 So I was asking, is there any difference between the vacation and break? Vacation and break, is there any difference? Tell me. The upcoming Dashara is a vacation or a break? If it is a vacation, tell yes, it's a vacation. So the 10 days which are there in front of you, that is an opportunity for someone who really want to get ahead and do better. Magic won't happen that you are sleeping and some magic will happen and suddenly you will be ahead of everybody else. That won't happen. So it is not compulsory. If you really want to do better than others, here is an opportunity because nothing will happen in school, nothing will happen in the center for probably 7 or 10 days. So there is an opportunity for you to focus on your week areas and get ahead. It's your choice if you want to do it. How if you are planning to go out station during the vacation? No one right? Everybody will be in the home only. No one will go. Corona effect. I think this Corona will be there for 2-3 years now. I mean the kind of attitude that is there right now. This change will there. People will adapt to it. Anyways, we will get ahead with our last physics class before your so called vacation starts. So last session we have discussed about the dark equation. So I will quickly recap all that. And today the agenda is that I will not teach very heavy concept. In fact, I am going to solve a lot of questions with you. And then I will be going a little slowly also so that you can understand in a much better manner. Here is an opportunity that if you haven't understood the dark concept, please ask doubts. If you don't ask doubts, I will not get a dream that you have certain doubts and you know something so I should tell that again. So if you get stuck somewhere, immediately let me know. We will discuss it. This is an opportunity for you. I think some of the students' vocations already started. 7-8 students haven't come in today's session. Alright. Write down. Today is also a focus on torque only. Torque equation. Alright. What is a torque equation? Can anybody quickly type in? Torque equation is what? Respond quickly. Okay. Type in the torque equation. F r sin theta. What else? Okay. That is not the torque equation. That is the definition of torque. Torque equation is tau is equal to i alpha. That is the torque equation. Okay. So when you say force into perpendicular distance, that is the definition of torque. That is not an equation. Alright. So the torque equation is tau equals to i alpha. When I can apply this? When I can apply this equation? How can I apply this equation? Where? What is the condition? Anyone? Only about the fixed axis? That's all. So this equation is valid for a rigid body. Okay. And i is the moment of inertia about a fixed axis. Alright. If i is, let's say fixed axis, this equation is valid or not? This equation is valid for the fixed axis? All of you agree? Now, moment of inertia is about the fixed axis. What about torque? Torque is about same fixed axis. Remember that. You find out torque about the same fixed axis and equate that to moment of inertia about that axis into alpha. Alright. And you know torque is equal to perpendicular component of force into distance or perpendicular distance into force. Right. So if you look at the diagram, I think we have made it here clear. So f r sin theta it is, you can see that perpendicular component of force is f sin theta. So f sin theta into r is torque. You can read it like that. You can also read it as f into r sin theta. What is r sin theta? r sin theta is the perpendicular distance on this line of force. So when you drop a perpendicular on this line. Okay. This angle is theta. This distance is perpendicular distance which is r sin theta. Fine. So you can write the torque equation as f sin theta into r or r sin theta into f. Both ways you get the same value. Okay. There is a lag. Is there a lag? When I speak, when I write there is a lag. Okay. Let me do some magic. Here goes the lag. Let me write something then only you will understand. Tell me. Is there a lag now? Lag is gone. Okay. Good. Fine. So you all understand, right? Torque is equal to perpendicular component of force into distance or perpendicular distance into force. Both ways. Fine. Now you can, no kinship there is no lag. Check your internet speed. So now torque is a vector quantity. One direction to take positive other direction is negative. Which direction typically we take as positive for the torque? Or tell me how you define the definition of how you define the direction of torque? How you define the direction of torque? What it is? Direction of torque is the direction in which it is trying to rotate. Getting it? Direction in which a torque is trying to rotate. Now same force can have clockwise or anticlockwise torque depending on about which axis you look at. Suppose you look at this force. Okay. With respect to let's say this axis. The axis that comes out of your plane with that axis this force will try to rotate clockwise or anticlockwise. About this axis. Clockwise all of you understand? Now about this axis which way it is trying to rotate? About the second one. Anticlockwise. The same force can produce both anticlockwise and clockwise torque. You have to see with respect to what axis. In this chapter everything depends on the axis. Fine. So be very sure about it. Now is this equation valid only for the fixed axis? No. This equation is also valid about the center of mass axis. Then I is equal to ICM alpha. Okay. Now tell me one thing. If I use let's say there is a rod. Okay. Suppose there is a rod like this. Okay. This rod is fixed at this point and it can rotate about that axis. So there is a fixed axis. Alright. So you have a force. You have a force. Let's say passing through the center of mass. There is a force. Fine. So the torque with respect to center of mass is how much? Sorry. With respect to the axis of rotation is how much? About the fixed axis what is the torque? This is what is the value of torque about the fixed axis? F into perpendicular distance L by 2. Right. This is equal to moment of inertia about the fixed axis which is how much for the rod? About one of the ends moment of inertia of the rod is how much? ML square by 3. By now you should be knowing it. ML square by 3 into alpha. Okay. This is writing the equation about the fixed axis. Now same equation torque equal to I alpha. I can write about the center of mass axis also. Center of mass is here. This is the center of mass. What is the torque about the center of mass? Tell me. Can you find out torque about the center of mass? Talk due to this force is how much? About the center of mass. Zero. No perpendicular distance. So which force is creating torque about this point? About the center of mass. Is there any force to create the torque about that point? The hinge forces is hinge forces. From where it is tied there will be some hinge forces which we do not know. Let's call it as FX and FY. Okay. If you find out FY where it is fixed that point the pin will apply force. We need to find out what is FY. Okay. So what is the torque due to FY? FY into L by 2. This is equal to ML square by 12. Now we have to use moment of energy about center of mass into alpha. Okay. So my question is for the same object if you use the equation about the fixed axis and about the center of mass axis will you get the same alpha or different alpha? Because torque and I are different about fixed axis and about the center of mass axis. What about alpha? Will alpha also depends upon which axis you are finding which axis you are writing the equation? Everyone type it. Will alpha also depend upon which axis you are writing torque equal to alpha equation? Quickly I am waiting. You don't need to solve something. What do you think? That's what I am asking. Don't tell me you don't think anything. Alpha remains same. Fine. Alpha will be same. Doesn't matter whether you write the equation about the center of mass or about the fixed axis. Alpha for a rigid body is fixed about any axis. Alpha, omega and change in angle they are same for all points on the rigid body. Remember that. This is with respect to torque equation. You can use a torque equation for a rigid body which is rotating. In a question you need to find out which object is rotating and for that object you can use the torque equation to find alpha. Now for a rigid body you can use net force along x axis is equal to mass times total mass of rigid body into expression of center of mass along x axis and similarly about the y axis and about the z axis. Whatever you have learned in Newton's second law chapter or laws of motion chapter everything is valid for the rigid body also but attrition is of the center of mass. Fine. Now we need to also find out relation relation between movement of different points on the rigid body. So this will help us to visualize better because Newton's second law equation for a rigid body will give me expression of center of mass and torque equation will give me alpha what if I am interested in finding the expression of some other point. What I am trying to say is this. For example if you have a rigid body like that this is a rigid body. It is rotating about the fixed axis. This is the axis about which it is rotating its angular acceleration is alpha. Fine. Total length is L. Now you have to tell me what is the acceleration of this point which is at a distance of x away. The distance of x away. What is the acceleration of that point this point. Suppose it just started rotating omega is almost zero. So what is the acceleration of that point in terms of alpha. No one knows how this point is moving everyone is this point moving in a circle. Tell me is this point moving in a circle or not with respect