Conservative and Non-conservative Forces

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. Carry a bag up to the third floor, by the stairs or by the lift: gravity does the same work both times. Now drag a box between two points on the floor, once straight and once the long way round. Does friction also do the same work both times? Not so sure now, right? Today we find out why some forces care about the path and some do not. This comes up often in JEE, so let's begin.

0:31Conservative forces: the path does not matter

So, here is the big idea. We call a force conservative if its work from a point A to a point B depends only on those two points, not on the path. On screen, that is W with a small A arrow B below it, okay? Now think. The particle goes from A to B on one path, and comes back to A on another. What is the total work done by a conservative force? Zero! Yes, exactly. Going, it does some work W; coming back, the end points swap, so it does minus W, on any path. So the work round every closed path is zero. Got it? Now, the conservative forces: gravity, the spring force, the electrostatic force, and every constant force. Okay? Also every central force. Here, f of r depends only on the distance r from a fixed centre, and r hat, the r with a little cap, is the unit vector pointing away from that centre. With me? And the non-conservative ones: friction, air drag, and other forces that depend on the velocity. Friction and drag always act against the motion, so a longer path means more work lost.

1:41Is friction conservative?

Right? Now, a classic JEE trap. Every constant force is conservative, and kinetic friction has a constant magnitude. So kinetic friction is conservative, right? No, no, careful! Its magnitude is constant, but its direction is not: it always opposes v, the velocity, the v with an arrow on top. Okay? So on every piece of a closed path, friction does negative work. The total, W with friction written small below it, is less than zero, never zero. Clear? So the exact rule: constant in both magnitude and direction, then, as the double arrow says, conservative. Constant magnitude alone is not enough. Got it?

2:24The test for a force in a plane

Now, the test for a force given as a formula. F vector, the F with an arrow on top, equals F x i hat plus F y j hat. F x and F y, with small x and y below, are the components, and i hat and j hat, with little caps, are the unit vectors. The force is conservative if partial F x over partial y equals partial F y over partial x, everywhere. The curly d means a partial derivative: differentiate in one variable, keeping the other fixed. Okay? In three dimensions, the condition is nabla cross F equals the zero vector. Nabla is the upside down triangle, the cross is the cross product, and the zero with an arrow is the zero vector. Right? Let's try it. Is y i hat plus x j hat conservative? Yes, it is! F x is y, so partial F x over partial y is one. F y is x, so partial F y over partial x is also one. One equals one: conservative. Clear? Now change one sign: y i hat minus x j hat. Partial F x over partial y is still one, but now partial F y over partial x is minus one. One is not equal to minus one, the equals sign with a slash through it. Not conservative! With me?

3:40Topper's trick: work from a potential function

Now a trick that toppers use. If the force is conservative, then F vector equals nabla phi, for some function phi of x, y and z. Phi is the Greek letter, a circle with a line through it, and nabla phi means: differentiate phi in x to get F x, in y to get F y, and so on. Okay? Then the work is just a subtraction: W from A to B equals phi of B minus phi of A, end minus start. No path needed, no integral along a curve. That's it! To find phi: integrate F x with respect to x. Then take only the terms of F y that have y alone, integrate them with respect to y, and add. The terms with x in them already came from F x. Clear? And the potential energy, capital U, equals minus phi. Watch that minus sign! We use this in Chapter three. Got it?

4:38Example 1: friction work on two paths from A to B

Now, our puzzle with the box. Let's settle it. Example one: a two kilogram block is dragged on a floor from A to B, five metres apart. Mu, the Greek letter for the coefficient of friction, is zero point one. Find the work done by friction, first directly, then via a point C, with A C three metres and C B four metres. Pause the video, and try it first. Done? Okay, let's check. Friction, small f, is mu m g, with m the mass and g the acceleration due to gravity: zero point one times two times ten, taking g as ten metres per second squared. Two newtons. Now, the displacement is five metres both times. So the work is minus ten joules both times, right? No, no, wait! That is the trap. Friction opposes the motion on every bit of the path, so it acts over the whole distance travelled. So W equals minus f times the path length. With me? The straight path is five metres: minus two times five, minus ten joules. Okay? Via C, the path is three plus four, seven metres: minus two times seven, minus fourteen joules. Right? So, puzzle solved! Same end points, but different work. So, as the double arrow says, friction is non-conservative. Beautiful, isn't it?

5:59Example 2: find a so that F is conservative, then the work

Now a harder one, JEE level. F vector equals small a x y plus three, times i hat, plus x squared plus two y, times j hat. Find small a so that the force is conservative, and then the work done from zero comma zero to one comma one. Pause and try. Hint: test first, then find phi. Done? Okay, let's check. Partial F x over partial y is small a times x. Partial F y over partial x is two x. These are equal everywhere only when small a is two. Clear? So F x is two x y plus three. Step one: the integral, the tall S, of two x y plus three, d x, is x squared y plus three x. Okay? Step two. The integral of the whole of F y, d y, is x squared y plus y squared. So do we add all of it to phi? Oh no, careful! We already have x squared y from step one; adding it again counts it twice. So take only the y-alone term of F y, two y. Its integral, d y, is y squared. Got it? So phi equals x squared y, plus three x, plus y squared. Quick check: its x derivative is two x y plus three, and its y derivative is x squared plus two y. Perfect! Now the work. Phi of one comma one is one plus three plus one, and phi of zero comma zero is zero. So W equals five joules, along any path! See how fast that was?

7:37Quick revision: the test and the potential function

So, quick revision. Conservative: the work depends only on the end points, and it is zero round any closed path. The test: partial F x over partial y equals partial F y over partial x; then the work is phi of B minus phi of A, end minus start. And friction and drag are non-conservative: constant magnitude is not enough. Now try the practice problems on this page. All the best!