 Yeah, no but like the one male is they send you a CD. So we have a whole set of power, one of those is mind-standing. I'm not telling you if you're sure with my option from ADIZ. Do you have any more cards? I just want to say it's Malyar, I think it's Malyar. I think it's Malyar, I think it's Malyar. Ah, I think it's Malyar. Malyar, I think it's Malyar. Malyar, I think it's Malyar. I'm just going to show you the two by the end of the day. This is the answer, it's ADIZ. It's ADIZ. Dude, I think it's a good one. He is not older than me. Oh, you should move to my ADIZ. You see the bottom is older than me. There's more to go there. There's more to go there. There's more to go outside the box. Yeah, Shaker is 5 days younger than me. Wait, Shaker is 5 days younger than me. Shaker is older than me. Does anyone count on me? 5 days younger than me. Yeah, that's not interesting. It's ADIZ. You know what I'm saying? You were only surprised. You know what I'm saying? You just be like... It's easy. You should see the... You should see the last one. I'll go with you. Hello. Hi. I'm sorry. Excuse me. All right, so let us start. Others will join. Keep your phones inside. What are you looking at? Which match? Good ball. We will start the talk. Heard of talk before? All right. Heard of talk before. Talk before. What is? Talk is something which will... Causes rotation. Causes rotation. Simple. Force is something which causes translation and the torque is something which causes rotation. Write it down. Disturbing. It is something which will take force. Or in this world. We need to just identify it. What it is. And we need to quantify it. Like best possible way to quantify it. And the way you should quantify it should be consistent with the observation. But exact quantification whether you call this length as 1 meter or not it should be consistent. Nobody is going to ask you why force equal to mass and acceleration. But it should be consistent that if velocity is increasing quickly the force has to be more. That's all. You could say why not force equal to m into a square. That also you can take it as force. But people have taken the simplest possible expression combined with mass and acceleration. So similarly let's try to see the observation related to the rotation. And then we can quantify what the torque should be. Fine. Now in order to have translation we need a force. Now if a rigid body is at rest if a rigid body is at rest do I need force for it to translate? Any object needs a force right? And is the force required for it to rotate? If the object is initially at rest I want that object to start rotating. Do I need a force yes or no? I need a force. And larger the force faster the rotation yes or no? Logical. So torque should be proportional to the force. Is there any doubt in this? So torque in physics is represented by letter tau. Have you seen tau before? Tau should be proportional to force. Exactly. You will see that in your project like for example see saw you might have played no? Yes. Rotation is rotation it need not be complete circle. So this is the fulcrum you apply some amount of force over here. This is the force. It will rotate yes or no? Now if you apply the same amount of force over here will it rotate faster or slower? Slower. The force is same. So torque is not just proportional to the force. It has to do its distance as well. So the greater the distance from the axis of rotation greater is the effect of rotation. Simple. So this is let us say distance d. So the torque should be proportional to the distance. Always find an axis of rotation. Right now there is a fixed axis of rotation. So you can easily see that there is a axis which is visible very clearly. Now if I throw this chalk in the air it rotates and then comes back do you see any axis of rotation? Not exactly right? So basically I mean this is a convenient way of writing. In your textbook it was written that torque is proportional to distance from the axis of rotation but in reality you can find torque about different different points. You can say torque with respect to a particular point is proportional to distance of that point from the force. Now I am finding torque with respect to this point. Now if I find torque of this force with respect to that point the torque will be zero because distance is zero. Getting it? So the torque is proportional to perpendicular distance from the point about which you are finding the torque. And when you see a fixed axis you naturally tend to find the torque about that fixed axis but there need not be fixed axis all the time. Now what if I am applying force like this will this seesaw apply like this what will happen? It will rotate it will rotate but not as well as this like this right? So if this is angle theta only the f sin theta component helps it to rotate fine? So it not just depends on force and perpendicular sense it also depends on angle and we have found out that if angle is 90 degree the torque is maximum and if angle is zero the torque is zero okay? So that is why we need to bring that angle also in the picture so all of them have agreed that let us call the torque as distance from the point into the force into sin of angle to determine this angle theta theta is what is the angle between force and force and no no it is the angle between the force and the distance from the axis it is the angle between the suppose from this point you are finding torque so you draw a line from this to the force this is your one vector and this is another vector getting it? So it is the angle between this vector and that vector your distance vector starts from the point about which you are finding the torque so it is rf sin theta can this torque be negative? you consider it to be positive right? it can be negative so right now and this force is the torque negative? what do