 Do anyone have that paper you have written on Monday? No, no, no, like he sent it to us on Monday. No, the hard copy he has given to us. No, no, no. I told you to find it. See, he has it. You don't like the paper, is it? Fan, switch off the fan. Switch on the fan? You don't like the fan at all? No, the fan versus AC. Oh, they get on the cold. Oh, so they are strategizing it. Fine, let us take one question from this. Put your bag on the desk. We'll take a question on the previous class concept and then we'll start the class, the new concept. Let us see whether you are well versed in this. Give it back. There is this rod. Stop talking. There is this rod. The rod mass is capital M and the length is L. There is another mass, a point mass, that is fixed at the end of the rod with mass small m. You are, this rod acquires some angular velocity. By the way, the rod is free to rotate about this axis. It can swing like this in a vertical plane. Fine, so you are giving some angular velocity so that one end of the rod acquires a velocity v0. You have to find out what is the value of v0 so that the rod just becomes horizontal. So the masses are different, right? Small m capital M. You have to write down this work method theorem. W is equal to u2 plus k2 minus u1 plus k1. So the work done is 0. Stop talking. Because k2 will be equal to u1 plus k1 as work done is 0. Sir, but there is displacement. Is work done 0 or not? 0 becomes 0. Which force will do work? Oh, minus mg of work done. See, mg is doing work, no doubt about that. But for mg you are considering potential energy. If you are considering potential energy for a force, you don't need to find a work done for that force. And there are forces from the hinge, but the point of application doesn't move of the hinge. So the work done by the hinge forces is 0. Between this position and that position apply this theorem. This question is very important. The hinge is the rod plus small m can be treated as a single rigid body. Assume this line to be 0 potential energy. It will be better than that. This line is what? u2. u2 is 0, let us say. U2 is 0. What about k2? k2 is also 0. It just goes to that position. Both u2 and k2 are 0. Let's try to write down the kinetic energy of this rod along with this mass. So you can write down the kinetic energy in different ways. Given you can write it as half moment of inertia of the rod about the fixed axis into omega square plus half m into v0 square. So you have written the kinetic energy of the rod plus the kinetic energy of the mass. You are writing it separately. And moment of inertia of the fixed axis is how much? mx square by 3. So half L square by 3 omega is what? v0 by L. This is the kinetic energy. Another way to write the kinetic energy is to find the moment of inertia about this fixed axis for the entire rigid body. For the entire rigid body moment of inertia is what? mL square by 3 plus small m into L square. So kinetic energy will be mL square by 3 plus mL square into omega square. This will be the total kinetic energy. This is k1. What about u1? U potential energy it is better to split. Otherwise you have to find out the centre of mass location for the combined system. If you can find out no problem. The potential energy will be m plus capital M into g into height of the centre of mass. Or you can split it into two parts. Potential energy of the rod plus potential energy for the small mass. This will be equal to what? Negative of capital Mg L by 2 it is down. The centre of mass of the rod is down. And negative of MgL. This is u1. This is k1. u2 and k2 substitute there. You get the answer. Any doubts? I did not know that we have can you show the notes what we have done at the end of the last class. We just started the energy. And we have written kinetic energy about the fixed axis as well as rotation plus translation. Right. So we have just introduced what is the kinetic energy and potential energy of our body. So we are continuing solving problems right now. So we have learnt that kinetic energy of a rigid body is half moment of inertia about the fixed axis into omega square. Fine. Which omega about which axis? Fixed axis. About the centre of mass. Or the fixed one. Which one? Omega. Omega is same for entire rigid body. That is why it is called the rigid body. Fine. If omega of one point is different from other point it will no longer remain a rigid body. It will distort. Okay. So this is the kinetic energy. If you are not able to find the fixed axis no problem you can write it like this. Half m into vcm square plus half moment of inertia about centre of mass into omega square. Two ways you can write the kinetic energy both ways you get the same answer for the kinetic energy. Alright. Let me sit here somewhere. All the four sides are occupied though. Okay. Potential energy mg into height of the centre of mass. Okay. W is equal to u2 plus k2 minus u1 plus k1 or you can remember it like this also. u1 plus k1 plus w is equal to u2 plus k2. So you can do some amount of work to increase