Young's Modulus of a Wire

Transcript

0:00Introduction

Hello everyone, here is a small puzzle. Hang four kilograms on a thin wire two metres long, and it stretches by only one point four millimetres. But a change of room temperature affects its length too, and the support can sag. So how can you trust such a tiny reading? The answer is a neat idea with a second wire. Questions on this experiment do come up in JEE, so let's begin.

0:28Young's modulus: stress over strain

Look at the left of the figure. The wire hangs from the hatched support, and capital L, with the long arrow, is its length. Capital M is the load, and small l, beside the little double arrow, is the extension. The label gives small d, the diameter, found with a screw gauge. The graph and the notes come later, okay? Young's modulus, written capital Y, is stress over strain. The stress is capital M times small g, with small g the acceleration due to gravity, over pi times r squared, the area, with r the radius. The strain is small l over capital L, the extension over the length. So capital Y equals capital M times small g times capital L, over pi times r squared times small l, clear?

1:16From the radius to the diameter

But the screw gauge gives the diameter, not the radius. So when you put r equal to d over two, what changes in this formula? Yes, exactly, a four comes to the top, because with r equal to d over two, r squared becomes d squared over four. So capital Y is four times capital M times small g times capital L, over pi times d squared times small l, with me?

1:43Searle's method: why two wires?

The experiment is called Searle's method. Two identical wires hang from the same support, okay? One of them, the reference wire, carries a fixed load. The other, the experimental wire, is loaded in equal steps. Right? The extension, small l, is read with a spirit level and a micrometer screw, or a vernier, and the length, capital L, with a metre scale. Now, we add load only to the experimental wire, so why do we need the second wire at all? The reason is simple, both wires are affected equally by a change of room temperature and by any yielding of the support. So the difference in their lengths shows only the extension due to the added load. That is our puzzle answered. Got it?

2:29The graph: Y from the slope

In the graph, extension, small l, is on the vertical axis, and load, capital M, on the horizontal axis. The dots are the readings, and they lie on a straight line. Delta, the little Greek triangle, means a change, so the slope is delta l over delta M, the sides of the dashed triangle, okay? Now suppose, for the same length and radius, the line is steeper, so each kilogram produces more stretch. Is capital Y then larger, or smaller? It is smaller, because the slope is extension over load, and the extension, small l, sat at the bottom of our formula. So capital Y equals small g times capital L, over pi times r squared times the slope. The formula in the figure says the same thing in terms of d, four times small g times capital L, over pi times d squared, times delta M over delta l. That last part is just one over the slope.

3:30Taking good readings

Clear? Now, we read while loading, and again while unloading, and we average. So why read in both directions? Because the mean removes any lag of the wire in responding to the load, and backlash in the screw. And if the loading and unloading readings agree, the elastic limit was not passed. Right? Now here is a trap. Suppose the load is so large that the wire does not return to its original length. What has gone wrong, and how does the graph show it? The elastic limit has been exceeded, so always keep the load within it. The graph bends away from a straight line, and the unloading readings do not retrace it. Got it?

4:11Errors in Y

Now let us look at the errors. Here is our formula, with d squared at the bottom. Which measured quantity counts double in the error? Yes, exactly, it is the diameter, because of that square. Here, delta means the error in a quantity. Delta Y over capital Y equals delta M over capital M, plus delta L over capital L, plus two times delta d over d, plus delta l over small l. All the terms add, okay? The diameter enters as d squared, the most important source of error, as the note on our figure said. Right? We measure it with a screw gauge, at several places, in two perpendicular directions. Why so much trouble? Because the wire may not be uniform along its length, or perfectly round. Since capital Y depends on d squared, the mean diameter must be known well. Clear?

5:08Example 1: Y of a 2 m wire stretched 1.40 mm by 4 kg

Now the examples, the easier one first. Here, a wire two point zero metres long and zero point six zero millimetres in diameter extends by one point four zero millimetres under a load of four kilograms. Find Young's modulus, with small g equal to nine point eight metres per second squared. Pause the video and try it first. Done? Okay, let's check. First, does zero point six zero millimetres go in as the radius? It does not, and this is a common mistake, because that is the diameter. So r is zero point three zero millimetres, which is zero point three zero times ten to the power minus three metres. Right? Now let us substitute. On top is four times nine point eight times two point zero, and this four is just the mass in kilograms. At the bottom is pi, times the radius, whole squared, times the extension in metres. With me? The top gives seventy eight point four, and the bottom about three point nine six times ten to the power minus ten, and the wavy equals sign means about. So capital Y is about two point zero times ten to the power eleven, in newtons per square metre, written capital N times small m to the power minus two. Got it?

6:30Example 2: Y from a micrometer reading, pitch 0.5 mm

Now the harder one. The micrometer of Searle's apparatus has a pitch of zero point five millimetres and one hundred circular divisions. Its reading changes by one rotation and forty divisions when the load is increased by five kilograms, on a wire one point five metres long and zero point eight zero millimetres in diameter. Find capital Y here. Pause the video and try it first. Done? Okay, let's check. The extension is hidden inside the micrometer reading, so how much does one full rotation add? One full rotation adds one pitch, zero point five millimetres, and not one millimetre. The least count, the pitch over the circular scale divisions, is zero point five over one hundred, which gives zero point zero zero five millimetres. Okay? So small l is zero point five, plus forty times zero point zero zero five, which gives zero point seven zero millimetres. Right? Now substitute, with the load as five kilograms, the increase that produced this extension. On top is five times nine point eight times one point five. At the bottom is pi, times the radius, whole squared, times the extension in metres, and the radius is half of zero point eight zero, so zero point four zero millimetres. Clear? The top gives seventy three point five, and the bottom about three point five two times ten to the power minus ten. So capital Y is about two point one times ten to the power eleven newtons per square metre. If you got this, very good work.

8:10Quick revision: formula, second wire, slope, diameter error

So, quick revision. Use this formula for capital Y, with the load within the elastic limit. The second wire cancels the effect of any change of temperature and any yielding of the support. From the graph, the slope goes at the bottom of the formula. And in the error, the diameter counts double, so measure it with great care. Now go and try the practice problems on this page. All the best!