you want to be negative and positive? rf sin theta tell me whether it is negative or positive r is the magnitude of distance f is the magnitude of the force we consider positive it is positive what is your negative? why does it no because you see length will be positive f will be positive and theta here will not give you a negative side okay sure you keep quiet then others what do you think? negative or positive? what is the angle theta? it is acute or obtuse here? then positive theta is greater than greater than theta this is one vector this is another vector this is the angle actually this theta is not that theta actually this is the tail this is the tail and here is the tail for the force this is the angle tail to tail or head to head both will give you the same thing vectors you will not watch the video also this is rf into sine of angle between r and f why it should be angle between f so if this is the definition right now the torque is what negative or positive? sine of obtuse is still positive sine of obtuse is still positive it tries to rotate like this it tries to rotate it in this way fine and then there should be some sine convention we usually take anti clockwise to be positive the sense of rotation in anti clockwise to be positive and sense of rotation in clockwise to be negative so over here r cross f is in which direction this is r where r cross f I have to find out r is like this and f is so I have to align my right hand in the direction the first vector and then curl in the direction of second vector r cross f is into so this is clockwise or negative torque curl in the direction of the force see if suppose this is vector a and this is vector b how do you find a cross b other and then curl in the direction of second vector so thumb is a cross b b cross a it depends on perpendicular distance sorry it depends on distance the force and the angle between the distance and the force is it clear I can write the torque f perpendicular alright and in this chapter remind this sine of the torque we just use our sense of rotation or just look at it if you take that direction to be positive one sense of rotation will be positive other sense of rotation will be on a single axis only two sense of rotations are possible perpendicular distance perpendicular component of force and then torque sine theta into f what it is r sin theta is what perpendicular component of distance perpendicular component of the distance from the force this is the torque so physically perpendicular distance of the force that causes the rotation so basically if the perpendicular distance from the axis then the torque will be 0 it will not rotate at all for example this force sine of force was passing through the axis its torque and perpendicular distance from the axis is 0 so it will not be any rotation are you getting it similarly if it says this let's say this is a door of that axis if I apply a force like that this force if I extend will cross the axis it will cut the axis what is this perpendicular distance from the axis perpendicular distance of this force from the axis tell me how much it is sit back what is this perpendicular distance from the axis so the breadth of the what you are saying this then how much the answer is 0 it is passing through the axis this line of force cuts the axis there is no distance from the axis are you getting it so it will not cause any rotation we write down a force passes through the axis line of force if passes through the axis creates 0 torque about that axis creates 0 torque about that axis any doubt still no if you have to find out the torque and extend it extend it and drop a perpendicular about which you are finding the torque that is the perpendicular distance that into the force is your torque and suppose when you component the force take the perpendicular component of force and multiply with the distance that again will give you the same torque now tell me if the net torque is 0 if the net torque is 0 will the object ever translate it will do can it no it can't it can't translate okay example so if I have something and I push it from both sides if I have an injured body and I push it from both sides okay one second now tell me suppose there is one rod like that you are pushing from like what Param is saying suppose and push it from both sides with F what will be the acceleration of center of mass 2F by M 2F is the net force on this system 2F is equal to M into ACM so ACM will be about the torque about this point about the center how much is the torque why both of the forces this force creates a rotating torque in this way that creates in opposite way so their rotating effect gets cancelled away so net torque is 0 so just because the net torque is 0 doesn't mean that net force is 0 getting it perpendicular distance from here into drop of perpendicular on this line of force so this is the perpendicular distance itself it is already perpendicular okay here now what if what if net force is 0 can the object or rigid body rotate no no no it can I can flip that force to that side then it will get a net torque but net force is 0 I have understood what he is saying okay so suppose you have a rod like this one force from top like this one force from below equal to opposite force will this object rotate when you open the there also you apply equal to opposite force you don't want center of the cap to move so S is from center of masses 0 force is not moving but entire object is rotating fine there is some rotation or there is an angular acceleration force causes acceleration torque causes angularization please write it down if net torque is 0 acceleration will be 0 if net torque is 0 angular acceleration will be 0 is there any doubt sir even if net torque is 0 it will be rotating with uniform angular velocity yes if the net torque is 0 the angular velocity