Moment of Inertia and Radius of Gyration

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Every body has just one mass. Weigh a cricket bat, and you get one number, right? But try to spin that same bat, and it can feel easy or hard, depending on the axis you turn it about. So tell me: how many moments of inertia does one bat have? One? Two? The answer is surprising, and by the end of this video you will see why. This idea comes up often in JEE rotation problems, so let's begin.

0:33Moment of inertia: the mass of rotation

So, push a box in a straight line. Its mass resists any change in its motion. That is inertia, right? Now we want the same kind of quantity for rotation about an axis. Let's build it together. Break the body into small particles. Capital I, the moment of inertia, equals the sum of m i times r i squared. The big Greek letter on screen is sigma, and it means add over all the particles. Small m with a little i below is the mass of each particle, and r with a little i below is its distance from the axis. Okay? Now tell me. One kilogram close to the axis, or the same kilogram far from it: which adds more to I? Far away, yes! And by a lot, because r is squared. Double the distance, and that particle adds four times as much. But careful: r is the perpendicular distance of each mass from the axis, measured at right angles to the axis line. Clear? For a continuous body, like a rod or a disc, the sum becomes an integral. I equals the integral of r squared d m. The tall curved S is the integral sign, and d m is a tiny element of mass. Got it? The unit is kilogram metre squared, and the dimensions, in square brackets, are M L squared: mass times length squared, just like the formula. With me? So what does I really measure? Rotational inertia: how strongly a body resists any change in its rotation. In rotation, I plays the role that mass plays in a straight line. Right?

2:05The trap: always state the axis

Think back to our formula. What was inside it? Only masses, and their distances from the axis. So I depends on three things: the mass, how that mass is distributed, and the axis. Okay? Now a question that catches many students. Imagine a wheel spinning faster and faster. Does its moment of inertia increase? No, not at all! Speed is nowhere in the formula. I does not depend on the angular velocity omega, the Greek letter on screen. Spin it fast or slow, and the same axis gives the same moment of inertia. Don't fall for this one! Got it? Now, our puzzle from the start. Any body has just one mass. But how many moments of inertia does it have? Infinitely many! One for every axis you can choose. Change the axis, and the perpendicular distances change, so I changes. So in JEE, never write the moment of inertia of a body without saying about which axis.

3:03Rotational kinetic energy and radius of gyration

Clear? Now, energy. For a rigid body turning about a fixed axis, the kinetic energy is K rot, with the small word rot below for rotational. It equals half I omega squared. Compare it with half m v squared, in the brackets: mass becomes I, and speed becomes omega. Same pattern! With me? Next, a neat idea. Suppose the whole mass, capital M, sat at one single distance k from the axis. What would I be then? M k squared, exactly! Every particle is at the same distance k, so the sum is just capital M times k squared. Turn it around, and the double arrow means so: k equals the square root of I over M, the total mass. This k is called the radius of gyration. Right? So what does k mean? It is the distance from the axis at which you could put the whole mass, as a ring, and get the same moment of inertia. Okay?

3:55Two particles: a handy shortcut

One more handy result. Two particles, of masses m one and m two, with the small one and two below, sitting a distance d apart. Right? Take the axis through their centre of mass, perpendicular to the line joining them. Clear? Then I equals mu d squared, where mu, the Greek letter on screen, equals m one times m two, over m one plus m two. Remember it for molecules with two atoms. Got it?

4:20Example 1: I and k for three point masses

Alright, Example one. Particles of one, two and three kilograms are at distances one, two and three metres from an axis. Find I and the radius of gyration. Pause the video, and try it first. Done? Okay, let's check. Each mass times its distance squared. One times one squared, plus two times two squared, plus three times three squared. One, plus eight, plus twenty seven: thirty six kilogram metre squared. Right? Now the radius of gyration: k equals the square root of I over M, as on screen. But which mass is capital M here? The total mass, correct! One plus two plus three: six kilograms. Not just the heaviest particle. Okay? So k is the square root of thirty six over six, which is root six metres. The wavy equals sign means about: roughly two point four five metres. So all six kilograms, as a ring of that radius, would give the same thirty six. Clear?

5:22Example 2: four masses on a square, three axes

Now a harder one. Four particles, each of mass m, sit at the corners of a square of side small a. Find I about three axes: part a, the axis through the centre, perpendicular to the square; part b, a side; and part c, a diagonal. Pause and try. Hint: for each axis, find each particle's perpendicular distance. Done? Okay, let's check. Part a. How far is each corner from the centre of the square? Half the diagonal, yes: small a over root two. So r squared is a squared over two. Four particles: four m, times a squared over two, and the dot means times. That gives two m a squared. Okay? Part b, the axis along one side. Two particles sit right on that side. What is their r? Zero! They are on the axis, so they add nothing. The other two are one full side away, small a. So I equals two m a squared. Got it? Part c, the axis along a diagonal. Two particles lie on it. And the other two are small a away, right? No, no, careful! Small a is the length of a side, not the distance at right angles to this diagonal. The two on the diagonal add nothing, and the other two are half the other diagonal away: small a over root two. With me? So I equals two m, times a squared over two, which is just m a squared. Right? Now look at all three answers: two m a squared, two m a squared, and m a squared. Same four particles, same mass, different axes, different moments of inertia. There is our opening puzzle, solved! Always state the axis.

7:05Quick revision: I depends on the axis, k = sqrt(I/M)

So, quick revision. I is the sum of m r squared, with r the perpendicular distance from the axis. I depends on the mass, its distribution and the axis, but not on omega, so always state the axis. And K rot is half I omega squared, while k, the radius of gyration, is the square root of I over the total mass. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!