the initial mechanical energy. But if you are doing the negative work your initial mechanical energy will reduce to u2 plus k2. Fine. So both ways you can remember take few questions now which will test you on these fundamentals. I think there should not be any confusion. We have done lot of questions in work by energy chapter already. Okay. Now we are going to use work by energy concept for a rigid body. It is exactly the same. The only difference is the energy. That is the only difference. So keep it simple in your head. It is very easy to complicate. A simple will take simple scenarios and we will bid on it. And length L. You have to find out the angle velocity when it has rotated by an angle of theta. It is rotating in a vertical anybody else absent today? To see where the center of mass is initially and where it is finally. The mass is moving like this. It reaches here. It travels vertically at distance of how much? This is L by 2. So this will be? L by 2 sin theta. L by 2 sin theta. This position and that position. Fine. So step number 1, write down w is equal to u2 plus e2 minus u1 plus k1 and then substitute. Work done is 0. Final potential energy is what? Initial potential energy. Where should you take initial potential energy? So my u2 becomes? Sorry u1 becomes 0. k1 is also 0. It starts from rest. So u2 will be minus of ngl by 2 sin theta plus kinetic energy will be fixed axis. So I can direct moment of inertia about a fixed axis. ml square by 3 into omega square. This is equal to 0. So from here I will get the value of omega. Theta by L. Under root of 3g sin theta? No. 3g sin theta by L. Time t. How much it will rotate? Then what you will do? For angle theta after time t. d theta by d theta. You write omega as d theta by and then bring here 1 by root sin theta d theta you have to integrate now. And that side will be dt. If you are able to integrate 1 by root sin theta you will get the answer. End of question we have done. Same problem we have done. We got the angle velocity then also. Calculated angle was checked whether it is the same thing. Torque we have calculated. We got alpha and then integrated alpha. You have got the same thing right? Now it becomes? 95. 3g cos theta is equal to omega d omega by d theta. And 3g by 2 L sin theta. And then you get omega is this. So you get the same answer same way here as well. Find out what will be the reaction force of the hinge. Did we do that? So quickly we I tell you. Let's say this is reaction forces. Let's say this is f y and this is f x. So f x will be what? f x will be equal to m omega square and f y will be equal to m alpha into L by 2. Because alpha into L by 2 is the tangential acceleration, radial acceleration. Why L by 2? Why not? L full L. Center of mass you could track. Center of mass is at a distance of L by 2. Entire rod behaves as if it is a point mass located at the center of mass. So what's the acceleration of the center of mass? How do you find that? Do you have alpha? Right? So alpha into L by 2 will be the tangential acceleration. There is a circle, right? With angular acceleration alpha. So alpha into L by 2 is tangential acceleration. We just differentiate omega. Differentiate or you directly have this alpha, no? Sir, I just understand why is it L by 2 with the center of mass. You have to see net fourth is equal to the acceleration of center of mass. Yes sir, but how did we get that? Using torque equation we have done this. Torque into I alpha. Right? Mg is a fourth. So Mg sin theta is perpendicular component. Mg sin theta into L by 2 is I alpha. Yes. You get alpha. We learn the torque equation. Then you can think of valence and forces. Alright, so do this now. It is in the sphere of mass M and radius R. Okay? Sphere of mass M and radius R. This angle of inclination is theta. From center the height of the ground is edge. Okay? There is a solid sphere, right? Yes, solid sphere. It is not mentioned what I said. You have to assume it is a solid one. Okay? When it comes down we have to find out the angular velocity. What is the angular velocity omega R when it comes down? There is friction sufficient for the pure rolling. Friction is sufficient so that pure rolling happens. Alright? There is no slipping. So it can't be measured? You want to shift it? No. Not right now. Incline plate is at rest. Moment of inertia about the center of mass for a sphere is 2 by 5 M R squared. Is friction doing any work? Yes. Friction is doing some work. How many of you say yes? Somebody asked me doubt in the group, right? You asked. I asked. In pure rolling case friction doesn't do any work. Please write it down in box. If it is pure rolling there is no slipping between the two surfaces. Right? Relatively they are not moving. So work done by the friction is zero. But that doesn't mean that friction doesn't apply force. Force is there to create acceleration and angular friction. But that doesn't do any work. What did you do? You caught the acceleration. I see I want you to apply work energy theorem. Alright? You