 Okay, so today we are going to take up rigid body motion previous your questions. Okay. Now rigid body motion is a vast, vast topic. Okay. So at times people, they consider center of mass concept and moment of inertia, all of that also part of rigid body motion. So there is nothing wrong with it. Okay. But in today's class, the focus is more towards, you know, the torque equation and angular momentum and work energy theorem related to rigid body. Okay. So that is what the main focus is. So I will quickly just list down the formulas that we are going to use. All right. So first of all, the torque is the cause of the rotation. The equation torque is equal to i alpha. Okay. Now this equation is valid only about fixed axis. Okay. So torque is equal to i alpha is about the fixed axis or about center of mass axis. Okay. Now, if you change the axis, things will change for torque as in torque will be different about different accesses. Okay. And moment of inertia will be also different about different accesses, but alpha is same. Okay. For entire rigid body, alpha is something which is a constant. So that you should take care of. Okay. Then, you know, at times you have to calculate moment of inertia as well. So you should be aware about parallel axis theorem, which is i is equal to ICM plus total mass of the rigid body into d square. Okay. These two accesses should be parallel. All right. So these are the two equations. And then you have for a planar object, IZ is equal to IX plus IY. IZ is moment of inertia, which is perpendicular to the surface of the object. Fine. So these things we should be taking care of. And when you calculate torque, it is not same as when you calculate force. Like when you calculate force, you just try to see which direction the force is acting and you just add up all the forces. But torque is little tricky to find for torque. There is a formula that r cross f is the torque. Okay. Now you say that torque will do i alpha and then torque is equal to r cross f. Okay. See, this is what the torque is. Okay. And i alpha is the effect of the torque. Fine. So cause is r cross f. You can equate the cause with the effect, which is i alpha. Okay. Now for every force, you need to find out the torque separately. Okay. And the torque could rotate the object in clockwise or anticlockwise direction. So the torque which is rotated in which is trying to rotate in one direction need to be subtracted from the torque that is trying to rotate the object in other direction. Okay. Now this is the vector representation. You can also simply find out the torque. You can also simply find out the torque by finding the perpendicular distance of the force from the axis. Okay. For example, if you have a rod like this and there is a force. Okay. This is the force and suppose this is the axis of rotation. Okay. Then how will you find perpendicular distance from the force? You just draw or you just extend the force line and then you drop a perpendicular on this force. Okay. So when you drop the perpendicular on the force, this is the perpendicular distance. Fine. So torque due to this force is this perpendicular distance into force. Okay. You can also find the torque by finding the component of force which is perpendicular to the length. For example, this is the force. You find its component along this line which is perpendicular to this distance. Let's say L is a distance L into f cos theta is also torque and f into r cos theta is also the torque. The magnitude comes out to be same. Okay. So you should be very familiar with how to find out the torques and then there is a concept of angular momentum. Fine. So angular momentum of a rigid body about the center of mass is ICM into omega. Okay. Now you will see that in this chapter many things can be like for example torque is there, angular momentum is there. So all these things can be different about different axes. Okay. So you should always think in that manner. Don't think that torque is same about each and every axis, angular momentum is same about each and every axis. Everything is with respect to some axis. Okay. So angular momentum of entire rigid body about a certain point is r cross MVCM plus ICM omega. So I am not getting into too much of derivation. So this is what the angular momentum of a rigid body is. And the angular momentum is not only for the rigid body but for the point object also. For a point object which is moving with velocity V, the angular momentum is r cross its linear momentum P. Okay. So it comes out to be r into M into V. Fine. So this you can also write as perpendicular distance of the velocity multiplied by mass into velocity is the angular momentum of a particle. This is the angular momentum of a rigid body. This is the angular momentum of a particle. Why I am talking about angular momentum of particle because in a system there might be particle as well as a rigid body. Okay. So I hope all you guys have come with a basic revision of the theory part. Okay. That is why I am going little fast. Is it the correct assumption I