 All right, so good evening to everyone after a good long break, people are still joining in, how was your school exams, can you type in your marks, at least for physics you can let me know, type in your marks, only I can see that, don't worry, out of 50, okay, okay good. So I can see that school you are able to manage, anybody got 50, full, all right, highest is looks like 49, right, 49 is what, teacher, fine, so I hope now you can understand, got practical marks also, does this include practical also, whatever you're telling me, practical marks I'm not that interested because I know practical, everybody will got full, right, everybody will get full, how the practical was done online anyways, how they do practical, lie, okay, anyways, so, all right, good. Good, so what is going on in your school, in NAPL, what is going on, and in Rajachinagar what is going on, fluids you are doing in NAPL, NAPL is fluids, and in R&R work by energy, fluid statics is removed, don't worry, I don't think it will be removed from complete exams, it will be there. So, I think we are, our class hasn't started anything yet. All right guys, so today we will complete this chapter, rigid body dynamics, fine, after completing rigid body dynamics, we are going to, we are going to do thermal physics, fine, so that's how, that is what the plan is, right. Session, so I hope all of you have had a reading of whatever we have done previous class before coming to this, because you know this is a chapter which is connected, all the things are connected, so I'm assuming all of you have come prepared, you know what is moment of inertia, you know how to find it, you understand the, the torque, how to find the torque, you know how to write torque equal to IL phi and all that, okay, so all of that will be used in problem solving, fine. Is my assumption correct, all of you have come prepared? So, the, all right, there you go, last session, our focus was the cause of rotation, which was torque, okay, we were mainly talking about the torque and the angular acceleration, fine. This class, we are going to discuss about the energy consideration and collisions, fine, so for example, if two masses are colliding, we have learned about the conservation of linear momentum, right, what conservation of linear momentum tells us, linear momentum conservation tells us what will happen to their linear velocities, in each velocity is 2 meter per second, final velocity will be what, after collision, so that you can find out if you can conserve momentum, if you know coefficient of restitution, all that will be used to find out their final velocities, okay. Now suppose two rigid bodies are colliding, suppose there is this body and let's say you have another object, okay, here is, here are two objects, two rigid body and both are colliding, fine. When they collide, because of the collision, not only linear velocity will be attained but there is a chance that the bigger object start rotating also. Don't you think so? When two objects collide, the bigger object can start rotating as well, right, so we need to not only understand what happens to the velocity, we need to also understand what will happen to the angular velocity, all right. So when it comes to, write down the collisions, when it comes to collisions of rigid body, we are tracking velocity of center of mass and the angular velocity, both we need to identify before and after collision, to define what happens before and after collision. If there is no rotation, let's say one mass comes here, other mass comes from there, they collide, rotation is absent, they don't rotate after the collision or during the collision, then you don't need to worry about omega. But when rigid body collide, then it might start rotating also. So for identifying, for dealing with velocity of center of mass, we can use all concepts of linear momentum. What does it mean? Suppose you are dealing with, you need to find out only the linear velocity, then forget about angular velocity, the concepts of angular velocity, how to find out that is different from concepts of linear velocity. Linear velocity you can get by just conserving the momentum or whatever you know already. For example, all of you do this question, suppose you have a rod like this, it is horizontally lying on a table, it is lying on a frictionless table and there is no fixed axis as such, fine. Then there is this mass of small m, a point mass of small m, coming with velocity of u, okay. This mass is capital M length is L, okay. This mass comes and hits this rod and immediately after hitting, immediately after hitting, after hitting, mass m stops, okay. And I want you to find out velocity of center of mass for rod, how much it is, everyone do this. Can I conserve linear momentum, everyone? Can I conserve linear momentum or not? I can, after the collision, will rod rotate also after the collision, is it correct? Will it rotate? It will rotate, right? So it will gain some angular velocity, let's say omega is angular velocity after collision and it will have velocity of center of mass PCM. Now, conserve the linear momentum, all of you, all of you conserve linear momentum. When we conserve linear momentum for rigid body, for rigid body, we can assume it to be, we can assume it to be a point mass with PCM velocity, okay. Just look at velocity of center of mass, you don't need to look at anything else with what velocity it is moving, okay. So, now tell me what is the answer? Gayatri got something, others? Correct, it's straightforward. Initial momentum of the system is M into u, final momentum is M into PCM, okay. Small m comes to rest, all right. This plus zero, zero plus this. So velocity of center of mass is M by capital M into u, all right. So with respect to the linear velocity, you don't need to do anything else other than the conservation of linear