Fine, so I think we can start now, all right? In the last session, we were doing rigid bird dynamics. We had quickly brushed up all the basic concepts as such, and we had done some problem practice, right? But that was not good enough. So I plan to do a little bit more today on rigid bird dynamics, the problems, and then we move to the gravitation chapter, okay? So without any further discussion, let us move forward. All of you, here is the first question. Start doing it. What is the first thing that comes in your mind which principle will be used here? Correct. Work energy theorem, or you can say conservation of mechanical energy also. Is angular momentum conserved? Is angular momentum conserved? About the hinge, is angular momentum conserved? Why? Why it is not conserved? Because gravity is the external torque about that hinge. So you can't do that. See, many times, you know, when we look at the question, we tend to correlate something which we have done earlier, right? You might have seen something similar previously. You're like, oh, there I have done conservation angular momentum. So quickly that is the first thing that comes in your mind. So when you look at the question, another similar question should not come in your mind. What should come in your mind is the concept. Understand the scenario. Then only start your solution. I hope there is no difference between I speaking without mic or with mic. Because when I speak with mic, I don't need to shout. I'm not in a mood of shouting. Anyone close to the answer? Aditya already got it. Hey, you did conservation of angular momentum or energy. Why? You realize mistake. What mistake? What is a mistake that you realized? What is a mistake? The mistake is rushing very fast. It is not that you ignored here. You know things, but you went into the wrong direction because you're ready to run. As soon as you look at the diagram, you're like, okay, fine. Let me use conservation angular momentum. When you do like this, you will make silly error every time, okay? So be very careful. I'm seeing that thing with you. So take a pause. Take a two second pause every time when you start solving a question. Understood? And you don't need to be first one to be wrong because if you're the first one, typically first one to answer will definitely make a silly error. Getting it? So this is not great fourth or fifth where you have to answer first. Take a pause. You'll get the right answer. I know. Anyways, so we need to find the angular speed when it becomes the vertical scenario. So according to work energy theorem, W is equal to change in potential energy plus change in energy. W is zero. Work done by all the forces is zero. Gravity is doing work, but for gravity we are considering the potential energy and the hinge forces. Are the hinge forces doing any work? Hinge forces doing any work? No, hinge forces are not doing any work because the point of suspension is not moving. Displacement is zero of the point of application of the force. So you can very well write U1 plus K1 is equal to U2 plus K2, like this. Okay, now you have to choose zero potential energy line. So you can say that this is my zero potential energy line. I think most of you have chosen that way only. All right? So when you write the potential energy, you can either consider it as one single rigid body or you can write potential energy plus rod plus potential energy of the particle. Separately also you can write and add them up. So whenever such kind of composite body is there, I am more comfortable in taking it separately for the potential energy. Kinetic energy you can take it combined. You get the same answer. So initial potential energy, you can see that the center of mass of the rod is also at zero line and the particle also at the zero line. So U1 is zero for the rod, zero for the particle plus kinetic energy is also zero. Everything is at rest. Then U2 is what? When the rod becomes vertical, what is the potential energy of the rod? U2 is what? Minus of MGL by two, it comes down. Then potential energy of the point mass is minus of MGL. Right? Minus, it is going down. Potential energy. Now kinetic energy. Kinetic energy can be written as if there is a fixed axis, you can directly write it as half I about the fixed axis omega square. Is there a fixed axis? Clearly there is a fixed axis. The hinge point is a fixed axis. So you can write it as half of moment of inertia body fixed axis. ML square by three for the rod plus for the point mass ML square into omega square. What do you get when you simplify this? Omega has option number one. I hope all of you are getting that. This, okay. Now option number one is the answer. All right, that is fine. Now if I, let's say, extend the question and if I ask you to find out the hinge force, tell me the hinge force when the rod is vertical. How much it is? Find out. You have to draw the free body diagram. Let's say hinge forces HX and HY, Harry Aaron and Pranav got some answer. Then what else? What are the force? It will have gravity force, okay. There will be gravity force from the center of the rod of how much MG and from the edge of the rod, another MG, okay. Is there any acceleration? Is there any acceleration? Any acceleration? There is acceleration. You forget that it is moving in a circle, okay. Centripetal acceleration. So this point is also moving in a circle and the center of mass is also moving in a circle. Fine. What you could do is find out the center of mass position for rod and the point mass from this edge. So that would be M into L by two plus M into L divided by M plus M, which is two M. So this is three L by four. So at a distance of three L by four, this is the center of mass for both the things. So the centripetal force would