is either 0 or uniform because angular acceleration is 0 right there constant omega just like a force is not 0 so if force is 0 there can be constant velocity where is Advik he came not he went back he was like no so you he could have gotten some shade or something so he came cycling why you prefer cycle so it empowers us he can use motorbike I am not waiting no you might have heard some story ok let's go it is getting recorded I like physics sir you are saying that if net torque is 0 it is 0 so this point let's say a particle is in equilibrium what we say net force is 0 translational equilibrium net force is 0 ACM is 0 translation won't happen but can rotation be still happening is it possible yes we have just seen it is still possible to have rotation are you getting it to have a rigid body in equilibrium not only net force should be 0 but also net torque should be 0 then only it will not have excretion of center of mass nor it will have angular excretion center of mass is alpha what will be the angular excretion of the tip of the circle alpha only it is a rigid body alpha of all the points will be ok finding the torque from this point the torque of this which is near by force from me ok now there is this special scenario of two equal and opposite forces which we have drawn actually these form couple so they are called couple they are special kind of torque I am not exactly sure spelling is this only but let's keep it like that forces are applied on it of this 3 minutes I have total length is at first find out how much is the torque about the center could you unbolo properly do it total torque total torque total torque will be total torque total torque total torque total torque total torque total torque total torque total torque total torque total torque total torque total torque total torque total torque total torque total torque total torque total torque which way like this about that for that force about this point is f l by 2 which direction seen like this right so they will add actually speaking it should be minus f l okay but which is at a distance of x from here because of this force is what x about that okay now find it can't be anything other than the same what is the torque because of this force about this point it will be f into l this way as well f into x so f into x like this what about that force torque is f l plus x but in which direction it will be like this so these two are in opposite direction sense of rotation about this point so till now this force was causing like that sense of rotation but that was about this axis it causes the rotation this way that is where this become in opposite this is now in opposite direction of that so you need to subtract this from that you'll get f l only right so you can take any point about any point you can take here also anywhere force is 0 will have the same torque about any axis anything is this for anybody yeah why not a couple is always opposite next force should be 0 couple is like two couple meant to write well female similarly here also two forces opposite in equal and opposite directions so what if we have three forces that couple is okay I said couple right with forces so we got it I'll edit it please write down condition mass expression should be 0 it should not translate with any exploration neither it should rotate with any angular simple conditions right so basically net force along x axis is 0 along y axis should be 0 axis should be 0 xs talk about any axis should be 0 dynamics of the rotational this thing in fact moment of inertia was a part of rotation dynamics only here we need to be very careful with where the force is applied alright what is its distance from whatever axis you're considering or whatever point you're considering taking with the distance so when you draw this direction is fine you need to not only take care of direction but also take care of about which point to find talk talk is perpendicular distance into force okay so going forward viewer diagram becomes little bit more involved okay and one more thing when you write net force 0 all the forces will feature in the equation all the force will be there will become 0 if you're finding about the point through which that force is applied and this is fulcrum suppose I'm finding talk about this point and equating it to 0 from here will not come in the equation yes or no target so you can make your life simple by choosing an appropriate point about which you equate net talk to be 0 it will itself the suppose there are three variables of equation itself only one variable is coming so you can tightly get the target question so net force equal to 0 there is no greater just write it down when you write down into 0 appropriately choose about which point you're writing the and you can read it about any point answer will not change but the amount of time you take will change so there are a lot of questions on rotational equilibrium this topic is called rotational equilibrium equilibrium of a digit body same thing okay first let's do a couple of questions on finding out this question is a heart that move over the heart I'm not making it up see it is there any other information you need where is the axis is it in this direction okay here is the question Calcutta total torque here so that arrow at the bottom is that also yeah that is also a force 20 newton so it is a 4 meter a centimeter hi there 4 meter long This is the angle 30, this is the angle 30. Which angle do you see? Is there any other angle that can be 30? That's at between 15 and 5. So around 90, right? 0.6275. That's the angle. 21, right? 21, no. What is the angle? You can't even talk. How are you going to make your mind? It's totally unsympathetic. It's made in Delhi? 