 So this shows a rough track, a portion of which is in the form of cylinder of radius are capital R. With what minimum linear speed should a sphere of radius small 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. So can you say that this radius is a lot smaller than that one? It is not small, you can't ignore smaller components. Where is the highest chance the sphere will leave the circle. So if the sphere doesn't lose the contact at the topmost point it will complete the full circle, that is the hint. Condition for the sphere to leave the surface is normal reaction becoming zero. Just do the force balance for the put normalization zero 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 just hardly three or four concepts. Once again topmost point or the force diagram equate normalization zero you get a relation between velocity and gravity, topmost point. Then apply conservation of energy between this point and that point. Okay? Try attempting it, do it your own way. You may be wrong, it's okay. So is it 20 into capital R minus small r by 700 root? Which under root 20? 20 by 7 g into capital R minus. You might have done silly or somewhere. Oh it's it's so. Root 27 by 7 g r minus r. Won't this just be the same thing as when we are using a string to push it out. What is string? What is string? String and the circle. Is there a difference? Difference is there or not? Sir, over here it's going to rotate. Force, string, the force, string applies is a pull force. The force, the surface applies is a push, normal reaction. Why are you quitting two scenarios? You are oversimplifying 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. It will keep it straight forward 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. Got it? Alright so normal reaction will be downwards. This is normal reaction. MG will be downwards and acceleration. How much into the acceleration? vcm square? No, by r minus small r. Center of masses radius is r minus small r. It is moving in this radius. This is capital R minus small r. So I will write n plus MG net force is equal to mass times acceleration of sensor 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. Leave a little other motion to write this equation. Net force equal to mass and acceleration. 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. r is the radius of the path. You have to check the radius of the path of the center of mass, which is r minus r. This equation and normal reaction will be 0 if it just leaves the top surface. So then 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. Now 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? This is 0. Mg into, this will be capital R, 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 1. 2 into r minus r. This will be U2 plus K2 is half icm 2 by 5 mr2 vcm2 by r2. This is half icm omega2 plus half m into vcm2. This is K2, this plus that. Minus U1 is what? 0 plus K1. Small r. U1 will be Mg small r. I am taking this to be 0 potential energy. Passing through the center of the sphere. Half into 2 by 5 mr2 initial velocity by r whole square plus half m into initial velocity of the center of mass square. From here you get initial velocity of the center of mass. You understood your errors? Your mistakes? Whatever you have done? Any doubt or any trouble? This chapter is difficult or Fluids was? It was not hard. I like this one more than Fluids. I like Fluids. Both are equally hard? No. This one is harder but it is more fun. Fluids. Fluids was not doable? Advanced level. Okay, whatever we are doing is advanced level only today. Now if you look at the main of the leg O's. All these see I am looking at the last questions of HCRM. Even those friends we have done already. So I am not able to find a new type of question. Alright, here is a question. It is a disk. It is a disk which can rotate about this axis. Alright. In fact it is rotating already. Omega naught. Alright. Now gently you have kept two masses. M1 and M2 at the edges. Okay. And M1 and M2 immediately when you keep it you have attached a favor quick on it so that when you place it it sticks there. 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. The mass is M, radius is R and it is a disk. It is like this. Disk is horizontal. Like this. Disk can rotate horizontally. Give me that worksheet Who had that? Solve this question. What I can use here? And you want them to use. But isn't there an external torque because of the weights of M1 and M2? Who said it is like like this? Torque is there or not? Right? Normalization because of M2 and M1? No, not normalization. The weight of M2 and M1? Torque because of M1, G and M2, G is about the axis that passes through the diameter of this disk. And this axis is perpendicular to it. So torque because of the weights of M1 and M2 M1 and M2 that torque 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 consider about this axis. Fine? So initial end momentum is what? I omega which is M R square by 2 I into omega this is initial angular momentum this will be equal to final end momentum which is what? M R square plus capital plus M1 R square plus M2 R square into omega. So you will get omega. Can I consider mechanical energy? No. You can't. There are more masses. What? There are 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. Okay do this. Stop talking. And listen here. Finish this question. Then we can all go home. A uniform rod of length L and mass M is fitted freely at one end as shown in the figure. This angle is theta. This angle is theta. This angle is theta. This theta. Whatever is drawn I am drawing exactly the way it is. Okay. Alright the rod was initially vertical. Nursed it and it start falling down. Rotating about this fixed axis. Alright. 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 the rod is like this. Initially it is vertical. Like that. At this theta where angle is theta find out alpha, omega and tangential acceleration of the free end. Same thing we have done. Similar thing on the second class of the rotational. Fourth. 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 12. You will not get time later on. When is the syllabus going to be bad? It will be two weeks. 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. Find all three and then answer. Next all chapters are 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. Okay. Done. Alpha all of you might have got. Talk about the 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 e2 is a torque. Mg sin theta into l by e2 is a torque. So mg sin theta l by e2 is a torque. This is equal to i alpha. Which is m l square by 3 times alpha. Torque into i alpha about the fixed axis. Alpha comes out to be 3g by 2l sin theta is alpha. How do you get omega? Multiply with no omega e. Other than that you can conserve energy also. That's better. Please find out using conservation of energy. Quick. So it's root of mg cos theta by m. Conservation of energy all of you please apply. Strange answer. Is there any answer? Conservation of energy? Energy should be conserved. Under root of 3g into 1l sin theta by 2l sin theta. I don't mind. I don't mind. It's not a different thing but it would be simplified. I'll do it now. For omega using this s by 0 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. So omega will be how much? 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. 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. All right? Tangential equation of free end is how much? So it's just 3 that alpha into l. Alpha into l? 3g sin theta by 2. 3g sin theta by 2. Okay now you are able to solve those kind of questions. We have couple of more minutes. Because I am a workshop. 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. 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. So open your eyes. Open your eyes. Accept the reality. That it is not straight forward the way it was in class 10. 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. When you are in mid of class 12 if you are not doing the homework it will destroy your chances completely. 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. And they are better prepared than you if you don't practice. So they will naturally beat you down. It is natural. Now you can choose to close your eyes and feel that I don't want to do it or I am very happy playing video games and chatting with you whatever time pause you are doing. That is completely foolish 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 mid of the class 12 and mark my words they will come with all crying and everything they will say that they are depressed whatever nonsense but we can't help that point in time even we can't help. We can console you saying it is okay and things like that but then 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 in rank 1 in J mains or J advanced if you deserve to be in 90s suratkal don't take admission in local colleges in bangalore you understand what I am trying to say if you deserve something better then you go and get it okay it will be sad that if you deserve let's say a college like like 90s suratkal and you land up in some third grade college in bangalore or 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 the world doesn't know what kind of students go where they know it okay so make your life easy by working hard in next couple of months and you have another one weeks time practice this chapter which is naturally very difficult I know very difficult because a 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 a score of 30 or 20 if you think you are at 100 maintain that later on it can go up to 120 or 130 whatever it is okay 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 more than 1 lakh right so probability of getting rank more than 1 lakh is what 19 out of 20 which is close to 1 competition is high alright that's it