 So, now we can take up questions from whichever topic we like and you can also start attempting questions anywhere from the chapter. So, let us take this question first, there is a rod of mass m and length l lying on a frictionless surface, stop the string. Now, there is a ball of mass small n coming with velocity u hits the rod at a distance l by 4 away from the center and after it hits the rod after it hits the rod, it comes back with the same velocity back with u by 2, u by 2. You need to find out after it comes back with u by 2 straight back, you need to find vcm and omega vcm the rod and omega. Again, use the frictionless surface or internal surface or horizontal external forces, 0 initial momentum is m into u, this should be equal to final momentum minus m into u by 2 as capital m into vcm. vcm will come out to be equal to how much? vcm straight forward conservation of linear momentum. At times whenever we learn something new, we forget that we can apply whatever we have learned earlier. So, this we have not learned in this chapter, this we already know, conservation of linear momentum. Now, can I use conservation of angular momentum about which axis? About any axis I can use, but center of mass will be the easiest one to apply because about center of mass angular momentum is I see a momentum, I can literally use about any axis, about any axis torque is 0. So, what will be the initial angular momentum of the system m into u into l by 4, yes or no will be equal to what? Minus of m into u by 2 into l by 4. Now, angular momentum has changed for the particle after collision, then what plus icm which is m l square by 12 into omega, how much omega will come from here? 9 m u by 2 capital m l. How many of you got vcm? u by 2 backward. Yeah, u by 2 backward, in hc or mind is given it comes to rest. Find out the point which will be at rest immediately after collision, find which will be at rest immediately after collision, similar to what we have done sometime back. Yeah, it is. How the rod will rotate after collision? It is l by 3. Rod will rotate like this and will move forward like that, so it has to be above only, below what will happen, omega into l is also this way and v is also this way. So, is it l by 3 above? Yes sir. Let us see, omega into d this way the velocity and this way it will be vcm, so total velocity will be vcm minus omega d, so t will be vcm by omega, vcm that and omega is this, so it will come out to be l by 3. So, at a distance l by 3 from the center above, that point will be at rest immediately after collision, any doubt? Sir, is there always a point at rest? Not necessarily. Sir, how do we know there will be one here? Because we have found out l by 3, if suppose d comes out to be more than l by 2, then that is not on the rod. Try the same thing, below the rod on the bottom side, you get a minus thing right from the center. Below is physically not possible, now both omega into d is also same way as v. Obviously, we get a distance is minus which means etc. Mathematically you will get minus, but people say that minus of 1 in terms of distance has no physical, it is an imaginary number itself, it does not have a physical significance. Alright, so let us take up next question, it is a sphere of mass m radius r, it is angular velocity 2r and the velocity of center of mass is given as v0 on the surface. Suppose coefficient of friction is given to you, mu is the coefficient of friction. Alright, you need to find time. Sir, it is rolling this way. You need to find time, you write down time after which pure rolling will start and you need to find vcm and omega at that point. The most of the questions you are solving today, you will not get it the first time and that is the whole idea, so that you will learn something. If you are able to solve all the questions, wasting your time. Just to better know that 5-3 how much will answer your field good about yourself, go back home happy, but you will not learn anything. This side v0 by 2 omega into r and that side v0, total velocity is v0 by 2 this way. So friction will act in backward direction and what is the value of friction? Mu times normal reaction or not? It is sliding, so this is mg and this is normal reaction, this will be mu times normal reaction. It is sliding right, the point of contact is sliding, so friction is mu times. It is not a static friction. You can do it, you find the acceleration, find out the angular acceleration using torque equation and then how much? Sir, v0 by 7 mu. Find out the velocity, v will be how much? 6 by 7 mu. 6 by 7, correct. How much will be the acceleration of this? Yes, this goes down. What is the force acting on it? Mu times normal