can solve it using talking question also. But I want you to use work energy theorem. I want to see that. Now under root 5 G is by R. 5 R is by R. 5 R is by R. 5 R is by R. 5 R is by R. 5 R is by R. 5 R is by R. 5 R is by R. 5 R is by R. 5 R is by R. Sir. Sir, we assume the bottom to be zero potential and potential at the top of the H plus R is by R. That's why I took it down. Yeah, so the H plus R is just H to the center. Yeah, but in the end it's R. Alright, so almost all of you did not get it. The final potential is not zero. It's MG minus R. Central mass is a high R above. Yes or no? So W is equal to U2 plus K2. This is the expression? K2 is what? K2 is half m vcm squared and gh plus zero. Did you get this? Yes. And since it is pure rolling omega is equal to vcm by R. Pure rolling on a fixed surface. That is why. Alright, so when you substitute vcm is equal to omega solve it to get the value of omega when it comes down the incline. Sir, if you think itself is moving then we are going to do that after we learn angular momentum and momentum. Because you have to apply condition momentum. So in that case vcm, even if there is no slipping, vcm will be different from omega. Yes. At the point of contact the velocity of the object will be equal to the velocity of the surface. Yes. Both tangential direction as well as radial direction. Guys, focus here. It is a disk. It is a disk mass m and radius r. And this is a horizontal line which is at a distance of r by 2 below the center. So if you drop a perpendicular from here on this chord this distance is r by 2. This entire disk can rotate about this horizontal line. Are you getting it? So it is like this that suppose this is a disk it can rotate like this. Are you getting it? It can rotate like this. I know. So it is like this. This can rotate about the horizontal axis. Are you getting it? Understood. So it can flip over like that. And it started rotating. You have to find out its angular velocity when it has flipped over completely. Understood. So it is horizontal like gravity makes it work? Yes. Also so it flip. So we just need to find angular velocity. So we have got it. It is under 4G. Yes. It is under 4G. Yes. It is under 4G. So you can use parallel axis still. So you can use parallel axis still. No, it is MR square by full glass MR square by full glass. Yes. We just take this. I don't remember the answer. But I do remember this came in J 1998. This is easy. This took us like a minute. It is supposed to be a very tricky question. But would you pass it? I am telling you now. You got it. Yeah, I don't. Somebody is teaching so well. I get all the credit. So focus here. Center of mass was here earlier. Now it has gone there from 0.1 to 0.2. I have to apply the conservation of potential energy sorry, conservation of mechanical energy between 1 and 2. The same thing W is equal to U2 plus K2 minus U1 plus K1. You can assume point number 2 to be 0 potential energy. So U2 becomes 0. U1 will be mg. 2 plus R by 2, R. So mg, R. K1 is 0. K2 is half moment of energy about the fixed axis into omega square. This is your K2. How do you find momentum about this axis? I have to apply parallel axis. Right. So about this axis I know it is MR square by 2. So about this will be MR square by 4. So this is the ICN which is parallel to that axis. You have to shift it M into D square which is M into R by 2 whole square. So this will be like this. Understood. And without total force the axis applies on the disc at this moment when it is at 0.2. Total force the axis applies on the disc when the disc is at 0.2. You have to use whatever how much omega comes by the way? 4G by R. How much alpha you are getting? Alpha is equal to Alpha about that axis is so much. Anybody got alpha? I got the force really. Alpha. How much is alpha? Isn't 0? Alpha is 0? Yes. The only force is mg that passes through the axis touches the axis of perpendicular distance is 0. So torque due to mg is 0 so alpha is 0. So there is no tangential acceleration. But is there a radial acceleration? Yes. Which direction? Yeah. No it's towards the axis from radius towards the axis. It is like this. How much is this acceleration? Omega square? Omega square R by 2. R by 2. See center is moving at a radius of R by 2. You have to see center is motion only. So that one is 2. Omega square R by 2. Are you getting it? All of you? Yes. Stop talking. So there will be a force from the axis in this direction let's say that force is n and there will be a force in downward direction which will be mg. So next force is 2 minus mg. This should be equal to mass time acceleration which is omega square R by 2. So you get n. Yeah. mg plus 2 mg. Yeah. Omega square R by 2 is how much? 2 mg. 3 mg. Omega square R by 2 is how much I am asking? 