momentum. Getting it, it's simple. Now, suppose I tell you there is this rod, okay, and there is a mass which comes and hits the rod with velocity u, same situation. But after hitting the rod, it doesn't come to rest. Fine, what is given to you is that the rod attains angular velocity omega and it rod starts moving with VCM. Sorry, VCM is not given, omega is given. Omega u, small m, capital M, L, these things are given to us, small m, capital M, okay. And this line of velocity, this line of velocity is at a distance of L by 2 from the center, okay. And it is an elastic collision, elastic collision. For elastic collision coefficient of restitution is equal to 1, fine. So you need to tell me what is the value of the final velocity V for small m and VCM for the rod, for rod in terms of everything which is given to you, okay. Do it. After collision, the velocity of mass small m is V, let us say. If you have written the equations, you can type in that you are done. You don't need to exactly, you know, simplify and get the answer. Just write the equation and tell me that you are done. Okay, one is done. Others, you write two equations, right? One is conservation of linear momentum. This equation you will write. How this equation will be? Can you type in what will be the equation of conservation of linear momentum? Alright, so initial momentum of, initial linear momentum of small m is m into u. Initial linear momentum of the rod is 0. This should be equal to the final linear momentum of the rod, which is MVCM plus m into V. Okay, so how many of you got this correct? m into u should be equal to m into VCM plus small m into V. Everybody understood this? Those who haven't typed anything, have you understood? Next equation is coefficient of restitution equation. Restitution equation. What is that? E should be equal to 1, which is equal to velocity of separation divided by velocity of approach. Now, tell me what is velocity of separation after collision? Small m is moving like this after collision with velocity V. So, velocity of separation is how much? VCM minus V, okay? VCM minus V is not correct. Now think, tell me what it is? Approach with the entire rod. Approach with the point of collision, separation with point of collision. The velocity of approach and separation, velocity of approach and separation should be, should be with the point of collision. Understood? Now tell me, okay, Prakul got something. What is r? r is not something which I have given you. After collision, tell me with what velocity this is moving in this way? What is the velocity of this point on the rod after the collision? Tell me, everyone? Velocity of the rod of this point. Center of mass is going with VCM and the rod is rotating with angle of velocity omega. Do you all agree that with respect to the center of mass, that point is moving in a circle? If you just look at the relative to the center of mass, it is going with L by 2 into omega. But that is a relative speed. You need to add plus VCM to it. Got it? This is the total velocity. Something like this we have done previous class also. Do you all remember? We have done it for the angular acceleration. Same thing for the velocity also. With respect to center of mass, omega into L by 2 is the velocity. But center of mass is not addressed. So we have to add center of mass as velocity also. So total velocity is VCM this way and omega into L by 2 also that way. Okay? So the velocity of separation is omega into L by 2 plus VCM minus V. And velocity of approach is U. Should be with point of collision along the normal. Along the normal. So normal is this line only. Do you all understand this second equation? The second equation is the important one or something new to you. Everyone understand the second equation? Okay. So this is how you solve collision with the rigid body. But over here, can I explain what again? This one. Second equation. So this is velocity of the numerator is velocity of separation. Velocity of separation. Now when I say velocity of separation, immediately first thing you will come in your mind is velocity separation with the center of mass. But that we should not do. Right? Velocity of separation should be with the point of collision where it is about to collide or where it has collided. Okay? When it is collided along that line, that point's velocity is omega L by 2 plus VCM. Omega L by 2 with respect to center of mass. Plus velocity of center of mass is the velocity of the rod at that point. So that point is separating with the mass with the velocity of omega L by 2 plus VCM minus velocity of the mass after the collision. That is velocity of separation. That is the velocity of approach which is U. Everything was at rest before the collision. So that divided by U is equal to 1. Okay? Now till now, we have assumed omega is given to us. Right? Omega need not be given to us. Omega suppose is not given. Then you will not be able to solve this question. Right? So there is something which is more than just conserving linear momentum which is going on which we need to understand. Because omega is the result of the collision only. So nothing else should be given to us. Only this situation should be given to us and we should have been able to find omega. Because there is nothing else I can give you to find out omega. Nothing else affects omega. Everything that affects omega is already given to you. But still we are not able to find omega independently. Omega I have to give you. So let us try to discuss something which is different from conserving linear momentum. I am telling you a concept of angular momentum here. Fine? So write down angular momentum. Can you try to define what is linear momentum? What does it represent? Linear momentum represents what? M into V is what? M into V is a formula. You could have said M into V square. Then formula can be anything. But what