be towards the center which will be omega square distance of center of mass, three L by four. This is the acceleration, centripetal acceleration. Is it clear to all of you? Type in. So net force towards the center, H Y minus MG, minus MG, this is equal to total mass times acceleration of center of mass, which is this. So from here, you get the value of H Y and H X is zero. As torque is zero, so tangential acceleration, which is alpha into R center of mass, that is also zero. Okay, so I just told you something little bit more than what question is asking because same question will not come in the exam, right? So something like this comes, I have seen people ignoring the centripetal acceleration as if it doesn't exist. But remember, it is rotational motion. Everywhere there will be centripetal acceleration. Everything is rotating. Don't forget, do this. Can you got something? Check, can you whether that is right or not? Are you able to see the mass, total mass behind my back? That's the corner we put all our things which we are not using. So unfortunately, that is seeing on my background. All right, many of you have answered. Any other answer? Any other answer? That's it? Okay, all right, let us discuss. Is there a fixed axis of rotation? Tell me, is there a fixed axis? Who is rotating? The, this thing is rotating, right? The cylinder is rotating. Now, is there a fixed axis? Is there a fixed axis of the cylinder to rotate? There is an instantaneous fixed axis of rotation, but that is fine. Even if you are not able to visualize, that's fine. What did I tell you in the last session? If you're not able to find out whether it is a fixed axis of rotation or not, and you want to use torque equal to I alpha. So about which axis you can use it safely about center of mass. So it is understandable that sometimes it is tricky to understand that this is the fixed axis. All right, so let us ignore that and solve it as if there is no fixed axis. All right, so consider reel as a hollow cylinder, acceleration of the center of the reel is what they're asking. So let's say this is tension T, gravity force would be from here, MG, this is alpha. Okay, will the center of mass accelerate or not? Center of mass will accelerate for the cylinder, right? It will have an acceleration of center of mass AC. Now, is this a pure rolling about this point about the thread? Is it sort of pure rolling? It's a pure rolling, right? And is thread is moving or thread is fixed? This entire thread is fixed. It is not moving at all. So you can say that since pure rolling sort of scenario is happening, acceleration of center is equal to alpha times r. Okay, that's a constraint equation because of the pure rolling. Then you have downward direction, MG minus T, net force in the downward direction is mass times acceleration of center. And torque about the center of mass is what T into r. Is there any other torque about the center of mass other than the tension torque? No, so T into r, this is a hollow cylinder. So m r square by two times alpha, okay? So one r is gone, so T is equal to m AC by two. What they're asking, acceleration, right? So substitute the value of T over there. So MG minus of m AC by two is equal to m times AC. So three AC by two is equal to G. And you don't need to now write that AC is two G by three. Just pick option four, fine? Got it? Is it clear to all of you? So while you are in the examination hall, in fact, there are some students who could write this and then directly pick the answer. You don't need to write all these steps in the examination hall. While practicing you should. Hollow cylinder moment of inertia is, oh, is it hollow cylinder? Real as a hollow cylinder. Yeah, it is MR square. I have considered it as like the solid cylinder. Sorry about that. So MR square alpha is ICM, okay? So I think answer will change now. AC would become equal to G by two, right? It'll be one, all right? It's the one in the Google it is, they have considered solid cylinder, I guess. They have taken MR square by two. If you search Google, all right? So let us move forward. Do it your own all these questions. I mean, we are towards the end of the session.  Brutally honest with yourself, okay? That is how you learn. That's the most important thing. Do this. Okay, I can see some of you have got the answer. Some of the students are joining now. So let me put them in after this question itself. All right, many of you got some answer. Good to see that. Let us solve this question. A solid sphere of mass M and radius R is pulled by a force F. Sphere does not slip over the surface. Friction force acting on the sphere. Now this is a routine type of question, isn't it? So let us say friction is acting on it, the backward direction, the friction, okay? Normal reaction is upward, gravity downwards and gravity down. So I want to write torque about the center of mass is equal to I center of mass alpha, okay? So which force is having torque of all the forces? Which of the forces have torque? F and the friction force, right? So the torque you do F and the torque you do friction force will get added up or subtracted. You get added up. Both are trying to rotate in this way only. I mean, I may be wrong. The direction of friction may be wrong, but whatever I have assumed, the torque will get added up. So this is alpha, A center of mass. Torque about this center of mass is F into R plus R into friction force. It's a solid sphere. So two by five MR square alpha. Then it does not slip. It is a rolling on a fixed surface. So AC would be equal to R alpha. And net force F minus small F should be equal to mass times friction of center. So we can now quickly solve it. F plus small F is equal to two by five M into A center. Mass, then you can subtract it. So you'll get minus of two F is equal to three by five M into ACM. Then you get ACM, substitute ACM, you'll get the value of F. What do you get the answer as? Can you do all of this? So number two, is the direction of force of friction correct or not? Friction, whatever we have assumed, it is coming out to be negative, right? Friction is coming out to be negative. F is equal to minus of three by 10 M into ACM. So that's why force of friction will be in this direction. Do this, this is another type of question which comes regularly. Someone is asking, do we ever need to be careful of choosing the direction of friction? Sometimes I have seen, they ask questions that which direction the friction will be, okay? So whichever direction the surface is trying to slide, trying to slide, opposite to that direction of friction is.  