 Now write down conservation of angular momentum. Just like conservation on linear momentum, there is something called conservation of angular momentum. Write down the statement. If net external torque on a system of particles and rigid body, if net external torque on a system is zero, the angular momentum of the system will be conserved. The angular momentum of the system will be conserved. So if we add up all the angular momentum, it will not change if net torque is zero. So you have to do what? If you have to apply the conservation of angular momentum, you have to find an axis about which net torque is zero. And then about that axis, initial angular momentum will be equal to final angular momentum. So basically the total angular momentum of entire system, let's say it has two rigid bodies and two point masses. So this is the total angular momentum. So if you differentiate it, you will get dL by dt will be equal to torque on 1 plus torque on 2 plus torque this, like that. There will be internet torque also and like the way you remember we have done internet torque also get cancelled away. So if we add all the torque of the system, you will get net external torque. External torque is equal to rate at which angular momentum changes about that axis. And if SNL torque is zero, rate of change of net angular momentum about that axis is zero. So sum of the angular momentum before will be equal to the angular momentum afterwards. Simple derivation, the way we have done for the linear momentum. Now let's discuss couple of scenarios where the angular momentum is there in our day to day life. Have you ever like spread your hands like that and rotated? I did that. No sir, like I let this thing in sixth grade or something. And then it said that if you take two really big textbooks and go on a swivel chair, and you start spinning and then you pull it together, then you spin fast. Yes. I did that. Have you ever done like spinning with your hands open like this and while spinning you do like that, the angular momentum increases. Anyone into dancing? Everybody rotates. So if you have to increase angular velocity, you decrease the momentum in Asia. Which dance? I know. I know. Which dance? Bollywood? Random. In classical in chakras, you take your hands inside once you start spinning. See, all the dance, dance, 50,000 dances spinning only about the fixed axis. And salsa, as you see, every, I mean 90% of the time, the girl is rotating. So that's translation class. I mean, yeah, but most of the time it will be the guy will be like this and the girl is spinning like that. What are you doing? No sir, like the pros doing salsa. Pros are different, like I am talking about you doing salsa. You guys are going to be like this. It's like they do something and the girl is spinning for some other guy. She can talk. While spinning, she goes too much. And see, if you have not done this, go home and do it. Not a lot of space here. If space would have been there, if I ask everyone to do one by one. Alright, listen here. Another example is, have you seen in the Olympics, someone jumps on top and goes in the swimming pool? Diving, it's called diving. So when the person is jumping from the top, his body is completely stretched like that. And in between, folds himself or herself. When he jumps, he or she has very little angular velocity. He was rotating like this slowly. And as soon as the person folds himself, what happens is that ICM decreases because all the mass comes near to the axis. So sum of MR square becomes very less. Moment of inertia decreases. But angular momentum is what? I into omega. If I decrease, omega has to increase because angular momentum has to be constant. So because of that, when he or she folds herself, angular velocity increases and then again when someone is about to touch the water, God knows how they find out they are about to touch the water. They again stretch themselves. Okay? Okay? Try not to do it. Fold it and put it in the water. And I have heard many stories where, not many, couple of stories, that someone jumped and fall on the stomach in the water. Very nice. Surface tension will not let you go inside very easily. So once the Russian spacecraft was there and inside it there were astronauts, and that point in time they have that disk which rotates. You know, I have seen the big disk. Oh sir, is there this middle axis theorem thing in which it just flips over? There is this other really cool thing. It's called the middle axis theorem in which there is this Russian guy. So he had this T script. Let me complete now. Middle axis theorem, I am not teaching. Listen here. So what I was saying was these Russians were fond of some music. So they went to space and started that gramophone and the disk was rotating. This has some angular momentum. Initial angular momentum of the spaceship was zero. When the disk rotates, it generates some angular momentum. So the spaceship slightly will rotate. Initially it was zero. Finally it should also be zero. So if this rotated like this, the spaceship should rotate in opposite direction. So there they found out the conservation of angular momentum properly. So even if the spaceship rotates by 0.001 degrees, it will go somewhere else. If you have to go to moon, you will land up in Mars. Because they are traveling extremely far distances. And there are so many examples of angular momentum. Angular momentum is in fact more generic than the linear momentum itself. So let's take couple of numericals on the conservation of angular momentum. Is there any doubt on angular momentum conservation? Stop talking. It's across two rods are welded. The mass of the rods M and the length of the rods L. At the end there are guns. And guns have bullets inside them. So they fire simultaneously. So this entire thing can rotate about this axis. It is horizontal. This thing is horizontal. It can rotate about this fixed axis. The bullets mass is small m. All the bullets mass is small m. Which is very less compared to capital M. So basically before firing and after firing, the rod mass doesn't change. Shouldn't the bottom be towards the lamp? No. He just didn't want it. I don't want it to be like God told me that this is how it should be. Put it like this so that you mix layers. So now you have to find out the angular velocity of the entire system. Immediately after the bullet is fired. The bullets velocity of firing is v. All the bullets are fired with velocity v. Can I use conservation of angular momentum? Use. Yes. There is no matrix to me. About which axis can I use? About this axis, about the fixed axis itself. Axis force as well as mg