 If you get this question correct you will get 2 minutes break. 5, 5. 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. So what is the hole inside? So should I speak? Yes sir. Okay, so when you look at it from this side it appears like that. Oh my god. Oh my god. It's a cylinder from the middle, another small cylinder comes out. Okay, this is a string which is wrapped around the smaller cylinder which is fixed on the bigger cylinder. Are you sure? Okay. Yes. So this is, what happened? How do you fix it up? Take a big log, wooden log. You remove this material from the wooden log, then you left with this kind of structure. Yes. This line is attached, now it will feel okay. Okay sir. Sir, both are not in order. Both are cylinder only see, total mass is given and complete radius 2R is given. So moment of nature is M into 2R square by 2. Okay, cylinder here. Oh 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, radius of gyration is, let's say K. What is that? Haven't taught radius of gyration? No. Nothing happens see, radius of gyration. Oh seriously. If radius of gyration is K, moment of nature is M into K square. Okay. So radius of gyration, for a disk it is R by root 2. Because M into K square should be equal to moment of inertia. So getting it. So radius of gyration is K, so moment of nature is M into K square. Okay. Okay. Okay. Okay. Moment of nature, hope this axis is given. Yes. So the radius of gyration is K, about that axis. You need to find out, you'll be at alpha, A and what else, force of friction. Just write out equation like this, that's all I want. It's not slipping. No slipping. No slipping. What strings had something over it. String is getting pulled. String is wrapped around here and it is pulled at an angle theta with the horizontal. What's happening in the horizon? If F is very large, but F is not that large. So can that thing slide on the motor? No. Can it slide on the motor? There is no slipping. There is no slipping, it is pure rolling. Rolling without slipping. So roll backwards now. You, attempt it. If you assume rolling without slipping, the bottommost point is to get rest. Just write down the equations. Done. Draw the free board diagram, represent forces, write down the torque equation, force equation. How simple is that? So you're pulling strings like this. Yes, with force F. Tension in 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? F only, that is a tension. Fx is friction. Oh, sorry. F r is equal to F cos theta. Why? Why should balance? Why should balance? There is skid night and not roll without slipping. You don't need this equation to vertical direction. Normal direction. It's correct. No one else. Why are you taking mu n as friction? It is not mu n. Friction becomes mu n. It is just sufficient for it to have to roll in. Mu n is when sliding happens. It is rolling friction we apply. But who else? No one else. I'll write down. Check compare books. So I just change mu n friction and it's okay. So because everything will remain the same. I didn't use mu n. No sir, the tension is wrong. Wait, no sir. Thought due to friction was wrong. Tension should be because it has to be rolled in. What I'm saying is that at times you'll worry about how it is rotating with the direction of friction and all that. What I'm saying is, don't be consistent. That's 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 emetration of that direction. Similarly, clockwise torque, you should not equate to I of alpha of anticlockwise. Are you getting what I'm trying to say? You can assume any direction of friction, any direction of what I'm telling consistency because torque here to alpha doesn't 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? Let's assume friction is this way. I don't know which way it is. So I'll assume any which way. This is the friction direction. If my assumption is wrong, friction magnitude will come out to be negative since the friction is. Here friction area, that's all. So if I write net force along x direction is equal to mass emetration. I write f cos theta minus friction is equal to what? m into a cm vertical direction I will not write because this is the normal reaction will come in place. Torque equation is what? f into cos f into r simply f into r because the entire force is perpendicular to this distance. To find the component of force you have to use s cos theta but in order to find torque force into r this is a cm 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? Only then this point can be at rest. Only then alpha into r this way and a cm 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 r 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 okay okay so after the break we will derive torque even to i alpha then we have angular momentum left center of energy left then let's see we can start angular momentum probably alright so let us talk about the derivation of torque equal to i alpha we will take a simpler case okay and then we will 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 to i alpha for the fixed axis okay suppose angular acceleration is alpha alright acceleration 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 so the acceleration of m1 is what alpha into r1 acceleration of m2 is alpha r2 okay acceleration are in opposite directions are you getting it that is understood acceleration is in opposite direction we will talk about it later on but right now the net force on m1 let's say that is f1 will it be equal to m alpha r1 is this the net force mass m acceleration m1 m1 is this the correct thing there is something wrong here tell me what it is so you are neglecting all the force we know I am taking 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 r1 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 this force will be so torque will be only because of this tangential force is it clear okay but torque because of the force f1 is what torque because of only this force which is how much m1 r1 r1 perpendicularity is r1 so torque is what m1 r1 r1 this is m1 alpha r1 into r1 so this is what m1 r1 square alpha and then you sum that below yes or no on this m2 although the acceleration is an opposite direction the torque will be in the same direction because both have same sense of rotation so if you add up all the torque you will get the total torque this is the total torque which will be equal to summation of y square into alpha this will be equal to summation of mi i square is what moment definition about that axis so i into alpha and what is the summation of all the torques this will be equal to sum of all external plus the internal torques