am making? Yes, sir. Those who are online, I have you guys come with a read of the chapter. Okay. So whenever like for example if you are stuck or you want to interrupt me or want to have any discussion on any particular topic, please feel free to do so. You can WhatsApp me on my personal number. You can message on the Skype. You can message on the group whatever is convenient to you. Okay. So this is with respect to the angular momentum. Okay. Now comes the work energy theorem. Now you might be remembering work energy theorem. We have used it so many times. At least 100 times we have written this formula. So work done on a rigid body is also following the same equation. Okay. Work done is equal to U2 plus K2 minus U1 plus K1. What is U? U is a potential energy and K is the kinetic energy. Okay. Now potential energy can be of two types. Okay. Potential energy usually in mechanics they are of two types. One is elastic potential energy which comes majorly due to the spring and other is gravitational potential energy. Okay. So when you divide the potential energy, you need to find out separately elastic and gravitational potential energy. Now elastic potential energy, there is no change to the formula that remains half Kx square. What about the gravitational potential energy for a rigid body? How you write gravitational potential energy? Gravitational potential energy for a rigid body is MGH center of mass. Because you know when the rigid body is there in space, the points are all over the place. When you say that this is the zero potential energy line, which points height will you take to write MGH as potential energy? You just find out where the center of mass is and find out its height from the zero potential energy line. Okay. Suppose this is HCM. So potential energy of this rigid body will be equal to MGHCM. Okay. So it's I think very straightforward. And kinetic energy, how will you write kinetic energy? It is half M into VCM square plus half ICM into omega square. Okay. It's like you first see what is the kinetic energy with respect to center of mass. You can think like that to remember the formula. With respect to center of mass, every point is rotated about the center of mass axis. You can say half ICM, omega square. And then you have to look at what center of mass is doing. So center of mass, if it is moving with VCM, its kinetic energy will be half MVCM square. Okay. And if there is a fixed axis for a rigid body, the kinetic energy will be simply equal to half moment of inertia about the fixed axis into omega square. Okay. So this is something which you can directly use. And if you're not able to find any fixed axis, you can anyway use this formula. You will get the same quantity. Getting it. So this is I have just dressed up all the formulas that are required in today's session. Fine. Any doubt with respect to any concept, just feel free to, you know, ask any one of you any doubts, any small concept which you want me to discuss. Fine. So we will start with, you know, I'll just give you a few questions and then let's see how it progresses. Suppose you have a square plate. Okay. The side length of the square plate is A. Okay. And you are applying forces on this square plate like this. One force is like that. Then another force like this. This is 30 degrees. Then there is another force F like that. And then there is a force in this direction. Okay. You need to find out the torque about this axis that comes out of this plane. Okay. About this axis, what is a torque? Let me come back to that in some time. Should I do it? Just a second. So is it F A into 7 plus root 3 by 8? So is it by 4? What are you saying? F A into 7 plus root 3 by 4. Whole by 4? Yeah. Okay. Those who are online, are you getting this? Sir by 4. Those who are on YouTube, are you able to get this answer? Okay. Now let's see. Here you have talked you to this force is what from this axis, what is the perpendicular distance of this force? The perpendicular distance of this force is this much, which is how much? A by 2. Okay. This force is also A by 2. And this force perpendicular distance is also A by 2. Okay. So these three forces, it is very easy to find the torque. Now, can you see that all these three forces are trying to rotate this object in same direction? And this force is trying to rotate in this direction, that force like that and this force the upper one also trying to rotate it like this. So all these three talks will be added up. Okay. So F into A by 2 is a torque due to one of them into 3. This is a total torque due to these three forces. Okay. Now let us see the torque due to this force. Okay. Now one way is to find out the perpendicular distance of this force from the axis. That will be little difficult in this particular situation. So what we will do, we will find out the component of forces in two direction and we will say that this force will be as if it is a combination of two forces. One is in this direction which is F sin 30 and one in that direction which is F cos 30. Okay. Now you can see that