2 mg. 2 mg. So the axis applies a force of 3 mg at this moment. Clear? No. An angular momentum. Because impulse in which chapter? Collisions came in work finance. So angular momentum and angular impulse both will do together. Okay? Remember linear is a mathematical expression for linear momentum. What it is? Momentum is what? Force is a flow momentum. Yeah amount of floating. Amount of angular motion. Linear momentum is amount of linear motion and angular momentum is amount of angular motion. Simple. If an object is going in a straight line does it have any angular momentum? No. Does it have? Sometimes every time I ask something like oh this might be something. Alright so here is Param right in the on the clip. Right? This is Ruchir trying to save Param. Thank you. So Param first checks what is the depth of this by dropping a ball. He is trying to see how far it is. Now when Param is seeing it from here he is looking it straight but Ruchir was not actually caring for Param. He was actually caring for the ball. So he is looking at the ball which he has dropped every time he looks at it he has to turn his head. See for the ball for the Ruchir it has angular momentum as well as linear momentum Param will say no. Fine? How about which axis you are looking at it? Alright? So quantification should be in such a way that if the distance of the perpendicular distance let us say this is the perpendicular distance and let us say this is the velocity. If this perpendicular distance becomes zero the angular momentum should become zero. Yes or no? increases. The angular momentum also increases into mass time velocity. Alright? We are not getting too much of detail into it. We are directly writing the expression. This is the angular momentum of the rectangular bracket for the point of mass. So for particular distance from the axis Yes. From wherever somebody asks you to find a torque what you should ask back about which axis? So whenever I am asking you to find the angular momentum the axis of point you are asking the angular momentum. Simple. Fine? So angular momentum about an axis which is perpendicular distance are from what? clockwise or anticlockwise? Right now it is in which direction clockwise or anticlockwise? Clockwise? So it is the angle is increasing like this. So it is trying to rotate like that. So it is clockwise. In vector rotation form angular momentum is written as r cross p. p is your linear momentum. Point mass. Let us now try to find out the angular momentum of a rigid body. Magnitude of angular momentum is r perpendicular mass into velocity. So dL by dt perpendicular is let's say constant. This will be m into dV by dt. Magnitude of velocity will change only because of the tangential acceleration. So this will be r perpendicular into mass into tangential acceleration. Yes or no? So mass into tangential acceleration is what? Tangential force. r perpendicular to tangential force is what? It is a torque momentum changes for a point mass is equal to torque about that axis. So if you have like that case that we had before you just had something like let's say a fixed velocity about some axis then there will be a change in angular momentum. So there will be a torque on it across that axis. But there is no external force on it. How can you say there is no external force? I just have some line and I have an axis like this. You have some object going like this. So this is the perpendicular distance perpendicular distance is fixed? No, no sir. It's an axis in the z dimension. This is an axis. No sir. It's not in the same dimension. This is your axis. This is your line of velocity. Your perpendicular distance is not changing. Your linear momentum is also not changing. So your angular momentum is also constant. It is not changing. So there is no torque. So perpendicular distance is fixed? Yes. So angular momentum is there but it is constant. Please write down. Sir, if I am standing somewhere I look at somewhere in front of me I kick a football. So it has angular momentum. But there is no external force after you kick it. It is like momentum can be there if velocity of object is there. But force will be there only when it changes. Something like that. Changes you can claim that there is a torque. It should change with time. Art changes then? Then also there can be momentum. There can be torque. If perpendicular distance itself changes it means it is not going in a straight line. Write down. For a rigid body we are trying to find the angular momentum. Fixed axis is translation. So if the entire rigid body is moving with the same velocity it is not rotating. It is as good as a point mass is 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. This is your rigid body. And this rigid body Stop talking. 