 Alright, so let's continue this. If you get the answer, let me know. Work energy on the rigid body, okay? So, we are going to apply work energy theorem, which is what? W is equal to, you remember, U2 plus K2 minus U1 plus K1. This I have to apply on the rigid body. So, that is what? Force into displacement of the point of application of the force, okay? And I need to understand how to write the potential energy and kinetic energy, alright? So, let's take one by one how to write the gravitational potential energy for the rigid body. Gravitational potential energy, okay? Suppose this is the rigid body, this is M1 at a distance of h1, M2 at a distance of h2. Like that, there are different even point masses. So, total potential energy will be what? M1 h1 plus M2 h2 plus M3 h3 and so on. I have to add up all the point masses potential energy on the rigid body, logical? So, potential energy total g h1, assuming g is constant, assuming g is constant, okay? M1 g h1, M2 g h2 and so on. So, it will come out to be equal to summation of Mi hi into g. This is the total potential energy, okay? Now, assume your y-axis in vertical direction. Let's say this is y-axis, fine? So, your hi is what? Yi, y-coordinate, okay? So, this is summation of into g. Potential energy of a particular point mass can be negative. Suppose, rigid body is still here. So, the y-coordinate will be negative. So, automatically consider the potential, negative potential energy also. This is my reference line, zero potential energy, which we all always draw her internal line, right? Wherever we write the potential energy, we draw her internal line for the gravitation. Now, I know that summation of y-central mass, let's call it as h-central mass, height of central mass. Got it? Central mass could be here. So, this will give me a summation of mass into y-cm, top talking M g h, center of mass, a potential energy. This equation is still, why we are doing it separately? Because rigid body can rotate. And when it rotates, it's in velocities, okay? So, I need to add up all the energy. So, in the course of this chapter, then I need one more class. So, we will spend one month on rigid body dynamics, system of 20% of a rigid body. This is, let's say, fixed axis. Please write down. I'll first find out the kinetic energy for the fixed axis. How to write it? Let's say, this is the rigid body. This rigid body is rotating about this fixed axis. Angular loss of omega. There are masses like this, okay? So, kinetic energy M1 v1 square plus half M2 v2 square is fixed for entire rigid body, right? This is right. So, v1 is what? This is what we get? M2 r2 square and so on into omega square. What is this bracket term? I about the fixed axis. This will be equal to half moment of inertia about the fixed axis into omega square. So, if there's a fixed axis, you don't have to do anything. Just write half I about fixed axis into omega square. It's a case one. What if moving with velocity v? Half a v square. Half a v square. Simple. It is as good as a point mass. That we are not considering at all. Any doubts? Now, case two. Axis is translating. Okay, rotation plus translation. This is the rigid body. Let's say it rotates with angular loss of omega and the center of mass is going with vcn. Okay? This is the case. Yes or no? So, velocity of a point which is here which is at a distance of r1 vcn plus omega r1. Yes or no? I am assuming my y axis to be this. This is let's say x axis and z axis comes out let us say. Center of mass is my origin. As simple as this, it will be a vector sum actually. r cross omega plus vcm vector. But we are doing it in a simpler way. So half m into vcm plus omega r1 whole square mass one. So like that there will be summation of square. This is equal to half icm omega square cm square. Prove it. No sir, but this is the vector sum. So that's not what it is. What? So this is vector sum of vcm and omega r1. So what? So it's up to vcm square plus omega r1 square. No, they are not perpendicular to each other. vcm and omega r1 are not perpendicular to each other. Okay sir, but it's still vector sum. Correct. So there will be cos theta factor coming in. Understood that. So there will be multiplied by cos theta. So like I said initially I am taking it in a simpler way. Okay. Plus summation of is not a distance. You have to consider it with sin. It's like a position vector. Fine. So r can be negative. So what is this? Summation of mi becomes 0. This is what? Total mass into vcm square. This is icm into omega square. Okay. So this is what the proof is. Half half. So this if you want to take 1 will be equal to vcm. What will be v1 square then? When you square it, how much it will become? v1 square will be what? vcm square mod of vcm square plus what? Mod of omega square. Then 2 vcm come inside the bracket dot with vcm. vcm is a constant. So summation can go inside vcm. So vcm dot of m1 r will be the simpler thing so that you understand the philosophy of how it comes. You can do it with the vector also. Translating only. It is not rotating. Then omega into r will be the axis. Is it it? Like if we take translation into consideration, won't there be any omega? So there won't be any omega. Which omega? I mean there won't be any. Not only translation. I am saying translation plus rotation. Both. So this is from the axis. The sine theta is actually the perpendicular distance. And in moment of inertia, summation of mi square sine square theta i perpendicular distance from the axis. So that is a little bit too much of detail. Otherwise I see, start seeing the blank faces. This is what it is. This is for the ith particle perpendicular distance from the axis. And that square into mass summation will be your moment of inertia. Not r i square. Getting it? Any doubts? So that's it for today. We will continue next week. We will finish up this chapter.