reaction which is mu and g, this is equal to m into a, so a will be mu g which were backside, so it is deceleration and then torque equation if you apply, torque because of friction will be mu mg into r. This will be what? It is a sphere, torque equal to i alpha, 2 by 5 m r square i times alpha, torque over center of mass is equal to icm into alpha. So alpha will come out to be mu g by 2 r. Angular acceleration or deceleration? Acceleration. Acceleration, alpha is acceleration. So velocity after time t will be equal to v0 since it is deceleration minus mu gt and omega is omega0 which is v0 by 2 r plus alpha which is 5 mu g by 2 r into t. Understood this v is equal to u plus 8t, omega is equal to omega0 plus alpha t. Suppose t is a time after which pure rolling starts then v by r should be equal to omega or v should be equal to omega r. Just multiply r with this and equate to that, you get t and once you get t, substitute here to get v and substitute here to get omega, any doubts? It is bound over it, you are pulling one end of the string with a force f. Friction is sufficient for pure rolling. The radius is r, it is a sphere of these things. A center of mass alpha and the value of friction. What is f? f is not given to you, you have to find what is f? Capital f. No, about center of mass f and friction, both are there. You might be wondering about direction of friction, it does not matter. You can take any direction. If it is wrong, it will come out to be negative. Since it is a pure rolling, there will be a static friction and static friction can range from zero to maximum value. It can be anything. It is not a constant friction. Only when sliding happens, you can say friction is mu n. The first step is to draw the free body diagram. Which side the friction will be? I don't know. So I will take anything. When they right or left, but many of you are saying left. It turns out to be right and is right. It does not matter. It is fr. This will have alpha and ACM. This is the free body diagram. You must start like this. The net force in the horizontal direction is what? F minus fr. This should be equal to m into ACM. This is your Newton's second law equation. And the torque equation for the object which is rotating. This object is rotating. So fr plus... Both are rotating in the same direction, isn't it? Both are clockwise. This is the net torque. This is equal to 2 by 5 mr square alpha. And since it is a pure rolling on a fixed surface, alpha into r is ACM. Three equations, three variables. Anybody solved it? Yes. Friction is towards the right. Friction equal to mu mg. Because there are four variables. No, no, no. Where are four variables? FR, alpha and A. What else? How much FR you are getting friction as? 3 by 10 mg. 3 by 10 mg? M, A. O, A is also... You are getting in terms of unknown. Solve it. Quickly tell me what is friction. Solve these equations. Tell me what is friction. Wait, sir. That might be a problem. Sir, it is 3 F by 7. Fixed 3 F by 7. 3 F by 7? Yes. Yeah. 3 F by 7. It will be 3 F by 7. Sir, 3 F by 7. Yes. Okay? Yes. See, it does not depend on mass friction. Yeah. Weird. Right? This is what a rolling friction is. It can be anything. See, it actually... mass has an effect of friction. Effect on friction. But friction has to be equal to this for pure rolling. It should adjust itself to this value. If let us say mass is lighter, the maximum value of friction may not be equal to 3 F by 7. Then it will never be having pure rolling. If mass is heavier, then maximum value can go beyond 3 F by 7 and that can pure roll. You are getting it. But for pure rolling, this much friction must be supplied. If friction, maximum value is less than this, it cannot have pure rolling. Alright? So this must be greater than or equal to mu and G. Let us do this. Let us take out this now. Why are you tearing your pages? What are you doing? Sir, he didn't bring his book. No, he brought his book. Now it is 80%. Why are you tearing your book? You are giving him pages. He has a book to tear off the pages. He is exploiting your friendship. Don't be friends with him. Bad person. Alright, so this is... It is a solid sphere. No, I have changed my mind. Should I keep it as this? It is a solid sphere. Stop talking. This side, friction is sufficient for pure rolling. This height is let's say h1. This surface is smooth. You need to find out till what height the sphere will reach. Over here. Same sphere. Now don't talk. Don't talk. Try this. Start from height h1. Burst to height h2. Find out the relation between h1 and h2. Sir, you mean both those arms should be straight, right? What? How do you want to know? Sorry, this is inclined plane. That is also inclined