0.26 or 0.026, whatever. I don't know. Whatever it is. I don't know. 0.26. No. 1.37. No. Yeah. It was my view on point. No. No. Yeah. No. I don't know which one is mine. Which one is yours. I don't know. I don't know. I don't know. Which one is mine. So where is the action? I'll do one thing. Find out the torque of all the forces one by one. Yeah. Yes. What you got? Yeah. Torque due to 5 Newton is? Wait. It's torque due to 5 is? 0. 0. 0. Passes through the axis. Yeah. Torque due to 15. 0.5. 0.5. 0.5. 0.5. 0.5. 0.5. 0.5. 0.5. 0.5. 0.5. 0.5. 0.5. 0.5. 0.5. 0.5. 0.5. 0.5. Guys. You can find the perpendicular distance. Or you get the perpendicular component. Yes. That's it. Shut up. It is. 0.5. 0.5. I am discussing here the solution. So you should pay attention. When I was asking you what is the answer you all said, something, something, right? I listened. Then I am telling you and now you are discussing among yourselves, perpendicular distance here is 6 into 6 times 37, 6 times 37 is perpendicular distance into 15, 500 because it is in centimeter. So minus 0, minus 0, minus 0, minus 0, minus 0, minus 0, minus 0, minus 0, minus 0, minus 0, minus 0. We can do plus minus 0. 10 into 4 divided by 100, what is this? 0.0, 0.4. 0.4. Yeah. Okay. Tower due to 20. What? Drop of pump bending a resolving force would be easier. This is 20 cos into 4 by 100. Force no rotation, this force like this, that force in this direction, this one in that direction. This plus that minus this. 0.5. 0.5. 0.5. 0.5. 0.5 is the answer. Okay. Sir, sir, for the tower we only write 0.5 for Newton meter. Newton meter, N dash M. That's it. You will get the whole number. Okay. In and out. Right. The right is cos. See cos or sign, don't remember like that. You need to find perpendicular component of the force which is perpendicular to this line. Okay. So, this is 60 degree. So, 20 cos 60 is along this line, right. If you consider this angle, 30, then it comes out to be sign 30. Sign 30 cos 60. If two lines are like that, angle is theta, the component is cos theta times. Okay. Any other doubt? The answer is 0.54. The surface has coefficient of friction mu. By the way, some object has mass M. Okay. There will be a force, gravity force. Which will be applied from where? Reason. Maybe the sum of the forces around each point. Okay. Let's first do the center of gravity, then we'll come back to this. Please write down center of gravity. Center of gravity is different from center of mass. Center of gravity is a point. Center of gravity is a point which can be considered as if, which can be considered as if entire weight of the object is, entire weight of the object is applied from that point. Earth is pulling the entire object from that single point. If you behave like that, the center of gravity should be 0. Why is the center of gravity not the center of mass? We said it's not always. That is what I was about to do. Any other doubt? Use a definition and find out where the center of gravity lies. This is an irregular shape and size of the object. Looks like a car. Let's say this is the center of gravity. According to definition, entire mg should act from here. A rigid body is made up of small, small point masses. Is there any doubt on that? Okay. So let's say this is the point mass of mass m1. So m1g. Yes, please come. But where do you sit? There. There you can sit. M2g. Multiple masses. This will be m3g. Now suppose distance. This is your r2. This is your r1. This one is your r3. The center of gravity should be 0. According to definition of center of gravity, gravity force about the center of gravity should be 0. Okay. So what it is? Let's say g is varying everywhere. It's a big object. Gravity is g1, g2, g3. m1g1, m2g2 and so on divided by... No, that's it. Equal to 0. The direction of g is constant of h. I can write it at like this. m1 direction of g is fixed. And if g is constant, let's say, if g is constant, then I can write it as equal to m2r2 and so on with vector g is equal to 0. This means what? The bracket term is equal to 0. So where the origin is? If bracket term is 0, then what? Summation of h by summation of mi is r center of mass. Okay. So it is a point, if I multiply and divide by summation of all the masses, this and here is total mass. I have multiplied and divide by the total mass. Okay. So this bracket term is basically, this should be equal to 0. And what is 0? 0 is the center of gravity as well as center of mass. Because we have assumed, when we started, we assumed, let's say this is the center of gravity. Total mg is applied, let us say. And according to the definition, net torque should be 0 about that point. That is what we have done. So it is a point which has m1r1 plus m2r2 and sum of all that should be equal to 0. Okay. And that point is actually the center of mass. All right. So if adjacent due to gravity, g is constant throughout the rigid body, then center of gravity and center of mass will coincide. Please write down. If small g is constant throughout the rigid body, if small g is constant throughout the rigid body, then center of gravity and center of mass will coincide. Then center of gravity and center of mass will coincide, will stop, gravity force, gravity force on a rigid body, gravity force on rigid body can be considered, can be considered as a single force, as a single force of m into g magnitude, m into g magnitude vertically down through the center of gravity. So, if there is a rigid body, capital Mg will be applied to the center of gravity and if nothing is said, you please assume that center of gravity and center of mass will coincide, because assumption of g being constant for entire rigid body is very common, usually it is same only, do not pull that string. Fine, let us solve few questions, all of you, the ladder has mass m and length l, it is placed like that, this is wall and this is floor, wall is smooth, floor has coefficient of friction mu, if I decrease the angle