 Alright, so let's take up next question. It's a sphere of mass m radius r. It has been given of angular velocity of v0 by 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. You need to find time after which pure rolling will start and you need to find vcm and omega at that point. Most of the questions you are solving today you will not get it the first time and that's the whole idea so that you will learn something. If you are able to solve all the questions you are wasting your time here. Now you are doing a 2 plus 2 but how can you know that 5 plus 5 minus 3 how much you will answer your field good about yourself, go back home happy but you will not learn anything. What is the velocity of the bottom most point? Omega into r this way which is what? 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 time 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 r. It is not a static friction you can do it you find the acceleration find out the angle of acceleration using torque equation and then how much? So v0 by 7 mu. Find out the velocity v will be how much? 6 by 7. 6 by 7 correct. Acceleration of this. What is the force acting on it? Mu time normal reaction which is mu mg this is equal to m into a. So a will be mu g which is back side 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 about 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? Angular acceleration. Is deceleration alpha is acceleration. So velocity after time t will be equal to v0 since it is deceleration minus mu gt and omega after time t is omega 0 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 pros a t omega is equal to omega 0 plus alpha t. Suppose t is a time after which pure rolling starts then at that time 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 will get t and once you get t substitute here to get v and substitute here to get omega. Got it? Any doubts? It is a sphere, it is bound over it. You are pulling one end to the string with a force f. Friction is sufficient for pure rolling. The radius is r, it is a sphere of mass and 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 o sorry. No. This 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 0 to maximum value. It can be anything. It is not a constant friction. Only when sliding happens you can say friction is mu n. How are you? Done? So first step is to draw the free body diagram. Which side the friction will be? I do not know. So I will take anything. You can take right or left but many of you are saying left. It turns out to be right and it is right. It does not matter. It is f r. So this will have let us say alpha and a c m. This is the free body diagram. You must start like this. So net force in the horizontal direction is what? f minus f r. This should be equal to m into a c m. This is your Newton's second law equation and the torque equation for the object which is rotating. This object is rotating. So f r plus, both are rotating in the same direction. Both are clockwise. This is the net torque. This is equal to 2 by 5 m r square alpha. And since it is a pure rolling on a fixed surface, alpha into r is a c m. Three equations, three variables. Anybody solved it? Yes. Also friction is towards the right. No, no, no, no. Where are four variables? f r, alpha and a. What else? How much f r are you getting friction as? 3 by 10 m g. 3 by 10 m g? m e. O A is also. You are getting in terms of unknown. Yes. Solve it. Quickly tell me what is friction. Solve these equations. Tell me what is friction. Sir, it is 3 f by 7. Fix it 3 f by 7. 3 f by 7. Yes. 3 f by 7. See, it does not depend on mass friction. This is what the rolling friction is. It can be anything. See, it actually, mass has an 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. So, this must be greater than or equal to mu m g. Let us do this. Let us take a test now. Why are you tearing your pages? What are you saying? Sir, he did not bring his book. No, he brought his book. Now it is 80 percent. 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. So, this is a solid sphere. No, I have changed my mind. Should I keep it as this? It is a solid sphere. Stop talking. Mass m radius r. This side, friction is sufficient for pure rolling. This height in let us say h 1. This surface is smooth. You need to find out till where, till what height the sphere will reach. Now don't talk. Don't talk. Try this. Start from height h 1. Burst to height h 2. Find out relation between h 1 and h 2. How do you want to learn? Sorry, this is inclined plane. That is also inclined plane. Sir, is it a disk? Sphere, sphere. Solid sphere. Sorry. h 1 is equal to… Function is different. But for pure rolling, friction does not work. All right. If you are getting h 1 equal to h 2, that is wrong. All right. That is the hint. Tell me which one will be higher. h 1 or h 2? h 2 is less. h 1 is higher. h 2 is higher. h 2 is higher. Should I solve? Mass answer. This is h by h 2 is equal to 7 by 5. No, I just made it up. h 2 is more… How can it be more h 1? Because it gains some rotation which it cannot use because it is smooth. Okay, guys. So, here is one final hint. See, what will happen is that when it rolls down, it will gain what kind of kinetic energy? Rotation as translation. Rotation is half ICM omega square and translation is half MVCM square. All right. When it is going in a smooth this thing, there is no torque about center of mass. Friction is not there. So, if you draw a free by diagram, there is MG normal reaction. So, the angular acceleration will not be there. So, omega will be constant throughout. So, even if it stops here, it will keep on spinning with whatever angular velocity it had acquired over here. 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? All right? Only translation energy, half MVCM square will be equal to MG h 2. Half i omega square you have to eliminate. Should I do it? I will write the expression down. There are the equations. This energy cannot be utilized. It will be hazardous. Okay? So, it says that if friction would have been there for it to pure roll, it will go higher. If it is smooth, it will not go higher. Sir, if there is friction on this side also, then would the height be the same? Yes. Then, if pure rolling both sides. Sir, I said that rotation energy can be utilized because omega will be constant. That is why rotation energy cannot be utilized. Sir, if it is smooth, it cannot go higher. Correct. Because if friction is there for pure roll, you would utilize this also. Then this plus this will be equal to MG h 2. But if friction is not there, this will not convert into potential energy. This will remain hazardous. Only this one will convert into MG h 2. But friction should be sufficient for pure rolling. Sir, what is the potential energy of friction when it is at the bottom? Yes, correct. I am assuming that radius is very less compared to h 2. Sir, but I did not like it. You did not like it? No, I did not like it. So, you can keep it there also. Sir, is it 7 instead of 5? Yes. Yes, whatever comes. This is h 1 minus r, you can say. This is h 2. h 2 minus r. What? Work done by the friction one or in the box. Where? Here. No, work done is 0 always. If it is pure rolling, work done by the friction is 0. Are you finding it tough? Okay.