Angular Momentum of a Rigid Body; Torque and Angular Impulse

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. There is a disc at rest on a fixed axle, with a string wound round it. You give the string one sharp jerk. The force lasts only a moment, and nobody knows how big it was at each instant. So, can we still find how fast the disc spins? Tau equals I alpha needs the torque at every instant, right? We don't have it. Today we learn a tool that does not need it: angular impulse. Questions like this do come up in JEE, so let's begin.

0:35Angular momentum of a rigid body

So, last time we found the angular momentum of a single particle. Now take a whole rigid body, turning about a fixed axis. Every particle moves in a circle about that axis, all with the same omega, the angular speed. Pick any one particle. What is its angular momentum about the axis? m v r, yes, exactly! On screen, L i equals m i v i r i; the small i below just numbers the particles. And the speed v i is r i times omega, so L i equals m i r i squared omega. Clear? Now add over all the particles. Omega is common, so it comes out. And the sum of m i r i squared, what is that? The moment of inertia, I, correct! The big Greek letter sigma means sum, so the bracket is I, and L equals I omega. This is the heart of today's lesson, for a rigid body about a fixed axis. Got it?

1:32Torque changes angular momentum

Now, let's find what changes angular momentum. First, recall the laws of motion: F ext vector equals d p over d t. The little arrow on top means a vector, and the small e x t below means external. So force is the rate of change of momentum, p. Okay? In rotation, torque takes the place of force: tau ext vector equals d L over d t. Tau, the Greek letter, is the torque, and d L over d t is the rate of change of angular momentum. Only external torques count. With me? Now, for a rigid body about a fixed axis, L is I omega, and I does not change. So, what does d L over d t become? I times d omega over d t, which is I alpha, with alpha the angular acceleration. Yes, exactly! So tau equals I alpha, the law from the last chapter, is just a special case of this one. Okay? Now multiply by d t, and add up over time; the tall S shape is the integral sign. The integral of tau vector d t is called the angular impulse, and it equals delta L vector, the change in angular momentum. Delta, the Greek triangle, means change. Clear? And if the torque is constant, the integral is simply tau times t. But even when the torque keeps changing, as in a sharp jerk, you don't need the torque at every instant, only the total angular impulse. That is the key to our puzzle. Right?

3:01Kinetic energy in terms of L

Now, energy. You know the kinetic energy of rotation: K equals half I omega squared. And from L equals I omega, omega equals L over I, right? So, put that in place of omega. What do you get? Half I, times L squared over I squared; the dot means times. One I cancels, and K equals L squared over two I, that's it! Got it? Now, a trap. Keep L the same, and make I half as big. K equals half I omega squared has I on top, so K becomes half too, right? No, no, careful! With L fixed, halving I doubles omega. So use K equals L squared over two I, where I sits at the bottom: K doubles. Clear?

3:45Moving and spinning: orbital plus spin

Last idea: a body that moves and spins at the same time. About a point O, L O vector equals r c m cross M v c m, plus I c m omega vector; the small O below L means it is taken about that point. The small c m below means centre of mass: r c m is its position from O, and v c m is its velocity. Capital M is the total mass, the cross is a cross product, and I c m is the moment of inertia about the centre of mass. With me? The first part is called orbital. Put the whole mass M at the centre of mass, and find its angular momentum about O, as in the last lesson. Okay? The second part is called spin: the rotation about the centre of mass itself, I c m omega. So, total angular momentum equals orbital plus spin. Right? Now tell me. When does the orbital part become zero? When r c m and v c m are parallel, exactly! The two upright lines on screen mean parallel. Parallel vectors have zero cross product. So if the centre of mass moves along a line through O, the orbital part vanishes, and only the spin is left. Got it?

4:52Example 1: constant torque on a wheel for 3 s

Alright, Example one. Here, a torque of five newton metres acts for three seconds on a wheel at rest. Its moment of inertia, I, is three kilogram metre squared. Find its final angular momentum, and its angular speed. Pause the video, and try it first. Done? Okay, let's check. We are given a torque, and a time. So, which idea do we use? Angular impulse, yes! The torque is constant, so delta L equals tau times t: five times three, fifteen. The wheel starts from rest, so this is its final angular momentum: fifteen kilogram metre squared per second, the k g m squared s to the power minus one on screen. Okay? Now L equals I omega, so omega equals L over I: fifteen over three, five radians per second. r a d s to the power minus one is just radians per second. Clear? Let's check with tau equals I alpha. Alpha equals tau over I: five over three radians per second squared, and in three seconds, omega is alpha times t: five again. Perfect! Got it?

6:06Example 2: disc given a sharp jerk of impulse J

Now a harder one: our puzzle from the start. There is a disc of mass capital M and radius capital R, free to turn on a fixed axle, at rest. The string wound on it gets a sharp jerk of impulse J, along the tangent. Find the angular speed the disc gets. Pause and try. Hint: angular impulse. Done? Okay, let's check. We don't know the force at each instant; we only know J, its total impulse. So, what is the angular impulse about the axle? J times R, correct! The jerk acts along the tangent, so at every instant its lever arm is the radius, and its torque is the force times the radius. Add that up over the jerk, and you get J times R, the angular impulse. With me? The disc turns about a fixed axle and starts from rest, so the change in angular momentum is just its final I omega. And for a disc about its own axis, I is half M R squared. So, J R equals delta L, equals I omega, equals half M R squared omega. Okay? Now cancel one R from both sides: omega equals two J over M R, that's the answer! See? We never needed the messy force at each instant, only its impulse. Puzzle solved! Clear?

7:28Quick revision: L = I omega, angular impulse, orbital + spin

So, quick revision. For a rigid body about a fixed axis, L equals I omega, and K equals L squared over two I, okay? External torque is the rate of change of angular momentum, so angular impulse equals the change in angular momentum. And a body that moves and spins has orbital plus spin angular momentum. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!