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. Fine? So I am giving you hints. You derive yourself just like we have derived the kinetic ratio for rigid body about a fixed axis. Same way, please do that. Let's say this is point mass m1 at a distance 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. I am omega. I am the fixed axis. I am omega. How many of you got that? Let us do this. The velocity of m1 in tangential direction will be how much? This velocity will be omega into r1. So the of point mass m1 is r1 into this will come out to be r1 is square omega. Similarly l2 will come out to be m2 i2 square omega. So if you add it up you will write total... No. So that is why I am putting you there. So if you talk next then what will... No, no, no. Something extra should be there. Right? What should we do? We should talk. We will sit here. Without... Right? What kind of punishments you get in school? Get out of the class. We don't. Yeah, they actually don't give us punishments. Is he? What? No punishment. My teacher used to hit me. No punishments. That's so... You will not have any stories to tell. Anyways. How should I punish? I am in a jail. That I will not do. Innovate. Embarrass. Being sarcastic. Then what? Shouting the routine thing. Then what else? Stand on the chair. With wheels on it. Or stand here and what are you talking? Explain to everyone. You are smiling. Come here and let everyone smile. Like that. Fine. So if you add up all the angular momentum, you get that total angular momentum of a rigid body. Fine. And when you talk about that total angular momentum of a rigid body, you use capital L for it. So capital L is summation of times omega. Right? Omega is constant, so comes out. This will be equal to i about fixed axis into omega. Angular momentum about the fixed axis. It is 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 alpha. Which is equal to? Talk. Getting it? Okay. Force is equal to 8 at which angular momentum changes. F is equal to dp by dt. That is more basic or f is equal to I mean to a. It is dp by dt. dp by dt. Even mass can change. There will be a force. V dm by dt is also force. Similarly here, torque is also equal to omega dI by dt. It can be omega into dI by dt also. Getting it? Case number one, fixed axis. Now we are trying to derive for the translational axis. Ignore that. It will be equal to icm into omega centre of mass cross m into vcm. That is total angular momentum. This is about which axis? This is the rigid body. It is a rigid body going with vcm 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 RCM vector. Getting it? Which one? RCM is a position vector of the centre of mass with respect to a point about which you are finding the angular momentum. Usually application of this will not be there. Even if it is there, you can always have momentum with respect to centre of mass. This will be how much? RCM becomes what? Zero. The position vector of centre of mass with respect to itself is zero. Moment axis always no matter what the rigid body is doing is icm omega. Sometimes centre of mass axis is your fixed axis also. Full mathematical derivation of this also. That is of no use. Sir, Mr. Kailam, what about the z axis if we just take icm omega? Yes. And it is a vector quantity. It is a vector quantity. Next topic, please write down. See, 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 it is equal to rate at which angular momentum changes for a rigid body. It is very similar to force equals to rate at which the linear momentum changes for a point of mass. These equations are similar. Fine. 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. Impulse. Leader impulse denoted by what letter? J. So J vector is equal to change of inner momentum. 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. 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 of this concept. Impulse from here. This is the impulse J. Impulse is given. You need to find out velocity of center of mass and omega stationary on the horizontal surface. Mass m and length i uniform dot. This is equal to rate of change of linear momentum over here. Sorry, it is not rate of change of linear momentum over here. So because it can translate and rotate. What are you have learnt for the center of mass? What are you have learnt with respect to linear impulse and everything? Everything is valid with respect to center 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? It is a but isn't J equal to m vcm plus i omega? No. Letting it J is equal to change of linear momentum if it is a rigid body for the center of mass. Simple. Now we have to also apply the angular impulse is equal to change of the right? Now is there fixed axis? No. No fixed axis? 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. But if you write about any random point writing angular momentum becomes difficult. It is a plus rcm into mvcm but if you write angular impulse with respect to center of mass it will be simply equal to icm omega. Yes or no? So I will calculate angular momentum with respect to center 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? The angular impulse is integral of tau dt tau is what? This is equal to l by 2 integral f dt which is l by 2 into J. Same thing. J into l by 2 is equal to the angular momentum change with respect to center of mass axis. So this will be what? ml square by 12 into omega. So you write omega also here. What do you get? 6 6 J by ml. Is it comfortable with this concept? Okay. So find out the point which is at rest immediately after impulse is given. That. 