plane. Sir, is it a disk? Sphere, sphere. Solid sphere. Solid sphere. H1 is equal to... But for pure rolling, friction is no good. Stop talking. I told you several times now. It was the same thing. Okay? Come on, let's go. Sorry, I won't do it anymore. Keep quiet. Shut your mouth. When I am requesting couple of times, you should respect that. H1 equal to h2, that is wrong. Alright, that is the hint. Okay, tell me which one will be higher. H1 or h2? H2. H1. H1. H1. H2. H2. No. H2 is lesser. H1 is higher. H1 is higher. H1 is higher. H2 is higher. H1 is higher. H1 is higher. What? Yeah, H1 is higher. Okay. Let's see how high it is. Should I solve? No, sir. H1, h2 is equal to 7 by 5. No, I'm just married off. H2 is lower. How can it be more than h1? Because it gains some rotation which it can't use because it's small. Okay, guys. So, here is one final hint. See, what will happen is that when it rolls down, it will gain what kind of planting energy? Rotation as translation. Rotation is half IC omega square and translation is half MVCM square. When it is going smooth this thing, there's no torque about center of mass. Friction is not there. So, if you draw a free word diagram, there is MG normal reaction that's it. So, the angular acceleration will not be there. So, omega will be constant throughout. So, it will, even if it stops here, it will keep on spinning with whatever angular velocity it had acquired over here. So, that's why h2 has to be less. So, why is being in between? Right? So, omega cannot be decreased. Fine? And hence, whatever rotation energy it has acquired here, cannot be converted into potential energy. Are you getting it? Translation energy, half MVCM square will be equal to MG h2. Half I omega square you have to eliminate. Write the expression. This energy cannot be utilized. It will be hazardous. So, it says that if friction would have been there for it to pure. If it is smooth, it will not go higher. Sir, if there's friction on this side also, then wouldn't the height be the same? Then give pure rolling both sides. I said that rotation energy can't be utilized because omega will be constant. That is why rotation energy can't be utilized. You said when it's smooth, it can't go higher. Correct. Yes. Because if friction is there for pure rolling, you would utilize this also. Then this plus this will be equal to MG h2. But if friction is not there, this will not convert into potential energy. This will remain as it is. Only this one will convert into MG h2. But friction should be sufficient for pure rolling. Sir, won't there be potential energy also when it's at the bottom? Yes. Correct. I am assuming that radius is very less compared to h2. Sir, but I did it like this. You did not like it. No, I did it like this. So, you can keep it there also. I mean, you keep it cancelled. Sir, is it 7 is to 5? No, whatever comes. This is h1 minus r, you can say. This is h2 minus. Sir, if the work done by the friction... What? If the work done by the friction... Where? Here. No, work done is 0 always. If it is pure rolling, work done by the friction is 0. Sir, I am finding it tough. Simple question, simple question. Alright, so this shows a rough track. It shows a rough track, a portion of which is in the form of cylinder of radius r, capital r. With what minimum linear speed should a sphere of radius r... It's a sphere of radius small r, solid sphere. It should be set rolling on the horizontal part so that it completely goes around the circle on the cylindrical part. You have to find out with what velocity you should project it so that it completes the full circle. Simple. It will roll without slipping, throw out. Can you say that this radius is a lot smaller than that one? It is not small. You can't ignore small or complicated. Where is the highest chance the sphere will leave the... The highest point. The highest point. So if this sphere doesn't lose the contact at the topmost point, it will complete the full circle. That is the hint. Foundation for the sphere to leave the surface is normal reaction becoming 0. Just do the full balance or put normalization 0 so you get something. And then apply conservation of energy between this point and that point. Simple. Same thing we are doing again and again. Conservation of energy, torque, equation. There is hardly three or four concepts. Topmost point, draw the force diagram, equate normalization 0. You get a relation between velocity and gravity. Topmost point. Then apply conservation of energy between this point and that point. Okay? Try to do it your own way. It's okay. So is it capital R minus small r by 700? How does noise come? Who created that noise? You. What did I ask you? Which underwrote 20? 