is the idea behind it? What it is? Linear momentum is what? What does it represent? Linear momentum is the amount of linear motion. Everybody agree to this? Amount of linear motion, quantity. If a heavier mass is moving with small velocity then also amount of motion is large. And if a very very light object is moving with high velocity then also amount of motion is lesser. So it does not only depends on velocity also it does not only depends on mass. It depends on both mass and velocity. So if you multiply both of them it will give you amount of linear motion. Or you can say loosely you can say it is like emotional inertia. With what inertia it is moving? The inertia of the rest and the inertia of motion. Both are combined here. Now similarly there is this amount of the angular motion also. That also can be quantified. So just like it is difficult to stop an object which is already moving with some certain velocity. You need to make an effort to stop it. Similarly a rigid body like for example fan. Fan even if you switch it off it will take time to stop. Why? Because of its rotational inertia. It wants to continuously rotate. But because of the friction and all it stops finally. So it doesn't have linear momentum overall. But it has an angular momentum. Like for example just to explain to you a bit more. Suppose your fan has just one wing. And it is rotating about this point. So all these points will have velocity this way. All these points will have velocity that way. The fan is rotating like this. Don't you think that vector being the vector quantity. Sorry, linear momentum being a vector quantity. All these momentum of the points will cancel out all the linear momentum of that. So linear momentum is zero. All of you agree to that? For a fan which is rotating about a fixed axis. Linear momentum is zero. So there will be one set of points going this way. Other set of points going in opposite direction. So when you add the linear momentum it will become zero. Do you all agree? Type in. Now even though linear momentum is zero. But there is certain amount of angular momentum. Which is still making it move. And that moment is rotation. So we are going to discuss about how to quantify the angular momentum. So first we will try to introduce the angular momentum for the point mass. Angular momentum of a point mass. So angular momentum is simple. It is like how fast or how quickly the angle is changing. Which angle? I'll tell you. For example, there is a mass, a point mass of mass m and velocity v. It is going like this. Okay. And you are standing here. You are standing here. You are looking at this. All right. Now after sometime it will reach here then it'll reach there. It'll keep on moving further and further. Now look at it. When you are looking at it, you will see that your head will turn. First it will look at it directly like this. Then you have to rotate your head like this. So because you are rotating your head. So with respect to you, the angular position is changing. Isn't it? Earlier the angle is zero. Then this much angle. Then that much angle. This is the angle. And that is the angle. So there is rate of change of angle for the mass with respect to you. Fine. And that is what we are trying to quantify. Fine. Now, if this distance, let's say this is the distance r. Fine. So if the value of r is zero, what will be the rate of change of angle? What do you think it will be? Will it be small or large? What it will be? If you are standing here, let's say r is zero. There is no perpendicular distance between you and the object which is moving. So what is the rate at which angle will change? Or angle won't change. Angle won't change. So angle won't change. So angular momentum L which is represented as angular momentum should be equal to zero. Okay. Now angular momentum will be large if both mass and velocity are higher. Fine. And the r is also large. If r is zero, angular momentum will be zero. Fine. So this r is what? r is a perpendicular distance. Are you getting it? So it cannot be any distance for what I'm trying to say is suppose there is a mass over here, then the angular momentum, this mass and velocity, you are here. Okay. Angular momentum depends on the perpendicular distance. It doesn't depend on your distance to the point. Otherwise, if you stand over here, if you stand over here with respect to you, the distance of this point is continuously changing, but angle is not changing. So angle depends on the perpendicular distance. Okay. So that is the reason why the angular momentum is quantified as perpendicular distance mass into velocity. Are you able to understand everyone? Now, just like torque, you can say that at any moment, if this is the value of r, what is this r? The point, sorry, the vector which connects, this is the vector which connects the point o with the mass that is moving. So with respect to this angle, what is the perpendicular distance? If I say that angle with r and velocity is theta, perpendicular distance is what? If this is r, this is theta, what is perpendicular distance? r sin theta, right? So r sin theta into m into v, this is the magnitude of the angular momentum, fine? So the angular momentum can be also quantized as a vector quantity, which can be written as r cross linear momentum p, where linear momentum is m into v. Okay. So this can be written as r sin theta into m into v or you can say r into m v sin theta. What is v sin theta? v sin theta is the perpendicular component of velocity. The velocity will have two components, this velocity, this will be v sin theta and this will be v cos theta. So the component which is along the r will not create any angular momentum but this component will create angular momentum. So you either take the perpendicular distance or take the perpendicular velocity. The formula is just like torque equation, which is r