Okay. So what is the condition? What is the condition? So that it can be pulled over this step. What is the condition for that to happen? Right, so torque which is trying to rotate in this way is more than the torque which is trying to rotate that way. I hope this is clear to all of you. And when it is about to pull over, which force will become zero? Normal force will become zero. So these are the conditions. If this condition is not clear to you, you'll not be able to solve such questions, fine? All right, so if that is a situation, we need to draw the free buoy diagram. And this is the normal reaction, passing through the center, normal reaction, then gravity force, is there any other force? Any other force? That's it, right? Friction from where friction will be there? It has been lifted up slightly from the ground. It has been lifted up from the ground, okay? Friction from the step. Oh, that is something you want to consider. Consider it. Although it is not given, no, it will not slip. Don't worry about it. You can assume friction though. Let's say this is the friction from the step, fine? This is the friction. Now, normal reaction and friction, which we have considered, the torque due to those two forces will be zero about this point or not. This is a fixed axis of rotation. So if torque about that fixed axis is just greater than zero, it is like tending to zero. Initially, what will happen is that torque due to gravity will be more than torque due to F. If F is less, then torque due to F kept on increasing. And after some time, the force of this force, which you are pulling, becomes so big that it will topple over. So when it is just starting, that point in time, torque due to F and torque due to MG will be equal and opposite, fine? So perpendicular distance of MG from this axis is given to us, right? No, it is not given to us. This distance is how much? Anybody calculated? This is R by two. This is R. So this distance is root three by two, root three by two. So MG into root of three by two is a torque due to MG. This should be less than or equal to torque due to force. Pappanian distance of this force is what? From this point, what is that distance? R by two, F into R by two, okay? So F should be greater than or equal to root three times MG. So option number three is there. Clear to all of you? Getting the answer is not important. Understanding that what is the condition that is important. All right, let's move. Don't worry, everything is getting recorded. I'm going to share the recording also. Condition is normal distance becoming zero and torque due to F is greater than torque due to gravity. That is the condition. Do this. Sir, why torque due to MG should be more? Initially it is more when capital F is very less. You just keep the sphere and capital F is very less. Then it will remain there only because torque due to gravity is slightly more than the torque due to F, about that point, fine? But it is not rotating. Even though torque due to gravity is more than F because normal reaction is also there, right? Normal reaction is MG. So torque due to a normal reaction and MG, they'll cancel out each other. But when it is about to topple, normal reaction disappears. Do this, this is about the collisions. When collision between the rigid body happens, can we conserve the linear momentum? Yes or no? Like the way we do with the point masses, can we conserve the linear momentum for the rigid body also? Yes. But we have to take the velocity of center of mass. What do you mean conservation of linear momentum about the point? Linear momentum conservation is not about the point. It is about the direction. I'm not talking about angular momentum. Do this, everyone, this question, all of you. First one is solid sphere and second one is identical sphere. The second one becomes solid or not? All right, so shall we solve this? Shall we solve this? 20 seconds, okay. I can see many of you giving the answers. Oh, I have to take the attendance also. I'll just quickly take the attendance. Take couple of seconds more. All right, so let us solve this. Hmm, huh. So can I conserve the linear momentum just before and after collision? Type in yes or no? But there's a friction, right? There is friction. It's a rough surface. So can I conserve linear momentum? There is external force horizontally. Can I conserve the linear momentum? Now tell me. First, this sphere is pure rolling, so what? Friction will be there or not? Friction is there. So can you conserve the linear momentum? Okay, somebody is saying it does not do any work. That is immaterial, whether it does work or not. There's, what is it?  