20 by 7g into capital R minus... You might have turned sillier somewhere. Oh, it's it's... The root of 35g capital R minus 23r by 700. Wow. Doesn't look that weird. It should be the same thing as like the string. 27 by 7g R minus R. Sir, but won't this just be the same thing as when I using the string to push it... What is string? What is string? String. String and the circle. Is there a difference? Difference is there or not? Sir, which unit is going to rotate? String. String applied is a pull force. The force, the surface applied is a push, normal reaction. Why are you quitting two scenarios? You are over simplifying it. There was no law that one scenario is equal to the other scenario. Okay, so see the thing is that many a times you don't have to learn, you need to unlearn. You need to throw away the assumptions from your head that, okay, fine. You have to keep on reminding yourself that these kind of assumptions should not come in your head. Otherwise, mechanics will really be a problem. Keep it straightforward, simple. The way it is. Just three or four equations. Same thing you have to do again and again. That's all. Read every question on its merit. Don't compare one question from the question which you have solved earlier. Compare it with the concepts. So easy to say like this. Like that. Got it? I just forgot about that. Alright, so normal reaction will be downwards. This is normal reaction. Mg will be downwards. And acceleration. How much will be the acceleration? vcm squared. No, by r minus small r. Center of masses radius is r minus small r, right? It is moving in this radius. This is capital R minus small r, okay? So I'll write n plus mg net force is equal to mass times acceleration of center of mass. How many of you got this equation? This is not this. We haven't learned in this chapter. We have learned in, not work by energy. Little log of motion to write this equation. Net force equal to mass and acceleration. Because the center is moving in the radius of r minus r. You have to see if it is a rigid body, you have to see So what happens is if it is a point mass, then you have to check the radius. But if it is a rigid body, it is a big object. So whose radius will you take? Radius of the center of mass. Center of mass is travelling a radius of r minus r. So r is not radius of the center of mass. You have to check the radius of the path of the center of mass which is r minus r. And normal reaction will be 0 if it just leaves the top surface. So when you put normal reaction 0, you get a limiting condition for it to leave the surface at top most point. If normal reaction is little bit higher, it will complete the full circle. So condition for it to complete a full circle is such that velocity of center of mass over here should be equal to how much? vcm should be equal to root over g times r minus r. This is the condition for it to complete the full circle. This velocity is velocity over here, not over here. I am going to use what is the theorem between this point and that point. Why I am using these two points? Because I have velocity at that point. Which is this? W is 0. U2 is what? Stop talking. Mg what? U2 is? Let's say this is 0. This will be capital R plus r minus r. Let's take this line to be 0 potential energy. It will be easier. So this will be Mg into? U into capital R minus r. 2 into r minus r. This will be your U2. Plus K2 is half icm 2 by 5 mr square vcm square by r square. This is half icm omega square plus half m into vcm square. This is K2. This plus that. Minus U1 is what? 0 plus K1. R right. Small r. What? U1 will be Mg small r. No, I am taking this to be 0 potential energy. Passing through the center of the sphere. Half into 2 by 5 mr square. Initial velocity by r whole square. Plus half m into initial velocity of the center of r square. From here you get initial velocity to center of r square. You understood your errors? Your mistakes? Whatever you have done? You are looking here, right? This chapter is difficult or fluids was? It's really hard. Both are hard. I like this one more than fluids. Both are equally hard. No, no. This one is harder but it's more fun. Flutes. Fluid was not doable. Advanced level. Okay, whatever we are doing is advanced level only today. Now when you look at the mains level question you will be like, oh. Okay, next question. All this, see I am looking at the last questions of HCRMA. Even those kinds we have done already. So I am not able to find a new type of question. All right, here is the question. It's a disk. It's a disk which can rotate about this axis. In fact it is rotating already. Omega naught. Now gently you have kept two masses. M1 and M2 at the edges. And