Surely I have something wrong with the physics though because if you had some material to make a parachute then a parachute with a bigger surface area would be more effective? However, since the parachute is roughly circular then doubling the surface area should double the weight?

 

[PeroK]

Hi. We actucally used a pair of scissor and made a single cut to the centre of the circle and gently twisted it into a cone. We then dropped the cone with the tip pointing downwards so that it fell nicely. Surely I have something wrong with the physics though because if you had some material to make a parachute then a parachute with a bigger surface area would be more effective? However, since the parachute is roughly circular then doubling the surface area should double the weight?

It doesn't double the weight of the parachutist, though!

 

[Jimmy87]

What about the coefficient of drag? Which is part of the drag equation.

Thanks. So would a paper circle with a bigger surface area have a different coefficient?

 

[Jimmy87]

It doesn't double the weight of the parachutist, though!

Why not? If it has roughly the same thickness then wouldn't a parachute with twice the surface area effectively mean you have twice as much parachute (so twice as much weight)?

 

[PeroK]

Thanks. So would a paper circle with a bigger surface area have a different coefficient?

What do you think? Does size matter?

 

[PeroK]

Why not? If it has roughly the same thickness then wouldn't a parachute with twice the surface area effectively mean you have twice as much parachute (so twice as much weight)?

The parachutist is the person hanging on the parachute. Their weight doesn't change.

 

[Jimmy87]

The parachutist is the person hanging on the parachute. Their weight doesn't change.

No, I meant that if you designed two separate parachutes - the second one with twice the surface area.

 

[PeroK]

No, I meant that if you designed two separate parachutes - the second one with twice the surface area.

I'd quite happily throw a toy soldier out of an aeroplane with a pocket hankerchief as a makeshift parachute, but I wouldn't bet on it holding my weight as I fell to Earth! Would you?

 

[JBA]

You should realize that generally the weight of a parachute itself is only a portion of the weight it is carrying; and, as a result, the small percentage of added material weight has much less effect than the increased area drag. For example, if the parachutist weighs 200 lbs and the parachute weighs 15 lbs., then, a 10% increase in the parachute weight is 1.5 lbs or 16.5 lbs total. As a result. the total % increase in the parachutist + parachute = 216.5 / 215 = 1. 007% as apposed to a 10% increase in drag.

Actually, the total packed weight of a parachute including an aerobatic harness with a 240 lb rated load is 15 lbs.

 

[Jimmy87]

I'd quite happily throw a toy soldier out of an aeroplane with a pocket hankerchief as a makeshift parachute, but I wouldn't bet on it holding my weight as I fell to Earth! Would you?

No indeed not. I understand that though because with much smaller parachute you would have a lot less air interacting with it as it fell so it would not work with someone who has a much larger weight force. I am more interested with object itself falling and not with a parachute with someone on the end. If a paper circle has four times the surface area then why does it fall slower because it should also have increased its weight by four times?

 

[PeroK]

No indeed not. I understand that though because with much smaller parachute you would have a lot less air interacting with it as it fell so it would not work with someone who has a much larger weight force. I am more interested with object itself falling and not with a parachute with someone on the end. If a paper circle has four times the surface area then why does it fall slower because it should also have increased its weight by four times?

Well, then we're back to coefficient of drag. Can you think of any reason why the coefficient might depend on size, not just shape?

 

[Jimmy87]

No not really. The way I see it is that an object with four times the surface area will experience four times more resistive forces as it interacts with the air molecules. Could you kindly explain?

 

[PeroK]

No not really. The way I see it is that an object with four times the surface area will experience four times more resistive forces as it interacts with the air molecules. Could you kindly explain?

Where do the air molecules go? Perhaps think about something flat falling.

 

[Jimmy87]

Where do the air molecules go? Perhaps think about something flat falling.

Where do the air molecules go? Perhaps think about something flat falling.

They would get deflected downwards as they collide with the surface?

 

[PeroK]

They would get deflected downwards as they collide with the surface?

And perhaps sideways as well?