cross f, r sin theta into f or r into f sin theta. Do you get this torque? You remember it is perpendicular distance into force or r into perpendicular component of force. Similarly, angular momentum can be written like this. Okay. All right. But then we have found out angular momentum for a point mass, which is r cross p. This is for a point mass. But then how does it help us? We have just randomly defined something. Now the physical meaning of the definition of the angular momentum will be clear like this. If you differentiate it, what you will get? dL by dt. dL by dt, take r as fixed, r cross dP by dt. What is dP by dt? Everyone? Force. So r cross f. This is what? Torque. Fine. So just like, just like rate of change of linear momentum is force. Similarly, rate of change of angular momentum is the torque. Fine. So we have just defined it for a point mass. I'm going to show you how it is valid for the bigger objects also like a rigid body. Okay. So I just thought I could quickly tell you how does it relate to what you already know. So now you can understand laws of motion chapter and rigid body motion chapter. They are conceptually they are similar. Okay. Now take a rigid body for a rigid body rotating about the fixed axis. Okay. So for example, there is this object, this object, this entire rigid body is let's say rotating about this fixed axis with angular velocity omega. Okay. So all the points on this rigid body, they start moving in a circle. Isn't it? They are moving in a circle keeping the center on the axis. Right. So let's say you have mass m1 at a perpendicular distance of r1. This is m1. So what is the angular momentum of m1? Everyone distances r1. Tell me angular momentum of the mass m1 about this axis or about this point. How much it is? What is the velocity? Velocity of the mass is how much? This point's mass velocity is how much? Omega r, is it perpendicular to r1 already? Is that velocity already perpendicular to r1? So r1 is a perpendicular distance r1, then m1 and velocity is what? Omega into r1. Understood all of you? So this is equal to m1 r1 square times omega. This is the angular momentum of one of the point mass on the rigid body. Okay. So like this, if I keep on adding all the angular momentum, then I'll get total angular momentum, which will be m1 r1 square omega plus m2 r2 square omega and so on for all the point masses. Now tell me what is this inside the bracket? m1 r1 square plus m2 r2 square and so on into omega. What is this inside the bracket? This is moment of inertia about what? About the fixed axis. We are talking about rotation about the fixed axis. Remember that. So we have got the angular momentum of the entire rigid body. If the rigid body is rotating about the fixed axis, angular momentum about the rigid body about the fixed axis is if into omega. All of you clear on this? Everyone type it. This is the angular momentum about the fixed axis. What is the difference between this and torque? Torque is I alpha. The formula you're talking about the difference. When you say what is the difference between this and the torque, the question is similar to what is the difference between force and linear momentum. Just like force and linear momentum, they are different. Similarly torque and angular momentum, they are different. By the way, when we talk about the angular momentum for the rigid body, we need to write it capital L. When we talk about the point masses angular momentum, we write small L. That's just a convention. For the fixed axis, the angular momentum is If into omega. Now, if you differentiate it, you get what? dL by dt will be equal to If into d omega by dt. Now, d omega by dt is what? If into alpha. What is If into alpha? If into alpha is what? This is the torque. Torque about the fixed axis. Whether you differentiate the angular momentum for a point mass or whether you differentiate the angular momentum for a rigid body which is rotating about a fixed axis, you're going to get a torque only. Now, let us see a rigid body which does not have a rigid body which is having rotation plus translation, both. Basically, there is no fixed axis. There is no fixed axis. For example, this object is moving and rotating also like that. This fixed axis example is like a fan. For this case, we can have angular momentum about the center of mass. We are not getting into derivation of this. It's pretty long. It will take a lot of time. We need to complete many things. That is beyond your curriculum. Angular momentum about the center of mass axis is ICM into omega. The ICM is along the angular velocity. LCM is ICM into omega. Just like torque is equal to If is valid only for the fixed axis or about the center of mass axis. Similarly, angular momentum is equal to I omega. That formula is valid only about the fixed axis or about the center of mass axis. Suppose you need to find out the angular momentum about any other axis. Then what you do? You first write it as ICM into omega then plus RCM. First write it down. Then I will tell you RCM into VCM. What you do is you first find out the angular momentum with respect to the center of mass, which is ICM into omega. Then you consider entire body as if it is a point mass moving with VCM. This formula RCM into MVCM is a formula for a point mass. You find the angular momentum of the rigid body with respect to the center of mass axis then you find angular momentum of the center of mass and then add them up. This formula is rarely used. I think it is not there in your NCIT textbook also but you should be aware of it. Maybe when you are done with the basic questions there are few questions on that as well. Remember, torque equal to I alpha just like that is valid only about the fixed axis and only about the center of mass axis. Similarly, angular momentum is equal to I omega is valid about the fixed axis and about the center