M1 and M2 immediately when you keep it you have attached every quick on it so that when you place it it sticks there. All right? You need to find the final angular velocity. It will attain after you have placed M1 and M2 over the edges. They are point masses. Not required. So the mass is M, radius is R and it is a disk. It's like this. Disk is horizontal. Like this. Disk can rotate horizontally. You may have a worksheet who had that. Solve this question. What I can use here? Consumption. And you want to use. But isn't there an external torque because of the weights of M1 and M2? Who says it's like this? Torque is there or not? Currently. M1 normalization because of M2 and M1. Torque because of M1, M2 and M2G is about the axis that passes through the diameter of this disk. This axis is perpendicular to it. So torque because of the weights of M1 and M2 doesn't pass through this axis. Understood? So torque about this axis is zero. But net external torque is there. But not about this axis. About that axis it is there. I can conserve about this axis. Fine? So initial angular momentum is what? I omega which is M r square by 2 I into omega naught. Initial angular momentum. This will be equal to final angular momentum. Which is what? M r square plus capital K. Plus M1 r square plus M2 r square into omega. So you get omega. Can I conserve mechanical energy? No. You can't. There's more masses. What? There's more masses. Understood? Alright, we take one final question. It's a paragraph type question now. From this only. So most of you got most of the questions wrong from this paper. We attempt that. Okay, do this. Stop talking. Stop talking and listen here. Finish this question then we can all go home. Uniform rod of length L and mass fit freely at one end as shown in the figure. This angle is theta. Whatever is drawn I am drawing exactly the way it is. Alright, the rod was initially vertical. Nursed it and it started falling down. Rotating about this fixed axis. There are three questions. On this scenario. You need to find angular acceleration when the rod makes angle theta like this. You need to find angular velocity when the rod makes angle theta. And you need to find the tangential acceleration of the free end of the rod. You should be able to solve it. Initially it is vertical like that. At this theta, when angle is theta, find out alpha omega of the free end. Same thing we have done. Similar thing on the second class of the rotational. How many class you missed? So it is sad that you are not in it. Sad for you, not for me. You are not looking sad. You have one more week to practice rotational. Mark my words, you will not get time later on. Tell your class 12th class. You can have one more problem solving class. It will get worse. One month we have spent on this already. In between like the J advanced one. How we can have it? I will call you guys. Tuesday we can have problem solving. Find all three and then answer. Next all chapters are pretty straight forward like the thermodynamics was. This was little tricky. Understand it takes lot of practice actually. You cannot sit here and think that. I will understand everything. In school what exactly they are doing in this chapter? They have done this. They have done that. We started today. What did you do? How much they have done? The difference question that is. They are straight forward. Alpha all of you might have got. What a fixed axis is how much? This is mg. Papandika component of mg is how much? Cos of sin theta. Cos of sin theta. Mg sin theta is Papandika component. Mg sin theta into l by 2 is a torque. So mg sin theta l by 2 is a torque. This is equal to i alpha. Which is ml square by 3 times alpha. Torque into i alpha about the fixed axis. So alpha comes out to be 3g by 2l sin theta. This is alpha. How do you get omega? I do omega d omega by d. Other than that you can conserve energy also. That's better. Please find out using conservation of energy quick. So it's l by 2g cos theta by l. Conservation of energy all of you please apply. First mobile phone when I was in second year of engineering. That point in time you were incoming you were charged. And it was all prepared. Do you have any answer? Do you have any conservation of energy? Sir energy should be conserved. Under root of 3g into 1 minus sin theta by 2. I don't know. Okay I'll do it now. For omega using this as my zero potential energy. So final potential energy is what? The height of sensor of mass from here is what? This one is l by 2 cos theta. Mg l by 2 cos theta plus k2 is half i omega square. So m l square by 3 i omega square. Minus u1 is mg l by 2 k1 is 0. Same thing we have been doing for entire class now. So omega