to this fixed axis point it is moving in a circle right. It is moving in a circle. And it is moving in a circle with angular acceleration of alpha. So what is the acceleration of this point. It will be tangential first of all right. It will be tangential. So what is that acceleration. Any object that is moving in a circle. Let's say this point. Suppose there is a point that is moving in a circle. This is let's say small angle d theta. This distance is rd theta by definition of angle. Okay. So my distance is rd theta we have done all of this when you divide by dt you will get speed of the particle which is r into omega all right. When you differentiate it again you will get du by dt is equal to r times d omega by dt. What is d omega by dt. What is this d omega by dt. Alpha so this is r alpha and this is acceleration all right. So acceleration of any point which is moving which is moving in a circle of radius x and direction alpha tangential direction of acceleration is alpha into x. Everybody understood this. If you find the center about which a point is moving in a circle then the acceleration of that point in tangential direction is alpha into x. Type it quickly. Have you understood this. Now tell me if angular velocity of this rod is omega so how much is the acceleration towards this center this acceleration is how much it is moving in a circle forget about the entire rod just focus on that point that point is moving in a circle point B although it is a part of rigid body forget about that fact so it is omega square x so any particle that moves in a circle will have an acceleration towards the center which is omega square r that is what it is r is x right and tangential direction it is alpha into x sometimes in uniform circle of motion alpha is 0 so only acceleration is centripetal but in this chapter alpha will be 0 alright but then here we assume throughout the chapter till now that we are talking about the fixed axis what if the object doesn't have a fixed axis for example you have this there is this rod lying on a horizontal table which is moving forward the acceleration of center of mass is given to us this acceleration of center of mass in this direction is ACM that is given and while moving it is also rotating with angular acceleration alpha fine now I want you to tell me what is the acceleration of a point P which is at a distance of x away at a distance of x away what is the acceleration tell me any one acceleration of point P which is at a distance x away the point is same as in the figure what what I was saying kinship as I like that yeah right now you tell me what is the acceleration nobody else only kinship got it okay now tell me relative to center of mass what is the acceleration of point P this you answer how much it will be and in which direction relative center of mass when we find out relative acceleration we make that point as if it is at rest you stand there and watch the rod with respect to center of mass how this rod is moving everyone relative to center of mass what this rod is doing relative to center of mass it is don't you think it is spinning about the center of mass if you just look at with respect to center of mass relative to it it is just spinning all the points they are rotating with respect to center of mass in a circular path do you all agree to this type in I am telling you something which is very very important these things are skipped when things are taught in everywhere else so relative to center of mass all the points are moving in a circle okay with what angular acceleration with what angular acceleration is there any angular acceleration of all the points are moving in a circle so with what angular acceleration they are moving is there an angular acceleration first of all if yes what it is center is going in a straight line or not everyone center is going in a straight line angular acceleration of center of mass? No. So, relative acceleration of all the points relative angular acceleration is alpha only. So, all the points are rotating with an angular acceleration of alpha. So, this point p relative to the center of mass is going backwards with what acceleration? This point p relative to center of mass which is at additions of x moving a circle with alpha. So, what is the tangential acceleration of that point? Alpha x alpha into x, right? So, relative to center of mass is this acceleration. So, absolute acceleration is what? Absolute acceleration how will you find? With respect to center of mass it is alpha x. So, how will you get the true value of it? For example, I will tell you a scenario then you will understand better. For example, this object is going with 5 meter per second. This object is going with 2 meter per second forward. So, relative to this object, relative to object 2, what is the velocity of object 1? Relative to object 2? 3? 3, right? So, you know the relative at velocity let us say is 3 meter per second. Relative to an object that is going forward with 2 meter per second. So, what you do to find the true velocity? If you already know the relative velocity, you add the velocity of an object with respect to which you are finding. Similarly here with respect to center of mass which is going that way, the acceleration is alpha x backwards. So, in order to find the total acceleration, you need to add the acceleration of center of mass to it in that direction. So, this is ACN. So, acceleration of point P is ACN minus alpha x in this direction taking that direction positive. All of you understood this. This is extremely important whatever I have just discussed. Everyone understood this? Type it quickly. Let me give you one more scenario. See, we are not doing theory. We are solving numerical like a theory. There is a difference. So, suppose you have a rod. This rod is making angle. Let us say this angle is theta. It is angle theta and the total length is L. Mass is M. This point is center of mass. This point is L by 2. Mass is uniformly distributed. The rod is rotating in this direction this way with alpha angular acceleration. The rod is also moving forward. The center of mass is accelerating with ACM. You need to tell me what is the acceleration of the tip. You need to find out what is the acceleration of the tip in the perpendicular direction of the rod. So, basically I am asking you what is the acceleration of the tip in this direction perpendicular to the rod in this. Do it. Okay, Archer got something. Sudhanava Kashyap. Is that the name of the student? First time and that is your dad's name. Brothers. Anyways, so over here what is the acceleration in this with respect to center of mass? Relative to center of mass. What is the acceleration of the tip? What is the acceleration of the tip? Relative to the center of mass? What is the acceleration of the tip? And in which direction? It is alpha into L by 2. Alpha into L by 2. Which direction it is? Everyone, which direction it is? Will it be perpendicular to the rod in this direction only? Along this direction only which is shown like this? Everyone? See, you need to understand that the tip is moving in a circle with the center at the center of mass. It is like that. Approximately I have drawn. So, this tip with respect to center of mass is going in a circular path. So, the acceleration of this tip will be tangential like this along the tangent direction of circle which is this. Now, total acceleration when you find you need to add the acceleration of the center of mass also which is this. You may or may not be aware that there is a centripetal acceleration also since it is rotating which is omega square L by 2. But I do not want this to come in. So, I am interested only in the component that is perpendicular to the radial acceleration. So, if this is theta, this is also theta. You can check that. This is ACM. No, sorry, that is not ACM. The above one is ACM. So, total acceleration is alpha into L by 2 plus ACM into cos theta that is perpendicular to the rod. So, these things you need to be very familiar with and there is no escape to it. You must understand. Otherwise, this chapter will give you a lot of trouble. You need to understand these things. Why centripetal acceleration is there? Centripetal acceleration is there because a point is accelerating with the centripetal acceleration of this point is there because it is going in a circle. That is the reason why. Only this point's acceleration I am telling. But if the center of mass goes in a straight line, there is no net centripetal acceleration on the rigid body as a whole, total. If center of mass moves in a circle of radius r, then there is a centripetal acceleration of omega square r for the center of mass. Here I am talking about the center of mass. If there is a point that goes in a straight line, if center of mass does not move in a circle, of course, there will not be any centripetal acceleration for the center of mass. But for the point there is, the point is moving in a circle. You need to detach the circular motion from the rigid body. So, now write down a very, very important topic rolling. Tell me what are the kind of objects that can roll, objects that can roll. Sphere can roll. Only sphere. Ring, disc, cylinder. Sphere, ring, disc and cylinder. These objects can roll. Will you use cone in your bike? No, right? So, of course, then anything can rotate. But then we are not interested in the irregular shape object. We are interested in an object which looks circular. If you look from the cross section, it should look circular. So, basically we are talking about rolling of a sphere, disc, ring and cylinder. Fine. Now rolling is a very, very important concept because rolling has actually changed our lives. If rolling was not there, we could