Conservation of Angular Momentum

Transcript

0:00Introduction

Hello everyone! Let's start with something you have surely seen on television. An ice skater spins slowly, with her arms stretched out. Then she pulls her arms in, and suddenly she spins much, much faster! Nobody pushed her. Nothing outside gave her a twist. So where did the extra spin come from? And here is something even stranger: her kinetic energy goes up too. By the end of this video, you will calculate exactly how much. Today: conservation of angular momentum. It comes up often in JEE, so let's begin.

0:38The law: no external torque, no change in L

So, recall the last concept. Tau, the Greek letter with an arrow on top, is the torque. The arrow means it is a vector, okay? The small word below it says external: torque from outside the system. And d L over d t is the rate of change of the angular momentum, L vector. So, the net external torque equals d L over d t. Right? Now tell me. What happens to the angular momentum if the net external torque is zero? It stays constant! Yes, exactly. If tau external is zero, then d L over d t is zero, so L does not change at all. The double arrow on screen means implies. And read the words carefully: zero torque about a point means L about that same point stays constant, okay? Point, or axis: we will come back to this. This is the law of conservation of angular momentum. Clear? Now, for rotation about a fixed axis, L equals I omega. I is the moment of inertia, and omega, the Greek letter, is the angular speed. The small subscript one means before, and two means after. So, I one omega one equals I two omega two. With me? So, a quick question. If the moment of inertia becomes smaller, what happens to omega? It goes up! Correct. The product, I times omega, must stay the same, so a smaller I means a faster spin. Now, when you pull your arms in, your muscles act inside your own body. Those are internal forces. And internal forces cannot change the angular momentum of the system. Got it?

2:21Same L, but not the same energy

Alright, now the energy. From the last concept, the rotational kinetic energy K equals L squared over two I, okay? So, if L stays the same, does the kinetic energy also stay the same? Most students say yes. What do you say? No, no, careful! Look at the formula. L is fixed, but I sits in the denominator. So, K two over K one equals I one over I two. Smaller I, larger kinetic energy. Right? So write this down: angular momentum conserved does not mean kinetic energy conserved. These are two separate laws. Mechanical energy can change while L stays the same, and the other way round too. With me? Two classic cases. One spinning disc is dropped onto another, and friction makes them turn together. L stays the same, but energy is lost. And a person walking on a turntable: again L is the same, but the energy changes. Clear?

3:21Trap: conserved about which point?

Now, the biggest trap of this topic. Angular momentum can be conserved about one point, and not about another point. Okay? Take a central force: a force that always acts along the line joining the body to a fixed centre, marked O, okay? The force may push or pull, but it always points through that centre. Right? So tell me, what is the torque of this force about O, the centre itself? Zero! Yes, exactly. Torque, tau, is r cross F, the cross product of r and the force. Here r, the position vector from O, and F, the force, lie along the same line. So the cross product is zero. But about other points, the force generally has a lever arm, so the torque is not zero. Got it? So, the Sun's gravity on a planet, and the tension in a string that passes through a hole at the centre of a table. In both, L is conserved about the centre only. In the exam, always ask: zero torque about which point?

4:24Example 1: skater pulls arms in, new omega and KE ratio

Clear? Now, remember our skater from the start? Let's finish her story. She spins at three radians per second, with I equal to four kilogram metre squared. In the units, the small minus one up on the s means per second, okay? She pulls her arms in, and her I becomes one point six kilogram metre squared. Find her new angular speed, and the ratio of the kinetic energies. Pause the video, and try it first. Done? Okay, let's check. Pulling her arms in uses only her own muscles, which are internal forces. There is no external torque about her spin axis. So, I one omega one equals I two omega two. Right? So, omega two equals four times three, over one point six. Twelve over one point six: seven point five radians per second. Two and a half times faster! And that answers our puzzle: no outside twist needed, just a smaller moment of inertia. Okay? Now the energy ratio. Angular momentum is conserved, so energy is conserved too, and the ratio is just one, right? Oh no, careful! That is exactly the trap we just saw. L is the same, so K two over K one equals I one over I two. Four over one point six: two point five. Her kinetic energy goes up two and a half times. Clear? And the extra energy comes from the work done by her muscles as she pulls her arms in. Internal forces cannot change L, but they can change the kinetic energy. Got it?

6:06Example 2: student on a stool flips a bicycle wheel

Now a harder one, JEE level, and it has a lovely trap. Here, a student sits on a stool that turns freely. Student plus stool together have I equal to five kilogram metre squared. The student holds a bicycle wheel, with I w equal to zero point five kilogram metre squared. The small w below means wheel, okay? The wheel spins at twenty radians per second, with its axis vertical. Now the student turns the wheel upside down. Find the student's angular speed. Pause the video, and try it. Done? Okay, let's check. The stool turns freely, so there is no external torque about the vertical axis. So L about the vertical axis stays the same. About which axis, always! With me? At the start, only the wheel spins. L equals I w omega w, where omega w is the wheel's angular speed. Zero point five times twenty: ten kilogram metre squared per second, pointing up. Right? Now the wheel is upside down, so its spin points down. So the student and stool pick up ten, and omega is ten over five, that is two radians per second, right? No, no, wait! That is the trap. The total must stay ten, up. But the wheel now has ten pointing down, that is minus ten. So student plus stool need ten plus ten: twenty, up! Angular momentum is a vector, so always watch the direction. Okay? So, omega equals twenty over five: four radians per second. And the student turns in the same sense as the wheel's original spin. Four, not two. That is why the direction matters so much. Clear?

7:53Quick revision: I1 omega1 = I2 omega2, but KE changes

So, quick revision. No external torque about a point means L about that point stays constant. For rotation about a fixed axis, I one omega one equals I two omega two, but the kinetic energy usually changes: K two over K one equals I one over I two. And always ask: zero torque about which point, or which axis? Now go and try the practice problems on this page. All the best!