both F sin 30 and F cos 30. They both are also trying to rotate in the same direction. So torque due to F sin 30 and F cos 30 both will be added up. Like you know for F sin 30 it will be F sin 30 into A by 2. Okay. And for F cos theta, F cos 30, the torque is F cos 30 into A by 2. All of you done this? Yes sir. Okay. So like this you can find out the torque in this particular situation. Okay. So do this question. You have a circle or let's say you have a disc. Okay. You have a disc which is free to rotate about this axis. Okay. So you know this disc is vertical right now. Okay. It is vertical. The axis and due to gravity is like this. Okay. This axis is horizontal and this disc can you know flip over. For example, when it flips over the disc will be something like this. Getting it. So this is the center. Okay. This distance is R by 2. The radius of this disc is R. The mass is M. You need to find out what is the angular velocity this disc gains when it flips over. All of you got this question? Yes sir. Solve it now. What will do here? Any particular equation comes in your mind? That work energy theorem? So what do we have to find in the end? Angular velocity when it flips over. Okay sir. So the question is you have this red horizontal line. You can see about this line this disc can flip over. Okay. So basically this disc is rotating about this line. You need to find out what is the angular velocity of the disc when it flips over. Okay. So is it a root of 8 by 3 into G by R? Root of 8 by 3 into? G by R under the root only. G by R. This is what you are getting? Yeah. Is this what others are also getting? So I am getting 2 root G by R. Okay. I am getting 8 by 3 root over 8 by 3 G by R. Oh then I did something wrong. Okay. Naman is also getting same thing. Okay. Those who are online, are you able to solve it? Okay. So what I will do, see every mechanics problem like this where you apply conservation of energy, you need to assume one horizontal line to be zero potential energy. Where do you think that is appropriate here? Red line. Red line, you can just, I mean, you know, don't spend a lot of time identifying it because that is something, you know, the answer is independent of it. So wherever you see some logical choice, just go with that. Okay. So you can see that this is zero gravitational potential energy. Okay. So initial gravitational potential is what? Initial is mg into R by 2. Right. Central mass is here, which is at a height of R by 2 from the zero potential G line. And final potential energy is minus mg R by 2 because the center of mass has gone below a height of R by 2. Okay. And K1 is what? K1 is zero. Initial kinetic energy of this entire system is zero. And K2 is what? K2, you can see, you can see that this red line of fixed axis, right? So you can write kinetic energy as half into I about fixed axis into omega square. Yes or no? Yes, sir. Okay. And also, you know, use that formula half MVCM square plus half ICM omega square. That will also give you the same answer. But whenever you have a fixed axis, you can write the kinetic energy of the fixed kinetic energy of the rigid body as half I into half I about fixed axis into omega square. That will easily give you the answer for kinetic energy. Now, the problem is that fixed axis is not something that passes through the center of mass. So you need to find out what it is. Okay. So we know that moment of inertia about this axis that comes out of the disk is what? MR square by 2. Okay. So after that, we use Papandika's theorem to find out moment of inertia about this dotted line. How much it is about this moment of inertia will be MR square by 4, right? Because for the disk about this is MR square by 2. So I'll say that that is IZ which is equal to IX plus IY. Now, both these axes, they are similar. Okay. So two times of moment of inertia about this axis is equal to MR square by 2. So about this horizontal axis moment of inertia is MR square by 4. Okay. Once I get moment of inertia about this axis, since this is passing through the center of mass and this red line is parallel to this dotted line, I can use parallel axis theorem to find moment of inertia about the fixed axis, which is what? MR square by 4 plus M into R by 2 whole square. Because I'm going a distance of R by 2, the distance between these two parallel axes is R by 2. Okay. So this will give you MR square by 2 as moment of inertia about this fixed axis. Fine. So this you have to use here and work done is zero. There is no force that is doing any work. Okay. So this is how you guys have done? Yes sir, but I made a silly mistake. So I took a wrong moment of inertia. Okay. So that is something which is very common. People make errors like that. Okay. So, you know, immediately you have to see about which, I mean, is there any fixed axis and about that axis only you have to write half I omega square as kinetic energy. And if you do not, if you're not able to find out any fixed axis, use the formula half ICM omega square plus half MVCM square. Okay. That way you'll be safer.