Potential-Energy Curves and Equilibrium

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Take two balls. One sits at the bottom of a bowl; the other is balanced on top of an upturned bowl. Both are at rest, and the net force on each is zero. So both are in equilibrium, right? Now I give each a tiny push. What happens? One rolls back. The other rolls away and never comes back! Same zero force, completely different behaviour. Why? Today one picture, the potential-energy curve, answers this. And it comes up often in JEE questions. So, let's begin.

0:37Reading the curve: the force points downhill

So, here is our picture. Along the bottom is x, the particle's position, and up the side is U of x, its potential energy there. The dark blue line is the potential-energy curve. Clear? Now, remember from earlier in this chapter: F equals minus d U over d x, and d U over d x is the slope of this curve. Because of that minus sign, the force always points downhill on the curve, towards lower potential energy. Right? So tell me. Where on this curve is the force zero? Some of you said: where U is zero. No, no, careful! The force depends on the slope of the curve, not on its height. So the force is zero where the curve is flat, where d U over d x equals zero: an equilibrium position. In the figure, the points A, B and C: a valley, a hilltop, and a flat stretch. All three are equilibria. With me?

1:31Stable, unstable or neutral?

Now, are all three equilibria the same? Put the particle at A, and nudge it a little to the right. Which way does the force point now? Back towards A, yes, exactly! Just right of A the curve goes uphill, so the force points downhill, back again. Nudge it left, same story. This is stable equilibrium, where U is a minimum. The test: d squared U over d x squared, the second derivative, is greater than zero. Okay? Now B, the top of the hill. Nudge it, and the force points downhill, away from the top, so it runs off. Unstable equilibrium: U is a maximum, and the second derivative is less than zero. Right? And C, on the flat stretch. Move it a little along the flat part, and there is no force at all, so it just stays in its new position. Neutral equilibrium: U is constant over a range. Got it? So, our two balls from the start. The ball in the bowl sits at the bottom of a potential well: stable, it comes back. The ball on the upturned bowl sits on top of a hill: unstable, it rolls away. Same zero force, but one sits in a valley and the other on a hill. Puzzle solved! Clear?

2:43Total energy E: where can the particle go?

Now, the orange dashed line is E, the total mechanical energy. When only conservative forces do work, E stays constant, so the line is flat. Now tell me: can the particle ever be where U is above E? Never! The kinetic energy, capital K, equals E minus U, and K can never be negative. So K is greater than or equal to zero, and U of x must be less than or equal to E, okay? The greater-than and less-than signs with a line underneath mean or equal to. Clear? Where U equals E exactly, K is zero: the particle stops for an instant and turns back. These are the turning points, x one and x two, that is x with a small one or two below it. The two-headed arrow between them marks the allowed region. With me? Now look carefully. Our particle starts between x one and x two. To the right of B, the curve dips below the dashed line again. So, can the particle get there? No, it cannot! To get there it must climb over B, and near the top of that hill, U is more than the total energy. So it is trapped in the well around A, going to and fro between x one and x two, fastest at A, where U is least. Right? So, to cross B, its energy must be more than U at B, the U with a small B below it. One comparison, and the graph question is done. Clear?

4:07Small oscillations about a stable minimum

One more idea. Say a stable minimum is at x naught, that is x with a small zero below it. Give the particle a small push, and it oscillates about x naught. These small oscillations are simple harmonic. Okay? So, what decides how fast it oscillates? Two things: how sharply the valley curves, and the mass. Omega, the Greek letter like a round w, is the angular frequency. It equals the square root of U double prime at x naught, over m; the tick shaped sign is the square root. U double prime, with two little dashes, is the second derivative, and m is the mass. Got it? And the period, capital T, equals two pi over omega. Topper's tip: no force diagram, no equation of motion. Just differentiate U twice, put in x naught, divide by the mass, and take the root. That's it!

4:59Example 1: equilibria of U = x^3 - 3x and their nature

Now, Example one. The potential energy is U equals x cubed minus three x, in S I units. Find the equilibrium positions, and their nature. Pause the video and try it. Done? Okay, let's check. Equilibrium means zero slope. U prime, with one little dash, is the first derivative: three x squared minus three, set equal to zero. Okay? So x squared equals one, and x equals one. Done, right? No, no, wait! x squared equals one has two roots. The double arrow means so: x equals plus or minus one, the plus sign over the minus sign. Students drop the minus one all the time. Don't! Got it? Now the nature. Differentiate again: U double prime equals six x. With me? At x equals one, U double prime is six, greater than zero. So U is a minimum there: stable equilibrium, a valley. Clear? And at x equals minus one, U double prime is minus six, less than zero. So U is a maximum there: unstable, a hill. One valley, one hill. See how fast that was?

6:11Example 2: period of small oscillations for U = 8x^2 - 4x

Now a harder one, Example two. Here, a point five kilogram particle has potential energy U equals eight x squared minus four x, in joules; k g on screen means kilograms. Find its equilibrium position, and the period of small oscillations about it. Pause the video and try it. Hint: differentiate twice. Done? Okay, let's check. U prime equals sixteen x minus four, set to zero. So x naught is four over sixteen: point two five metres; small m is metres. Okay? Next, U double prime equals sixteen, greater than zero. So it is a stable minimum, and small oscillations about it are simple harmonic. Got it? Now omega equals the root of sixteen over point five: root thirty two. The wavy equals sign means approximately, so about five point six six radians per second, written rad s to the minus one. So the period is about five point six six seconds, right? Oh no, careful! That is omega, not the period. T equals two pi over omega, two pi over root thirty two: about one point one one seconds; small s is seconds. So, equilibrium at point two five metres, and a period of about one point one one seconds. Clear?

7:34Quick revision: slope, stable or unstable, turning points

So, quick revision. The force points downhill on the curve, and equilibrium is where d U over d x equals zero. Minimum, stable; maximum, unstable; flat, neutral. The particle can be only where U is less than or equal to E, and near a minimum, omega is the root of U double prime at x naught, over m. Now go and try the practice problems on this page. You can do them, I promise. All the best!