not imagine the, you know, modern day travels by using bikes, the cars, the trains. All those things are not possible. They are possible because rolling friction is less. It is very less compared to sliding friction. Fine. So, because rolling creates lesser amount of friction, it is a preferred way of, you know, going from one place to other. All right. How much friction and all that, we will discuss that. So, hold on. Now, I will show you, for example, you have, let's say, a disc. All right. You have a disc. You can say a sphere, ring, whatever you may want to call it. There is this disc. Fine. This disc is rolling. What is component, lesser? Compared, compared. Even I got confused. This is compared. Fine. So, you have, let's say, a disc of mass M and radius R. All right. It is moving in such a manner that the center of mass has the velocity as VCN. All right. And it has an angular velocity like this. Now, tell you, when it rolls, angular velocity will be like that or like this. Which one do you think will happen when it rolls and moves forward? That way it will roll or this way it will go? What do you think? Omega 1. Right. That you're telling from your experience. Right. So, let's say this is a rolling scenario. All right. Now, there can be two kinds of movement for the disc when it is rolling. One is rolling without slipping. And second is rolling with slipping. So, have you experienced them when you, let's say, ride a bicycle? Have you ever experienced rolling with slipping? It is rolling also and slipping also. Have you experienced that? Many times, Kinshukh still experiences that. So, now, rolling without slipping and rolling with slipping, both can happen. But rolling without slipping is a constrained motion. This is a constrained motion. Now, I want you to tell me what is the constraint when that happened? What is the criteria? Forget about omega and V and everything else. If you crack this, it is good enough. Tell me, what is the criteria for rolling without slipping? Slipping should not happen. So, what should be there actually? What should be happening? What is the condition for it? Friction. No, friction is actually required. Friction, if it is not there, do you think your bicycle will roll if there is no friction on the roads? What is the condition? Sudhana. Tell me, what should happen at point of contact? Rotation of the edge is equal to velocity of body. What does it mean? Also tell me, forget about the rotation also. Here are the two surfaces. When one surface slides on the other surface, what is the condition that one surface slips on the other surface? What is the condition? No friction. Are there friction between the two hands? Still it is sliding. Still it is sliding. Not about the force. I am not asking anything related to force. I am asking related to the movement. What should be the condition? Tell me, relation with respect to their velocities, their accelerations. Is there any relation between velocity and acceleration? Velocity of one hand is more than the other. Clear? There should be some relative velocity. All of you agree to that? There should be some relative velocity. Between both the hands, velocity of one on the hand should be more than the other. Then only sliding will happen. All of you agree to this? Quickly tell me. Now, here is a point of contact. Here is a point of contact where this point of contact at the point of contact, if velocity of the disk is equal to velocity of the surface, then tell me whether there will be slipping or no slipping. Simple, no slipping. Now tell me, using VCM and omega, what is the velocity of point of contact of the disk? Tell me, what is the velocity? Just like we found out that total acceleration, similarly find out the total velocity. What it is? Along the direction, horizontal direction, what is the net velocity? VCM and omega, in terms of that you find out. With respect to center of mass, it goes backwards with what velocity? Omega into R and you have to add the velocity of the center of mass this way. So, this is VCM. So, total velocity is VCM minus omega R. This should be equal to what? Velocity of the surface, which is how much? In this case, how much? Zero. It is at rest. It is at rest. So, from here, you will get VCM is equal to R omega. You got a constraint equation. This is a constraint equation. If I differentiate it, what I will get? If I differentiate this, what you will get? Which time you differentiate? dVCM by dt is equal to R d omega by dt. What is that? dVCM by dt is equal to R. What do you get? This is what? This is ACM is equal to R alpha. Please remember this constraint. This is a constraint equation. All right? And this will come again and again. They will tell you that how they will tell you that rolling without slipping is happening. They will either write that it is rolling. They will write like this, rolling without slipping. They will write or they will say pure rolling. Okay? Like this, they will say. If they write like this, you have a constraint equation for the rolling. Conditions of pure rolling you have to write. Okay? Now, there is a thumb rule that if there is a rolling going on on a fixed surface, so if there is a pure rolling on a fixed surface, does not matter orientation of the surface. Whether surface is like this and you have your disk like that, whether the surface is like this and your disk is like that. As long as the surface is fixed, you can write blindly what ACM is equal to R alpha. Surface is fixed. All right? But surface can move also. For example, in this case, suppose you have a case like this, this one. All right? This is going this direction. Let me move it in other direction. It is going in this direction with acceleration a1. This is a cylinder. Let's say sphere. It actually doesn't matter. It's a sphere whose expression of center of mass is ACM and it is having angular acceleration like this alpha. All right? Tell me what is the condition so that pure rolling is happening over here or no slipping is there at the point of contact? Tell me that we got something or chip thus second object center of mass doesn't matter. Ritu Shashank helping you right now. All of you try it. Get the relation. Now, here with respect to center of mass, what is the acceleration of this point? Which direction? Alpha r in which direction? Right or left with respect to center of mass? How is left? Left case. It is right. It's rotating like that. Look at the sense of rotation. It goes like this and comes back like that. Right? All of you understand that this is alpha r. Understood? Type in quick. Now, why what happened? This plus ACM is a total acceleration right hand side, but the acceleration of the block is a1 and I'm saying that there is no slipping. So, what should happen is that alpha r plus ACM should be equal to negative of a1. They are in opposite directions. All right? So, alpha r plus ACM plus a1 should be equal to 0. If a1 would have been in this direction, then these two are equal. I don't need to put negative sign, but they're in opposite direction. Now, how can this be true? Everyone tell me. How can this be true? Adding three quantities. How can they add up to 0? Don't waste your time. We are going to do a lot of questions. This thing will come again and again. All of you answer me. How can three quantities get added and equal to 0? The answer is 1 or 2 quantities in different directions than shown. So, what will happen is there's nothing wrong with the equation. There's nothing wrong with the equation. You can imagine it like this. When you solve the question, you will probably get alpha to be equal to minus 5 like that. You will not get a positive number for one or two quantities and probably a1 you'll get minus 2 like that. And ACM, you'll get a positive quantity, maybe plus 10 or something like that. So, you have to maintain that consistency in the equation. That's it. Do not start thinking that how can three quantities be added up to 0? One or more maybe negative. Negative means in the opposite direction that's it. But you have to maintain that at a point of contact, the velocity of both the objects should be same for no slipping. Fine. Anyone has any doubt? Anyone has any doubt? Clear? Okay. So, now we are good to solve questions. So, here is the first question. You have an inclined plane that is fixed. This inclined plane has angle of theta. Then you have a sphere. You have a sphere of mass m and radius r. So, there is no slipping or you can say that friction is sufficient for no slipping. Friction is large enough for pure rolling. Fine. So, if this happens, you have to tell me these things. Acceleration of center of mass, alpha and the force of friction between the cylinder is solid sphere. So, it was a solid sphere. By the way, whenever in the question, it is just written sphere, take it as solid sphere. There is some time may not mention that it is a solid. The solid sphere is that find out ACM, alpha and Fr. All of you. First step is draw the free body diagram. Have you drawn the free body diagram? Done with the free body diagram? All of you. Those who are done, type in free body diagram. Done. Okay. So, I'll also draw the free body diagram. If you think that you are done with the free body diagram, continue. So, mg force from the center of mass, you need to be very particular about the point of application of the force also when you draw the free body diagram of a rigid body. This is mg. Okay. Normal reaction from the surface like this. Friction. Now, sometimes you will wonder which direction the friction will act. Okay. So, don't worry about it. Just assume any direction. Don't break your head by which direction the friction will be top or bottom and all that. Doesn't matter. If you