 Alright guys, first item, system of particles. So if ever you find, if ever difficulty in physics, this name will come first. System of particles. So why students find this chapter difficult? Because there are multiple things here. In this chapter they have a number of things. So we have this habit as just mugging up or trying to do it last minute of the exam preparation. So what happens is that if lot of information is there, you do not understand the link between the information or you do not do enough problem practice. Then you will get confused in this chapter which comes up when you might be still confused in the other chapter because of lack of problem practice. This chapter will double it if you do not pay attention. If you ever miss, let's say 30 minutes of the class, you are daydreaming about something or someone, you are thinking about something. And you do not pay attention. The entire chapter is gone. This is like a chain. Why I am telling you all this? Because the basic chapters are over now. So these will form a basics. And then here is a chapter studying a special case about the motion. For example, you always have patient and some velocity. It is moving like this. You never consider that while moving, for example, masses and a pulley system, you draw free body time, you get the answer and you feel happy about it. But then going forward, you are considering multiple particles together. For example, if there are like zillion molecules in this room, you are not going to study one molecule at a time. What is this molecule doing? It becomes senseless. So you need to understand how collective behavior is what? As a collection of all the particles, what are they doing collectively. So this is what this chapter tells you and this is closer to the. So we have learned the basics from special cases. Now we are making it slightly generic. And when it comes to each and everything is a system of particles. This theory itself, multiple point masses, you can put small, small, small masses, you can go to the atomic level. The atom is, let's say, smallest particle of this. You can say electron and this thing also. So there are zillions, millions of those particles are there of which the TV is made up of. So this is one entire, it will look like a single object, but this is actually a system of particle. But this is a special kind of system of particle. There can be system of particles like this cap and the pen. This is also a system of particle. When I am studying cap and pen together, two particles I am studying together. So this is also system of particles only. But they are moving. The motions are not point, will move along with that point. There is a definite relationship between these two. So their motions are related in some way. So studying this kind of system is easier because knowing the motion of one particle, you can predict the motion of the other particle. But if system of particles have loose particles like this, then you can predict knowing one particle's motion. This kind of system of particle which doesn't get deformed or there are some relation, this is called rigid body. So we are going to study a special kind of system of particle which is a rigid body later on. Right now I am going to just introduce a concept of system of particle. How to study any system of particle? Whether it is rigid body or collection of particles like this, they are moving dependent on each other. I think we have already started studying the system of particles. Can't be theory is one. We have studied the entire, all the gas molecules together, what they do. So we have already started studying the system of particles. But here we are going to study it in a different way. We are going to, we are trying to apply the neutral second law of motion. Then we are trying to find out the velocity of the system as a whole and so on. So this is slightly different but we have already started understanding the system of particles. Whereas have we studied the system of particles as in multiple particles together? System of particles here, right? You have found out that we have already studied the system of particles. And flutes, what is flutes? Is it a single particle? No, multiple particles are moving. So even flutes system of particles, so we have already started studying system of particles. First time we are studying it exclusively. So when we are studying system of particle exclusively, we don't know anything. So we will start from the scratch, fine? So when we were, suppose I study, let's say particle. Like you remember, what is the first thing about the motion that we should study? It's a point object. In kinematic, no, see that is different. You study and displacement. But before even, that is an assumption. How you start studying it? The first one is the location. Which location will you tell? Location means starting from 0, 0 to 2, 0 and 2, 2 and 0, 2. So you can't tell me that, right? Location is a coordinate, right? So first thing we are going to learn here is how to get that system. Suppose there are multiple masses scattered everywhere, how can a system, right? So please write down location of a system. Center of mass. Have you heard of the center of mass? This is called center of mass. It's