theta, what will happen if I keep on decreasing, there will be a point at which will not be nucleogram, you need to find minimum theta for which it will not slide, as in it is about to slide, there is the answer, real body diagram, theta is always dimensionless, when you get sin theta to be equal to something, it has to be dimensionless, if it has dimension, your answer is wrong, you can check your answer whether it is right or not, if mu increases theta should decrease further, it should allow lesser angle if theta is, if mu is large, there will be a normalization from here or not, this will be normalization, let us say n 1, there will be a friction, which direction, it is about to slide, so this way, what magnitude, since it is about to slide mu n 1, there will be gravity mg, what else, friction from here, this will let us say n 2, there is no friction at the top, so next we can write net force along x axis to be 0, then along y axis to be 0, by x and y axis are these, along x axis, what are the forces mu n 1 minus n 2, this will be equal to 0, along y axis we have n 1, then I can write net torque is 0, about any point, about which point should I write, why, if I write about this, torque because of these two will become 0, so this is more sensible to write about this point, so what is the torque about this point because of mg, what is this perpendicular sense, l by 2 cos theta, so mg into l by 2 cos theta is a torque because of mg, what about torque into n 2, you have to drop a perpendicular on the line of n 2, this, this is what l sin, so minus of, why minus, because sense of rotation is different, n 2 into l sin theta, so this is equal to 0, now n 2 is what, mu times n 1, n 1 is mg, so n 2 is mu mg, mg l by 2 cos of theta minus mu mg l sin theta is equal to 0, mg l, mg l, theta is 1 by 2 mu, theta is tan inverse 1 by 2 mu, any doubt, when is k d p y of mass of mass of mass of mass of mass M and length L can rotate, this distance is l by, here a point mass is kept small m, you need to find out how much mass you should keep here, what is the point mass? You should keep here, simple, simple questions, m by 4, ready? Which is wrong? One is small m and three is capital M. Which m by 4? Small m by 4. Small m by 4. Okay, so capital M is a mass of the rod. Capital M is a mass of the rod. You probably ignore the gravity force on the rod. Small m by 4. Correct? Is it 2 m minus 3 m by 8? Yeah. 2 m minus 3 m by 8. By what? 8. Yeah. Yeah, yeah. 2 m minus 3 capital M by 8. Yeah. 2 m minus 3 capital M by 8. Okay, nobody is getting 2 m by 8? 2 m minus 3 capital M by 8. So 1 m. Always getting that? Yeah. 2 m minus 3 m by 8. In the law of river diagram, you need to also draw capital MG force from where? L by 2. L by 2. Distance. This is your normal reaction, let's say N1. And actually it can assume that MG force is acting on the rod here. Actually what is acting? Normal reaction between small m and the rod. And since small m is at rest, normal reaction is equal to MG. Okay, so that is why this force can be considered as MG. But it will not be equal to MG if rod starts rotating with angular acceleration. Are you getting it? Then it's right, N minus MG is equal to M into A. And from there you get the normal reaction between small m and the rod. Right now rod is fixed. Okay. Let's say this mass is N1. So N1g. So about which? About which I should equate, talk to this 0. About this point. Then N1 will not come in the equation itself. Right? So I will be able to write it as MG into L by 5 minus capital MG into 2 minus L by 5. Then N1g into what? L minus is like this. One last question. A uniform rod of length L rest again the smooth roller. This is roller. Okay. And it is smooth. Find coefficient of friction between the ground and the lower end. If minimum angle is theta between the ground. So mu is what? If a degree is that theta, it is like. The cross. I don't have. And what about the roller? No it is like this. I need a normal roller so you don't balance across the roller. Then you don't need a machine. Balance start the roller. So I am not balancing the roller. I am just balancing the roller. Because of the roller, perpendicular to the rod, that is the difference from the previous question. This is N2, this is N1, cos square theta y, no it is a complicated expression actually. The rod has mass, okay. If rod does not have mass then theta n is equal to N2 sin theta by mg. I will write the answer first. No one. Okay, let us solve this. There will be mg force, there will be new times, now the question. Okay, I can take component of N2, horizontal and vertical. This is theta, this is theta, theta, this is theta and that is theta. So this is N2, cos theta, this is N2 sin theta. Stop talking. So N2 sin theta, minus mu n is equal to 0, this all of you. Mu n1. Mu n1, okay. Vertical direction, N1 plus, both these equations you got, still you are getting it wrong. Then talk about this point, let us equate that to 0. What is the torque because of mg? You will have to find. mg into, what, mg into l by 2, no l by 2, it is a weird expression. What is this about? Yeah, but then the normal thing will be solved. Stop talking. This l by 2 cos theta, okay. It is N2 l. How do you know what is this length? It will be l minus h cos theta. L is still here. It is l minus h cos theta. So N2 into, this entire force is perpendicular to the rod. So you do not need to worry about taking its component when you are finding the torque. So N2 into what? L minus h cos theta. h by sin theta. These are equations. How many variables? N1, N2, mu. So we take mg into l by 2 cos theta. We assume that mg is being applied at one end of the rod. No, here, center of the rod. Yeah, and then it will rotate around the center of mass, right? No. There is no fixed axis. It depends on, suppose this rod I fixed one end, it will