assume it to be a wrong direction, friction will come out to be negative and you will know at the end. Don't worry much. Then you should show the acceleration also because they're asking you to find out that. So, this is ACM and there will be angular acceleration. Let's say this is my angular acceleration. Use a different color. This is my alpha. How many of you got this correct free body diagram? Different direction for friction. Now, can you actually tell me that is there a way to without even solving the question, can you tell in this direction, which direction the friction will be? Up or down? Without solving, can you tell why it will be up? Why it has to be up? The reason is what? Now, that is one way. Another way is that torque about center of mass should not be zero. You can see mg and normal reaction, both of them passes through the center of mass. So, friction is the only one which is creating torque and torque is going to be i alpha. So, torque should be able to create alpha in that direction. Then a little roll down. Do you understand that? Do you understand that? Friction should create a torque in this way so that the sphere can rotate like that and go down. Now, I want you to write down the equations. Let's say this is my y-axis. This is x-axis, x and y. So, I want you to write net force along x-axis to be equal to mass times equation along the x-axis and net force along y-axis, mass into equation along the y-axis. Let me know once you're done writing these two equations. Acceleration of the center of mass you have to take. You might have to take components of mg. Okay. Only Archit has done others. Anybody else? To get this along x-axis, mg sin theta minus friction is equal to m into Acm and along the y-axis n minus. Do you think that this is any different from whatever you have been doing till now? It's the same thing, right? You got this equation, right? Now, you have to write torque equal to i alpha. Is there a fixed axis here? Is there a fixed axis? Is there a fixed axis? What is the velocity? If it is pair rolling, okay? What is the velocity of this point at the bottom? If it is pair rolling, what is the velocity of that point? This point velocity should be zero or not. Surface is at rest. Do you all agree? That point of contact velocity should be zero, right? So that point is at rest. That point is at rest. So that is the fixed axis. But that is a different type of fixed axis. It is instantaneously at rest. Instantaneous fixed axis, okay? So this point, I don't know how to describe that. For example, you have something rolling, you know? This object is rolling. This point right now is at rest. As soon as it rolls a little bit, this point is replaced by that point. This point is replaced by that point. Now earlier, this point was at rest. Now, this point is at rest. And then this point is at rest as it rolls forward and forward, okay? But then, you know, I'm just telling you for your knowledge, later on, once you get comfortable with a lot of problem solving, you can use this. But right now, let us assume that there is no fixed axis. And anyways, if you are not sure, if you play safe and write down this equation about the center of mass axis, okay? So write down this equation, all of you. And let me know once you're done. Torque equation about the center of mass, write down those who are done typing. What is a torque about center of mass? How much it is? How much is a torque about the center of mass? Torque, in terms of force, how much it is? You have to find torque, right? Torque is equal to force into perpendicular distance. What is that? How much it is? Did you calculate the torque in terms of all the forces acting on it? How much it is? No one? Which force is mg creating torque about center of mass? Which force is creating torque? How much? Torque into friction is how much? As a radius is r. What is the perpendicular distance of friction from the center? Pappan's distance of friction from the center is how much, right? r is a perpendicular distance. So torque is r into fr, perpendicular distance into force. That is the definition of the torque, right? Now, this should be positive or not. That depends on alpha. Direction of alpha, you take positive, not clockwise, anti-clockwise. Okay? Just like you write net force is equal to mass and acceleration, direction of acceleration, you take positive. Similarly here, r into fr, icm is what? I think most of you got this, 2 by 5 mr square into alpha. Okay? So friction is equal to 2 by 5 mr alpha. Any other equation other than these? Any other equation? Consider an equation for pure rolling, right? So alpha r should be equal to acm, pure rolling on a fixed surface. So friction is equal to 2 by 5 macm, which you can put over here. You will get mg sin theta minus friction of minus 2 by 5 macm is equal to macm. So acm is equal to 5 by 7, right? No, 5 by 7 g sin theta. This is acm. All of you understood this, type in quick. Is this clear? Go through this entire thing. Tell me what is the frictional force? How much friction is required? What is the friction required? Friction we already know. 2 by 5 m into acm, which is 5 by 7 g sin theta, right? So this is equal to 2 by 7 mg sin theta. All right? This is the rolling friction. This is the amount of rolling friction that is applied on it. Getting it? Only this much friction will be applied, this much. And if coefficient of friction is mu, maximum friction is how much? f max is equal to mu time normal reaction, which is mg cos theta. So mu mg cos theta should be greater than or equal to 2 by 7 mg sin theta. Otherwise rolling won't happen because for rolling this much friction is must. If this much friction is not there, rolling won't happen, okay? I hope things are clear. Clear, right? Friction is this. You have to substitute acm there. You'll get the value of friction. Acm you got it, right? Now you can take the questions or this one we just did. Do this. Both the tensions are equal because of the symmetry. Anyone close to the answer? Kinship got something. From the side, how does it look like? It looks like this. A circle, a string over the top. It'll look like this. All of you agree? Right? One string will be front, one string will be behind. So it'll be t and t, two t's. There will be mg force like this. Any other force? Any other force? No. Only tension and mg are there. So do you think that it will roll like this? Everyone? Do you think it should roll like that? The cylinder will, the wrap will open up. Everybody? Anger equation alpha and it will accelerate like that. Acm down, okay? So now just write down the equations. Nothing is left now. Vertical direction net force is mg minus 2t is equal to m into acm, okay? This is the force equation. You have to write the torque equation also. Now to write out the torque equation, you can say if you're not very sure about this being the fixed axis which it is, you can say torque about the center of mass is equal to icm into alpha. So what does it talk about the center of mass, everyone? What should I write instead of this? T into r, no, 2tr. Two strings are there, right? Which are creating torque on the cylinder. icm is how much? How much is icm? For the cylinder icm is what? mr square by 2, right? That into alpha. Any other equation you can think of that I'm missing here. Anything else? t is equal to mr square mr by 4 times alpha. Any other equation that you think we should write over here? What is the relation between alpha and acm? Is there any relation? This point on the string, this point on the string is at rest, right? So it is like rolling without slipping only. That point is at rest, so that point on the cylinder should also be at rest. So acm should be equal to alpha into r. So tension will come out to be macm by 4. So then you write down mg minus two times of tension macm by 2 is equal to macm. So 3 by 2 acm is equal to g. So acm is 2g by 3. If acm is 2g by 3, the tension value is m by 4 into 2g by 3. So tension is mg by 6. Now go through this question. All of you, it's a very important question. Step by step go through it and realize that we have not done anything new. All right? All of you spend maybe a minute on it. Once you're clear, type in that it is clear. Clear? Yeah, r alpha is equal to acm. That's what we did. Type it quickly. Is it clear? Entire question, solution now clear? They go. See, this thing is, you're asking me how I am calculating. You should be just happy that I have written these equations and now you solve it yourself. These are the equations. Do it yourself. Why you're asking how it is 2 plus 3, what you have done, how you're substituted, this and that. You solve these linear equations, right? I'm not here to tell you how to solve linear equations. Understand that. You have to do it. Otherwise, the same thing will happen in the exam also. Fine? So all I have done is I have substituted r alpha as acm. So t is equal to macm by 4. Okay? Do it yourself. Do it yourself. Then only it'll be good for you. All right? Otherwise, the same thing is written in the book also. You can as well copy it from the book. Solution book, right? If you consider string to be surface and should be also considered frictional force. Tension is there now. That's like a friction force only. Tension is due to the friction only, you can say. Let's go to next. Write down the equations and solve them. First, nicely draw the free word diagram and write down the equations. All of you. What about b? What about b? What about it? Okay, b. What is the speed of cylinder as it falls through a high-tech? Tell me how will you do the b part? Why don't you do the a part? What do you do with the b? How will you get it? Now it is accretion is this constant accretion. So