Elastic Potential Energy

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Pull back the rubber cord of a catapult and hold it. Nothing is moving, right? Now let go, and the stone shoots off, at about twelve metres per second in one of today's examples. So where was that energy hiding? Inside the stretched rubber, as elastic potential energy. Today we learn how to calculate it, and we will get that twelve metres per second ourselves. Questions on this come up often in JEE, so let's begin.

0:33Stretching a wire stores energy

So, a quick recap from the last lesson. Within the elastic limit, Hooke's law holds, and a stretched wire behaves like a spring. The force F equals k times small x, the extension. And k, the force constant, equals Young's modulus Y, times capital A, the area of cross section, over capital L, the length. Okay? Now, stretching the wire takes work, and that work is not lost. It is stored in the wire as elastic potential energy, capital U, okay? And here is the key idea: it equals the area under the graph of force against extension. Clear? Now think. At the end, the force is F and the extension is delta L; the little triangle, delta, means change in. So the work is simply F times delta L, right? No, no, careful! The force was not F all the time. It starts at zero, and since F equals k x, it grows steadily. So the graph is a straight line, from zero up to F, at x equal to delta L, with me? So the area is a triangle, with base delta L and height F: one half, base times height. Capital U equals one half F delta L, okay? That half is there because the average force is only F over two. Forget it, and every answer comes out double. Got it?

1:58Three ways to write U

This same energy can be written in three ways. The first is the one we just found: one half, times F, times delta L, okay? For the second, put F equals k delta L; the double arrow on screen means so. Then U equals one half k, times delta L squared; the small two above the bracket means squared. The same one half k x squared you met for a spring. Right? For the third, use the elongation formula from the last lesson: delta L equals F L over A Y, okay? Put this into the first form. What do you get? Yes, exactly! One half F, times F L over A Y, which gives F squared L, over two A Y, the same energy, now written with the load. Clear? Topper's trick: if the question gives the load, use F squared L over two A Y; if it gives the extension, use one half k delta L squared. Pick the right form first. With me?

3:01Energy per unit volume

Often we want the energy in each cubic metre, not in the whole wire. That is small u, the energy per unit volume: capital U over capital V, okay? And the volume of the wire is capital A times capital L, area times length. So divide one half F delta L by A times L, and look at the pieces. Which two familiar quantities appear? Regroup: one half, times F over A, times delta L over capital L; the dots mean times. F over A is the stress, and delta L over L is the strain! Very good. So small u equals one half, stress times strain. Same pattern: one half force times extension became one half stress times strain. Got it? Within the elastic limit, Hooke's law says stress equals Young's modulus Y, times strain. Put that in, and small u also equals stress squared over two Y, and one half Y, times strain squared. Clear? The unit is joules per cubic metre; the minus three on the metre means per cubic metre. Careful: if a question asks for energy per unit volume, do not multiply by the volume. Right?

4:17Example 1: energy stored in a 2 m wire stretched by 1 mm

Now, Example one. Here's a wire of length two metres, area one square millimetre, and Young's modulus Y equal to two times ten to the power eleven pascals. It is stretched by one millimetre. Find the energy stored. Pause the video and try it. Done? Okay, let's check. One millimetre is ten to the power minus three metres. So one square millimetre is ten to the power minus three square metres, right? No, no, wait! The square applies to the unit too, so the area is ten to the power minus six square metres. And delta L is ten to the power minus three metres. Okay? The extension is given, so we find k first. Y A over L: two times ten to the power eleven, times ten to the power minus six, over two. Ten to the power five newtons per metre; the minus one on the metre means per metre. Got it? Now capital U is one half, times ten to the power five, times ten to the power minus six, which is delta L squared. Zero point zero five joules. That's it! Quick check with the first form: the final force, k delta L, is one hundred newtons. One half, times one hundred, times ten to the power minus three: zero point zero five joules again. Perfect!

5:45Example 2: speed of a catapult stone from elastic energy

Now Example two: our catapult. The rubber cord has length forty two centimetres, area six square millimetres, and Young's modulus five times ten to the power six pascals. It is stretched by twenty centimetres and releases a twenty gram stone. Find the stone's speed, assuming all the elastic energy becomes kinetic energy. Pause the video and try it. Done? Okay, let's check. Everything into metres first: capital L is zero point four two metres, and capital A is six times ten to the power minus six square metres. The stretch, small x, is zero point two metres. Okay? Which form of the energy fits here? Is the load given, or the extension? The extension, yes! So one half k x squared, with k equal to Y A over L, gives Y A over two L, times x squared. Five times ten to the power six, times six times ten to the power minus six, over two times zero point four two, times zero point two squared: zero point zero four. Right? On top, the powers of ten cancel: five times six is thirty. At the bottom, zero point eight four. Thirty over zero point eight four, times zero point zero four, is about one point four three joules; the wavy equals sign means approximately. With me? Now the speed. This energy becomes kinetic energy, one half m v squared. The stone is twenty grams, so m equals twenty, right? Oh no, careful! The mass must be in kilograms. Twenty grams is zero point zero two kilograms, so small m equals zero point zero two. Clear? Now one half m v squared equals U, so v equals the square root of two U over m; the tick-shaped sign is the square root. Two times one point four three, over zero point zero two, is one hundred and forty three, and its square root is about twelve metres per second. There it is: our catapult stone, at the twelve metres per second from the start!

8:03Quick revision: energy stored and energy per unit volume

So, quick revision. Capital U equals one half F delta L, or one half k delta L squared, or F squared L over two A Y: the area under the force against extension graph. Per unit volume, small u equals one half stress times strain, or stress squared over two Y, or one half Y strain squared. All within the elastic limit, and never forget that half. Now go and try the practice problems on this page. All the best!