now summation of all internal torques will be 0 some of all forces will be 0 this will be Newton's third law summation of all torque you have extended from your side and net force is 0 is it necessary that net torque will be 0 no so why net torque is 0 why net internal torque will be 0 let's say there is a force between two particles one will generate a torque in one direction the other will be in opposite direction prove it mathematically that mathematics prove that sum of all internal torques will be 0 just take two masses a red two mass internal forces 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 0 origin this is r1 that is r2 prove it sum of all torque will be 0 no one got it there you go because of this force is that force it is r1 yes or no so if you add these two equal to r2 minus these two points torque will be 0 right so if you can have pairs of all the internal forces two at a time they will cancel that is where net internal torque will be 0 so from there it comes out that net external torque to a rigid body will be equal to i into alpha of the fixed axis what is 0 you are not there in vectors you have not seen the videos where is r2 you know how to subtract the vectors r2 is this r1 is this where is r2 there will be no video or while having your dinner or lunch copy pen you should have it make the notes treat as if it is in a class r2 minus r2 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 then it will make 0 angle will be 0 with f net external torque will do i alpha x axis of the center of mass axis also but then it will become very involved this formula torque is equal to icm into alpha this is also true but torque is the center of mass axis I will just give you overall logic of it if you are using any other axis other than center of mass axis other than center no fixed axis then there will be a pseudo force there will be a pseudo force which is going to act about the center of mass this derivation assumes that this axis is fixed so what you will access and look at the end 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 axis passes through the center of mass then torque because of that pseudo force will be 0 derivation of all that but if we keep on doing that 15-20 minutes will be gone and we will 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 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 always fixed axis you may or may not find any doubts? I have 5 minutes for the questions sir for the derivation of that the torque internally is 0 we said that like the two masses 1 with force f and minus f each other but what if we have more mass on this side of the axis so we have more particles here more that the internal torque see more particles here doesn't matter actually because I am this point doesn't assume that both the particles will be both the particles can be on the right side of this point it doesn't matter if we have like 10 things with force minus f and 5 with force f it cannot happen cannot all the internal forces should add up to 0 does u agree? so how can it be happen ok if somebody is applying f on that minus f will be applied that is a Newton third law itself they will always I am center of mass so you said that you have to apply something extra if you are using about center of mass then you don't need to worry about torque into pseudo-force fix axis fix axis anyway you don't 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 it is 0 so that 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 can't use this formula so basically this formula you can use only about fixed axis and about center of mass let's not debate on the thumb rule assume it to be like the rule which you should follow it will be like simpler many of the students don't even know by the time they come in class 12, about to pass out also nobody tells them that you can use torque in the alpha only about 2 axis okay I interviewed he also did now how? how do I know? that's like forgetting I mean you grow old doesn't mean that for everything even if like from I also don't know everything if you give me a question probably I may not be able to solve that and there okay 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 still I have couple of questions from some J advance batch there is a question which still it has been a month now we are not able to solve it and it is from kinematics so simple I will give you 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 IC river frictionless IC river IC means frozen glacier is there wait is it frozen frozen? IC river okay no frozen this is frictionless no friction mu is equal to 0 here coefficient of friction is mu on the surface okay here is Param standing he has to cross the river at the shortest time possible he smells law what? he smells law the answer is not there okay 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 frictionless now there is some optimum distance if you go in a straight path you will get the wrong answer there should be a reason, logic and final answer I will tell you final answer I have seen this question before have you seen that between my three to one drama he is done this no no no maybe like this I want to be self which be logic one though no problem no problem if you have incidences zero if you have incidences zero then you have incidences zero and if you have incidences zero then it will be zero where zero between them no problem no problem incidences zero please copy down the answer it's not that straight forward I think it came in some a physics olympiad we write down if you think it is simple two step you are wrong sir why won't it be in a straight line even I am wondering right now I will give you if you assume it to be a straight line the answer come 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 I am 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 on this ground so maybe that is the reason why expression 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 constant velocity is not possible constant velocity is not possible constant velocity is only in the IC river here he is moving because of friction sir how much can you accelerate with if we need how much how fast he can accelerate with 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 but if he himself is running what do you mean by how can I walk at different speeds so how can I walk at different speeds and friction is always the same friction is acceleration speed is velocity how can you accelerate in instant acceleration 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 I will stick and give it I should not have brought this up I am using this ok