 Alright, let's go to the next question. Shall we move on? Let's go to the next question. We keep it simple only. No difficult questions today. Of mass M and length L. Okay, two strings are there. Everything is symmetrical. Properly the method is there. Alright, now this string is cut suddenly. Immediately after cutting, what is the tension on this string? Immediately after cutting. How I got the answer? That's not correct. No, no, no. Immediately after cutting it will be exactly the way it is before cutting. Is it 7 mg per second? Yes, 7 mg per second. How? Something like this you have to do. It will be 3 mg by 2 L. It will be what? 3 mg by 2 L. That's not correct. I said tell me, immediately after cutting, will there be an expression of center of mass? No. Correct? It's alpha. Immediately after cutting, is there a torque? Yes. No. So when I cut it, tension becomes 0 this way. So this is a fixed axis of rotation immediately after cutting. There is a torque because of mg. How much is a torque? mg L by 2. mg L by 2. This should be equal to i is mL2 by 3 into alpha. Alpha is what? 3g by 2 L. So center of mass is what? Alpha by 2. This into L by 2. So this is 3g by 4. There is ACM. This is 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 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? That is axis of rotation. This is not force. It's axis of rotation. Force is equal to mass of rotation I am writing. So we know the concept knowing the concept and applying the concept are two different things. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay. Okay, but I am asking immediately after you give me. No doubt right... Okay. All right. Next question. It's a cylinder which acts like a pulley m1 m2 stop talking mass m and it is R. This is the cylinder. The pulley is a cylinder and there is a friction friction between the cylinder and the thread. Thread does not slip on the cylinder. Thread does not slip on the cylinder. You have 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 it will rotate? This is the real thing. What we have been doing was the ideal scenario where pulley is mass less, friction less, pulley does not 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. Will tension both side be same or not? Same thing, same tension without friction. That is a complete sentence. The friction on the string. 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. Oh, he is selling like. Cylinder is mass per by 2. Why are you selling mass per? It is solid cylinder. This is solid cylinder. How will the pulley be? Why can't it be a... It can be hollow, correct. But what is it last class? If I just say sphere, it means cylinder. Acceleration of the masses and the tension in the string. Should I do it? On the cylinder, 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 makes T1 and T2 different. And let's say it is coming down with acceleration A. This moves up. What is the acceleration of this block? Because of constraint. Equationally, you don't have to solve. It moves, it rotates. Sender rotates. We know, sir, there is something wrong with one side. Sir, there is something wrong. Yes. I'll change it. I'll change it. I'll change it. I'll change it. I got it. What is the force applied on the cylinder? T1 plus T2 is an engine. What is... What is... Suppose if I cut this. I cut it like this. What is the force applied on the cylinder? T1 and T2 considering cylinder plus that string on it as one system. I am 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 we are assuming to be 0. So net force has to be tension only on the string. Equal and opposite it will be. So the equation of M1 will be what? M1g minus T1 is equal to what? M2... T2 minus... Minus M2g equals M2a. Minus M2g equal to M2a. M2a, right? Now this cylinder rotates. Is there a fixed axis? Yes. Fixed axis as well as center of mass of axis both coincide. So it will be... It takes sense of rotation like this to make it consistent with A. If A is down, it has to rotate like this. It cannot rotate in other direction. That way you have to maintain the consistency. So T1 is like this. T2 tries to rotate in opposite direction. So T1r minus T2r is equal to i alpha. i is what? M1 square by 3. 2. M1 square by 2 into alpha. And 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. Any doubts? The friction is sufficient for it to not to slip. Pure 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? Not sure. It's... Do we need to find coefficient of friction or... Sir, coefficient of friction or full friction force? Friction force. Acting on it. It should be what? So that it rolls without slipping. Is it 5G or 2R? More than Mg. Mg r is like... Sir, is it 5G or 2R? Yes. Is it okay? No. So tau is equal to... Which force? Friction? 2 by 5Mg is equal to... No, no, no, no. No, no, no, no. No, no, no, no. No, no, no, no, no. Sir, is it... Mg is the sine theta minus 5 cos theta. Alright, I'll do it now. There will be Mg force. There will be normal force. Friction will be in which direction? Upwards. So that it start rolling like this. You have to look at the rotation as well. This is the force of friction. Okay? Any other force that we have missed? No other force, right? So which force... Are you seeing any fixed axis? Bottom-house point is instantaneously at rest. But at times, we don't feel comfortable trading at this as a fixed axis. So we'll apply torque into IHF about the center of mass. So which force has a torque about center of mass? Friction... Mg passes through the center of mass or not? Yeah. So torque due to Mg is? Zero. Torque due to normal direction? Zero. Only friction will have torque. So if friction is not present, alpha can never happen. Alpha can never be there. It will just slide down. It will not rotate. Because torque is zero about center of mass. Okay? So this should be equal to what? 2 by 5 to alpha. So FR on a fixed surface, alpha into R is ACM. We have done that. Right? So friction will be equal to mass. Okay? ACM. So if I take component of Mg, Mg sin theta minus FR is equal to what? Friction value or I get Mg sin theta minus 2 by 5 is gone. 7 by 5 by 7 g sin theta. Understood? Friction to be equal to 2 by 7 g sin theta. 2 by 7 Mg sin theta. This is what they call rolling friction. Only this much friction is required for it to roll without slipping. Friction is mu into normalization which happens because of sliding. This much will be the friction acting on it. Which is very less. Any doubts? So we can also do this by balancing torque about that point at the bottom. No balanced torque. Yeah, so it's over there. About this you can write a torque Mg sin theta into alpha. This axis into alpha. Okay. You can't use about that axis at the moment of motion. That's why. Okay. But then don't do like that. Because it is a fixed axis but it doesn't appear to be a fixed axis. So every time when you do like that you will not be feeling in control of the torque. Any of your doubts? So what is the moment you can't parallel axis theorem? Yeah parallel axis theorem. 2 by 5 Mg square plus Mg square. That's okay. Any no doubts right?