will be how much? Okay you can integrate this also and get it. You write alpha as omega d omega by d theta. This is equal to 3g by 2l. Probably when you are integrating you are forgetting that limit is from 0 to theta. So when you put cos 0 it is 1. Integrate sin theta with minus cos theta. So when you put 0 as theta cos of 0 is 1. Alright tangential equation of free end is how much? So it is just 3 that alpha into l. Alpha into l? 3g sin theta by 2. Okay now you are able to solve those kind of questions. Okay we have couple of more minutes. Because of what? So let's leave it. Try attempting the homework questions. Those who have not done the homework probably was feeling totally out of sync today. Okay and this is going to grow as you move slowly and slowly. Class 11 syllabus will be over then class 12 will start coming up piling up. So it becomes very difficult to catch up. Okay so open your eyes accept the reality that it is not straight forward the way it was in class 10. Okay already damage is done and probably it is permanent damage you can't do anything about it. But if you keep ignoring it that permanent damage is going to increase only and you cannot do anything. Okay when you are in middle of class 12 if you are not doing the homework it will destroy your chances completely. Alright so just because you are not doing the problem doesn't mean that others are not doing it. Others are solving many difficult questions and you have chosen to sit with them in an exam J or whatever you are writing. Okay and they are better prepared than you if you don't practice. So they will naturally beat you down. It is natural right. Now you can choose to close your eyes very happy playing video games and chatting, chatting whatever time pause you are doing. That is completely foolish. Okay because soon you will realize that after class 12 and I am seeing it every year. Every year I am seeing it for past 10 years I am seeing this. Every time I say this there are people who take it seriously. There are people who ignore it and then same people who ignore they come back middle of the class 12 and mark my words. They will come with all crying and everything they will say that they are depressed or whatever nonsense. But we can't help that point in time even we can't help. We can console you saying that it is okay and things like that. But then if you don't wake up now time is going to run away and destroy it completely permanently fine. Waking up means that at least get where you deserve. I am not saying get more than what you deserve. I am not saying anyone that be your rank one in J main or J advanced. If you deserve to be in NIT Surat Kal don't take admission in local colleges in Bangalore. You understand what I am trying to say? If you deserve something better let it go and get it. Okay it will be sad that if you deserve let's say college like NIT Surat Kal and you land up in some third grade college in Bangalore somewhere else then you have to naturally prove yourself to the world and listen even though I have land up in some not so good college but I am good. For that you need to prove to the world otherwise world will not accept you. You have to work hard some point in time. Better to do it now otherwise once you are in some third grade college all of it is known to the world. Do you think the world doesn't know what kind of students go where? Okay so make your life easy by working hard in next couple of months and you have another one week's time to practice this chapter which is naturally very difficult I know very difficult because lot of concepts are there so it requires practice. Practice and get what you deserve. I am not saying get something more. Understand the difference right? I am not saying get let's say 360 or 360 but if you can get 150 don't get 50 and if you can get 200 don't get 100. At least be around and if you are around 100 there might be. Okay there a score of 30 or 20. If you think you are at 100 maintain that. Later on it can go up to 120, 130 whatever it is. So don't make it worse. Just maintaining your status also requires lot of effort. Otherwise you may think that you are there only but others are going ahead of you so not doing anything doesn't make your place stable. It makes you go down not doing anything because others are getting ahead of you and it is all relative. Ranks you are going to get right? You don't want to see your ranks in lakhs but somebody will get it. 20 lakhs it will write the exam. 4 fifth of them will get a rank less than sorry more than 1 lakh. So probability of getting a rank more than 1 lakh is what? How much it is? 19 out of 20 which is close to 1. Competition is high. Alright that's it.