rotate about its, from its top. It does not rotate from center of mass. It does not need to rotate about center of mass. So sir, how can we calculate torque if we do not know what the axis is? There is no rotation at all. It is an equilibrium. So N talks about any end point is 0. Yeah, but for the equation, we still need force and distance from the… You write, see I have written torque about this point is 0. You can write torque about any point to be 0. You get the same answer. But if I write about this point, my equation will be simpler. Because of these two, torque into 0. Sir, why? There, torque from N to y to b, right? L minus x times theta, right? x times… Yeah, so why is it not… x times… Correct. Yeah, why it is not h by sine theta only? It is. No, I just wrote it. This distance is h by sine theta only, right? Yeah. So, it is simply N to N to… h by sine theta. Sir, why does the roller make it perpendicular to the normal reaction? See, perpendicular to a point is not defined. So, roller makes it a smooth contact. That is it. And suppose it is not given it is a roller or not, still you take perpendicular to rod. Okay, it is a reasonable assumption. But in reality, it may not be perpendicular to rod. Because the point will not only apply a normal force, it will also be a little bit clear inside the rod and then it will try to apply along the rod also some force. Any other doubt? No doubt, no doubt less. Okay, so this is about… Sir, we take it perpendicular to the rod only when it like extends beyond that wall, right? The rod. See, the normal reaction we have discussed in Laws of Motion, normal reaction force should be perpendicular to both the surfaces of contact. Understood? If both surfaces are plane, multiple points of contact will be there and perpendicular plane is same anywhere you go. And if the shape changes, there will be a point contact. They cannot be a surface of contact. One sphere and the surface also has a single point of contact. If it has more than one point of contact, it is not a sphere. It may look like a sphere. It has to be a point of contact. No doubts, right? Alright, so next we will write down the torque equation. Should we derive it? Directly we will write it. We will derive it. See, what will happen is that you will get mathematically so much bombarded with all the equations and everything that you will not be able to enjoy the numerical stuff. Okay? First, let us do everything. You will feel good about yourself and then I will tell you how that torque equation comes. Okay? Make sense? So torque is equal to I times alpha. This is equal to mass of acceleration. Torque is equal to I times angular acceleration. What is I? Moment of inertia. Moment of inertia. Torque we anyway know how to find out and alpha is angular acceleration. Okay? Can I use this equation about any axis at about any point? No. I cannot use it at any axis. There is a restriction. The restriction is, we write down, this can be used only about fixed axis, any axis that passes through center of mass. This cannot be used about any other axis. If the object is rotating, if the object is not rotating interchangeably, then every point is a fixed axis in a way. I have to find torque and I have to find moment of inertia about the fixed axis. I will get alpha about the fixed axis. Isn't it? Alpha of the rigid body is fixed. Okay? Alpha of every point is fixed about every other point, every any other fixed point it is. Alpha only. Right? It can change, but the ratio between torque and I is no other theory. Okay? Torque is equal to I alpha. Okay? And then if suppose there is an external force applied on a rigid body, of course there will be external force because there is a torque, right? Most of it will be an external force. So net external force, texas, will be mass into acceleration along x axis. This acceleration is whose acceleration? Central. Central mass acceleration. So now you have alpha as well as A, ACM. And using the kinematics, you can find out the total acceleration now. First you observe with respect to central mass, then you add what central mass is doing. Okay? Similarly, Fy will be equal to M into Az. Okay? Do you know equations you already have learned? Newton's second law is acceleration. But whose acceleration? Central mass is acceleration. That you should apply without hesitating. Apart from torque equation also, which will give you alpha. All right? And suppose you have a relation between alpha and A. If you know that it is rolling without slipping, alpha is equal to A by A, then you can connect the equation with that equation. There is a connection between these two equations. Are you getting it? But if suppose there is no relation, no connection, is that clear? Okay? So let's take couple of simple questions then you will understand. Can't rotate about this fixed axis? Okay? Let's say it rotates by an angle of theta. We need to find out the angular acceleration at this moment. 12 cos theta by 1. So 3g cos theta by 1. So 3g cos theta by 1. 3g cos theta by 1. 