Things are pretty simple. If external force is there, you can't conserve momentum. Is it written anywhere that it has to do the work, then only you can't conserve the momentum? No, no, no, no. You are creating rules. So clear that, clear it from your head. All those things are not valid. But yes, you can conserve momentum. You can conserve momentum even though there is external force. The reason for that is that it is not impulsive. It is non-impulsive force. So just before and after, immediately before and after you can conserve momentum. If you leave it for very long, you can't conserve momentum. Let's say five hours before and two hours after the collision. The friction will change the velocity and angular velocity, okay? So yes, you can conserve the angular momentum because it is non-impulsive. And if you do not know what is impulsive, non-impulsive, get in touch with me after the class. I will send you a video. I can't teach you what is impulsive, what is non-impulsive in this session, okay? Otherwise, half an hour or 45 minutes will be gone there itself. Isn't it impulsive on the second sphere? No, it is not impulsive on the second sphere. The force of collision is this way, right? The first sphere comes and hits the other sphere this way. So this is the normal reaction between these two. So why friction has to be impulsive? The force of collision is horizontal. Friction at max will be mu times this normal reaction. Will this normal reaction change due to the collision or it remains mg? Normal reaction from the ground remains mg or not? Stays mg, right? So friction doesn't change during the collision. So just before and after you can conserve the linear momentum, they are this thing. Identical elastic collision happens, head-on collision happens. So can I say that the first one comes to rest and second one starts moving with the same velocity? Can I say that everyone? They exchange velocities in the elastic collision. So the second one is moving with velocity of v0, v0 velocity and the, okay. Coincident friction is given? Plus your second wave when it starts pure rolling. No, it is not given to us. What will be omega of the second one? Initial omega, immediately after collision. Is there an impulsive torque about the center? No, impulsive force is there, but torque due to that is zero, okay? So the initial angle velocity is zero, but friction will be there, right? What kind of friction will be there? Kinetic or static initially? Kinetic, right? Will be kinetic, sliding is happening, omega is not there, it'll be this way. Whatever it is, anyway, coefficient of friction is not given to us. Let's say friction is f. The best thing about the kinetic friction is it remain constant. So if it is a sliding happening, the deceleration will be constant. So you can use equations of motion. Okay, someone is asking, didn't the first sphere already have angle velocity? How is initial omega zero? Yeah, first one has angle velocity, right? First one has it, but why second one will have? Why second one will have? You tell me that. Is there any torque due to the collision about the center of mass? For second sphere to have omega, you need impulse torque. Impulse torque is not there. Friction between two is very zero. It is smooth, clearly written. Read the question. Okay, anyways, so the acceleration of the sphere of mass m is f by m. And it is the deceleration in backward direction. There's a deceleration. So the velocity after time t is v naught minus f by m into t. And omega, angle velocity will be there because there is alpha. Omega is omega naught, which is zero plus alpha into t. Okay, now how will you find omega? Sorry, alpha, how will you find? Friction is creating torque f into r is equal to two by five m r square alpha. Okay, so alpha is five f by two m r. This alpha you have to use here. And if pure rolling starts, then v should be equal to omega r. And this will give you time t when the pure rolling will start. And once you get the value of t, substitute it here, you'll get the value of velocity. Okay, what do you get the answer as here? Everyone, what do you get on here? What do you get answer here as? Good, so just do this calculation and find out. Okay, someone is saying, we can conserve angular momentum. Yes, there are some textbook and in the Google also, people have conserved the angular momentum to find out the answer. But for that, you need to understand the principle of instantaneous axis of rotation. This is a fixed axis of rotation actually. About this point, net torque is zero. So you can conserve the angular momentum initially and finally. You can get the answer like that also. Okay, but then if you have not done it till now, I would say avoided. When you look at the solution somewhere, they do talk about the other method also. But this is the regular one, which you can apply anywhere. So I'm talking about a generic method. Understood, Siddharth? Okay, someone is asking, what happens to the first sphere? Who cares about the first sphere? First sphere stops, good. Are you saying first spheres, linear velocity becomes zero. What about the angular velocity? What about its angular velocity? Angle velocity remains unchanged, non-zero, it will be that only or not? It'll remain that only whatever it was earlier. Okay, so if there is this sphere, linear velocity becomes zero and it has the angular velocity omega. So what will happen? This point is trying to go back. So friction is acting this way. Fine, so now after some time, even this will start pure rolling. Omega will decrease, linear velocity will increase. And then pure rolling will start after some time, even for the second one also. Clear to everyone? Okay. Good that you're asking intelligent questions like that only. We should probe all the situations. Our motive here is not to get the answer for one question. What is our motive? To get the clarity of the situation because the same thing can come in a different manner in the exam. Same question will not come exactly. Do this. This is on linear impulse and angular impulse. Okay, some of you got the answer already. Any other answer? Linear impulse is equal to what? What is the impulse equation? Linear impulse equals to? Linear impulse is change in the momentum. Okay. So linear impulse to the system is M into V. These are linear impulse.  