if you use v square equal to u square plus 2as, use that. v is equal to root of 2a into h. That is the answer to this. Have you drawn the free word diagram? Okay. Have you written the equations also? Did you get the answer as well? All of you do this. Which direction of friction will act? Right or left? You can't say. Looking at the question, you can't say anything. Just assume any direction. Let's say you assume that direction to be friction. This is friction. The downward direction will be mg. This is the normal direction. All right. So acm is this way. This is acm and alpha is this. How many of you got the free word diagram? Correct. Okay. Great. Many of you got it. So write down the force equation along the x-axis. Force equation along the x-axis will be what? F minus force of friction. This will be equal to m into acm. This is the first equation. Vertical direction n minus mg is equal to 0. Second equation. Now torque equation. What is the torque about the center of mass? I have to write torque about center of mass to be equal to icm into alpha. Tell me what you will get torque as about the center of mass. What will be the torque? So which are the forces that are creating torque? Is mg creating any torque? It passes through the center of mass. No perpendicular directions. Normal reaction? No. What about small f? Yes. Capital f? Yes. Both small f and capital f, they are creating torque. Are they creating torque in the same direction or in opposite direction? Torque, I'm asking. They are trying to rotate in the same direction. Right. So when you find the torque, you have to add them up. Right? So f into r is a torque due to capital f and small f into r is a torque due to the small f. They will both add up because in the same direction. This is equal to icm. It's a sphere. So 2 by 5 mr square into alpha. This is the third equation. Right? And it is rolling without slipping. Written. So I can say that acm is equal to alpha r. Now all of you solve these four equations, get the answer quick. I'm not here to teach you how to solve these four equations. Physics is still here only. All of you solve this linear equation, get the answer quick. Okay. Weibo already got it. Others? Okay. Gayatri got something. Archit, how can you subtract 5r from 2? Don't you think? Dimensionally, it doesn't make sense. Small f is not given. Okay. You have to find in terms of capital f. In fact, even r is not given. Radius is also not given. You can see how pathetic the situation is. You guys are not able to solve the linear equations. So you should be doing these calculations yourself. Otherwise, you'll never learn. Fine. Nobody will come and suddenly magic will happen and you start doing the calculation of your own property. You can see you're not able to solve. All the equations are in front of you. Any other answer? So capital f plus a small f to be equal to 2 by 5 m into ACM. Fine. And we have capital f minus a small f is equal to m into ACM. So if you add them up, you'll get 2f is equal to 7 by 5 m ACM. So ACM is 10f by 7. Okay. So Gayatri and Weber got it. This is ACM. Now just substitute the value of ACM in any of these two equations. You'll get the value of friction also. Everybody understood? Small f is not given. Small f is your assumed variable. Look at the question. Only capital f is given and small m is given. Go through it. All of you tell me, is it clear? Friction comes out to be negative. So friction is in the same direction as capital f. Opposite of what you have. Clear to everyone? Now we'll move to the next question. Read the question and tell me with the tension on both side of the string, both side of the pulley, tension on the string will be equal or different. Pulley has mass and friction both. It is not an ideal pulley. It is rotating. It has an angular acceleration. Fine. So the tension will be different. Otherwise, the net torque about the center of the pulley will be zero. It will not be able to rotate then. This is let's say T1. This is T2. Different tensions. Now solve it. Get the answer. When we assume same pulley, same string, same tension, we also assume pulley is frictionless and has no mass. Write down the equation and let me know once you're done. Equations written? No worries. I'm not hurrying up. Even when you guys don't respond, then I go with my pace. If you say that, I'll spend some more time. I'll spend more time. Okay. I want you to speak up. Animated done with the equations? Kind of. What is kind of? Kind of. So I hope you have drawn the free body diagram at least. This is mg. This is a normal reaction. This is T2. This is T2. T1. T1 mg. How many of you got this free body diagram? Correct. So now let's say acceleration of this block is small a. What is the acceleration of that block down? How much it will be? Is it a only? The reason is what? What is the reason? String has fixed length, constraint. It has nothing to do with friction or force or anything like that. Forget about that. It is a constraint equation. All right. Then this cylinder or the pulley will rotate. Angular acceleration is let's say alpha. Fine. What is the relation between a and alpha? About this fixed point, the entire pulley is rotating. So what is the relation? If radius is r, let us say small r given. Do you all understand that ACM? There is no ACM here. Okay. It is a only. A is equal to alpha r. This point should move it alpha into r. Pulleys not rolling without slipping. You have to understand this is different. The tip of the pulley is moving in a circle of radius is small r. So if alpha is the angular acceleration, tip has an angular acceleration of alpha into r and that tip is in contact with the string which is moving with a. So that is why both of them are equal. A is equal to alpha r. Otherwise slipping will happen. Okay. Now we can write down the equations. Right. So we have for capital M, mg minus t1 is equal to m into a. Okay. For small m, we have t2 is equal to small m into a. And for the pulley, what is the torque? Torque in terms of t1, t2. What it is? In terms of t1 and t2. Tell me the torque. Torque into t1 is t1 into r. t1 into small r. Okay. t1 into small r. Torque into t2 is in the same direction or in opposite direction. Opposite. Right. So t1 into r minus t2 into small r. This is a net torque. This is a net torque in the direction of alpha. This net torque is equal to i times alpha. Okay. So we have one equation. Two, three and four. All of you solve this. Get the answer quick. Just getting the equations is not enough. Solving them is also important. Oppose. Anybody got the answer? Any guesses how to solve this? t1 minus t2 is this. You can see if I use these two equations, I'll directly get the value of t1 minus t2. Okay. So mg minus t1 plus t2 is equal to capital M plus small m into a. Okay. So from here, t1 minus t2 is equal to mg minus capital M plus small m into a. So substitute this in place of t1 minus t2 and use alpha is equal to a by r. You'll get the value of acceleration. Okay. So let me go to the next question now. Okay. Here the rod is in equilibrium. Alpha is zero. This is similar to what we had done last class. Draw the free by diagram and let me know once you're done. Done with the free by diagram. No one. What is the direction of, let's say I'm draw a rod like this. There is a normalization over this point. Which one it will be? How? How the normalization over here will be at this point. It will perpendicular, right perpendicular to both the planes of the contact. Now there is no plane of contact for the wall. There's a roller there. So perpendicular to the rod you have to take like this. This is your n1. Okay. This is your n2 from here. This is your n2 and friction force this way. How much friction be acting over there? It is about to slip. What is the friction? How much? How much should be the friction force? Mu mg. No. Mu n2. Okay. There'll be mg, mg from where? Center of mass. Okay. mg from center of mass will be acting down. So this is mg. M is not given. This is mg. This force. All right. Now the net force is zero and net torque is also zero about every point. So see I want you to solve this question. Get the answer yourself. Then only you'll feel confident. If I keep on solving, you keep on noting down on your notebook. It is not going to help you. Okay. It is waste of your effort. Okay. It's a donkey's job to just copy whatever is written word by word. Okay. Do it yourself. Even if you get it wrong, doesn't matter. Try this. Distance between forces and center of mass you can find out, right? The center of mass is at the distance of L by 2. H is given. This H is given. Use it. Geometry. Let me tell you what I want from you. I want these three questions. Net force along x-axis should be equal to zero. It is an equilibrium. Net force along y-axis should be equal to zero. And I want torque about this point P. This torque point P is equal to zero. These three questions I want. All right. Let me know once you're done with these three questions. No one got it. Equations. No one is able to write. No, I don't want to see the equation. Just tell me that you're done. It may be right or wrong. Doesn't matter. Do it. Complete it. Done. Archive is done. Others. Shraddha is done. Others. Done. Kinshok is also done. The two is done. All right. So net force along x-axis is how much you need to resolve N1 into two components. Which component is this? N1 cos theta or sin theta? This angle is how much? This angle, how much? N1 theta, right? So I will get mu N2 minus N1 sin theta. That is cos of N1 theta. That is sin theta is equal to zero. The net force along x-axis is zero. Along the y-axis, we have N2 plus N1 cos theta minus NG. All right. This is equal to zero. All right. Now the torque equation