 Alright, so next we'll write down the torque equation, should we derive it or directly we'll write it? Derivation. Derivation we can take later, see what will happen is that you'll get mathematically so much bombarded with all the equations and everything that you'll not be able to enjoy the numerical stuff, okay. First let us do everything, you'll feel good about yourself and then I'll tell you how that torque equation comes. Did it make sense? So torque is equal to I times alpha. Just like force is equal to mass times acceleration, torque is equal to I times angular acceleration. Moment of inertia torque and alpha is angular acceleration, okay. Can I use this equation about any axis at about any point? No, I cannot use it at any axis. There is a restriction, the restriction is we write down this can be used only about fixed axis or any axis that passes through center of mass. This cannot be used about any other axis. If the object is rotating, if the object is not rotating it is stationary then every point is a fixed axis in a way. I have to find torque about the fixed axis and I have to find moment of inertia about the fixed axis. I'll get alpha about the fixed axis, isn't it? Alpha of the rigid body is fixed, okay. Alpha of every point is fixed about every other point, every any other fixed point it is, alpha only, right. Torque and I can change but the ratio between torque and I will remain same only. What else? There is no other theory, okay. Torque is equal to I alpha, okay. And then if suppose there is an external force applied on a rigid body, of course there will be external force because there is a torque, right. Most of it will be an external force. So net external force will be mass into acceleration along x axis. This acceleration is whose acceleration? Center of mass at acceleration. So now you have alpha as well as A, ACM. And using the kinematics you can find out the total acceleration now. First you observe with respect to center of mass then you add what center of mass is doing, okay. Similarly Fy will be equal to m into Ay, Fz is equal to m into Az, okay. Do not hesitate to use all these equations you already have learned. Newton's second law is center of mass is acceleration. That you should apply without hesitating. Torque equation also which will give you alpha. And suppose you have a relation between alpha and A. If you know that it is rolling without slipping, alpha is equal to A by R or A is equal to alpha, then you can connect the equation with that equation because alpha is equal to A by R. So there is a connection between these two equations. Are you getting it? But if suppose there is no relation between alpha and A, there is no connection. Is it clear? Okay. So let us take couple of simple questions then you will understand how to come up with this. There is a rod fixed axis. Okay. Let us say it rotates by an angle of theta. We need to find out the angular acceleration at this moment. So 3G cos theta by R. So 3G cos theta by R. 3G cos theta by R. Yeah, that is right. So 3G cos theta by R to R. Yeah. See MG will act from the center. MG forces on the right. Okay. Let us take forces like this. There will be one force this way and other force like that. Let us say. This you can say NY and this you can say NX. I mean I do not know which direction the reaction from the pin is exerting. So I will just say that it will be in a single plane only. Otherwise the rod will come out. It will remain in a single plane. The reaction force on the pin will be on that plane only. So assume it has two components. One in X and one in Y. Are you getting it? Right. So now I have to find out the angular acceleration. Is there a fixed axis? First step is to find out if there is a fixed axis. Have fixed axis? All right. This is here. This is MG. This is a fixed axis or not? This comfort of MG is what? Why I am finding that comfort? Because that is perpendicular to the distance. How much is that? MG cos theta. MG cos theta? Yeah. So this is MG cos theta. This is 911 theta you guys are saying. So this is MG sin theta. So torque because of MG is what? Is there any other force which can give torque about the fixed axis? No. Only MG can give. So torque because of MG is MG cos theta into L by 2. This is equal to moment of inertia about fixed axis into alpha. Which is what? M L square by 3 into alpha. So how many alpha? 3G cos theta. 3G cos theta. 3G by 2L cos theta. Okay. Now you have to find out what will be its angular velocity when it has reached a vertical position like that. You know. Angular velocity. Omega is what? Omega when the rod has become vertical. How are you? Sir. 3G cos theta. 