 Next is solving this fear. In the school, the syllabus, they will not ask you to derive it, but these methods, you must understand how to do it, because Jee likes to test you on the basics. They will mask changes with length and find out the moment of inertia of the rod. So if you know how to exactly solve it, you will apply the same method to solve there as well. Without knowing how it is coming, you will not be able to solve many such kind of questions. Solidity is fear, but our fear, solidity is fear is made up of hollow spheres. In hollow spheres, we will expand the solid sphere. In hollow spheres, this is r and the width is dr. Can you find out integral of d? Hollow sphere is dm. So, dm is r squared dm and hollow sphere. So, you have to integrate 2 by 3, this you have to integrate. It is hollow sphere, right? For hollow sphere, moment of inertia is 2 by 3 and the mass is dm half. That is exactly same which we did to find the side drop mass of the solid hemisphere. Just refer it back, do it yourself. How you find dm is same? You find dm is same. And it is same not only for this chapter, it will be same for about 4 years. And current and everything else, same way. In the volume of the hollow spheres, mass. There is some significant answer. 2 by 5 and mass squared is the answer. How many of you find it? 2. 3. Do it a bit. Then, total volume is into what? 5 r squared dr, which is surface area you can substitute. 5 by 5. So, you will get some basic integration to find out the moment of inertia of the some regular shape objects. The issue is for the same object, for the same object for different different axis and find the moment of inertia. So, there has to be a way to find out moment of inertia of other axis knowing moment of inertia of one of the axis. So, there should be a relation between the two axis. Fine. For that, we are going to learn two theorems which will relate moment of inertia of one axis with respect to the moment of inertia of other axis. Very straightforward, simple theorems. Please write down. I am going to end this topic. Otherwise, just 5 minutes topic, I should not leave it to the next class. Let me finish this off. Please write down parallel axis theorem. So, draw a rigid body like this. This is a, need not be a regular shape. Fine. This is a, this is an axis that passes through the center of mass. This is an axis that passes through the center of mass. And moment of inertia above the center of mass is ICN, let us say. There is another axis. By the way, the axis need not pass through the object. So, this, about this axis also the entire object can rotate like this. Okay. So, about this axis, another axis, both axis are parallel to each other. In fact, ICM and IE are parallel to each other and the distance between them is D. IE will be equal to ICM plus total mass of the rigid body into free square. Parallel axis theorem. Can I say that ICM is the least? So, moment of inertia, ICM plus a positive number. So, moment of inertia or inertia against the rotation is least about the axis that passes through the center of mass. And how many such axis are possible? How many center of mass axis possible? Infinite. Infinite. Infinite. Okay. And there can be different center of mass axis. Right? But one thing is sure, if you are finding the least moment of inertia of all the axis possible, it must pass through the center of mass. Okay. There is a small proof of this. Let us assume this is your Z axis. X. And let us say this is Y. Let us assume that there is a mass here whose coordinates are X, Y, Z. Let us say mass is small m i. X i, Y i, Z i. It is perpendicular distance from the Z is what? Root of X square plus Y square. So, moment of inertia, Z axis which is also ICM is sufficient of right now, X i square plus Y i square. Yes or no? Properties distance is root over X square of Y square. M i R square is M i into X square, X square plus Y i square. Okay. This is the parallel axis to the Z axis. This axis is parallel to the Z axis. So, moment of inertia about that axis. What do you think this will be summation of M i into? What will be its perpendicular distance from this axis? The axis is displaced along the X square. What do you think? It is X coordinate will be reduced by D. Yes or no? So, it is X coordinate relative to this line will be X i minus D. Very good. If D is exactly equal to the X coordinate for this line will become 0. So, it will be summation of X i minus D whole square plus Y i square. It is displaced only along the X axis and you can create such an area. You can rotate your coordinate axis in such a way that it got displaced only around the X axis. Okay. Now, expand this summation of M i. X i square plus Y i square X square plus Y i square plus D square summation of M i minus 2D summation of M i X i. Did you get this? How this comes? Understood. I have just expanded and taken summation inside. This constant comes out of summation here and there both. What is this? ICM i is equal to ICM plus what is this? D square into summation of all the masses is total mass. So, M into D square. Now, what is this? Summation of M i X i divided by summation of M i. This is what? X center of mass. Did you remember that? And what is X center of mass over there? Zero. X center of mass is at the origin. So, this is equal to zero. So, summation of M i X i becomes zero. So, this term goes. This is the proof of parallel axis theorem. Stress. This is next theorem, the last topic for today. Just two steps. The restriction in parallel axis theorem. One axis must pass through the center of mass. Are you getting into this between the two axes and none of them pass through the center of mass. The last topic for today is perpendicular axis theorem. What are the planar objects tell me? Square plane, a ring which entirely in a single plane are called the planar objects. Okay? This can also be irregular shape. X, Y, Z. Moment of inertia about Z axis will be the sum of the moment of inertia about X and Y. The restriction that it should pass through center of mass. Please write down Z axis. Z axis must be perpendicular to the plane. I z is perpendicular to the plane. I x and I y can be in the plane. But I z must be perpendicular to the plane. And there is a reason why we have called I x, I y and I z. The reason is perpendicular to each other. X, I, Y, Z. For mass and Y. What is perpendicular distance from the Z axis? Zero ks of x. This is Z axis and this is the point X, I, X. So root over X as Y square. We have been using it till now. G nox solid. Okay? I z is perpendicular to the formation of M I. And then talk to me. As soon as this topic gets over, all of you will die. Come back here. What is happening? Focus. We should help your idea of knowing parallelism. Go ahead and tell your friends. Just watch that comment. Stop talking. What is this? Tell me. What is X i perpendicular to the distance of the point from which axis is X i? Y axis. The formation of an X i star is I y. Why I connect from which axis is perpendicular to the distance? X axis. So this is your I y. And that is your I z. So this is the proof of perpendicular theorem. Please write an oblique theorem. See, guys, it is used quite often if I x is equal to I y due to symmetry. For example, if you do this, the moment operation of any diameter should be same. So I x is equal to I y for a test. And I z is M R square by 2 anyways, you know it. So I s plus I y becomes 2 times I x is M R square by 2. So about the diameter moment definition is M R square by 4. Because this is M R square by 2 and these are 2 equal. So 2 times of this is equal to M R square by 2. This is equal to M R square by 4.