of mass axis. This thing is very clear that if you differentiate the angular momentum you are going to get torque. That thing should be very clear. All that is fine that if we differentiate angular momentum I will get torque but so what? How can I use it? Do you remember when we discussed the linear momentum we had a law of conservation of linear momentum? Is there something like that here also which will be like conservation of angular momentum? The answer is yes, there is something like that. Let us see what it is and how we can use it in problem solving. Write down conservation of angular momentum. Whenever we conserve the linear momentum we have more than one mass because collision has to happen. Similarly, when we talk about conservation of angular momentum, again there will be multiple objects which will be colliding and exchanging the velocities angular velocities and all those things will happen. Let us consider a system, a hypothetical system which has multiple masses in it. This is the boundary of the system. Inside it I have let us say two point masses one and two rigid bodies. Randomly I have picked up this is M1, M2, this is M1, M2. Now the angular momentum of the entire system can be written as angular momentum for the mass M1, angular momentum for mass M2, angular momentum for first rigid body plus angular momentum for the second rigid body. This is the total angular momentum. So if you differentiate it dLt by dt you will get this. Now if you see that right hand side becomes the torque on the mass M1 plus torque on mass M2 plus torque on mass M1 plus torque on mass M2. Now when I add up all the torques will it include the torque due to the internal force also? Because when they collide they will apply force on each other will that torque is also included here? This torque which is experienced by M1 do I have to include the torque which is given by other masses on M1 when they collide? When I write like this, everyone type in of course it will include, right? It is rate of change of total angular momentum. It will include all the torques for that mass. So your thing is this M1. So M1 will get torque from let's say M2, M1 and all those things can give it force or torque whatever you want to say, right? So all of these four terms will have some of the external torques plus some of the internal torques. Now this external and internal is external to the system and internal to the system, okay? But when this let's say M1 is experiencing a force because of collision from M1 capital M1 is getting a torque because of which it rotates. So when I just talk about M1 then there is a torque because of small M1 when it hits it, getting it? But when I say external internal I am mentioning, I am saying torque is external only when it is outside the system, okay? Now look at the nature of the force which is internal. Suppose you have there will be one force like this and there will be another force like this, okay? So what happens is the internal force is along the line joining the two masses. Are you able to understand till here? Is this clear? I guess some of you are confused. Till here is it clear? Any doubts type in till here it is clear, okay? What I am trying to say is suppose you have two masses, okay? So from outside there is a force. There is a force, okay? But the internal force will be equal and opposite and because of this internal forces the torque will be zero because the force is along the line joining. There is no perpendicular distance. So it is the one and the same thing, okay? Let me show you vectorally. Suppose you have to find torque about this point O. For the entire system you are finding the torque about point O, okay? Just look at the internal force F. This is F, this will be minus F, okay? This is R1, this is R2, okay? M1 and this is M2. So torque on M1, torque on M1 is R1 cross F. Torque on M2 is R2 cross minus F. I am just looking at internal forces right now. So when I am adding the internal torque, torque on M1 plus torque on M2 I am getting R1 minus R2 cross F. What is the value of this? R1 minus R2 cross F, everyone? Where is R1 minus R2? Can you tell me where it is? Vector minus R2 vector. Where it is? Only Archit knows, is it? Nobody else? R1 minus R2. Don't you think this is the R1 minus R2 vector? R1 minus R2 is this? According to triangle law of addition. Minus R2 should be equal to this? Third vector, okay? So R1 minus R2. Do you all understand R1 minus R2? How it is that? Tell me quickly guys. Do you all understand that R1 minus R2 is this vector? This is R1 minus R2. Okay? R1 minus R2 is parallel to F. Okay? So when you add the internal torques you are having a cross product between two parallel vectors if you add it. What it will be? Cross product will be how much? If these two vectors are parallel? Zero. Right? Two parallel vectors cross product is zero. So if you differentiate the total angular momentum you are going to get the external torque. Okay? Because internal torques are zero. So DL by DT is equal to all the external torque on the system. So if external torque on the system is zero. If external torque is zero. Can I say that angular total angular momentum is a constant for the system? Do you all agree? If total external torque is zero. Can I say that angular momentum of the system is a constant? Because its derivative is zero. Right? And because it is a constant. I can say that before after sum of all angular momentum should be equal. Make sure. And they are equal only about an axis zero torque. You know sometimes you will find about one axis there is a torque about another axis there is no torque. Right? So you need to pick the axis about which torque is zero and then use the conservation of angular momentum about that axis only. Alright? Now let us take some examples. Otherwise it will remain a theoretical concept which will be