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# The Physics of Paper Helicopters

Extracts from this document...

Introduction

Frost ma 7137                                                                                        Physics Coursework

Michaelmas 2001

## Introduction

In my physics coursework I intend to study ‘paper helicopters’.  A paper helicopter is a piece of plain A4 paper that has been folded, so that when dropped, it spins whilst falling to the ground.  Below is a picture of one of the helicopters used.

The top half of the piece of paper is cut down the middle, creating the two wings, and then the rest of the piece of paper is folded up to make the body of the helicopter.

The reason for this choice was that there are many variables that I can change on the helicopter.  I intend to have different experiments, with the wing size, mass of helicopter and the height from which it is dropped all being varied.  I have done some initial measurements, and found that when dropped down some stairs of height 6m, the helicopter takes about 5 seconds, which is long enough to measure relatively accurately.  Although I originally intended to measure the revolutions per second of the helicopter, my initial experiment showed that it would be too difficult to do accurately.

## The basic physics of the paper helicopter

As the helicopter falls through the air, it spins.

Middle

3.05

3.09

1.13

2.21

10.56

2.87

2.72

2.68

2.76

1.01

2.36

11.97

2.41

2.41

2.43

2.42

0.88

2.48

13.38

2.22

2.13

2.22

2.19

0.78

2.59

14.79

1.91

1.96

2.03

1.97

0.68

2.69

Evaluation and conclusion

If we look at the graph of total mass against time (Graph 3 in appendix), we can see that there is a curve.  This means that the relationship between the two is not linear.  I expect the relationship to be:

Time = K x Massn

⇒ln(time) = ln(K) + nln(mass)

So when we plot ln(time) against ln(mass), the gradient will be n, and the y-axis intersect will equal ln(K).

Gradient = (δln(time)) / (δln(mass)) = ((-0.48/0.5) + (-0.88/0.9)) = -0.97

The graph is not big enough to see where it crosses the x-axis, but we can work it out.  Taking a point on the line: (1.5, 1.24):

ln(time) = m ln(mass) + c  (but ln(time) = 1.24, ln(mass) = 1.5, and m = -0.97)

⇒ 1.24 = -0.97 x 1.5 + c

⇒ c = 2.695.

Therefore, n = -0.97 and 2.695 = ln(K)

⇒ K = e2.695  = 14.806

Therefore, the relationship that we have derived is:

Time = 14.806 x Mass-0.97

## Evaluation and conclusion

Another variable that I think might effect the time taken for the helicopter to fall is the placing of any extra mass on the helicopter.  As the helicopter falls, it obviously spins.  When an object is spinning, it has momentum just like an object moving along a plane.  So, when you start to spin a playground ride, for example, it is much harder to spin it to begin with than once spinning, just like pushing a car.  This is because it takes energy to give the object its momentum.  When something is spinning, we call this rotational momentum.  When the helicopter starts to spin, it reduces the air resistance in a vertical direction, as the wings are effectively moving in the direction of that they are pointing.  Below is a diagram concentrating on one of the wings, and its movement:

Because the wing is both spinning and falling with the whole helicopter, its net direction is diagonal, as shown by the red arrow.  This will mean that the leading edge of the wing, which is what causes most of the air resistance, will be the very thin edge of the paper.  This will mean much less air resistance than if it was not spinning, as then the leading edge would be the entire area of the wing.

So we have deduced that when spinning, there will be less air resistance.  One way of increasing an object’s rotational momentum is to put most of its mass as far from the center of rotation as possible as this will maximize its speed and therefore give it more momentum.  If a spinning object has more momentum when its mass is far from the center of rotation, then it must require more energy to make such an object go the same speed as one with its mass in the center of rotation.

In this experiment I will place two paperclips (one either side of the center) at different distances from the center of rotation.  I expect the helicopter with its mass on the outside to take longer as it will take longer for it to spin quickly.  The readings will be taken with a constant height of 6m, and mass of 2.84g.

 Length from middle Time (sec) Velocity (cm) 1 2 3 Ave (m/s) 100 2.73 2.43 2.75 2.64 2.28 80 2.75 2.78 2.69 2.74 2.19 60 2.93 2.95 2.51 2.80 2.15 40 2.87 2.61 2.80 2.76 2.17 20 2.78 2.65 2.73 2.72 2.21

Conclusion

(m/s)

90

5.01

5.09

5.10

5.07

1.18

80

5.18

5.15

5.13

5.15

1.16

70

4.20

4.32

4.70

4.41

1.36

60

3.83

3.96

3.83

3.87

1.55

50

3.13

3.42

3.51

3.35

1.79

40

3.07

3.00

3.19

3.09

1.94

30

2.54

2.41

2.50

2.48

2.42

20

1.99

1.98

2.10

2.02

2.97

## Evaluation and conclusion

As we can see from the graph, there is a straight line, showing that the wing angle is proportional to the time.  When the angle is 90 degrees, the wing is offering its maximum air resistance as it has a much higher cross-sectional area facing the direction of motion.  When the angle is 20 degrees, there is a very small cross-sectional area, so there is less air resistance.  Another observation that I made is that as the angle got smaller, the helicopter span much faster.  If we refer back to the introduction, where I explain why the helicopter spins, we can see that it depends on the horizontal component of the force exerted on the wings by the air.  When the angle is 20 degrees, the air is exerting a force on the wings in an almost horizontal plane.  This means that the horizontal component is far greater, making it spin quicker.

#### General conclusion

Although my results have been quite consistent and predictable, I think that there has been one weakness consistent throughout the coursework.  This was the flexibility of the paper used.  The smallest of touches could deform the paper, and subsequently alter the wings.  I would have liked to have been able to measure the revolutions per minute of the helicopter, as I think this would have led to some interesting results.  However, the helicopter revolved too fast to get any accurate measurements.

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