We are to investigate the factors that determine the terminal velocity of a 'helicopter'.

Richard Venables 10W

Helicopter Investigation

Planning

We are to investigate the factors that determine the terminal velocity of a ‘helicopter’.

We will be making helicopters with two controlled variables and dropping them from the Great Hall balcony, a distance of 3.6 metres. We will time how long they take to drop from half way up (or some distance like that) to the bottom so that we are only timing them when terminal velocity is reached.

When released from rest, the helicopter usually starts to spin after a short while. Once it has started spinning it has reached terminal velocity. We are to measure what the terminal velocity is and will be using two variables and seeing how they could affect the terminal velocity.

Possible variables are:

Length of wings
Number of paper clips
Width of wings
Colour of card
Shape of wings
Length of stem
Angle of wings

The two variables I will use will be the length of the wings and the number of paper clips.

This is what my helicopters will look like:

Prediction: I predict that the terminal velocity squared is directly to the mass of the ‘helicopter’.

Prediction of what the graph will look like:

I have predicted this because as the mass increases the ‘helicopter’ will speed up so the terminal velocity squared will increase. The graph will be a straight line instead of a curve because the terminal velocity has been squared.

Prediction: I predict that the terminal velocity squared is inversely proportional to the area of the wings of the ‘helicopter’.

Prediction of what graph will look like:

I have this because as the greater the area of the wings, the more air resistance is applied on the ‘helicopter’ so the slower it will drop.

Background Information

Weight = Air resistance at terminal velocity
The weight is affected by the number of paper clips and the amount of card used. The dye used in colouring the card could also affect the weight of it.
The air resistance is affected by the area of the card perpendicular to fall. Thus angled blades may reduce this, but also have a ...

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I have this because as the greater the area of the wings, the more air resistance is applied on the ‘helicopter’ so the slower it will drop.

Background Information

Weight = Air resistance at terminal velocity
The weight is affected by the number of paper clips and the amount of card used. The dye used in colouring the card could also affect the weight of it.
The air resistance is affected by the area of the card perpendicular to fall. Thus angled blades may reduce this, but also have a propeller affect.
The air resistance increases as the square of the speed

Aspects of Safety

Aspects of Safety include:

When using scissors to cut the card we must be very careful to use them in a way that would not cause any danger to ourselves or someone else.

When dropping the ‘helicopters’ off the balcony in the great hall we must ensure that there is no one beneath us that the ‘helicopter’ might hit.

On the balcony no one should run or lean over the edge as they could fall off.

Results

Table for results when changing the surface area of the wings. All ‘helicopters’ with a weight of 0.8g on the bottom.

Table for results when changing the weight on the bottom of the ‘helicopter’. All ‘helicopters’ with surface area of the wings at 10 cm2.

Analysis

My results have shown that terminal velocity squared is inversely proportional to the area of the ‘helicopter’ wings, is directly proportional to the mass, and one over terminal velocity squared is directly proportional to the to the area of the ‘helicopter’ wings.

The graph to show the terminal velocity squared in relationship to the area of the ‘helicopters’ wings clearly shows through the line of best fit that there is inverse proportionality and all the results for when the variable was area of wings look accurate. However when you see the graph for one over the terminal velocity squared in relationship to the area of the ‘helicopters’ wings, there is direct proportionality as there should be, however there seems to be an outlier in the graph for one over the terminal velocity squared when the area of the wings was 32cm2.

With the results I have obtained I think that it would be appropriate and possible to come up with some conclusions in relation to area of the ‘helicopters’ wings and the mass of the ‘helicopter’ in relation to their terminal velocity.

Conclusions

The terminal velocity squared is inversely proportional to the area of the ‘helicopters’ wings.

This links in and replicates what I predicted at the beginning with the inverse proportionality shown between the terminal velocity squared and the area of the ‘helicopters’ wings, which shows that I have obtained good and accurate results.

The reason that terminal velocity squared is inversely proportional to the area of the wings is that as the wing size increases so to does the air resistance. And the air resistance is affected by the area of the card perpendicular to fall. So as that increases and the air resistance increases terminal velocity squared increases.

The terminal velocity squared is directly proportional to the mass of the ‘helicopter’.

This links in and replicates what I predicted at the beginning with the direct proportionality shown between the terminal velocity squared and the mass of the ‘helicopter’, which again shows that I have obtained good and accurate results.

The reason that terminal velocity squared is directly proportional to the mass of the wings is that as the mass increases the ‘helicopters’ speed will increase so the terminal velocity will increase. However the air resistance will also increase because

air resistance=mass at terminal velocity

One over terminal velocity squared is directly proportional to the area of the ‘helicopters’ wings.

This does not link in to a prediction I made however looking at the graph it clearly shows direct proportionality. As the graph for terminal velocity squared and area of wings is inversely proportional it is logical that one over terminal velocity will be directly proportional to the area of the ‘helicopters’ wings.

Evaluation

There are several ways in which I could have improved my investigation:

For when the variable was area of wings, when the area of the wings was increased more card had to be used to increase the size of the wings. As more card was used the mass of the ‘helicopter’ increased, and the mass of the ‘helicopter’ affects the terminal velocity. So to improve this aspect of the experiment I could have taken card off the bottom of the stem of the ‘helicopter’ when I added to the wings.
Another aspect that could have been improved is the place where we dropped the ‘helicopters’ off to measure their terminal velocity. Some of us dropped them off down a stairwell where there could have been an updraft affecting the air resistance, others (including me) dropped them off from the balcony in the great hall. There were probably no updrafts here however there would have been other drafts of some sort.
Another thing worth considering is that I took my results over two different lessons and dropped the ‘helicopters’ off two different parts of the balcony where the drafts might be different. I also used two different stopwatches, which might have been very slightly different.

Looking at the graphs for one over terminal velocity squared to the area of the ‘helicopters’ wings, and the terminal velocity squared to the area of the ‘helicopters’, the result for when the area of the wings was 32cm2 looks to be slightly inaccurate compared to the others. Apart from that there don’t seem to be any ‘faulty’ results and the investigation went successfully showing that the terminal velocity squared is inversely proportional to the area of the ‘helicopters’ wings, the terminal velocity squared is directly proportional to the mass of the ‘helicopter’, and one over terminal velocity squared is directly proportional to the area of the ‘helicopters’ wings.