useless. Okay? So let us take one example then we will proceed to see what else we can do. Okay? So suppose you have suppose there is a rod. Okay? There is this rod which is fixed here. This rod is fixed. And this rod can swing like this. Okay? So you can swing freely about this axis. Fine? So there is a mass m that comes with a velocity of u hits this rod. Rod having mass capital M and length L. It hits it and then its velocity becomes zero. After hitting velocity becomes zero and it drops. After hitting it drops. We need to find out the angular velocity of rod immediately after hitting. Find out. Can I conserve the linear momentum here all of you? Can I conserve linear momentum? Some of you are saying yes. Is it correct? Can I conserve linear momentum? All of you type in can I conserve linear momentum? The answer is no. We cannot. Now can you tell me why we cannot conserve linear momentum? All of you. Let me tell you a linear momentum not angular momentum. Because it is fixed so what? What is the condition for conserving linear momentum? What is the condition? No external force right? Is there an external force? Horizontally is there an external force? Horizontal direction. I am taking small m and capital M both in my system. Where is an external force? Horizontally from here there is an external force. Where it is hinged that will create external force and that will be impulsive in nature. Not only external force but impulsive also very large amount of force. So you cannot conserve the linear momentum in horizontal direction. Clear to all of you? Everyone clear? That you cannot conserve the linear momentum? Okay. Now can I conserve the angular momentum? Can I conserve angular momentum? Can I conserve angular momentum about the fixed axis? Can I do that? About the axis that is passing through the point of hinge? Can I conserve it? I have pole you know I can use the pole. Tell me can I conserve angular momentum about the axis that passing through the hinge? Can I conserve the angular momentum? Okay this is over. This is what you can see. See there is an external force by the hinge. But what is the torque of that force about the axis that passing through the hinge? How much is the torque? External force is there from the hinge. But torque due to that force about the axis that passes through the hinge is zero. Because perpendicular distance is zero. So only external force which was bothering was the hinge force and torque about that force is zero. So there is no torque about the axis which passes through the hinge. So you can conserve the angular momentum about the axis that goes inside the hinge like this about that axis. Can I conserve angular momentum about the center of mass axis? Can I conserve? Not conserve. Why? Because now this external force has torque about the center of mass. F into L by 2. So because there is a torque about the center of mass axis, I cannot conserve angular momentum about the center of mass axis. But I can conserve the angular momentum about the fixed axis. All of you clear right? Now find out what is the value of omega. Everyone tell the value of omega. We cannot do it about the center of mass axis because you can draw the freeway diagram. Now you can see that from the hinge there will be forces. There will be forces like this. So the hinge force will have torque about the center of mass. This affects into L by 2. Because there is an external torque you cannot conserve the angular momentum. Condition for conserving angular momentum is net external torque should be zero. But about that axis external torque due to Fx and Fy will be zero. Because their perpendicular distance is zero. Anybody got omega? So angular momentum of the mass m plus angular momentum of the rod initially should be equal to angular momentum of mass m plus angular momentum of rod finally. Initially whose angular momentum is zero? Initially whose angular momentum is zero initially? Rod. So this is zero. Angular momentum of small m is how much? Small m is a point mass. Its angular momentum is how much? About the fixed axis? You remember angular momentum is perpendicular distance into linear momentum perpendicular distance from the velocity. This is the angular momentum for a point mass. How much it is? From where omega came for point mass? Omega is for the rigid body. What is the perpendicular distance? How much is the perpendicular distance? L right? This is a line of velocity. This is the line of velocity. Its perpendicular distance from the axis is L. So perpendicular distance is L into m into v. This is the angular momentum for this small m. This is equal to what is the final angular momentum for small m? Everyone? This is how much? This is zero. For the rod, how much? For the rod, how much? Let's say omega is angular velocity. How will you write? M L square omega. The formula is what? Moment of inertia of a fixed axis into omega. This is the formula, right? So what it is? I f is how much? This is how much? For the rod. M L square by 3. You have forgotten many things. Revise it. All of this you should be clear. So 3 m v by capital M into L. This is the angular velocity. Anyone has any doubt here? Anyone has any doubt? Quick. All right. Let's proceed. One more question. Then we'll proceed further as in the next concept. Suppose there is a rod lying on a horizontal table like this. This rod is not fixed anywhere. Rod is free to move on a frictionless horizontal surface. Free to move anywhere. A mass comes from here. Capital M, a mass comes from here. Hits it with velocity v. Total length of the rod is L. From the center, the distance of line of velocity is L by 4. So after hitting, let me keep it simple only right now. After hitting, M comes to rest. You need to find out part a velocity of center of mass for the rod. Part b omega of the rod. There is a part c also. But first find out a and b. About which axis can I conserve the angular momentum? Everyone, can I conserve about that axis? Can I conserve about the axis that goes inside from here? From there. These are 1, 2, 3, 4, 5. Tell me about which all axis can I conserve the angular momentum? For small m and capital M together. All of you tell me. Tell me about which axis can I conserve the angular momentum? For both small m and capital M. Tell me what is the external force for small m and capital M? What is the external force? Normalization from the surface and mg. Both are non-impulsive. They can't suddenly change anything. So is there an external force to be considered just before and after collision? As in that there is no hinge. So about any axis you can conserve the angular momentum. 