 Alright, so do this now, it's in a 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, alright. There is a solid sphere, right? Yes, solid sphere. When it is not mentioned, what is it, you have to assume it is a solid one, when it comes down, we have to find out the angular velocity, what is the angular velocity, omega, when it comes down, there is friction sufficient for the pure rolling, friction sufficient so that pure rolling happens, alright, there is no slipping, you want to shift it, not right now, inclined plane is at rest, moment of inertia about the center of mass for a sphere is 2 by 5 m r square, how you have done, is friction doing any work, friction is doing some work, how many of you say yes, somebody asked me doubt in the group, right, is friction doing, you asked, what did I say, in pure rolling case, friction doesn't do any work, okay, 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 0, but that doesn't mean that friction doesn't apply force, force is there to create acceleration and angular equation, but that doesn't do any work, what did you do, you got the acceleration, I see, I want you to apply work energy theorem, alright, you can solve it using torque equation also, but I want you to use work energy theorem, I want to see that. Sir, we assume the bottom to be 0 potential and potential at the top will be h plus r, that's what I have done, this is from center, yeah, so the h plus r, no, no, no, it is 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 0, it's mg minus r, yeah, center of mass is a height r above, yes or no, okay, so w is equal to u 2 plus k 2, this is the expression, alright, w is 0 plus k 2 is what, k 2 is half m v c m square i minus mg h plus 0, did you get the answer, yes, and since it is pure rolling, omega is equal to v c m by r, pure rolling on a fixed surface, that is why, alright, so when you substitute v c m is equal to omega r, now you can 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, okay, because you have to apply conservation of momentum. Sir, but in that case v c m, even if there is no slipping, v c m will be different from omega, right, yes, at the point of contact the velocity of the object will be equal to the velocity of the surface, yes, okay, both tangential direction as well as radial direction, guys focus here, this one, it is a disc, it is a disc mass m and radius r and this is a horizontal line which is at a distance of r by 2 below the center, got it, so if you drop a perpendicular from here on this chord, this distance is r by 2, getting it? This entire disc can rotate about this horizontal line, are you getting it? Yes. So it is like this, suppose this is a disc, it can rotate like this, are you getting it? It can rotate like this, okay, I know, so it is like this, disc can rotate about the horizontal axis, are you getting it? Understood. So it can flip over like that, so when it flips over, how it looks like? To listen here, listen to this axis, it nursed it a bit and it started rotating, you have to find out its angular velocity when it has flipped over completely, understood? Sir, it is horizontal and gravity makes it work. So you can use parallel axis still, no it is MR screw by 4 glass MR screw by 4 glass, I don't remember the answer. But I do remember this came in J 1998. This is easy. This too comes like a minute. See it is supposed to be a very tricky question. I am telling you. Yeah, I don't. Alright. Somebody is teaching so well. I get all the credit. Alright, so focus here. Centre of mass was here earlier. Now it has gone there from 0.1 to 0.2. I have to apply the 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 mg. This distance is r by 2 plus r by 2, r. So mg, r. K1 is 0 plus K2 is half moment of energy about the fixed axis into omega square. This is your K2. If I find moment of energy about this axis, I have to apply. So about this will be mR square by 4. So mR square by 4 is your ICM, 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 square by 2. Okay, like this. Understood. Fine. Let it out. Find out the total force the axis applies on the disk at this moment when it is at 0.2. Sir, the what? Total force the axis applies on the disk when the disk is at 0.2. Sir, here we have to use torque. If we to use whatever. How much omega comes by the way? Sir, under root 4G by r. 4G by r. 4G by r. How much alpha you are getting? Alpha is equal to? Sir, alpha is equal to? Alpha about that axis is how much? Anybody got alpha? Sir, I got the force really. Alpha, alpha. How much is alpha? Isn't 0? Alpha is 0? Yes. The only force is mg that passes through the axis. That's the axis of perpendicular distance is 0. Okay, so torque due to mg is 0. So alpha is 0. So there is no tangential acceleration. Which direction? Yeah. No, it's towards the axis. From the radius towards the axis. It is like this. How much is this acceleration? Omega square? Omega square r by 2. r by 2 is perpendicular. The center is moving at the radius of r by 2. You have to see center's motion only. So that one is 2. So omega square r by 2. All of you? Yes. Stop talking. 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 n 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. Omega square r by 2 is how much? 2 mg. So the omega square r by 2 is how much I am asking? 2 mg. 2 mg? So this is 3 mg. So the axis applies a force of 3 mg at this moment. Clear? Shall I proceed? Write down angular momentum. We discussed impulse in which chapter? What's the most important? Collisions came in work-pinage, right? Yeah. So angular momentum and angular impulse both will do together. Remember linear momentum, mathematical expression for linear momentum. What it is? Momentum is what? Fourth element of momentum. What do you think angular momentum is? Amount of angular motion. Amount of angular motion. Linear momentum is amount of linear motion and angular momentum is amount of angular motion. Simple, right? It is going in a straight line. Does it have any angular motion in this scenario? I have asked you so many great questions. Every time I ask something like, oh, this might be something. Alright, so here is Param. Right in the clip. This is Ruchir trying to save Param. Thank you. So Param, first checks what is the depth of this by dropping a ball. He's trying to see how far it is. Now, when Param is seeing it from here, looking it straight like that, but Ruchir was not actually caring for Param. He was actually caring for the ball. So he's looking at the ball which he has dropped. Every time he looks at it, he has to turn his head for the Ruchir, but not for the Param. Ruchir will say that it has angular momentum as well as linear momentum. Param will say no. It depends on about which axis you are looking at it. You have to quantify it. 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's say this is the velocity. If this perpendicular distance becomes zero, the angular momentum should become zero. Yes or no? It increases. The angular momentum also increases. The angular momentum is basically perpendicular distance into mass time velocity. We are not getting too much of detail into it. We are directly writing the expression. This is the angular momentum of a point object. Please write down in brackets for the point of mass. So perpendicular distance from the axis. Yes, from wherever should you find a torque? What you should ask back about which axis? So whenever I am asking you, find the angular momentum. You should ask me back about which axis or point you are asking the angular momentum. So angular momentum about an axis which is perpendicular distance r from what? It can be clockwise or anticlockwise. Right now it is in which direction? Clockwise or anticlockwise? Clockwise. The angle is increasing like this. So it is trying to rotate like that. So it is clockwise. In a 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. One more thing. Angular momentum is magnitude of angular momentum is r perpendicular mass into velocity. Magnitude of 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 perpendicular to tangential force is what? It is a torque. It is torque. At which the angular momentum changes for a point mass is equal to torque about that axis. So if you have that case that we had before, let's say I had something moving with 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. I have an axis like this. You have some object going like this. Yes. So this is the perpendicular distance. The perpendicular distance is fixed? No, no sir. It's an axis in the z dimension. This is like this. Now this is the axis. No sir. It's not in the same dimension. Yeah, yeah. 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. The perpendicular distance is fixed, no? Yeah, yeah. All right? Yeah. So angular momentum is there, but it is constant. Yeah. All right. Please write down. No sir. I'm standing somewhere. I look at somewhere in front of me. I kick a football. I have to turn my head to see the football goal. So it has angular momentum. But there is no external force after you kick, etc. Yeah, no external force. Angle momentum, it is like momentum can be there if velocity of object is there. But force will be there only when momentum changes. Something like that. You can claim that there is a torque. It should change with time. Art changes, sir. If art changes because someone is working in a state. 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.