3g cos theta by 1? Wrong. Yeah, that's right. So 3g cos theta by 2. Yeah. Check. Why 2? Because gravity does not work. See, mg will act from the center. Forces. Forces on the right? Okay. Let's take forces like this. There will be one force this way, another force like that. Let us say. This you can say ny and this you can say nx. I mean I don't know which direction the reaction from the pin is exerting. So I will just say that it will be in a single plane only. Otherwise the rod will come out. It will remain in a single plane. So the reaction force on the pin will be on that plane only. So assume it has two components, one in x and one in y. Are you getting it? Right? So now I have to find out the angular acceleration. Is there a fixed axis? First step is to find out if there is a fixed axis. Is there a fixed axis? You can think force like that. What happens? I just took it as 4 by 7. I think it is the angle. And it is the angle. And it is the angle. And it is the angle. And it is the angle. I am also doing an answer. I am not doing something like this. Here you are. I am seeing this. I am doing it. It is correct. It is correct. It is correct. What happens? What happens? It is three. It is three. It is three. And it is connected to the T. You can think of it like this. It is three. Check what? Check what is the point of this. No. So, now we can find out what the problem is. We might just go to a particular company, and that's how it all starts. We have a really easy way to write it, you're all going to see what it says. This is the problem. This is the problem. So, buddy, there is a fixed access or not? Fixed access, yeah. Fixed access, yeah. Okay? What happened? Why are you talking? So, what's the problem? What's the problem in a group? Wherever you look, there's a chopper. What's going on? I think he was talking and I caught you. Isn't it? You're innocent. Isn't it? This company of MG is what? Why am I finding that company? Because that is perpendicular to the distance. How much is that? MG cost theta. MG cost theta? Yeah. So, this is MG cost theta. This is 911 theta, you guys are saying. So, this is MG sin theta. So, torque because of MG is what? Is there any other force which can give torque about the fixed access? No. Only MG can give. So, torque because of MG is MG cost theta into L by 2. L by 2. This is equal to moment of inertia about fixed access into alpha. Yeah. Which is what? M L square by 3 into alpha by alpha. MG cost theta. MG by 2L cost theta. Okay. Now, you have to find out what will be its annual velocity when it has reached a vertical position like that. Angular omega when the rod has become what? Where is the center stop? Where is it like? AMC. AMC. AMC. Annual maintenance cost. Where is it? 3D cost. Will you go? Will you go? Is it very far? It is far, but I have to go there. Some people don't go there. It's too far. A lot of money, right? Oh my God. Who will go there? It's in the exam. In the exam. Admission cost. I got a good file. I got a good file. One of the students forgot to with the form as I came here. I went to postpone, I guess. I went to the mountain. I got a constant number of things. So I made them from starting on, starting on, starting on. I mean, it's here. Yeah. How are we? Sir. 3G cost. 3G cost. Yeah, C divided by minus theta. Okay. From where are you getting theta? I am releasing from here. Theta is 0. And here I am finding the angular velocity. Theta is 90. From 0, theta is 90. What is omega? And what is theta? It's 3G over 5. 5 by 2. 5 by 2. Theta is 5 by 2. So theta, 5 by 2, who will put it in that expression? Sir. What is the answer? Blue 3G cost. Blue 3G 5 by 2. I like that. Blue 3G 5 by 2. Basically, it's under root 5 alpha. Possibly. Why? Not correct. Wait, sir, like from this position till the vertically. It is released from here. That's right. This road is like this. If you go like this, what is the angular velocity? Oh. Look at this. I will tell you in the game. Nothing is going on. Let's see how it goes. Look at this sector. Sir, because you do that. So I thought you meant from that angle. Oh. Oh. Oh. Oh. Look. My mistake. The root of 3G by 2. Sounds correct. What is it? Not sure. How did you get it? Alpha is omega into d omega by 2. Yeah. Just omega by 2. Here you need to apply kinematics. This is alpha. Alpha is what? Omega. d omega by d theta. 0 to omega. Theta goes from 0 to 5 by 2. 5 by 2. Okay? Like this, you get the answer. Sir, why can't we conserve energy? Because I haven't taught yet. So like, can I just say that it starts off with potential energy? No, no, no. Not now. I don't want you to use kinematics. All right. I'll teach then you can use. Sir, wouldn't the answer with that? No. You have done it wrong. I'll check that during the breaks. Okay? Are you getting it? Sir, but the side is minus that. What? Minus? Zero minus cos. Yeah. 5 by 2 minus cos. And this is a square minus 5 by 2. Theta, which is 0 to 90. Minus 5. Minus 5. Minus 0. Sir, it's coming as 0. Sine. Sine. What is sine? Sine. Sine. Sine. Sine. Sine. Sine. Sine theta. It is not minus sine theta. Yes. Okay? Sine is minus cos. It is reverse. Sir, I'm going to... What happened? Fall, fall. Sir, I thought that if the central mass moves from that top position to that position over there. Yeah, top. Right? So, no sir, the bottom. So the change in height of central mass is L by 2. No, L by 2. MgL by 2 is loss in potential G, which is equal to half i omega square. i is mL square by 3. Oh, half i omega square. Because half mv square. No, why half mv square? It is rotating. Then you can write half icm omega square plus half mvcm square. And vcm is omega into L by 2. That's what I'm saying. Wait for me to teach. Otherwise, you'll not learn right now. Both you'll not learn. Shit. Then, do you have to find the reaction from the Parangu itself? Let's say this is our y and this is our x. We'll have four. When it's rotating, one force will act in one and out the other. They should be two. Latching? Yes. Sir, you can take centripetal acceleration. Centripetal acceleration. So, our x will be equal to mv square by centre of mass is moving. Are you getting it? Is equal to mass into vcm square divided by L by 2. Here is one. Keep it this way. This is r y. So, r y is equal to m into x is now centre of mass. Which is alpha into alpha is 3g by 2 L cos theta. So, substitute here you'll get r y. And vcm is beginning 0 to theta. Total reaction will be what? So, root over this square plus that square. Sir, have some centripetal acceleration. Got it? Yes sir. I shut. It's out. It's out. You don't know why you can't be conservative. No, you can conserve energy. That