1, 2, 3, 4, 5, whatever axis you want to pick anywhere. You can conserve the angular momentum. There is no external force which can affect the conservation of angular momentum. Those who are saying only one of the axis. Is it clear? You have any doubt? You can conserve about any axis. But about which axis you know the formula? About which axis you know a convenient formula? About the center of mass axis. You know about the center of mass axis. The angular momentum is ICM omega. I will pick center of mass axis for conserving angular momentum. Because it is easier for me. But you can conserve about any axis. And there is no fixed axis here. Now tell me all the answers. A and B. Gayatri got something. Others? Okay. Good. Many of you got the answer. For part A simply you can conserve the linear momentum. M into V is equal to capital M into VCM. This is you can say VCM. And you can represent angular velocity omega like this. Okay. So from here you get VCM is equal to M by capital M times V. For part B you can have conservation of angular momentum about ICM. Fine. What is the initial angular momentum of the system? How much it is? Everyone? Papandika distance of the velocity from the axis is L by 4. So MVL by 4 is the initial angular momentum. Angle momentum of rod is zero before the collision. This is equal to angular momentum of now mass is zero. That plus the rod's angular momentum which is ICM into omega. What is ICM for the rod? ML square by 12. All of you? ML square by 12 times omega. Okay. So 1L is gone. This is 3. So omega is 3MV by capital M into L. Okay. This is the answer. Understood all of you. Those who got it wrong. Is it clear? If you take moment of inertia wrong, you definitely get a negative mark then. Such is the exam. Now you need to tell me the point on the rod which is at rest immediately after collision. All the points will have different different velocities. All of you agree? All the points will have different different velocities. There will be a point on the rod which will have zero velocity. Where it is? Midpoint is going with VCM. It is written. How can you say it is zero? Center of mass is going with VCM velocity. It's not zero. Tell me the location and how far and all that. Will it be upper point or a lower point? Will it be upwards or downwards from the center? Up or down from the center? All of you clear it is upward. Because lower, there will be omega into whatever distance plus VCM down. Okay. But upwards because of omega velocity and because of VCM, the velocity is there in opposite directions. That's the reason why. Okay. So let's say at a distance of y, the velocity is zero. So there the omega into y minus VCM is the velocity. Omega into y this way and VCM that way. Subtract. That should be equal to zero. So y should be equal to VCM divided by omega. So VCM by omega is how much? Is it L by 3? Can you check your calculations? It's L by 3. Clear to all of you. Move ahead now. See, you take a point at a distance of y away at this point with respect to center omega into y on the left-hand side is the velocity. You need to add the velocity of center, which is VCM. But VCM is this way, omega into y is that way, so you need to subtract them. That is a total velocity. When that becomes zero, if that becomes zero, that point is out in the rest. So this is what we have done. Okay. So there is a small thing about the angular impulse right down. Have you, do you remember linear impulse? Do you remember that linear impulse f is equal to dp by dt. Do you remember we writing integral f dt to be equal to change in momentum, final momentum minus initial momentum. Type in quickly. Do you remember this? Only Aryan remembers it. Okay. This is the linear impulse. Okay. That we have discussed in detail. We have spent an R also on the linear impulse. So there is something called angular impulse also. For example, we know the torque is dL by dt. Okay. Angular impulse is tau dt. Tau dt is delta L, which is final angular momentum minus initial angular momentum. Okay. So we will understand this concept by using a numerical. Okay. Theory is only this much. Clear. So just like there's a linear impulse, which is because of a force, because of a large amount of force for a very short amount of time, the momentum can change suddenly. Okay. So let's take one example and I will explain this concept. Every collision, every collision impulse also have angular impulse. What you need to have a linear impulse, you need to have impulsive force. Impulsive force is needed. Okay. In order to have the angular impulse, you need to have torque due to impulsive force. It is simple. Okay. So now let us take one example. We'll just take one example for this. It's a concept which you will be able to master only by problem solving. I'm just introducing to you that listen, this is something which exists. Okay. So for example, let's say there is a sphere, which is thrown at certain velocity in this direction. This ball is not very far away. Okay. Ball is very near to surface. So in this direction, it comes and collides the surface. This is what I'm trying to say