 Write down for a rigid body, we are trying to find the angular momentum. Two cases we will consider like always fixed axis and rotation plus translation. So if the entire rigid body is moving with the same velocity, it is not rotating. It is as good as the point masses moving and you can consider the entire rigid body as if it is a point mass located at the center of mass. So that case we already considered. Now we are considering case number one fixed axis. This is your rigid body. And this rigid body. Stop talking and focus here. This is your rigid body which can rotate about this fixed axis. I want to find out the angular momentum of the entire rigid body with respect to this axis. So I am giving you hints. You derive yourself just like we have derived the kinetic ratio for a rigid body about a fixed axis. Same way please do that. Let us say this is point mass m1 at a perpendicular distance r1 away and angular velocity is omega. This m2 at a perpendicular distance r2 away. Now start doing it. Try doing it your way. Anybody? I omega. I omega. I am about to fix axis into omega. How many of you got that? 2, 3, 4. Alright let us do this. The velocity of m1 in tangential direction will be how much? This velocity will be omega into r1. Right? So the angular momentum of point mass m1 is r1 into m1 omega r1. This will come out to be m1 r1 square omega. Similarly, l2 will come out to be m2 i2 square omega. So if you add it up, you will get total angular momentum. That is why I am putting you there. So if you talk next then what will happen? Something extra should be there right? Then what you will do? You will still talk. You will sit here. Without. Right? What kind of punishments you get in school? Get out of the class. We don't give punishments. Yeah they actually don't give us punishments. What? No punishments? They shouted at us and then that's all. My teacher used to hit me. No punishments. That's so... I will not have any stories to tell. How should I punish? So then you can hit them. I am in a jail. That I will not do to innovate. Embarrassing, sarcastic. Then what? Shouting to the routine thing. Then what else? Stand on the chair with wheels on it. That's just hitting me first time. All stand here and what are you talking? Explain to everyone. Like some joke you are cracking. You are smiling. Come here and let everyone smile. Tell everyone what you are talking. Fine so if you add up all the angular momentum you get that total angular momentum for reject body. When you talk about the total angular momentum of a reject body you use capital L for it. So capital L is summation of r i square times omega. Right? Omega is constant so comes out. This will be equal to i about fixed axis into omega. This is the angular momentum about the fixed axis. It's coming out so simply. Just like kinetic energy came out simply for the fixed axis as half i about the fixed axis into omega square. Angular momentum also comes out very nicely like that. Right? If you differentiate it, what do you get? i f into i alpha which is equal to? Talk. Getting it? Okay. Force is equal to rate at which linear momentum changes. f is equal to dp by dt. That is more basic or f is equal to m into a? It's dp by dt. dp by dt. Even mass can change. There will be a force. v dm by dt is also force. Similarly here it is not just i into d omega by dt. It can be omega into di by dt also. Getting it? Yes. Case number one fixed axis. Now we are trying to derive for the translational axis. Alright? It will be equal to i cm into omega r center of mass cross m into v cm. This is the total angular momentum. Fine? This is about which axis? This is suppose this is the rigid body. It's a rigid body going with v cm like this and let's say angular velocity is omega. I am trying to find out angular momentum of the entire rigid body with respect to this point which is such that this vector is your r cm vector. r cm is a position vector of the center of mass with respect to a point about which you are finding the angular momentum. Usually application of this will not be there. And even if it is there, you can always have momentum with respect to center of mass. This will be how much? With respect to center of mass is how much? i cm. r cm becomes what? Zero. Poichet vector of the center of mass with respect to itself is zero. So moment mass is always. No matter what the rigid body is doing is i cm omega. And sometimes center of mass axis is your fixed axis also. It's a full mathematical derivation of this also. We are not getting into that. That is of no use. Sir, I am always thinking about the z axis if we just take i cm omega. Yes. And it is a vector quantity. It's a vector quantity. Next topic, please write down. Soon we are about to finish this chapter's theory. Then we will solve only numericals. We are towards the end of the theory of this chapter. Write down angular impulse. So I know that torque is equal to rate at which angular momentum