Year 12 Young's Modulus of Wool Experiment

 I tied the wool around something heavy. The heavy object has been heavier than 500 grams because we'll be hanging 500 grams of water on the other end. I did consider tying it around the table leg on the far side. That would work. The length of wool has to be as long as you can make it.

 It helps if the end of the table is smooth. There must be as little friction as possible where the wool rubs over the table edge. Tie a knot in the wool at about 25cm from where it goes over the edge of the table. This will be our marker for measurements.

A bottle needs to be tied to the loose end of the wool and I'd advise putting a washing up bowl underneath to catch any drips.

 Once this is done, write down the position of the knot on the ruler. Get a measuring jug and fill it with 100ml of water. This is 100 grams.

Pour it carefully into the bottle. Write down the position of the knot on the ruler scale and calculate the extension of the wool (how much it has stretched). Repeat until you have done 200 grams, 300 grams, 400 grams, 500 grams. You need a table that records mass, weight and extension.
Finally carefully pour the water from the bottle back into the jug trying to disturb the wool as little as possible. Take a reading of the unloaded position again. You should find that it hasn't gone quite back to its original position. There is a slight permanent deformation in the wool.
You need to measure two other things:
1. The diameter of the wool.

Calculate the cross-sectional area A of the wool in square metres.

 2. The total length of the wool from the bottle along. I included all of the wool that was tied around my dumbbell because all of that wool stretched as well.
This is L, the length of the wool. Write it in metres.

Here are some bits of the formula sheet

Plot a graph of force on the y-axis against extension on the x-axis. It should curve at the beginning but draw a best fit line through the straightest section further on.

Calculate the gradient and explain why the gradient is the spring constant k.

Change the units so that the spring constant is in Newtons per metre

I then combined four of the above equations. 

Use your gradient along with L and A to calculate a value for the Young Modulus of wool in Newtons per square meter. It will be a very big number.

Calculate the biggest %U for the measurement of the water volume. 

Explain why this gives us the biggest %U for force F.

Calculate the biggest %U for extension.

Calculate the %U for the diameter. (What does it mean if it comes out as 100%, which it might here?)

Explain why we have to double it to get the %U for A.

Calculate the %U for length L.

Now explain why we have to add %U for F, extension, A and L to get %U for the Young Modulus.

How many significant figures can we justify for our answer for the Young Modulus given the total %U?

Calculate the +- absolute error in the Young Modulus.

Comment on any problems in the experiment and how you overcame them.

State any control variables and their values. Comment on whether it is realistic to say that they actually remain constant during the experiment.

Give URL and date accessed for any web pages used.

Hooke's Law experiment

I hung a spring from a clamp stand. I put ruler from the bottom of the loop on the spring across to a metre stick behind. I did this because I needed to be sure that  although the spring wasn't touching the metre stick, I had the spring and the metre stick lined up properly. If you don't do this and end up looking across at the wrong angle, you'd get a different reading. Looking at a scale from the wrong angle is called PARALLAX ERROR.

 I made sure that when there was nothing on the spring, the bottom of the loop lined up with 0cm. It means that this week we can read extension directly. We don't need to subtract the original length. I added 100ml of water at a time from a measuring jug in the kitchen. I was adding 100 grams of water every time - that is 1 Newton of weight.

The next picture shows 100 grams of water -  that is 1 Newton of weight. You need to make a table for the following pictures where you have a column for Weight of water and a column for Extension. You must include units in your column headings. Read the extension each time from the photograph. Enlarging the photograph will help. The first extension is 3.0cm because you are reading to the top of the ruler.

Here is the reading for 200 grams of water which is 2N weight of water. The picture is a bit blurred but the extension is 6.1cm.

Next picture is 300 grams of water. You need to come up with the correct weight in Newtons and read the extension from the photograph.

Below is 400 grams of water.

Below is for 500 grams of water.

Before we go, here is a quick pause. After 500 grams I took the bottle off and found that the spring went back to 0 cm extension. This will be important later.

Then it was 600 grams of water.

700 grams of water.

800 grams of water.

900 grams of water

1000 grams of water

1100 grams of water

1200 grams of water

1300 grams of water

1400 grams of water

After 1400 grams, i took the water off and found that the spring did not go back to 0 cm. We had added too much weight and we had permanently stretched the spring.

Here's my graph. Notice that it is a straight line through the origin for a long time. We say that the extension is directly proportional to the force. This is called Hooke's Law. The the line starts to curve. The point at which the line starts to curve is called the LIMIT OF PROPORTIONALITY. If you remember, for 5N I took the weight off and the spring went back to its normal original length. We'd say that it is still elastic. In the end, at 14N, when I took the spring off it was permanently stretched by 2cm. That's what happens if you go beyond the limit of proportionality.

The points are not all perfectly on the line. They are close to the line but not perfect so we say that there is random error. Here is a picture of my measuring jug to help you to work out why.

Has the sky always been blue?

Posting about the polarised sky the other day and about the scattering by oxygen and nitrogen causing the blue colour, I started thinking about whether this had always been true. The atmosphere has changed over time since the beginning of the planet. There is a theory that the sky used to appear orange due to methane. https://physics.stackexchange.com/questions/277544/orange-sky-3-7-billion-years-ago-because-there-was-little-oxygen The first answer refers to Titan, the moon of Saturn. Here is a lovely film of the visit to Titan by the space probe Huygens https://www.youtube.com/watch?v=HtYDPj6eFLc (I've always liked Beethoven's 4th Piano Concerto!) In the film, there is a definite blue layer visible at the beginning. That must be the nitrogen. Then everything goes orange. I suppose I'd assumed a red dust storm such as happens on Mars but apparently not. For more details on the atmosphere https://en.wikipedia.org/wiki/Titan_(moon)#Atmosphere

Polarised sky

The last time there were so few aeroplanes in the sky over Wigton was when the volcanic ash cloud grounded flights in 2011. It was then that I learned that the sky is naturally polarised in certain directions. When I say the sky, I mean the scattered light from nitrogen and oxygen molecules. 