force are applied passing through this axis itself. So net torque about axis is zero. So before the bullet was fired whatever the angular momentum will be equal to the angular momentum after the bullet is fired. What is the initial angular momentum of the system? Zero. Zero. It should be equal to final angular momentum. Add up all the angular momentum of the particles which are bullets and the rods. Equate that to zero. Angular momentum of a bullet is how much? m into v into l by 2. These two bullets angular momentum are in same direction. Same direction. What about that? Opposite. So total angular momentum of the bullets will be m v l by 2. Three bullets in the same direction minus m v into l by 2. This is the total angular momentum of the bullet. Yes or no? Okay. Angular momentum of the rigid body. How many rigid bodies are there? So icm which is m l square by 12 omega. Both the omegles of the rod will be same because they are welded as a single rigid body. So omega of entire rigid body becomes same. This will add up to zero. Final angular momentum is equal to initial angular momentum. Okay. This came in j-ad-1 2014. So in that was it this way or that way? It was j-ad-1. Got it? Any doubts? Anything? So if you wanted to find the net momentum of the system then you'll have to add the translation along. What do you mean by that? Like even the setup will move forward because both the two... No fixed axis. Extra force is there. You can't apply congestion momentum. Extra torque is not there. Extra force is there. Okay. Actually it's applying force, right? Yes. But torque because of that force is zero. So let's say this axis is not fixed. So it will start moving forward. So the question is if I put that and fired it the other way then both ways the change in momentum will be the same. Which? So like... Oh then it will not rotate. If this is like that then the sum of the linear momentum is zero. Yes. So it will not move. But then right now it will move. It will have some VCM. Yes. As well as omega. Okay. This... Find out if suppose axis is not fixed. If suppose axis is not fixed find out VCM in this case VCM and omega. Remember I can use conservation of linear momentum whether it is a point mass or a rigid body. If net external force is zero. Can I conserve linear momentum? Yes. No external force. Can I conserve angular momentum? External force. MG is there. MG force is there. Axis. The torque because of MG is zero. So about central mass axis I can always conserve the angular momentum. Okay. In fact torque because of MG is never about that axis. Whichever axis you take. Okay. Torque because of MG is trying to rotate like this. Sorry. It is trying to rotate like this. So horizontally the torque is there because of MG. Not in this direction. So you can literally use any axis. But it is always better to apply conservation of angular momentum about the center of mass axis or about the fixed axis. Your expressions will be simpler. Then VCM you got. So net angular momentum. Sorry, net linear momentum about the x axis should be zero because initially it was zero. So about the x axis you have 2M into V minus of that plus total mass of the rod 2M into Vx is equal to zero. This all of you got. So these two bullets are going in the negative x direction. So for that the linear momentum is minus of 2MV plus let's say that Vx is the velocity of center of mass along x axis. Plus total mass M plus M two rods are there 2M into Vx is equal to zero. This will give you Vx. So Vx is what? Vx is M by capital M times V. Vy will be what? You can see that along the y direction the momentum of this bullet cancels away from that. So momentum bullet is zero plus momentum of the rod equal to zero so Vy is zero. So it will only move in the x direction with Vx. But won't they move in a circle? Why a circle? If you constantly shoot bullets. If you constantly shoot bullets. Now it will not move in a circle. Yes sir because the horizontal rod will begin to rotate. But then with respect to this point it is moving in a circle. Otherwise it is a weird path. It goes forward as well as rotates. No sir but if this can also move then those things itself will shift so then it will start to do that. Not circular motion. Curved motion you can say. Sir but it will spend the same amount of time firing this way at each angle. So it will be a circle. Vx is this so the entire structure is moving forward as well as rotating with that omega which you have found out. So which is doing circular motion? Which point? That is only at this point. Once it moves a little bit now it will go in this direction. Correct. So what will happen is it will move little forward and this will happen. Yeah. That's all. But it is not a circle right? Sir but if your tangential velocity is the same and your angle. I know what you are saying. Yeah. But you should also understand what I am trying to say. With respect to the frame these are doing the circular motion. But if you are on the ground you will not see the frame doing circular motion. And anyways in the rigid body one point is always moving in a circle with respect to any other point. That is a generic statement. But if you talk about specific scenario here it is not a circular motion exactly. But it will be some sort of weird curve. Which point you are talking about? This is the central mass. Central mass goes straight. No circle. Yeah. Central mass velocity is Vx. Central mass doesn't rotate. Okay. Central mass will follow this path. It doesn't have Vy. Vy is zero always. Vx is this always. While the entire structure is moving forward the structure is rotating. Sir once it is 45 degrees. On the height of the center to the ground V like L by 2. You have not understood the problem. It is the horizontal. On the desk it is kept. And desk on the desk it is lying forward. Okay. Any other doubt? Conservation of angle of momentum along the y axis. That is only at this instant. At only at this instant. So at any instant the some of the angle some of the let's say momentum along y axis will be zero. Yes or no? Yes sir. Because there is no force between any instant later on and any instant now. Okay. So it will be always zero. Vy comes out to be zero. Initially and always. I am considering this is after time t whichever time you consider. And this zero is at equal to zero. So Vy is zero at any point in time. And Vx is this at any point in time. So the frame will move forward like this and rotate. Center of mass will go in a straight line. Okay. But with respect to center of mass everything is moving in a circle. Any other doubt? Come out to be same which you got earlier. Okay.