 Alright, so let us talk about the derivation of torque equal to i alpha, okay. We'll take a simpler case, okay, and then we'll try to extend it to a generic case. Suppose, this is the axis of rotation, it spins, okay. It is a fixed axis of rotation. I am going to prove torque equal to i alpha for the fixed axis, okay. Another axis is alpha. This perpendicular distance of the point mass m1 is let's say is r1 for let's say m2, the perpendicular distance is r2, okay. The acceleration of m1 is what? Alpha into r1, acceleration of alpha r2 in opposite directions. Are you getting it? Acceleration is in opposite direction. We will talk about it later on. But right now, the net force, let's say that is f1. Mass m acceleration? Yes, acceleration 1. Is this the correct thing? There's something wrong here. Tell me what it is. Sir, you're neglecting all the force. Any acceleration as alpha r1. Is that the only acceleration? There will be centripetal acceleration also. It is moving in a circle, right. So centripetal acceleration will be omega square into, so there will be centripetal force as well to provide centripetal acceleration. Centripetal force will be, this is what tangential force. There will be centripetal force fc, which is m1 omega square r1. Yes or no? Two forces are there, tangential and centripetal. But the good thing is that centripetal force passes through the axis. It touches the axis. So the torque because of this force will be? Zero. Zero. So torque will be only because of this tangential force. Is it clear? Okay. M1 r1 square alpha. And then you sum that with r1. Yes or no? This m2, although the acceleration is an opposite direction, the torque will be in the same direction, same sense of rotation. So if you add up all the terms, even the total torque, this is the total torque, which will be equal to summation of into alpha. This will be equal to summation of omega i square is what? Moment of inertia about that axis. So i into alpha. And what is the summation of all the terms? This will be equal to sum of all external plus the internal torque. Summation of all forces will be zero. This is Newton's third law, Newton said. Summation of all torque you have extended from your side. And net force is zero. Is it necessary that net torque will be zero? No. So why net torque is zero? Why internal torque will be zero? Because one will, let's say there is a force between two particles, one will generate a torque in one direction, the other will generate it in the opposite direction. Prove it mathematically. Prove that sum of all internal torque will be zero. Just take two masses, okay. You write two masses. Internal forces will be acting towards each other, equal and opposite. If one is f, other will be minus f. You need to prove that torque about this point will be zero. Origin, this is R1 that is R2. Prove it, sum of all torque will be zero. Force is that force. At these two, if a total torque is equal to either which is connecting these two points. So cross total will be zero. So if you can have pairs of all the internal forces two at a time, they'll cancel out. That is when net internal torque will be zero. So from there it comes out that net external torque to a rigid body will be equal to I into alpha without the fixed axis. What is zero? You are not there in vectors. You have not seen the videos. Do you know how to subtract the vectors? R2 is this, R1 is this. Where is R2 minus R1? Copic pen, you should have it, make the notes, treat as if it is in a class. R2 minus R1, sorry, R2 plus this vector x should be equal to R1. R2 plus x is R1, this is the, x will be equal to R1 minus R2. It makes zero angle or 180 degree. Both ways, the cross product will be zero with f. So the net external torque will go I alpha about the fixed axis. But then it will become very involved ICM into alpha. This is also true. But torque you have to take here about the center mass axis. I'll just give you a basic logic or overall logic of it. If you other than center mass axis, other than center mass fixed axis, then there will be a pseudo force which is going to act about the center mass. If this derivation assumes that this axis is fixed, so what you will do? No problem. But if this axis has an acceleration, you will have to apply the pseudo force. And that pseudo force point of application is center of mass. So then you have to use torque due to the pseudo force as well over here. But if your axis passes through the center of mass, then torque because of that pseudo force will be zero. But if you keep on doing that, 15-20 minutes will be gone and we lose all the track of it. It is not going to be used anywhere. Just remember these two equations. You can use one about fixed axis, other about the center of mass axis. So if there is a fixed axis, it is easiest to use. Center of mass axis is safest to use because center of mass axis is always valid. Fixed axis you may or may not find. I have five minutes for the while. Sir, for the derivation of that torque internally is zero. We said that the two masses, one with force F and minus F, they cancel each other out there. But what if we have more mass on this side of the axis than this? So we have more particles here than here. What will that affect the internal torque? See, more particles here and there does not matter actually because this point does not assume that both the particles will be, both the particles, it does not matter. Sir, if you have like 10 things with force minus F and 5 with force F. It cannot happen. It cannot. All the internal forces should add up to zero. Do you agree? So how can that