Moments of Inertia of Standard Bodies

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Take a solid ball of clay, and spin it about a diameter. Now melt that same ball, recast it into a flat disc with twice the radius, and spin it about its own axis. Same clay, same mass. So it should be just as easy to spin, right? Hmm, not so fast. By the end of this video, you will know if it gets harder or easier, and by how much. Today: the moments of inertia of standard bodies. They come up often in JEE, so let's begin.

0:35The thin rod: centre or end?

Quick reminder first. The moment of inertia, capital I, equals the sum of m r squared. This big Greek letter, sigma, means add up over every particle; small m is the particle's mass, and r is its distance from the axis. So mass far from the axis counts much more. And today, every body is uniform, with total mass capital M, okay? Now the simplest body: a thin rod of length capital L, with the axis through its centre, perpendicular to the rod. Then I equals one twelfth M L squared. With me? Now you tell me. Shift the axis to one end of the rod. Does the moment of inertia get bigger, or smaller? Bigger! Yes, exactly. Now the far end of the rod is a full length L away from the axis. So I equals one third M L squared, four times the value about the centre. Same rod, new axis, new answer. Clear?

1:35Ring and disc: axis or diameter?

Next, a ring of radius capital R, about its own axis: the line through its centre, perpendicular to its plane. Tell me, how far from this axis is each bit of its mass? Exactly R, every single bit! So the sum of m r squared is simply M R squared. About a diameter, a line across the ring in its own plane, it is half M R squared. Right? Now fill the ring in, and you get a disc. Its mass spreads from the centre to the rim, so about its axis, I is only half M R squared. And about a diameter, one fourth M R squared. Oh no, careful: many students use half M R squared for the diameter too. Check the axis, okay? Now cut a hole in the middle: an annular disc, with inner radius R one and outer radius R two, written with a small one and two below. Mass was removed, so do we add R one squared and R two squared, or subtract? Plus, not minus! I equals half M times, in brackets, R one squared plus R two squared. The minus is hidden inside M, the mass left after cutting the hole. Quick check: put R one equal to zero, and you get back the full disc, half M R two squared. Got it?

2:50Cylinders, spheres and a cone

Now, the cylinders. Thin hollow cylinder about its axis: M R squared, the same as the ring. Solid cylinder about its axis: half M R squared, the same as the disc. Hold on to that. But turn the solid cylinder about a perpendicular axis through its centre, and now its length L matters. I equals M times, in brackets, R squared over four, plus L squared over twelve. Okay? Now spheres. Take a solid sphere and a thin hollow shell, same mass, same radius, both about a diameter. Which one is harder to spin? The shell! Yes, exactly. All its mass sits out on the surface, far from the axis. Solid sphere: two fifths M R squared. Thin spherical shell: two thirds M R squared, the bigger one. With me? And last, a solid cone about its axis: three tenths M R squared, with R the radius of its base. Right?

3:49Plates, and the topper's stretching trick

Now a rectangular plate, with sides small a and small b; that is what a cross b means here. About the axis through its centre, perpendicular to the plate: one twelfth M times, a squared plus b squared. Okay? Now turn the plate about the axis through its centre, parallel to side b. Does the length b still matter? No, it does not! Mass spread along the axis never changes its distance from the axis. So I equals one twelfth M a squared. Look, that is exactly a rod of length a! Clear? And this is a topper's trick. Stretch a body along the axis, and its moment of inertia stays the same. With me? So the disc and the solid cylinder: both half M R squared. The ring and the thin hollow cylinder: both M R squared. That is why they matched earlier! And the thin rod and the plate: one twelfth M a squared, each about the matching central axis. Fewer formulas to learn. Got it?

4:52Example 1: sphere and shell with equal I, radius ratio

Alright, Example one: a solid sphere and a thin spherical shell have the same mass, and the same moment of inertia about a diameter. Find the ratio of their radii. Pause the video and try it first. Done? Okay, let's check. Call the solid sphere's radius R s, that is R with a small s below, and the shell's radius R h, with a small h, for hollow. Same moment of inertia, so two fifths M R s squared equals two thirds M R h squared. Okay? Now, M cancels from both sides. So R s squared over R h squared equals two thirds over two fifths, which is five thirds. Okay? So the radii are in the ratio five is to three, right? No, no, careful! Five thirds is the ratio of the squares, not of the radii. That is the trap. So take the square root; that is the root sign on screen. R s over R h equals the root of five thirds. The double arrow means so: R s is to R h is root five is to root three. And see, the solid sphere needs the bigger radius, because its mass sits closer to the axis. Makes sense, right?

6:02Example 2: sphere recast into a disc of radius 2R

Now, remember our clay ball from the start? Let's finish it. So, a solid sphere of radius R and mass M is melted and recast into a disc of radius two R, okay? Compare its moment of inertia as a disc, about the disc's axis, with its moment of inertia as a sphere, about a diameter. Pause the video and try it. Done? Okay. Many students stop here: the thickness of the disc is not given. So we are stuck, right? Not at all! Melting does not change the mass: the disc still has mass M, the same as before. And by our stretching trick, the thickness never changes a disc's moment of inertia about its axis, so we do not need it. Clear? First the sphere: I sphere, that is I with the word sphere written below, equals two fifths M R squared. Right? Now the disc: I disc equals half M times, two R, the whole squared. Oh no, careful: square the two as well! That gives two M R squared. With me? Now divide: I disc over I sphere is two over zero point four, because two fifths is zero point four. Five! So the flat disc is five times harder to spin than the ball it came from. Same mass, but now spread far from the axis. Puzzle solved!

7:21Quick revision: I of rod, ring, disc, sphere and shell

So, quick revision. Mass far from the axis means a larger moment of inertia: ring, M R squared; disc, half M R squared. Rod: one twelfth M L squared about its centre, one third about an end. Solid sphere, two fifths M R squared; shell, two thirds. And always check the axis; stretching along the axis keeps the moment of inertia the same. Now go and try the practice problems on this page. You will enjoy them. All the best!