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. No, no, no. No. It is not that. What is the condition of having a point at rest? 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 1 by 6 from 1 by 6 from? Left and side. Right. Right. Right of center. Stand up. Stand up. So it is not that. No, actually it is not. Every criminal say that I haven't done it. VCM is like that. So if you take a point over here which is at a distance of d, what will be its total velocity? Omega into d like this. This is omega VCM. Yes or no? It rests 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. You have to find out a point where VCM minus omega y becomes 0. So y will be what? VCM by omega. Omega. Pro VCM by omega. You get l by 6. So a point which is l by 6 on the left-hand side will be addressed immediately after collision. Make collision the impulse. By the way impulse comes off collision. Getting it? This is slightly tricky concept but it requires little bit of problem practice and then refine with that. Not lot of varieties will be there with angular impulse. So you can just solve couple of numericals and refine. So this is the angular impulse now write down conservation of angular momentum. Just like conservation on linear momentum there is something called conservation of angular momentum. If net torque if net external torque on net external torque on a system of particles and rigid body if net external torque on a system is 0 the angular momentum of the system will be conserved. The angular moment of the system will be conserved. So if I add up all the angular momentum it will not change if net torque is 0. 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 0. And then about that axis initial angular momentum will be equal to final angular momentum. Getting it? So basically the total angular momentum of entire system lets 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 it 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. You will get net external torque. External torque is equal to rate at which angular momentum changes about that axis. Fine? And if SNL torque is 0, rate of change of net angular momentum about that axis is 0. So sum of the angular momentum before will be equal to the angular momentum after. Simple derivation, the way we have done for the linear momentum in doubt. Now let's discuss couple of scenarios where the angular momentum is there in our day to day life. Okay? You have seen, have you ever like spread your hands like that and rotate it? I did, yeah. No sir, like I let this thing in 6th grade or something and then it said that if you take two really big textbooks and want a swivel chair and you start spinning and then you pull it together then you spin faster. Have you ever done like spinning with your hands open like this and while spinning you do like that the angular velocity increases? Anyone into dancing? Everybody rotates? Everybody rotates? So if you have to increase angular velocity you decrease the momentum in Asia? Which dance? I know. Which dance? In classical and chakras you take your hands inside once you start spinning. See all the dance dance, 50% dance is spinning only about the fixed axis. And salsa as you see every I mean 90% of the time So that's translation, classification I mean yeah most of the time it will be the guy will be like this and the girl is spinning like that What are you talking about? What are you talking about? What are you talking about? What are you talking about? No sir, like prose doing salsa I am talking about you doing salsa You do it like this No but it's like they do something and the girl goes spinning to some other guy to take a dog Oh, while spinning she goes It's too much And See if you have not done this go home and do it Not a lot of space here Space would have been there if I had asked everyone to do one by one She can do it What? You can do it. Yeah, just come here Why you are telling him to do So what do you do? I can't because I don't have a spinning chair You don't need a spinning chair You can just do it I don't know whether I am putting the same amount of force there It's no integrity of the girl Who cares about physics then Alright, listen here Another example is, have you seen 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 Right? So because of that when he or she folds herself angular velocity increases 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? Try not to do that I have heard many stories where not many, couple of stories that someone jumped and fall on the stomach Very nice Surface tension will not let you go inside very easily What happened? You are telling your story? No, no stories to share Nothing Nothing So they 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 What is happening? There is this other really cool thing It is called the middle axis theorem in which there is this Russian guy So he had this T screen Let me complete Where did the middle axis come from? We are 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 The disk has some angular momentum The spaceship was zero When the disk rotates it generates some angular momentum So the spaceship slightly will rotate The momentum initially was zero Finally should also be zero The spaceship should rotate in opposite direction So there they found out the conservation of angular momentum properly Even if these 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 travelling 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? Two rods are welded The mass of the rods and the length of the rods At the end there are guns and guns have bullets inside it inside 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 Getting it? 