Theorems of Parallel and Perpendicular Axes

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle before we start. Take a long ruler and spin it about its middle. Now hold it at one end, and spin it again. Much harder, right? But how much harder? About the end, the moment of inertia is how many times the value about the centre? Two times? Three? By the end of this video, you will get it in one line, with no integration. These two theorems come up often in JEE rotation problems, so let's begin.

0:32Parallel-axis theorem

So, look at the left half of the figure. The orange dot, marked C M, is the centre of mass of the body. The dark line is an axis through the centre of mass, and the moment of inertia about it is I c m: capital I, with small c m below. The green line is a second axis, parallel to the first. The theorem says: I equals I c m plus M d squared, where capital M is the total mass. And as the brackets say, this works for any body, flat or not. The right half comes later. Okay? Now, d is the distance of the new axis from the C M axis, the one through the centre of mass: the double-headed arrow in the figure. So, one condition: you must always start from the C M axis. Right? Now think. M d squared can never be negative. So, out of all these parallel axes, which one gives the smallest moment of inertia? The axis through the centre of mass, exactly! There, d is zero, so nothing is added. Move the axis away, and the moment of inertia only grows.

1:36The trap: always go through the centre of mass

Clear? Now, a trap that catches many students. Suppose you know I one, with a small one below, about an axis at distance d one from the C M axis, the parallel axis through the centre of mass. And you want I two, about another parallel axis, at distance d two. With me? Many students write this: I two equals I one, plus M times the bracket d two minus d one, whole squared. Looks natural, right? But is it correct? No, no, careful! That is the trap. M d squared is added only when you start from the centre-of-mass axis, and I one is not about that axis. So first subtract M d one squared, to come down to I c m. Then add M d two squared, to go out to the new axis. Got it? So the rule is simple: down to the C M axis first, then out to the new axis.

2:27Perpendicular-axis theorem

Clear? Now the second theorem. Look at the right half of the figure. This is a plane lamina: a flat body, like a thin plate, lying in the x y plane. Okay? The x and y axes, the dark arrows, lie in the plane of the plate. And the z axis, the green arrow, stands straight up, perpendicular to the plate. Right? And all three axes meet at one point, the orange dot. Clear? Then the perpendicular-axis theorem says: I z equals I x plus I y. I z, with a small z below, is the moment of inertia about the z axis, and I x and I y are about the two axes in the plane. So, the axis standing up equals the sum of the two lying flat. Got it? Now tell me. Can we use this theorem for a solid sphere, or a cylinder? No, not at all! This theorem is only for flat bodies, like discs, rings and plates. Not for spheres, and not for cylinders. Don't fall for this one! With me?

3:28Quick results from the theorems

So the perpendicular-axis theorem cannot handle a sphere. But what about the parallel-axis theorem? Can we use that one for a sphere? Yes, of course! It works for any body, remember? Take a solid sphere about a tangent. The tangent is parallel to a diameter, at distance capital R, the radius. About a diameter, the sphere has two fifths M R squared. Add M R squared, and you get seven fifths M R squared. Right? Now a ring, about a tangent perpendicular to its plane. That tangent is parallel to the ring's own axis, again at distance R from it. About its axis, a ring has M R squared. So, M R squared plus M R squared: two M R squared. Clear?

4:12Example 1: rod about its end from I about the centre

Alright, Example one. Obtain the moment of inertia of a rod about an end, from the value about its centre. Take its mass as capital M, and its length as capital L, okay? Pause the video, and try it first. Done? Okay, let's check. About the axis through the centre, perpendicular to the rod, we already know I c m equals one twelfth M L squared. That axis passes through the centre of mass, so we can start from it. Right? Now the axis through one end, parallel to this one. How far apart are these two axes? Half the length, yes! d equals L over two. With me? So, by the parallel-axis theorem: I equals one twelfth M L squared, plus M times L over two, in brackets, whole squared. Okay? L over two, squared, is one quarter L squared. And one twelfth plus one quarter is one third. So I equals one third M L squared. Clear? Now look: one third M L squared is four times one twelfth M L squared. There is our puzzle, solved! That is why the ruler felt so much harder to spin about its end. One line, no integration. Got it?

5:27Example 2: disc about a diameter and a tangent

Now a harder one. Use the perpendicular-axis theorem to find the moment of inertia of a disc, of mass capital M and radius capital R, about a diameter. Then find it about a tangent in its plane. Pause and try. Hint: start from the axis you already know. Done? Okay, let's check. The disc is flat, so this theorem is allowed. Now, which axis of the disc do we already know? Its own axis, exactly! Through the centre, perpendicular to the disc. That is our z axis, and I z equals half M R squared. Right? Now take two diameters at right angles, lying in the disc. These are our x and y axes, and all three axes meet at the centre. Every diameter of a disc looks the same, so by symmetry, I x equals I y. Call it I d, with a small d below, for diameter. With me? Now apply the theorem: I x plus I y equals I z. So two I d equals half M R squared. The double arrow means so: I d equals one quarter M R squared. That is our first answer. Okay? Next, the tangent in its plane. It touches the rim, and lies flat in the plane of the disc. So which theorem now: perpendicular, or parallel? Parallel, yes! This tangent runs parallel to a diameter, at distance capital R, the radius. And that diameter passes through the centre of mass, so we can start from it. Clear? So, I equals one quarter M R squared, plus M R squared: five fourths M R squared. Oh no, careful here! As the bracket says, do not add M R squared to half M R squared. That axis is perpendicular to the disc, not parallel to this tangent. Got it?

7:14Quick revision: I = Icm + M d squared, Iz = Ix + Iy

So, quick revision. Parallel axes, for any body: I equals I c m plus M d squared, always starting from the C M axis. Perpendicular axes, for flat bodies only: I z equals I x plus I y. And among parallel axes, the least moment of inertia is about the one through the centre of mass. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!