over here. Okay. So this comes with velocity of u. All right. And at the time of collision, it has the angular velocity omega naught. Fine. The coefficient of friction is mu. You need to find out what is the... Okay. Let me keep it simple so that we focus only on concept not on mathematics. Suppose it is... The ball is dropped. Okay. From a height of H from the center. Okay. From height H, it is dropped. Okay. Mu is a coefficient of friction. Coefficient of restitution is E. E is also given to you. You need to find out omega after the collision. Immediately after the collision, what is the value of omega that you have to find out. Okay. Get it. Just write down the equation and tell... Once you have written, tell me that you have done. You need radius of the ball. Take it as capital R. Yes. Mass of the ball is M, if you want. During the collision, there will be normal reaction from the surface. There will be gravity. Any other force? All of you? Any other force? Friction. How much is the friction? Will it be mu time normal reaction or something which we don't know? Is there slipping over here or not? There is a slipping or not? Over here. Slipping is there? All of you? Yes. Slipping is there. The velocity is in this direction for that point. So, there is a slipping. Okay. Because of the angular velocity, that point has omega R velocity that ways. Left hand side. So, there is a sliding. So, there is a friction of mu times normal reaction. Okay. Now, tell me which of the forces are impulsive? Is MG impulsive? Can MG suddenly change the momentum? No, it can't suddenly change the momentum. So, in a very short amount of time, MG force doesn't matter. Right? So, I don't care about MG when it comes to suddenly changing the momentum. Okay. Now, linear impulse is normal reaction on it. So, integral of N dt should be equal to change in the linear momentum. So, let's say final velocity of the ball is V. Initial velocity is how much? With what velocity it hits the surface? Everyone? How much it has travelled down? How much centre has travelled? Centre has travelled H minus R. Okay. If you're looking at this, this also has travelled H minus R. Every point on the ball has travelled H minus R. Distance, not H. Okay. So, the velocity, if you use the kinematics, will be root of 2g H minus R. This is the velocity with which it will hit the surface. Clear? This is the velocity of approach. Okay. So, integral of N dt is final momentum minus initial momentum. Final momentum minus initial momentum. So, M into V minus of minus U. U is in other direction. So, it is negative. I'm taking upward positive. Is this equation clear to all of you? Equation number one, is it clear? Now, this is the linear impulse, linear impulse equation. We also know that the coefficient of restriction E should be equal to V by U. Isn't it? So, from here, you'll get V is equal to E times U. Even that, you will get. Okay. So, you know both E and V. Now, U is this. E times of that is V. So, linear impulse equation. Now, let's try to write down angular impulse equation. Anyone has written it already? You need to first pick about which axis you are writing angular impulse. Right? So, the safest axis is center of mass axis. So, is there a force which creates angular impulse about the center of mass axis? Any force? Friction force? All of you agree? Friction force is there. So, the impulse is torque dt. Torque is how much due to the friction? How much it is? Mu n into, mu n into R. This is the torque. Tau dt. Tau dt is a torque. This should be equal to ICM, which is 2 by 5 MR square. Final angular velocity. Now, this direction is positive. Direction of torque is positive. Are you getting it? So, minus of, what should I write here? 2 by 5 MR square into? MR square into omega naught or minus omega naught? What should I write? This is final angular momentum. This is initial angular momentum. This is angular impulse. What should I write here? Minus omega naught plus omega naught. The direction of the angular impulse is clockwise or anti-clockwise. Which direction the torque is clockwise or anti-clockwise? Anti-clockwise. So, anti-clockwise take positive. I am assuming final angular velocity in anti-clockwise direction. Initial angular velocity is clockwise or anti-clockwise? It is clockwise. So, you have to take minus here. Minus omega naught. Getting it? So, mu is a constant comes out of integral. R is a constant comes out of integral. So, you can write it as mu times R integral n dt. Isn't it? Integral n dt, which is m into v plus u. This is equal to 2 by 5 MR square omega plus omega naught. So, using this impulse equations, you can correlate linear motion with the angular motion. Because the same impulse force is creating linear impulse and the mu times that is also creating angular impulse. So, you can correlate this equation with that equation. Getting it? Is this clear to everyone? 70 to 80 percent, is it clear? That much clarity is enough right now. 100 percent clarity will never happen by just sitting in a class. 100 percent of clarity will happen only when you solve problems, your own. Otherwise, it will just be like 30, 40 percent and that too will be evaporate. That too will be becoming zero if you don't do any questions. So, I will send you few questions on the collisions, which will also include few impulse related questions. Nowadays, I have seen that in computer exams, they are asking more and more questions on impulse. So, this is about collisions with the rigid body. So, a point mass can come and hit the rigid body. Two rigid bodies can collide with each other. In both cases, you can use conservation of linear momentum as well as conservation of angular momentum. But about the axis through which net torque is zero. And sometimes you have to use impulse equation depending on what kind of question it is. All right. So, this is...