about point P. Why have to choose it? Why have to choose point P? Because torque due to N2 and mu N2 will become zero that way. Fine. So it will become easy to handle. You have to find torque due to lesser number of forces. So torque about point P, if I write it to be zero, I'll get, what is the torque due to N1 about point P? You can find torque by N1 cos theta or N1 sin theta separately. You can also find the torque by entire N1 together. That will be better because entire N1 is perpendicular. So torque due to N1 is N1 into, what is this distance from P to this distance? This is the perpendicular distance, right? This is the perpendicular distance from here to here. How much is this? H by sin theta. H divided by sin theta. There are torque due to N1. Which direction the torque is? The torque is in this direction. Just trying to rotate that way. Okay. So MG torque is in opposite direction. That is MG into what? It's perpendicular distances. How much? From the point. This is MG's line. It's perpendicular distance. It will be this. This is how much? This is the perpendicular distance, right? You drop a perpendicular on line of MG. That is a perpendicular distance, which is how much? From here to there, it is L by 2. So that is L by 2 cos theta. So MG into L by 2 cos theta, right? Is equal to zero. This is your torque equation. How many of you got this? Torque equation. MG's torque is MG into its perpendicular distance from point P. Pappantical distances you have to drop a perpendicular on line of MG. So which is this? So there is a right angle triangle. This distance is L by 2. This is theta. So this perpendicular distance is L by 2 cos theta. Okay. Fine. Now from here, you can get the value of N1. N1 is equal to MG L by 2h tan theta. No, not tan theta, sin theta cos theta. This is N1. So substitute the value of N1 over here. You'll get N2. Okay. Then put both N1 and N2 there. You'll get the value of mu. That is what they're asking. Okay. Everyone just go through this and tell me whether clear or not. Okay, then we'll take a break now and meet after the break. Is this mains level? I don't know. You can't say something is mains level or advanced level because in mains level, there will be 30 to 40% questions in the mains paper will be advanced level and in advanced paper, 30% of the questions will be mains level. So how can you just look at a question and say this is mains level advanced level? But then yes, in advanced paper, more questions will be difficult. In mains paper, lesser questions will be difficult. And in CT, even lesser number of questions will be difficult. But we are taking those questions which will help us understand the chapter better. Right now our entire focus is understanding the concept, right? Because you guys are facing it difficult. That is where that is the reason why you're learning. If you don't face the difficulty, if you're able to solve it quickly, if I ask you what is two plus two, what is five plus two, five minus two, you'll be very happy of the answers you'll get. But you're not learning anything if you get everything. Fine. So learning happens only when you don't get when you are during the learning. So you will start getting these questions, right? Once you practice a lot of questions yourself after the class, during the class, nobody can master the concepts just sitting in the class, not doing anything after the class. That will never ever happen. It's not like your class 9th and 10th, okay? Fine then. So we'll meet after a couple of minutes and we'll meet at 6.24. Okay, here's the question. Kinshok got something. What do you think, Kinshok? Is it correct? Anybody else? Any other answer? What is the acceleration of the string? See, it cannot be A because it is dimensionally incorrect. Alpha into R can be A. Alpha by R dimensionally cannot be A. Tell me what is the acceleration of this? Let's say acceleration of this point is lower disk is alpha, right? And ACM is the center of mass of the lower disk. This is ACM. So if alpha is the acceleration, let's say alpha, this is alpha, this is let's say alpha 1. So, acceleration of this point P will be what? Alpha into R, alpha 1 into R, right? All of you agree? If alpha 1 is the angular acceleration of the upper cylinder, then this point P's acceleration down will be alpha 1 into R, okay? Now, with respect to this point, with respect to this point, this point is going down with an acceleration of alpha into R. You need to add this point's acceleration also, alpha 1 into R and then say this is equal to ACM. Okay, with respect to this point, it is going down by alpha 1 in alpha into R with respect to that point P, okay? You need to add the point P's acceleration also to get the total acceleration of the center of mass. Everybody understood this equation? Is it clear to everyone? Now, what is the relation between alpha 1 and alpha? What is the relation? Why do you think it is equal? Why? Why they're equal? Because it is the constraint between these two points, all right? The constraint equation is that they're both alpha for both the rigid bodies to be equal. So, actually answer B is correct, not C or A, all right? See, the thing is that I'll tell you another reason why their alphas are equal. For example, if you look at the above one, they are identical deaths, okay? Above one, tension T is this way. So, with respect to center, the torque is what? T into R, this will be equal to I into alpha 1. For the below one, with respect to center of mass, you have again same torque, T into R. This is equal to I into alpha. So, torque is same, T into R with respect to center of mass. There'll be MG, but MG is passing through the center of mass. Because tors are equal and moment of inertia are also equal as they're identical, their alphas are also equal. This you can say CT or mains level. Something like this we had done earlier, but now also we should practice this, do this, okay? I think this question is a little ambiguous in the wordings. I will change the wordings of the question. Just ignore this. Sorry about that. I'll just change the question if you want. Don't do this. Something like this we have already done earlier, but I thought we'll do it again. There is this rod of mass M length L. It is hanging by using two strings, symmetrically placed, these two strings and also there is a mass small m attached from the center. Okay? And this center from the center with a string mass M is attached. Yeah. So, what you're doing is you're cutting this string immediately after cutting, immediately after cutting, you have to find the tension on the other string, how much it is. Immediately after cutting, by the way, you can say that it acts like as if this point P is the fixed axis. Immediately after cutting, it will try to rotate like that, that you can say. Let's do step by step, get the value of angular acceleration of the rod, immediately after cutting the string. More basic than that, first draw the free boy diagram. Then we'll write the equations. Let me know once you are done with the free boy diagram and the equations. That's all. How many of you have done the free boy diagram? Free boy diagram. There will be a tension from here. There will be a force of capital MG. What else? Any other force? What should I write? Small MG. Small MG down. Small MG. Small MG force acts on small M or on the rod. Who is applying MG force? Who applies MG? Earth is applying MG. Is it applied on the small M on the rod? On the small M. What force the rod feels? Tension. Tension need not be MG every time. Let's say there is a force T1. I can show that there is this direction rotation will happen. This is alpha. What will be the acceleration of center of mass immediately after cutting in terms of alpha? Can I say that? In terms of alpha, what is the acceleration of center of mass? Immediately after cutting, alpha into L by 2 is ACM. Now vertical direction, what is the equation in vertical direction? Force equation. MG plus T1 minus T is equal to mass times acceleration of center of mass. All of you understand this equation? Have you understood this equation? Clear? Okay. Now there is a torque equation. Torque equation you can either write about the fixed axis or about the center of mass axis also. All right. So it is up to you. So we can write about the fixed axis. It will be T1 T1 plus MG. This into L by 2. How much is the moment of inertia? What should I write here? M into moment of general rod. About this point. How much should I write? ML square by 3. I into alpha. Second equation. But there are three variables. T1, T and alpha. How to get the third equation? Everyone, what is the third equation? T1 is equal to MG. Why? Is a small m accelerating or not? Everyone, is small m accelerating or not? This is small m that is kept there. Is it accelerating or not? How much is the acceleration of that? It is connected with the center of mass only. So whatever is the acceleration of center of mass will be the acceleration of small m. Right? So the equation will be small MG minus T1 is equal to mass times acceleration. This is the third equation. Now you have to just solve them and get the answer. Go through these equations. Let me know. Is it clear or not? Now you can see that you can't put a tension as MG every time because there is an acceleration also. Type in. Is it clear? Giving you one minute to go through. No, I'll check. If you write about center of mass torque you'll get the same equation which is a linear combination of 1, 2 and 3. It is not a new equation. If you write about center of mass. Clear to everyone. Type it quick then we can go to the next one. Can you answer or you are here to just sit? I'm asking something. All of you. Everyone should answer. I'm