 

[Jimmy87]

And perhaps sideways as well?

Yeh I guess some of them would get deflected sideways.

 

[PeroK]

Yeh I guess some of them would get deflected sideways.

How far sideways?

 

[Jimmy87]

How far sideways?

To the extent of the surface? So are you trying to say that a bigger surface area would have greater friction with air molecules being deflected sideways along the surface?

 

[JBA]

You should repeat your experiment using a shallow cone. Any flat paper disc will oscillate as it falls and this results in both effective area and form Cd changes that are both too unstable to predict or calculate. In addition, the oscillations allow lateral flow that nothing to do with Cd.

 

[Jimmy87]

You should repeat your experiment using a shallow cone. Any flat paper disc will oscillate as it falls and this results in both effective area and form Cd changes that are both too unstable to predict or calculate. In addition, the oscillations allow lateral flow that nothing to do with Cd.

I'm very confused now. Ignoring these oscillations should doubling the radius of a paper disk cause a slower falling time or should it remain the same for the reasons outlined in my first post?

 

[PeroK]

To the extent of the surface? So are you trying to say that a bigger surface area would have greater friction with air molecules being deflected sideways along the surface?

For a larger surface area, anything that has to get pushed out of the way, must be pushed further. And if the larger area is trying to move at the same speed, then the things will need to be pushed faster.

 

[JBA]

I suspect you will see close to an equal falling time; but you must be careful to insure that the cone angle is exactly same on every cone size because any variations in that angle will definitely affect the Cd of test piece.

 

[haruspex]

For a larger surface area, anything that has to get pushed out of the way, must be pushed further. And if the larger area is trying to move at the same speed, then the things will need to be pushed faster.

I agree.

Jimmy, try this calculation: if a disc radius r descends at speed v for time t, what volume of air is displaced? What must its sideways velocity be to be displaced in that time? How much KE does that take?

 

[Jimmy87]

For a larger surface area, anything that has to get pushed out of the way, must be pushed further. And if the larger area is trying to move at the same speed, then the things will need to be pushed faster.

So for a larger surface area the air molecules being deflected sideways along the surface will travel further and fas

For a larger surface area, anything that has to get pushed out of the way, must be pushed further. And if the larger area is trying to move at the same speed, then the things will need to be pushed faster.

If I understand you correctly are you saying that for a larger surface area the air molecules being deflected sideways along the surface will travel further and faster resulting in more friction at that surface and this effectively increases the coefficient? So accoding to the physics as you described then should a paper circle with twice the radius fall slower despite having a four times increase in weight? This is what I get from what you have said but not the same as what JB has said as he said that the falling time should be the same for a larger object?

 

[Jimmy87]

I agree.

Jimmy, try this calculation: if a disc radius r descends at speed v for time t, what volume of air is displaced? What must its sideways velocity be to be displaced in that time? How much KE does that take?

Sorry didn't see this before my last thread. Just to understand everyone correctly are we saying that a paper disk with double the radius has four times the weight but it would have more air resistance due to a different drag coefficient? I will think about what you just said and post a reply after I have thought it through.

 

[Jimmy87]

I agree.

Jimmy, try this calculation: if a disc radius r descends at speed v for time t, what volume of air is displaced? What must its sideways velocity be to be displaced in that time? How much KE does that take?

The volume of air displaced would be volume of the cylinder of air so: pi r^2 h = pi r^2 (vt). For this cylinder to move out of the way in time 't' would need a sideways velocity of 2r/t. So the kinetic energy to move it out of the way would be 1/2 m (2r/t)^2. I think I see your point. If we double the radius then we get four times the area so four times the weight. However, in terms of knietic energy required to move the air out of the way we get 8 times more energy lost to air resistance which is why it would fall slower?

 

[mfb]

For a larger surface area, anything that has to get pushed out of the way, must be pushed further. And if the larger area is trying to move at the same speed, then the things will need to be pushed faster.

No, the overall scale and therefore the time the air has increases as well, the same speed will still work. You scale everything up by the same factor.