 I stood with my shoulder lined up with the Sun. I took my polarising clips and rotated them. You'll notice that the picture below is darker than the one above. Hence the light coming in when I face this way is polarised. What struck me this time was that the polarisation was not horizontal and vertical. In the top picture, the clips were tilted to be along a line from the Sun to me.

 Next I stood with my back to the Sun. You can just see my shadow to make the point.

 Looking into the clip made no great difference, though in the pictures this might not be apparent because the bottom picture shows more of the sky. The closer you get to the zenith, the darker blue the sky appears.

Rainbows around the Sun

It's been a great day for seeing a halo around the Sun. It started with this upper arc that was showing the full spectrum. Later there was a full halo with the requisite two sun dogs to the right and left of the Sun.

Today I noticed for the first time that the sun dogs weren't quite on the halo itself: they are just beyond. All of these phenomena come from the refraction of light through ice crystals very high in the atmosphere. Start here: https://en.wikipedia.org/wiki/Halo_(optical_phenomenon) and here https://en.wikipedia.org/wiki/Sun_dog This halo is 22 degrees from the Sun https://en.wikipedia.org/wiki/22%C2%B0_halo
I was even more excited by two phenomena I hadn't seen before.

The first was a small section of an outer 46 degree halo. It is just by the right hand side of the washing line. https://en.wikipedia.org/wiki/46%C2%B0_halo The second is just visible in the first two photographs. There is a hint of a horizontal, perhaps slightly upcurving bar tangent to the top of the 22 degree halo. This is the upper tangent arc. https://en.wikipedia.org/wiki/Upper_and_lower_tangent_arcs It is interesting the way in which each phenomenon is dependent on a different orientation of the ice crystals.

Year 10 Density experiment

I filled the ice tray with water and weighed it. It came out at 159 grams.

 Then I put it into the freezer overnight. When it came out, it was still 159 grams.

 But if you compare the picture below with the the picture of the water in the tray, you'll notice that the ice is taking up more space. In the freezing process, the ice has expanded and has increased its volume.

Both the water and ice have the same mass, so you can't say that water is heavier than ice. However, because ice has a bigger volume, we say that ice has a LOWER DENSITY.
You calculate density by doing density = mass/volume.
Density is measured in grams per cubic centimetre      or      kilograms per cubic metre.

It looks like it must be scattering

Here's another lovely Super Moon photograph taken through a net curtain that I have been sent by our friend. When I did the maths on my own photograph, the angular spread was not enough to justify a conclusion of diffraction as the cause of the spread. Thinking about what I did with the laser yesterday, the situation with the Moon is different. The laser was close to the curtain with the screen distant. The Moon was far from the curtain with the screen (camera) close. With the weave about 0.5mm wide, the screen needs to be several metres from the curtain to get observable diffraction. So it is likely that the effect with the Moon could be scattering. I don't really know enough about the maths there to be able to work it out. Next I need to try the laser far from the curtain with the screen close and also see what happens when I fire the laser at individual threads.
The weave for the picture above looks tighter than mine and the material looks shinier.

Net curtain diffraction

 I fired the laser though the net curtain and got a pattern on a screen 4 metres away. At first I thought that it was just showing light shining through the holes because the beam is a couple of mm wide and covers several holes. But there does appear to be some kind of diffraction extending beyond the central blob.

I tried to apply the diffraction grating formula to the line that goes through 12 on the ruler. So the first fringe n=1 is roughly 5mm from the centre. Small angle approximation means sin(theta) = 5/4000. Slit separation through close inspection of the weave with a 10x magnifying glass shows gap to be 0.5mm. So wavelength = 0.0005 x 5/4000 = 625nm. That's almost perfect. But it seems at odds with some of the contributions here https://www.researchgate.net/post/How_to_explain_the_diffraction_pattern_through_a_net_curtain Maybe it is the use of a laser and not an LED.

L6 Diffraction grating lab

I fired a laser through a diffraction grating which turned a single laser dot into 14 dots (fringes). The central fringe is n=0. There is a first order (n=1) on either side etc. I've labelled some of them on the photograph below.

 Your job is going to be to measure the distance for each order from one side of the centre across to the other side. I have shown below for n=1 and I am going to call that distance y1. You will then be asked to HALVE it to get x1. The reason for measuring twice the distance that we actually need is that it halves the percentage uncertainty in the reading.

 I also measured the distance D from the fringes to the grating.

 Here is the reading for D.

On the photograph below, x1 is half of the measurement y1 made above. When you have found x1 and D, you calculate angle theta by using trigonometry as shown below. Eventually you will be calculating sin(theta)

 The next set of photographs will allow you to make measurements for y1, y2, y3, y4, y5, y6 and y7. Make a table with columns for n, y, x, theta and sin(theta)

 I used the 100 lines per mm grating. Remember that in the equation n.lambda = d.sin(theta), d is the slit separation. d is NOT the same as lines per mm. There are 100 lines in 1mm. You can then calculate how far apart 2 lines are. That is d.

In the end, you plot a graph of n on the y-axis against sin(theta) on the x-axis. The graph should be a straight line through the origin. Gradient = d/lambda so you can use the gradient to calculate a value for the wavelength of this red laser light.