changes for a rigid body. It's very similar to force equals to rate at which the linear momentum changes for a point mass. These equations are similar. So torque dt integral is equal to integral of dl which is equal to change in angular momentum. Similarly here, this is what we have already learned. This left hand side is what? This one? Leader impulse denoted by what letter? J. So J vector is equal to change of similarly the angular impulse. There is no letter as such assigned for angular impulse. So it is integral of torque dt is your angular impulse. This is your angular impulse. This is equal to change in the angular momentum. And how do you get angular impulse? There should be torque for a very short span of time. Any doubts here? No doubt that we can take a question on this because this is a little vague, this concept. Impulse from here. This is the impulse J. Impulse is given. We are going to find out velocity of centre of mass and omega. First stationary on the horizontal surface. Mass m and length i. Uniform dot. Can I use impulse is equal to rate of change of linear momentum over here? Sorry, not rate of change of linear momentum over here. Change of linear momentum plus angular momentum. So because it can sound slate and rotate. Whatever you have learned, everything is valid but for the centre of mass. Whatever you have learned with respect to linear impulse and everything is valid with respect to centre of mass. So you can say that J is equal to change of linear momentum which is m into vcm minus 0. vcm is what? Right m. Wait sir, but isn't J equal to m to vcm plus i omega? No. Reading it. You have derived it, right? J is equal to change of linear momentum. If it is a rigid body for the centre of mass. Simple. Okay. Now we have to also apply the angular impulse is equal to change of the, right? Now, is there a fixed axis? No. No fixed axis, right? Alright. So, no it is not fixed. It is placed on the horizontal surface and impulse J is given at the end. It is not fixed. Which point angular impulse is there? Impulse is there with respect to this? No. About that? Yes. About everywhere else there is angular impulse but if you write about any random point writing angular momentum becomes difficult. It will be icm omega plus rcm into mvcm, right? But if you write angular impulse with respect to centre of mass it will be simply equal to icm omega, yes or no? Right? So I will calculate angular momentum with respect to centre of mass. How much that will be? This divided by It will be simply linear impulse into L by 2. How it comes? J is what? Integral of F dT, right? Angular impulse is integral of tau dT. tau is what? F into L by 2 equal to L by 2 integral F dT which is L by 2 into J. J, same thing. Okay? J into L by 2 is equal to the angular momentum change with respect to centre of mass axis. So this will be what? M L square by 12 into omega. So you get omega also here. What do you get? 6 6 6 J by mL. Are you guys comfortable with this concept? Okay. So find out the point which is at rest immediately after impulse is given. No, no, no. Find out location of a point which is at rest immediately after impulse is given. And if you are getting the answer immediately it is wrong. Do something. Write it down. It has to be that N. No, no, no. That N might move back. So what is the condition for there being a point at rest and there being no points at rest? Net velocity is 0 for the point at rest. Velocity is 0, right? For an object with point which is at rest. So it is not necessary that there will be a point at rest. But there is... Sir, it is 1 by 6 from that point. 1 by 6 from? Right hand side or left hand side? It will be that way. Good. That's right. Right of center. Stand up. Me? Yes, center. So are you showing that you have not done it? No, actually not. Every criminal say that I haven't done it. Alright, listen here. So angular velocity is omega like this. Vcm is how? Vcm is like that. So if you take a point over here which is at a distance of let's say d. What will be its total velocity? Omega into d like this. This is omega d plus Vcm. Yes or no? So no point can be at rest over here because the angular component and the linear component they are adding up. Over here though, it will be omega into let's say distance of y. So there is omega y and there will be Vcm. So total velocity will be Vcm minus omega y. So you have to find out a point where Vcm minus omega y becomes 0. So y will be what? Vcm by omega. Omega? Do Vcm by omega. You get l by 6. So a point which is l by 6 on the left hand side will be at rest immediately after collision. Let me collision the impulse. By the way impulse comes off collision only. Getting it, this is slightly tricky concept but it requires little bit of problem practice and then you are fine with that. Not lot of varieties will be there with angular impulse. So you can just solve couple of numericals and you are fine. Alright so this is the angular impulse