be happened? If somebody is applying F on that minus F will be applied. That is the Newton third law itself. Sir, I have center of mass. So you said that you have to apply something extra, some pseudo force or something like that. If you are using about center of mass, then you do not need to worry about torque into pseudo force. Go into worry. For the fixed axis we should. Fixed axis, anyway you do not need to worry about anything because you are standing on the fixed axis and looking at everything. But when you are using about center of mass, you are standing on center of mass axis. Even though center of mass accelerates, torque due to the pseudo force will be zero. So pseudo force torque will not come in the equation if you are using torque about center of mass is equal to ICM alpha. But if you are using about some other axis which is neither fixed nor passing to center of mass, then you cannot use this formula. So basically this formula you can use only about fixed axis and about center of mass. Let us not debate on it too much. Just take it like a thumb rule. Take it as in which you should follow. It will keep your life simpler. Many of the students do not even know. By the time they come in class 12, they are about to pass out also. Nobody tells them that you can use torque due to alpha only about two axis. From Fiji came. I interviewed. He also did now. How? I mean, I know everything. That is like forgetting me. I mean, you grow old doesn't mean that I also don't know everything. If you give me a question, probably I may not be able to solve that in there. Because the mechanics is vast and it is not about how much questions you complete. It is about then and there if you are able to think a couple of questions from some j advance batch. There is a question which still for it has been a month now we are not able to solve. And it is from kinematics. It sounds so simple. Please attempt it at home. I keep on thinking about that like while teaching you also I might be thinking about this problem. This is a icy river. Frictionless. I see river. I see water frozen. Wait, is it frozen? Frozen? What is icy river? I think we are river with ice. This is frictionless. No friction. Mu is equal to zero. No friction is on the surface. Here is Param standing. He has to cross the river at the shortest time possible. He snows long. What? He snows long. You have to find out the shortest time it will take to go to the other side. Are you getting it? Now he can't start from here. Reason? He is taking time. There is some optimum distance. Assume that it will go in a straight path it will get the wrong answer. There should be a reason, logic and answer. Final answer I will tell you. Sir I have seen this question before sir. I have seen that between my friends. He has done this. I have seen that. He is like this. I want to be like this. He snows long. What logic is that? No problem. He has seen the zero. He has seen the zero. He has seen the zero. No problem. Please copy down the answer. It is not that straight forward. I think it came in some physical olympiad somewhere. Please write down. If you think it is simple. Two steps you are wrong. Sir why won't it be in a straight line? Even I am wondering right now. I will give it to you. If you assume it to be a straight line the answer comes out to be to under root L by Muji. Which is actually more than this. So walking in a straight line is not even possible. So I mean that is little strange. So then I started to do some research. Then I found out that when a runner is running. He is not putting his, I mean it's not a wheel that is rolling. So he is putting his feet like this. So there will be some angle. At which he will push the ground. So maybe that is the reason why the acceleration will not be able to Muji all the time. So maybe you have to go little bit in detail but just little bit. He is moving at constant velocity. What is constant velocity? Constant velocity is not possible. Constant velocity is only in the IC river. Here he is moving because of friction. So how much can you accelerate with? We need how much he can accelerate. Acceleration is friction force Mu Mg divided by mass Mu G is the acceleration. That is straight forward. But that itself is wrong actually. Only acceleration force on the person is friction. What else acceleration force is? No but if he himself is within. What do you mean by that? How can I walk at different speeds? Sir how can I walk at different speeds and friction is always the same? Friction is acceleration, speed is velocity. What are you saying? How can you accelerate and accelerate? You can accelerate but you don't need to. Static friction can vary from 0 to maximum friction. You are not speeding while running right? You can give it. I should not have brought this up. Alright we have 20 minutes. We will continue solving problems. No point talking about a new concept. What should we talk? Sir I was just talking. Alright so let's... What? Right now. You have that person right in your class. He left. He left. He left? Why? He left. He left. Give it to Kota. Give it to Kota? Give it to Kota. He is an alien. Now you are an alien. Give it to Pranav. I will give it to Pranav. This is a part in 12th grade. His name is Pranav Asigar. He left. No he is in 12th. When they said he left they meant 12th and like songs. Alright let's stop the talk. Continue. Continue this. If you get the answer let me know. Take it. Energy consideration. What kind of energy? One energy to apply in the rigid body. You are considering energy whether to apply or not? You are wrong. You are wrong. Work energy body. We are going to apply work energy theorem which is what? W is minus u1 plus k1 rigid body. Force into displacement of the point of application of the force. 