is what I said. I am saying you are using it in the wrong way. And I can't teach everybody spending everybody 15 minutes how to conserve energy and that tell you. Are you getting it? Yes. You are using the wrong formula for counting energy. Either you write half icm omega square plus half mvcm square. Or you write half i about fixed axis into omega square. You are using neither. You are saying just counting energy is half mvcm square which is not correct. So, that's not the correct counting. That's not the correct counting energy. When you write half mvcm square you are assuming that entire mass is translating without rotating with vcm. Let's go to the next question. Simple only. No difficult questions today. Mass m and length l. Two strings are there. Everything is symmetrical. Properly the message is there. Now this string is cut suddenly. There is tension on this string. You got the answer? That's not correct. Why? It's all the direction. No, no, no. It will like that only. Immediately after cutting it will be exactly the way it is before cutting. Is it 7mg by 2? Yes. 7mg by 2. Yes sir. I wrote on the way. Like in a capital T or something like this you have to do d minus this. Mg minus this will be equal to mg by 2. Sir, it's a 3mg by 2l. It will be what? 3mg by 2l. That's not an answer here. That's not an answer here. I said tell me immediately after cutting. Will there be an expression of center of mass? No. Correct? Yes. It's alpha. Is there a torque? Immediately after cutting, is there a torque? Yes. Before cutting, when I cut it, tension becomes zero this way. Yes. Okay. So this is a fixed axis of rotation immediately after cutting. There is a torque because of mg. mg l by 2. This should be equal to 2l square by 3 into alpha. So alpha is what? 3g by 2l. 3g by 2l. So extension of center of mass is what? 3g by 4. This tension, this is mg and there is acm. So I can write mg minus T is equal to m into 3g by 4 is equal to mg by 3. It should be minus mg by 4. It should be minus. So minus m into 3g by 4. Because it's the same direction. mg and mg minus T is m into acm. So they are in the same direction. What direction? mg and acm. That is the axis of rotation. This is not force. mg is acm. It's acm. It's acm. Force is equal to mass and acm. I am writing. So we know the concept. Knowing the concept and applying the concept are two different things. After this, something weird will happen. But I am asking immediately after that. No doubt, right? No doubt. Stop talking. It's recording. All right. Next question. It's a cylinder which acts like a pulley. m1, m2, stop talking. Mass m and radius r. This is a cylinder. The pulley is a cylinder. And there is a friction between the cylinder and the thread. Thread doesn't slip on the cylinder. Thread doesn't slip on the cylinder. You need to find out everything. Acceleration and tension. Friction is enough for no slipping. Acceleration and tension you have to find. System is rotating. Why not? What we have been doing was the ideal scenario where pulley is mass less, friction less, pulley doesn't even rotate if it is friction less. Pulley will rotate only when there is a torque. If it is friction less, there is nothing to apply friction. Both sides will be same or not? Yeah. Same tension without friction. That's a complete sentence. It's a friction on the string. All right. So please remember that if the cylinder has mass and it has moment of inertia and it is rotating, tension will be different at both sides. Why is the cylinder moving forward? Why is the cylinder moving forward? It's a solid cylinder. This is a solid cylinder. How will the pulley be? Why can't it be a... It can be hollow, correct. But what I said last class, if I just say sphere, it means solid sphere. Acceleration of the masses and the tension in the string. It will be T1 this way and T2 that way. From here to here, the same tension. From here to here, the same tension. But because the friction is there, it will be T2 different. And let's say it is coming down with acceleration A. This moves up. What will be the acceleration of this block? A. A only because of constraint. Equationally, go forward for now. You don't have to solve. I got it. It's A0, right? Huh? A0. What? It moves. It rotates. Center rotates. We know so there's something wrong. I'll change it. I'll change it. I got it now. What is the force applied on the cylinder? T1 plus T2 plus NG. What is... Suppose if I cut this. I cut it like this. What is the force applied on the cylinder? Plus that string on it as one system. What is the force applied on the cylinder? The force applied on it as one system. I'm not removing string. If you remove the string, friction will come as an external force. And friction will be equal to the tension only. Because friction will be equal to mass immigration of the string. And mass of the string will be zero. So net force has to be tension only on the string. Equal and opposite. Equation of M1 will be what? M1G minus T1 is equal to what? T2 minus M2A. M2A cylinder rotates. Fixed axis? Yes. Fixed axis as well as center of mass of axis. Both coincide. First thing. So it will be consistent with A. If A is down for a direction. Then again that way you have to maintain the consistency. At this T2 twice rotated in opposite direction. So T1R minus T2R is equal to I alpha. I is what? Into alpha. Alpha into R is why? Because this point is moving in a circle with respect to that fixed point. So alpha into R is the expression of that point. So alpha into R is A. That's all. That's all the question. The friction is sufficient for it to not to slip. We are rolling. Find out the force of friction between the inclined plane and the sphere. So that it doesn't slip. It is rolling without slipping. What is the force of friction? Mu is not given. But acceleration. Fine friction.