3G cos theta by 2L cos theta. Okay. From where you are getting theta? I am releasing from here. Theta is 0. And here I am finding the angular velocity. Theta is 90 from 0. Theta is 90. What is omega? Sir. What is theta? 3G by 2L. Phi by 2L. Phi by 2L. Theta is pi by 2. So theta, pi by 2. Who will put this expression? Sir. What is the answer? Under root 3G cos theta by 2L. I like that. Under root 3G pi by 2L. Basically it is root pi. Under root pi alpha. So what? Under root. Wait sir. From this position till vertically. It is released from here. This rod is like this. It has angular velocity. Oh. The rod is like this. Look at the game. Tell me what it is. It is not explaining anything. Tell me how it is. This sector has angular velocity. Sir. Because you do that. So I thought you meant from that angle. Oh. That is correct. Oh. Adu. Okay. My mistake. This is root of 3G by 2L. Sounds correct. What do you say? Not sure. 3G by 2L. How did you get it? I took alpha as omega into d omega by 2L. Yeah. Just wake up man. Here you need to apply kinematics. This is alpha. Alpha is what? Omega plus theta d theta into d. Omega goes from 0 to omega. Theta goes from 0 to pi by 2. Okay. Like this you get the answer. Sir. Why can't we conserve energy? Because I haven't talked yet. No. So like can I just say that it starts off with potential energy. No. Not now. I don't want you to use kinematics right now. All right. I'll teach then you can use. Sir. But then the answer with that is? No. You have done it wrong. I'll check that during the breaks. Okay. Are you reading it? Sir. But the sign is minus that. What minus? 0 minus. Yeah. Pi by 2 minus positive. Yeah. Sir it's minus pi by 2 sir. It's going this way. Sir theta which is 0 to 90. Sir it's minus pi by 2 minus positive. Sir it's minus pi by 2. Sir this is theta. All the sign. The integral quantity is what? Sign. Sign is that. The integral quantity is what? Sign. Is that what I was talking about? Sign theta. It is not minus sign theta. Yeah. Okay. The sign is minus cos, it is reverse, any other term. Sir, I'm not, but I'm just, What happened? Fall, fall. Sir, I thought that if the central mass moves from that top position to that position over there. So, no sir, the bottom, so the change in height of central mass is L by 2. No, L by 2. Sir, we will do half i omega square. i is mL square by 3. Oh, half i omega square. No, why half mv square? It is rotating, it's not rotating. Sir, you can see that the loss of the central mass is some v. Then you can write half icm omega square plus half mvcm square. And vcm is omega into L by 2. That's what I'm saying. Wait for me to teach. Otherwise you'll not learn what I'm going to teach and you'll not learn what I'm teaching right now. Both you'll not learn. Oh, I forgot. Then you have to find out the reaction from the pin. Find out where it was. This is Parang. Let's say this is r y and this is r x. Sir, you'll use, like, you'll have a pole. When it's rotating, one force will act inward and outward and they should be equal. Now check. Sir, you can take centripetal acceleration. Centripetal acceleration will be around the rod. So r x will be equal to mv square by L by 2. The radius with which the center of mass is moving. Are you getting it? Is equal to mass into vcm square divided by L by 2. So center of mass alpha into L by 2. L by 2 taken in opposite direction. So let's keep it this way. This is r y. So r y is equal to m into x is now center of mass, which is alpha into L by 2. Finding alpha is 3g by 2 will cos theta. So substitute here, you'll get r y. For r y, you need vcm and vcm is... So this will come out from here while integrating 0 to theta. Total reaction will be root over this square plus that square. So we'll have some centripetal... Got it? Yes sir. Get out. You don't know why you can't... What? ...conserve energy. No, you can conserve energy. That is what I said. I am saying you are using it in the wrong way. And I can't teach everybody spending... Everybody's 15 minutes how to conserve energy and then tell you... Are you getting it? You are using the wrong formula for kinetic energy. Either you write half icm omega square plus half mvcm square. Or you write half i about fixed axis into omega square. You are using neither. You are saying just kinetic energy is half mvcm square, which is not correct. So that's not the correct energy. That's not the correct kinetic energy. When you write half mvcm square, you are assuming that entire mass is... Translating without rotating with vcm.