waiting. I'll see who is not answering. Do something. There you want me to take your name then you'll answer something. Do you think I should keep on calling your names then only you'll speak up? All of you answer. Is it clear? If it is not, what is the doubt? Almost everybody has answered. Let me check who hasn't. Sometimes I wonder whether you guys are sitting there or not. Throughout the class, if you just keep quiet, it is as good as you watching the recording. Why are you attending live classes? Go and see the recordings. Okay, still I think some of you haven't answered my question. Who is it? Rishika, please answer. Ruchita, you're quite throughout the class. You don't speak a thing. Why? Vignesh Madan in caps. Should I ask you separately? Ruchita, there or not? Vignesh Madan there or not? Looks like you're not there. Vignesh. Vignesh. Sinchan. Sinchan is there. Vignesh is not there. Okay, Drutty you're there. All right, so I'll have to call Vignesh. See why he's not there. All right, let's move forward. Because just sitting, it is very easy to hide. All right, and it is very easy for me to ignore also. Why should I care whether you're hiding or not? It is so easy for me to just ignore. Okay, fine. Let them hide. I don't care. And I'll just continue. But ultimately you'll be at loss. If I ignore or you don't speak up in the class, it's a complete loss for you, not to me. So I don't know how many times I have to say that. Okay. Or I should just start ignoring that some of you are not speaking at all. All right, do this. All of you. It's a very similar question, the one which he had just done. Right? So all of you should get the answer. In fact, this one is simpler than the previous. Dev Rajan already got some answer. Others, Shambhu got it. But both of you are getting different answers. So one of you is wrong. Both of you checked your calculations. All right. Others, have you drawn free bird diagram? See, I don't remember the answer. And I'm doing this question for the first time myself. I solve questions with you as a student, whenever I do that. So so that you also think like it, we can discuss the difficulty phase during the problem solving. So the free bird diagram, this is the free bird diagram. This is T, this is MG. Continue solving. I'll just write down the equations. If I release it, center of mass will have acceleration or not? Center of mass will have acceleration or not? How much? How much? If angular acceleration is alpha, how much is the acceleration of center of mass? ACM is what? Alpha into r by 2, right? Alpha into r by 2 is ACM. So what is the equation you will get? Net force will do mass and maturation in vertical direction. What you will get? Tell me what kind of equation you will get? Anyone wants to type it? Equation in vertical direction. Can you type it? What you have written? Net force MG minus T is equal to mass times ACM. ACM is alpha into l by 2. Okay, this is the force equation. Do you all understand this? Everyone? Mass same as center of mass is equal to net force in vertical direction. This is a Newton's second law equation from laws of motion. This is not something which we have learned in this chapter. We have learned this already in laws of motion. Now we'll write torque equation. That is what we have learned in this chapter is equal to i alpha. About which axis should I write this equation? Is there a fixed axis? Is there a fixed axis? P is a fixed axis. Should I write about P? Which one would be better or easier for you? Writing torque equation about O or writing torque equation about P? Which one will be easier for you? The torque equation which one will be easier? P? Do you know you need to find moment of inertia about P? If you write torque equal to i alpha about P, you need to find moment of inertia about P. Yes or no? Do you know that? You have to use parallel axis theorem to apply moment of, to find out moment of inertia about P. So, unnecessarily why to get in trouble? Write down torque equation about the center of mass. Okay, so it will be T into r by 2. This is equal to i alpha. i is what? m r square by 2, m r square by 2 into alpha. This is gone. That is gone. So, T is equal to m r alpha. So, substitute that. You'll get mg is equal to 3 by 2 m. What is this? l by 2. This is r by 2. 3 by 2 m r alpha. So, from here alpha is 2g by 3r. This is alpha. We need to find tension. Tension is m r alpha. So, 2 by 3 mg. Anybody got this answer? See, good. Madhumati got it. Shraddha got it. Riltu got it. Only girls are able to solve. Anybody else? Kinshok, you got anything? All right, you got b. Now, you understand, right? All of you understand your silly errors? Go to the next question. This exact question will never come in the exam. All right? Don't try to be in the mode where you just try to understand every question, how it is solved and then you will not be able to do an exam. You should be understanding how to use a concept in any scenario. Do this. Length from pin A. It is 1 by 4. This is l by 4. Again, a repetitive question, right? This is the third question which is similar continuously. Don't worry about your school exams for this chapter. School questions will be a lot simpler than this. You will find it very simple now. You have done many different varieties and types of questions. Anybody for part A, did you get the value of alpha? How much will be the angular acceleration? You can write torque about the pin is equal to i alpha now because a force from the pin is not known. That pin is a fixed axis. So, even if there is a force from the pin, torque due to that force will be zero about the pin axis. Should I wait or solve? Anybody about to get the answer? Any other force on the rod? Any other force on the rod? Tell me. Any other force should I show on the rod? I have drawn a free-body diagram for you. Do you think any other force should I show in the rod immediately after cutting the cable? What do you think, Arjit? Shraddha? Madhumati? Do you think any other force other than these two? Force due to the hinge and force due to the gravity. Any other force other than these two? Why I have taken vertical direction force? Why not there will be a horizontal direction force? Why not from the pin a force horizontally? What is the problem? I can see there are two components of forces from the hinge F1 and F2. Why not? It's not hinge about the end point. So what? From wherever it is inch, no problem. Why you have taken vertical force only? Why not oriental force on the hinge? It's not a normal reaction. It is not a normal reaction. It's a pin. That pin can apply force in any direction. You can say one component vertical, one component horizontal. Yes or no? Everyone? But F2 is zero. But F2 is zero. What is the reason F2 is zero? What's the reason? That's the only force horizontally. Next force along x direction is F2 only which is equal to mass and maturation of center of mass along x direction. What is the expression along x direction? Everyone? Over here, what is the expression along x direction? It is zero. That is why F2 is zero. But suppose it gains some angular velocity, omega would have been there. Then F2 would have been equal to, okay, F2 should be in opposite direction actually. So minus of F2 would have been m omega square L by 4 because omega square L by 4 is a centripetal acceleration of the center of mass. Do you all understand what I have just told you? There is no acceleration because omega is zero right now. There is no centripetal acceleration. But if omega would not have been zero, then there would be a horizontal component of the force also from the hinge. All of you are able to understand the reason why only F1 is there. So if omega is there, there will be centripetal acceleration. So there will be F2 then. Right now F2 is zero. All right, so if after releasing alpha is the angular acceleration, what is the expression of center of mass? What it is? ACM is what? Acceleration of center of mass. How much? Alpha into L by 4. Okay. Right? So vertical direction I can say that mg minus F1 is equal to mass times acceleration of center of mass. And now I can write the torque equation. Which torque equation will be better suited? You have fixed axis also and center of mass axis also. Which one do you think will be better? Center of mass. Otherwise you have to find moment of inertia about this fixed axis. Right? So F1 into L by 4. But you get the same value of alpha both ways. Remember that this is equal to ML square by 12 into alpha. Okay? This is 3. One of the L is gone. So F1 is equal to ML alpha L by 3. All right? So mg is equal to M alpha L by 3 plus 1 by 4. So M is gone. So alpha is equal to 12G by 7L. This is alpha. Right? So what is the expression of NB immediately after cutting? In terms of alpha tell me how much it will be? In terms of alpha what is the expression of B? Alpha into in what radius this from here to here the distance is 3L by 4. Right? So this point B is moving in a circle of 3L by 4. So alpha into 3L by 4. So what answer you are getting when you do that? Which answer? Maybe say that much at least you can tell. No. This into 3L by 4. You get option A. Right? Reaction at pin is what? F1. Right? So once you get that value you can write down the value of once you know this is the acceleration. Alpha is known to you F1 is M alpha L by 3. Right? So 12 by so you get 4 by 7 up. What happened Kinshukh? Last question for today. Not today. Last question for the entire week. We unfortunately will not be able to meet for the next week because of the Bashira holidays. So last question for two weeks now. This one. Second part they're asking you what is F1. Okay in the previous question they're asking you what is the reaction at pin. Reaction at pin is