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 No I don't want it to be like God said I will put it like this so that you mix layers So now you have to find out the angular velocity of entire system immediately after 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? Can I use? Yes, there is no net 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 0 So before the bullet was fired whatever the angular momentum will be equal to the angular momentum after the bullet is fired Notice the momentum of the particles which are bullets and the rods Equate that to 0 How much? m into V into L by 2 These two bullets angular momentum are in same direction What about that? So total angular momentum of the bullets will be m V L by 2 3 bullets in the same direction minus m V into L by 2 This is the total angular momentum of the rigid body How many rigid bodies are there? ICM which is m L square by 12 both the omegles of the rods 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 0 final angular momentum is equal to initial angular momentum This came in J-advance 2014-2013 So that was it this way or that way? It was this way We are doing a lot better Got it? Any doubts? Anything? If you wanted to find the net momentum of this system then you will have to add the translational law What do you mean by that? Even this setup will move forward because both the two... No fixed axis External force is there except torque is not there external force is there Actually it's applying force but torque because of that force is 0 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? Then it will not rotate If this is like that then some of the linear momentum is 0 But it will be more angular But then right now it will move It will have some vc as well as omega As well as omega 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 0 Because it constantly follows its own axis and it moves forward both ways Comfortable Can I conserve linear momentum? Yes, no external force Can I conserve angular momentum? Torque because of mg is 0 About central mass axis I can always conserve the angular momentum In fact torque because of mg is never about that axis whichever axis you take Torque because of mg is trying to rotate like this Sorry 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 Done vcm you got So net angular momentum Sorry, net linear momentum about the x axis should be 0 because initially it was 0 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 0 All of you got So these two bullets are going in the same direction So for that the linear momentum is minus of 2m v plus let's say that vx is the velocity of center of mass along x axis plus total mass m plus m 2 rods are there 2m into vx is equal to 0 This will give you vx So vx is what? vx is m by right direction the momentum of this bullet cancels away from that So momentum bullet is 0 plus momentum of the rod equal to 0 so vy is 0 So it will only move in the acceleration with vx Fine Why is that? If you constantly shoot bullets No, it will not move in a circle Yes sir because the horizontal 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 rotate 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 curve motion you can say It will spend the same amount of time firing this way at each angle So it will be a circle vx is this The entire structure is moving forward as well as rotating with that omega which you have found Yes sir So which is doing circular motion which point? This point Correct so it will what will happen is it will move little forward and this will happen Yes But it is not a circle right? If your tangential velocity is the same and you are I know what you are saying 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 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 Central mass velocity is vx Central mass does not rotate ok Central mass will follow this path It does not have vy No no no vy is 0 always vx is this always While the entire structure is moving forward the structure is rotating Why 45 degrees On the height of the center to the ground p You have not understood the problem It is the horizontal on the desk it is kept and desk on the desk will slide forward Ok any of the doubt I do not know why vy The conservation of angular momentum along the y axis That is only at this instant That is only at this instant Only at this instant So at any instant the sum of the some of