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. Suppose this is the rigid body at a distance of h1, m2 at a distance of h2. Like that 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 g h1 assuming g is constant. Assuming g is constant. m1 g h1 m2 g h2 and so on. So it will come out to be equal to summation of into g. This is the total potential energy. Now assume your y axis in vertical direction. Let's say this is y axis. Fine. So your h i is what? y i y coordinate 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 also. This is my reference line. Zero potential energy. Which we always wherever we write the potential energy. We draw our internal line for the gravitation. Now I know that summation of centre of mass h centre of mass. Height of centre of mass. Got it? Centre of mass could be here. So this will give me mass into ycm. Of talking m g h centre of mass. Potential energy. This equation is still but I don't know how to write the potential energy. Why we are doing it separately? Speeds and velocities. So I need to add up all point masses. The way we have done for the gravitation potential energy. Chapter. Then I need one more class. So we have spent one month on rigid body dynamics. System of 20,000. Or like someone here. We take different cases. Let's say fixed axis. Please write down. First find out the potential energy for the fixed axis. How to write it? Let's say this is the rigid body. The rigid body is rotating about this fixed axis. Angular loss of omega. There are masses. R1. M1. Like this. Kinetic energy. 1v1 square. Plus half. M2v2 square. For entire. I started remembering my mobile phone. She is the like whenever she wants to use calls. Whatsapp also came. Start sending stickers and all that. Take care. So v1 is what omega into? R1. This is what you get? Okay. So this is half. M1 R1 square. Plus M2 R2 square. And so on. Into omega square. What is bracket term? I about. Fixed axis. This will be equal to half. Moment of inertia about the fixed axis. Into omega square. So if there is a fixed axis. You don't have to do anything. Just write half I about fixed axis. Into omega square. What if it is just moving with velocity v? Half. Half and v square. Simple. It is as good as a point mass. So that we are not considering at all. Any doubts? Now case 2. The axis is translating. Translation. It say it rotates with angular velocity omega. And the center of mass is going with. Vcm. Okay. The case. Yes or no? So velocity of a point. Which is here. Which is. Which is at a distance of R1. Vcm plus. Omega. 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. Okay. 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. Square. So half m into. Vcm plus. Omega R1. Whole square. Mass 1. So like that there will be. Square. You need to prove that. This is equal to half. Icm omega square. Cm square. Prove it. But this is the vector sum. So that's not what it is here. What? So this is vector sum of Vcm and omega R1. So what? So it's until Vcm square plus omega R1 square. No they are not perpendicular to each other. Vcm and omega R1. Okay so plus. Yeah sure. But it's still the vector sum. Correct. So there will be cos theta factor coming in. Understood that. So there will be multiplied by cos theta. So life inside initially I am taking it in a simpler way. Cm square 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 or positive also. So what is this? Summation of Mi becomes 0. Cm square. This is Icm into omega square. Okay. So this is what the proof is. Half. Half C. Enough. I have to do something about you. And I'll do it soon. You can't smile like a fool. And I'm telling you something. So when I turn my back you start talking. You talk to him then talk to him then. So you can treat it like a vector also. So this if you want to take it like a vector. V1 will be equal to Vcm. Take V1 like this. That will be what? Vcm square. Mod of Vcm square. Plus what? Come inside the bracket. Dot with Vcm. Vcm is a constant. The summation can go inside Vcm. So Vcm dot will take a simpler. You can do it with a vector also. Okay. Translating only. It's not rotating. Then omega to access. So R want to access. Isn't it? Okay. So this term. Sir, it's not. Like if you take translation to 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'm saying translation plus rotation. Both. And see guys. R1 cross omega square. What it is? It is from the access. When you do R1. R1 sine theta into omega. R1 sine theta is actually the perpendicular distance. And in moment of an i square, sine square theta i, perpendicular distance from the access. Getting it? So see that too much of detail. Otherwise, I'll see. Start seeing the blank faces. This is what it is. This is for the ith particle, perpendicular distance from the access. And that square into mass summation will be your moment of inertia. Not r i square. So that's it for today. We'll continue next week. We'll finish up this chapter. And is optics a part of you? Last year it was? Again changed it. I think we don't have optics. Not there. See this. See this. So optics is not there. So after this what should we do? What is going on in school? What is going on in school? Gravitations. It's in one class. What's the other class? Gravitations? You're doing levitation after work, right? Yeah. You have two teachers in school? Different classes. Different classes? Same teacher. One teacher per class, right? Yeah. It's a chemistry.