 We need to find coefficient of friction or full friction force. Friction force acting on it. This should be what? So that it rolls without thinking. More than mg. mg r center is it 5g center? r speed up by 2r. Take care. Tau is equal to? Which force friction? 2 by 5 mg center. No, no, no. All right, I will do it now. There will be mg force, there will be normal force. Friction will be in which direction? upwards so that it start rolling like this. You have to look at the rotation as well. This is the force of friction. Any other force that we have missed? No other force, right? So which four, are you seeing any fixed axis? Bottom most point is instantaneously at rest but at times we do not feel comfortable treating at this as a fixed axis. So we will apply torque into iso about the center of mass. So which force has a torque about center of mass? Mg passes through the center of mass or not? So torque due to Mg is 0. Torque due to normal reaction? 0. Only friction will have torque. So if friction is not present, alpha can never happen. Alpha can never be there. It will just slide down. It will not rotate because torque will be 0 about center of mass. So r into fr is a torque about center of mass. This should be equal to what? 2 by 5 into alpha. So alpha into r is ACM. So friction will be equal to torque mass, ACM. Mg sin theta minus fr is equal to what? Friction value over there. So I get Mg sin theta minus 2 by 5 m into a 5 by 7 g sin theta. Substitute over there alpha friction to be equal to 2 by 7 mg sin theta. 2 by 7 mg sin theta. This is what they call rolling friction. Only this much friction is required for it to roll without slipping. The maximum possible friction is mu into normal reaction which happens because of sliding. This much will be the friction acting on it which is very less. Any doubt? You can also do this by balancing torque about that point at the bottom. They do balanced torque. Yes, over there. About this you can write the torque. Mg sin theta into r is equal to moment of inertia about this axis into alpha. You can't use about that axis moment of inertia. That's right. But then don't do like that because it is a fixed axis but it doesn't appear to be a fixed axis. So every time when you do like that you will not be feeling in control of the power. Any doubt? So what is the moment? You can't parallel axis. Yes, parallel axis will be 2 by 5 m r square plus m r square. That's okay. Any no doubt right? If this question correct you will get 2 minutes break. 5 minutes break. Let's see. 1 person gets 1 minute break. 5 person gets it 5 minutes break. 10 minutes, 20 minutes. I will know if somebody copies. What is the whole thing inside right now? So should I speak? Yes sir. Okay. So when you look at it from this side it appears like that. My god. It's a cylinder from the middle. Another small cylinder comes out. This is a string which is wrapped around the smaller cylinder which is fixed on the bigger cylinder. So this is, what happened? How do you fix it up? I took a big log, wooden log. You removed this material from the wooden log. Then you left with this kind of structure right? Yes. It's a, if you take this line, now it will feel okay. Okay sir. So where are you? You put it on the cylinder, put it on the call. Sir, both are like non-call. Both are cylinder only. See, total mass is given and complete radius 2R is given. The moment of it is M into 2R square by 2. Okay. So you don't need to add up each house. Actually, yeah, it will not be M into 2R square by 2. You can say that, see here, let me tell you the radius of gyration. The radius of gyration is let's say K. Haven't taught radius of gyration? No. Nothing happens. See, seriously. If radius of gyration is K, moment of gyration is M into K square. So radius of gyration of a ring is what? R only. For a disc it is R by root 2 because M into K square should be equal to moment of gyration. Okay. Okay. Okay. Okay. So, I should be scared about that axis. You need to find out the far A and what else? Force of gyration. Sir, it's impossible. Just write out equation like this. That's all I want. Sir, it's not slipping. No slipping. No slipping. We just watched string said something about it. String is getting pulled. String is wrapped around here and it is pulled at an angle theta with the horizontal. What's going on? If F is very large but F is not that large. Can I slide on the motor? There is no slipping. There is no slipping. It is pure rolling. Rolling without slipping. So roll backwards. Yeah. Attempt it. Okay. If you assume rolling without slipping the bottom most point is to get rest. Just write down the equations. Write down the torque equation, force equation. How simple is that? You're pulling string like this. Yeah. With force and tension. Okay. Okay. Okay. How simple is that? You're pulling string like this. Yeah. With force and tension this string is F. Oh, tension in this string is F. Are you pulling string with F? So string will pull you with what force? Yes. F only. That is a tension. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Fx. Fx what? Fx is friction. Both are F. If R is equal to F of course there is a Y. Why should balance? Why should balance? There is script now it does not roll without slipping. You do not need this equation to vertical direction, normal direction. It is correct, no one else. Mu n, why you taking mu n as friction? It is not mu n, friction is not mu n, it is just sufficient for it to have to rolling. Mu n is when sliding happens, it is rolling friction we are taking. But who else? No one else. Sir, sir, I just changed mu n to friction and it is okay. So, because everything will remain the same, I did not use. Tongue to friction is wrong. Wait, no sir, I thought due to friction was wrong. Tongue to friction is wrong. The tension should be, because it has to be rotating. See, at times you will, listen, listen, listen, at times what will happen is that you will start wondering how it is rolling, where is the friction applied. What I am saying is that at times you will worry about how it is rotating, where is the direction of friction and all that. What I am saying is, do not worry about consistency. That is all. You can assume any direction. You can assume any direction of force or friction. But when you write down the equation, you should not equate force in this direction to mass emfration of that direction. Similarly torque, clockwise torque you should not equate to I of alpha of anti-clockwise. Are you getting what I am trying to say? You can assume any direction of friction, any direction of maintaining consistency. Because torque here to I of alpha does not tell you that friction direction should be this and that. All it tells you that it should be consistent with applying the equation. Are you getting it? So, let us assume friction is this way. I do not know which way it is. So, I will assume any which way. This is the friction direction. If my assumption is wrong, friction magnitude will come out to be in the friction. That is all. So, if I write net force along x direction is equal to mass emfration, I will get f cos theta minus friction is equal to what? f into a cm vertical direction I will not write because this is the normal direction will come in place. Torque equation is what? f is what? f into cos f into r simply f into r. Because entire force is perpendicular to this distance. To find the component of force you have to use as cos theta. But in order to find torque force into r. I will just throw you out. There is a counter in my head. I will slowly and slowly become sarcastic and then there will be a limit. I am not that sarcastic. There is a limit up to which then I have to ask you to. While I am telling here you should maintain silence. When I am asking you to solve problem then you can discuss. But when I am teaching something here you should be quiet. This is a cm. This angular acceleration alpha. Why I should take angular acceleration alpha like this? No, sorry. It will be like that. It should be like this. Why should I take like that? The reason? Only then this point can be at rest. Only then alpha into r this way and ACM that way. Getting it? Everywhere there should be consistency. So f into this way. But alpha is in opposite direction. So minus f into r. Getting it? Friction into 2r. This is the total torque about the center of mass. This should be equal to ICM which is m into k square times alpha. So you can solve this and alpha is equal to a by r. This is also constraint equation rolling without slipping. 2r is equal to alpha into 2r. Sorry. So after the break we will derive the argument by alpha. Then we have angular momentum left, center of energy left. Then let's see we can start angular momentum probably. Come in 10, 15 minutes.