a point. Have one mass here, which is 1,082 masses, only these two masses. So please write it down. I'm using the content, okay? I have my own way of. So don't match it up with, okay, this we have learned, this we haven't learned. We cover everything when the chapter gets over, then you can see. Yes. So the location is closer, this is more on the location, right? So rather than finding, suppose this is my coordinate axis, this is the other coordinate axis, x and y, let us say. And the position vector of this mass is r1 and position vector of that is r2. I am trying to find the position vector of center of mass, right? So position vector of center of mass will be the weighted average so that the position vector comes closer to the average mass. So it will be m1 into r1, m2 into r2 divided by m1 plus m2, okay? Does it make sense? Suppose m don't count the number of numbers you have. Similarly here you are finding the weighted average, now the weight of the position vector r is its weight, getting it? So this is the center of mass location and this is for only two particles. Suppose there are multiple 10 particles then going to the formula m1 r1, are you getting it? So I have particles and then there will be infinite particles also. So do you know this? This summation, summation of mi of mi. This is the center of mass location. Make sense of these particles. So let us try to see whether we can find, given the x and y coordinates of 1000 kg in 1 gram, what will be the x and y coordinate of the center of mass, okay? So if the position vector x i cap plus, similarly that is what I do, r, c, u and vector will be x center of mass, i y center of mass. This will be equal to summation of xi i cap plus y i j tan divided by summation of mi, getting it? So this I can write as summation of mi xi divided by summation of mi i cap plus summation of mi y i divided by summation of mi j cap. So what will be x center of mass? You just equate the i component, y component also. So x center of mass will be x1 plus m2 x2 and so on divided by m1 plus y center of mass will be equal to m1 y1 like this divided by the total mass of the system, getting it? Is it straight forward? Okay, so you must know what is x center of mass, y center of mass, because we are going to use this question. Do this question. This is 1 kg binary square. The silent of the square is 1 units. You need to find out the center of mass of this entire system. The system has 5 particles. Simple. 1 into 2 into 1 plus 3 into 1 plus 4 into 0 plus 5 into 0.5 divided by 15, some of all the... And similarly y c n. Y c will be what? 1 into 1, 2 into 1 plus 3 into 0, 4 into 0. 5 into 0.5 divided by... Isn't it simple calculation? Identifiable locations. Particles. Point masses. Are you getting it? It is a big circle which is not located. And you can't say a bigger mass of mass m1. Then I will say that this m1 into different point masses. Get it? If I divide it into small point masses, then this will give me that summation equal to capital M1 mass. m2 let us say mass m2. So m1 x1, m2 x2 and so on divided by m1 plus plus m2 x2 divided by m1 plus m2. Now tell me what is the first term in the numerator? This is what? 1 into center mass of m1's location, m2 into x2 divided by m1 plus m2. This will be total center of mass. So if there is a bigger mass, you multiply its center of mass plus m2 into its center of mass and so on divided by... ...points to be noted. It is uniformly distributed. Then the center of mass is location. You can see that x is equal to m1 x1 divided by m1 plus m2 and so on. If mass is uniformly distributed, all the components are distributed. If mass is uniformly distributed, n of symmetry, portion of mass, the intersection is center of mass. Because it should... If mass is uniformly distributed, it will not have to be made up of whole... ...where the center of mass, which is not regular shape and size. Then I go because there will be infinite particles, time of symmetry to help me out. Then how to find the center of mass location? The summation will integrate that. And this is how it will happen. So x center of mass is summation of m1. Now I am saying that suppose there is a big mass, a regular shape object. Here up this big mass is there. I assume that the entire big object is made up of small, small masses of mass dm. dm is the small mass and its location is x. The location of dm is x. So the numerator will become what integral of... So first of all it is x and dm. For a particular... Similarly here, this is a formula which you have to solve for different objects. Any doubts? No? Okay, now do one thing. Take a rod. This is a uniform mass distribution rod. Mass is n, length is n. Find out the center of mass using this formula. We know that in the next scenario, the mass locations, but they expect you to do in all the ordinary exams. That's the answer. Because you have not... So you should pay attention here. You should know something complicated here. So integral of s is going from here to here. The masses are distributed as constant. 0 to l, x dx cancels, l by 2. kg per meter. It will be center of the rod. So where... We tell. That into length is the mass of location. It will not be like simple state part. This is my y coordinate.