 Above this axis is a solid cylinder mass and radius R. Is all the mass look like this? No. No. M R square with 2 because you can imagine it as it is made up of this star on top of each other. Fine. So for solid cylinder also the moment of inertia is M R square by 2. The solid cylinder moment of inertia of this rod about the axis which passes through the tank and perpendicular to the rod. The axis is this axis. So how it rotate like this or it rotate like that. This is also about which it rotates. Fine. Why not moment of inertia? Like that exactly same the way we have here in the center of mass for the rod. So if you get answer or you put that in the square then that is wrong. Next let us say this is dm by the distance of x. The width of this is? It is right. How do you please do it? Get it yourself. You feel lot more confident. You are in university in chair and doing this. But as you learn it why to waste time with nonsense. What is the moment? x square dm. This I have to integrate to find the total moment of inertia. Yes or no? Because dm is not different in terms of x. So I have to write dm in terms of dx. So x dx length mass per edge length into dx. Okay. In the case of this the mass was distributed on the surface. So that is why in the case of this you have taken mass here by area. Because it was distributed in area. Now it is distributed in the length. That is the mass per edge length. Okay. So when you substitute it you will get integral of x square into m by L is constant comes out of integral. What is the integral x square ds? Sq by 3 0 to L it will be? Lq by 3. Lq by 3 or L and L. 1 L from denominator will cancel the rest. m L square by 3. Nodes are hung from one of its ends. So that is why this moment of inertia is important. Now find out the moment of inertia about an axis that passes through the center. And per particular point. How about this axis you have to find? Same mass and total length is L. Okay. So here you are considering a small mass addition of x. So this is also integral of x square dm. Which is x square m by L into dm. Everything remains same except the limits. What are the limits? If you take which will be at the distance same distance. Listen here. You are changing orientation when you call L by 2 this side and then L by 2 that side. So your limits are minus L by 2 to plus L by 2. To cover the full round. What is the answer now? Have to apply integral calculus in physics. Everybody faces difficulty. It is not somebody learns by work. But if you face difficulty and then resist it over this is not something I want to do. Now you will never learn from scratch right? What is the answer? L by 2 is? L by 2 is? L by 2 is x cube by. So you will have m by minus of minus L cube by 8. So you are going to get m L square by. So whatever you are doing you must remember this. Nobody is going to tell you that moment of an issue of rod about this and this and that. We have to remember it when you solve problems. But anyway when you solve other problems you tend to remember this automatically. So moment of inertia for the rod you must remember from the end and from the center. The moment of inertia of the rod about this axis. The axis passes through the rod. It will be 0. Moment of inertia about that axis 0 because all the mass is at a distance of 0 from the axis. Fine that is perpendicular to the. How many axis is passing through the center exist which are perpendicular to the rod? How many? How many such kind of axis exist? And where is r? Is the answer ever? Some answers closer to the axis. Okay. You find out moment of inertia of this without details. Okay. This is your z axis. This is your x axis. All of you please draw this. Suppose there is a small mass dm. This is dm mass. Where the mass and let's say its coordinates are x, y and z. x comma y comma z. The origin is the center of the sphere. Okay. Now let's say I want to find out moment of inertia about z axis. I said. Okay. I said for this small mass should be perpendicular distance of the small mass. Square times dm. Yes or no? Now tell me what is the perpendicular distance of this mass from the z axis? Since of this point having x, y, z coordinate from the z axis. This is z axis, x and y. Whose coordinates are x, y? z axis is distance from the flow. z axis. I want the perpendicular distance from this line. So all you find out from the z axis is root over x square plus y square of the dm mass. Let's call it as di because it is for the small mass. So di z will be equal to root over x square plus y square the whole square. So x square plus y square dm. Okay. And down here. Now there will be multiple such particles. Right? So add up. i z is summation of xi square plus yi square. This is the moment of inertia of the entire sphere about the z axis. Any doubt? Now tell me what should be ix? ix will be what? Summation of? yx squared plus zx squared times i. Why it will be? x squared plus zx squared. Summation division are same. I want to tell you how to get without integration also. Now is i z, ix and iy all are equal or not? It is a sphere. Any line passing to the center is same in all aspects. Right? So ix is equal to i z is equal to i z. So if you add it up. So ix times i z is equal to what? Two times of summation. Please do it your own. x square plus y square plus z squared times dm. Do you get this? Right? And can you tell me what is x square plus y square plus z square? Distance of the surface from the center what it is? Radius. Radius squared. Right? So this entire private thing is r squared. Which is a constant for whichever x, y, z you take. Getting it? So you get three times of i z is equal to two times summation of r squared dm. So r squared is constant. So comes out three times i z is two r squared summation of dm. What is summation of dm? Total mass. So i z is two. So moment of inertia of the hollow sphere of mass and radius r is two by three m r squared. The axis should pass through the center. This is r squared. This is the distance of the point of the center. For these three i's. See if it is a sphere. Moment of inertia about any line passing through the center should be same. Because this axis is no different from that axis. It is a sphere. It is three dimensional symmetrical sphere. That is why I have taken i z, i x and i y all things equal. Due to symmetry, all of this comes. Now you can choose to integrate out the center of mass, location of a hemisphere. Similar way you can follow and get the answer after integrating. That is the homework. You write it down. The homework is to determine the moment of inertia using integration for the hollow sphere. But this is tricky actually. It is straight forward. When you learn it, it becomes simple. So the homework is to find out moment of inertia of the hollow sphere using integration. How do you integrate? You assume a ring below. Here you assume a ring like this. The way we have done it for the hemisphere. And this is the theta. You remember, the radius of this ring is r sin theta. And the width of the ring is r d theta. So mass per unit area of the ring. Why do you do it?