the let us say Momentum along y axis will be 0 Yes or no Because there is no force between any instant later on and any instant now So it will be always 0 vy comes out to be 0 Initially and always I am considering this is after time t whichever time you consider and this 0 is at equal to 0 So vy is 0 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 ok So center of mass everything is moving in a circle same which you got earlier ok Next question It is a rod of mass m and length l There is a bullet that comes from here bullet of small m comes from this side it comes hits the this rod comes out from the other side Should I simplify this or it is ok no Simplification bullet does not come out it gets embedded inside Ok bullet becomes a part of the rod it does not come out alright it got embedded inside the bullet embedded inside the rod of mass m and length l This rod along with the bullet swings by an angle of theta of mass It gets shown by an angle theta only You need to find the velocity of the bullet So where does it hit Sorry Where does it hit This is l by 4 l by 5 color So there is small and insignificant comparison to capital L No No it will move still move velocity could be high Anyways do it now Just one question Can I use consumption of energy between this point and that point No This collision loss of energy Once it collided Once it collided after that this point or that point you can use between this point and that point Because the collision some energy will be lost It is an inelastic collision of linear momentum just before and after collision I cannot use because there is a force generated at the hinge So you just find i and n I can use Because there is a linear momentum No because there is an access force from this angular momentum I can use about which access about this access reason there is no torque about that just before and after collision All of you here I will use conservation of angular momentum to find the angular velocity just after the collision and from that point onwards I use conservation of energy What is the angular momentum of the bullet m into u 4 l by 5 perpendicular distance Yes or no This should be equal to initial angular momentum should be equal to the final What is the moment of inertia about the fixed axis capital ML square by 3 plus small m into Like I got all of them The answer is like Have you got this one I am asking only that And anybody else got this equation What fixed axis into omega is the total angular momentum of the rigid body Now rigid body has bullet also a part of it That is I have to add its moment of inertia Ok, so from here with omega and once you get omega forget about the fact that collision has happened Now it is conservation of mechanical energy So which you can write like this w is equal to u2 0 I can assume this to be my 0 potential energy Potential energy Initially See initial critical energy is what I into omega square where I is this u1 is what You can write down the potential energy of the bullet and the rod separately So you can say For the rod it is minus of m g l by 2 plus for the bullet minus of m g into 4 l by 5 Understood and then k2 is what k2 is 0 u2 I have to write now u2 is the final potential energy for the capital m it will be this distance which will be what l by 2 cos theta So m into g into by 2 cos theta plus minus potential energy is negative m g into 4 l by 5 and how to solve this rather than just noting down So guess what the chapter is over and now I can give you the mixed question multiple concepts together Whenever you solve problem you know almost every time if you solve properly you will feel that oh this concept was not done with this little thing was not told So like that there can be many different things and these small small things are your own learning when you solve problems Do not rely only on the theory which is taught in a class I spent around a month teaching this chapter still there are a lot of small small little things which you will understand when you solve problems and it's not that we are not solving problems we are solving problems but in the size of a chapter is as big as combining entire kinematics 1d 2d, laws of motion and bhoppa energy all these 4 chapters together the size of that is the size of this single chapter So we can't do each and every variety of the problem in the class even if we spend one month like what we have done but then it is up to you to practice that Fine All right let's start we'll start with the simple ones and then go to the complicated ones rather than directly jump into it There is this rod of mass capital M and length L Sir, did we have break? After this Is break time has come? 6.30 6.30 is a break time? 6.50 6 is a break time? 6.30 Is the official break time? No Chapter will be done after we are done with this Now is a good time for the break We'll take 10 minutes break 10 minutes break Then we'll take much bigger break I know Go and take 10 minutes break and come back and don't make noise because there is a class early night