F1. Okay and F1 is M alpha L by 3. Alpha you already got put the value of alpha here. Okay do this. Tell me the normal reaction should act from where? Should act from B by 2 upwards? Is it required that it should act like this? That's not required. When it is about to tip over I'm giving you a hint to solve the question. So when it is about to tip over normal reaction will act from where? Leftmost point edge good corner because the entire surface will be slightly lifted up when it is about to topple. Isn't it? It will be as if from here to here the surface lifted up and everything is trying to get rotated about that point. So normal reaction will shift this side. That is the condition for it to be tipping over started to tipping over. Normal reaction shifting to the corner that is the condition. Okay now do it. No one got the condition. No one got the answer. Let me do it. Minimum force is that force at which alpha is just greater than 0. Tipping over means what? Tipping over means alpha is more than 0. It started rotating about the edge. About this point P it has started rotating. So alpha is there. So the condition for the minimum this thing would be that the alpha is almost 0. About point B all the talks are balanced. If you increase force little bit torque will be greater than 0 and then it will just topple over. Okay and normal reaction will be shifted here anyways. I hope all of you have drawn the freeway diagram. Only force left is MG. MG will be always acting from the center of mass. Doesn't matter tipping over or whatever it is. For MG it doesn't matter. Now friction. Friction is also there. So we will be taking friction like this. This is how much friction will be there? Everyone will it be maximum friction? Yes because it is already sliding moving with concept velocity. Right? Even to find greatest height the force will be applied to let's say this is y. This is let's say y. So torque about point P should become equal to 0 and if you increase the force it should become greater than 0. That's how it will be. So torque about point P for force F is how much? Force F what is the torque with respect to P? F into what is the perpendicular distance of F from P? Extend the line of force and drop a perpendicular on it. It will be y only. Right? So F into y is the torque due to MG. Sorry due to F. Due to MG will be what? MG into what is the perpendicular distance of MG from P? From P MG's distance. Y by 2? No. It is B by 2. Extend the line of force of MG and drop a perpendicular on it. This is a perpendicular. Right? This is a perpendicular. That is B by 2. So MG into B by 2. Will they add or subtract? They're in the same direction or opposite directions. MG is trying to rotate it clockwise. F is trying to rotate in which direction? Anticlockwise. So this minus that is equal to 0. All right? So you will get Y is equal to MGB by 2F. Fine. Now we have to substitute the value of F. So horizontally there is no extension of center of mass. Right? Testing the constant velocity. All right? It is a constant velocity. So F minus mu times normal reaction is equal to 0. Alpha is also 0. So extension of center of mass is 0. Fine? So F minus mu n is 0 and vertically normal reaction minus MG is equal to 0. All of you able to understand what I'm doing here? So F is equal to mu times MG. These are the force equations. So substitute here the value of F as MG mu MG. So MGB divided by 2 times mu MG. So you'll get B by 2 mu. All of you understood this question? Everyone type it. Is it clear? Clear? Fine. Let us do one advanced level question now. Last question. So there is a, this is fixed. Okay? Don't worry about that. This is fixed. There is a plank. This plank. See you don't need to solve. We just need to write the equation. All right? I'm just looking for the equation. Nothing else. A force of F is applied. An angle theta. The mass of the cylinder is M. Radius is R. It is a solid cylinder. This is mass small m. Okay? There is sufficient friction everywhere for no slipping. All right? So I want you to write down the equations only. Assume that this is alpha and the expression of center of mass is ACM. Fine? With respect to these, just draw the free by diagram and write the equation. Don't need to solve. Okay? Assume that the force is not very large. This plank remains horizontal. Assume that. You just need to write down the equations of motion, force equation and torque equations. All right? The plank is not rotating. So you don't need to write torque equation for the plank. Just write down the force equation for the plank. You need to write force equation for plank and force and torque equation for cylinder. This is what you have to do. Clear? Yes. Can show friction is there everywhere. Friction won't leave you. Should I write down? Have you drawn the free by diagram? At least free by diagram. All of you. Continue doing it. Meanwhile, I'll start. Some of you might be in a hurry for your applications to start. Don't want to hold you back. How many of you got this free by diagram correct for the plank? It'll accelerate also. Let's call it as activation A1. All right? Then we'll draw free by diagram for the cylinder. Here are cylinder. This is a cylinder. So we'll show the pair of force and one pair of friction, then one normalization from the bottom, one friction from the bottom. Then what else? Is that it? Yeah, that's it. So this is the friction 1, friction 1, normal 1, normal 2, FR2. Yes, yes. MG, gravity. Gravity should be the first force. How can I forget that? Maybe class timings are over, so I'm started forgetting things. Gravity is MG. All right? You have a CN. Don't worry if you're not getting this at the first go. This is a advanced level. I have to show you some advanced level. Right? So that you should not ask whether it is advanced level or not. The other questions. This is alpha. All right? So now this cylinder is rolling on a fixed surface. So I can quickly write down alpha into R should be equal to ACM because it is rolling on a fixed surface. Right? That is there. Put this. And then I can write down the force equation for the cylinder. FR1 minus FR2 is equal to capital M into ACM. Torque equation for the cylinder. Torque will get added up. Both the frictions will try to rotate in the same direction. This will be equal to I into alpha. These are the force equations for the cylinder. Force equation for the blanks will be F cos theta minus FR1 is equal to small m into A1 and vertical direction F sin theta plus n1 minus mg is equal to zero. Vertical direction equation may not be that useful. You may not be interested in finding n1. So that's right. Now we need to write. Now count the variables. Okay? Count the variables. You have FR1, FR2, ACM, alpha, then n1. What else? Is that it? 1, n2? Yeah. It is not n2. Let me write down the variables here. ACM, A1, FR1, FR2, then alpha, FR2, ACM, alpha and n1. So we have six variables and only five equations. Where's the sixth equation? We need six equations for six variables. What do you think it will be? What do you think it will be? Vertical direction for cylinder, another variable will come. n2 will come. Are you getting it? If you write vertical direction n2 minus n1 minus mg. Although you got one more equation, but you got one more variable also. You need one more equation. Tell me. What will be the other equation? No slipping condition over here. Why you forget that? No slipping there. No slipping condition over there. Okay? Now tell me what is the no slipping condition over there? What is the equation for that? What is the condition for no slipping between cylinder and the plank? What is the condition? What is the attrition of this point? Just look at the cylinder. Attrition of this point is what? Tell me. This point's attrition is what? With respect to center, it is alpha into r this way plus ACM. All of you agree? ACM is also this way. Do you all agree that this is the attrition of that point on the cylinder? Quickly type in. Okay, this should be equal to what? This should be equal to A1, attrition of the plank because there is no slip. Okay? And ACM is anyway equal to alpha r. So two times ACM is equal to A1. Attrition of the plank should be two times attrition of the center of mass of the cylinder. Okay? All right. So today we have comprehensively covered almost every type of question that can exist. But then having said that, there will be small, small tweaking here and there and you can still find some new varieties. And I'm sure that you are not, you know, you still need a lot of practice. You're not as smooth as you should be before the exam. Right? So you need a lot of practice. So, so vacations are in front of you. So you can either choose to enjoy vacation. All right. Or you can choose to work towards your weakness and come out very strong the other side of the vacation. The choice is up to you. Okay? So I'm not telling, I'm not strictly telling you to do anything. Just that I'm telling you that here is a chance for those who have fire in their belly, but they have lag behind because they did not get the head start at the beginning. They might have taken probably class 11, not so seriously. Now they realized, okay? So you have 10 days to look back upon and work on your strength or you can enjoy vacation, close your eyes as if nothing is happening around you. Okay? But then I'm also not saying that you should 24 hours, you just study. Find out time to meet your friends, spend some time with your family in free time. But at the same time, do not just close the book for the next 10 days. Okay? It'll be not cool. All right. So happy the shower to all of you. We'll meet on the other side of the break. And I hope you'll come not more prepared after the break. Bye.