The rod is massless. They're going to change in momentum. So two M into velocity of center of mass minus zero. So the velocity of center of mass is V by two. It comes directly like this. Then, but nobody is interested in finding velocity of center of mass. What they're asking is angular velocity. This impulse also creates an angular impulse. This linear impulse has an angular impulse about the center of mass. So I can write the angular impulse, which is linear impulse multiplied by the perpendicular distance from the center. This is the angular impulse. This is equal to change in the angular momentum about the center of mass. Change in the angular momentum about the center of mass is ICM, final angular velocity minus initial angular velocity. Initial one is zero. ICM is M into L by two, whole square, that two masses, so into two omega F, omega I is zero. So what you get omega F as from here, option one, this is how you do this. You cannot conserve the energy here. How can you conserve energy? Clear to all of you? Type in, can we, somebody asking, can we consider second mass as an instantaneous center of rotation? I'm not using instantaneous axis of rotation to solve any of the questions as of now, because it is not required for J main at least, fine? You can solve it without that. Someone is asking in these situations, it is not absolutely necessary to do anything about center of mass, even if suppose we wrote angular momentum about the left mass, does it still works? Are you telling me or asking me? Are you telling me or asking? See the angular momentum, no, no, no, Siddharth, that's incorrect, totally incorrect. See torque is equal to I alpha. This equation is valid about the center of mass or about the fixed axis. Similarly, the angular impulse, the angular impulse is equal to change in the angular momentum. This is also valid about the center of mass or about the fixed axis. In this question, this point happened to be, sorry, not that point. The other point, this point happened to be instantaneous axis of rotation. So that is why doing it like that, you are getting answer in this question. But if suppose this doesn't turn out to be instantaneous axis of rotation, then whatever you did will give you the wrong answer, fine? So always use this equation, angular impulse equation, either about the fixed axis or about the center of mass axis. If it is not clear where is the fixed axis, use it about the center of mass axis.  Clear to that? Okay, so whenever you have such misconceptions or something which you are not very clear, voice is out.  Many times we have something in our head, we want to use it again and again, just clarify it once. You don't need to shy away asking such things. Okay, good that you asked. Solve this question. I think VCM you can get easily. Can you tell me what is the loss of center of mass? Zero, it will be zero. So it is between two and three. Conservation of linear momentum. Initial linear momentum, two M into V minus M into V, this is equal to zero. That should be equal to two M, three M, eight M, that is 11 M into VC. So velocity of center of mass is zero. Okay, people are getting different answers. Two particles mass M and two M strike the rod and stick to the rod after collision, velocity of center of mass and the angular velocity about the center of mass. See, when they say angular velocity about the center of mass, it doesn't mean anything. They should have just said, what is the angular velocity? They don't need to write that. Angular velocity is same about any point. So let's say angular velocity is this, final angular velocity. So can I conserve the angular momentum? Tell me, can I conserve the angular momentum? People are saying, yes, but then when these particles are moving and hitting the rod, they will apply force on the rod, right? That will create external torque. That will create the torque, yes or no? So how can you apply the conservation of the angular momentum? They are the internal forces because, because your system is all of it together. About which axis can I apply conservation of angular momentum? About which axis? I'm asking about which axis I can apply. I may or may not apply that axis. Only about center of mass, only about center of mass or about any axis. Let's say I'm applying about this axis. Is external torque is there about that axis? No, right, about any axis. But when I'm using about any axis, then it is difficult to write this thing. Difficult to write the angular momentum equation about any axis. If I'm using the center of mass, it is simply I into omega is the formula. Otherwise, the angular momentum expression becomes little lengthy and complex. So that is the reason why, even though I can write about any axis, but I'm using center of mass axis to write the conservation of angular momentum. Conservation of angular momentum about the center of the rod. So let us do that. Initial angular momentum is what? M into two V into two A. Now angular momentum of M and two M will get added up or subtracted. It'll get added up. Plus two M into V into A. This is the initial angular momentum. This should be equal to the final angular momentum. Okay, so how will you write the angular momentum of the rod and the, okay. Here, when it gets stick over it, the problem is center of mass location changes. The center of rod no longer remains the center of mass. So we have to be very careful, okay. So what you can do is that after the collision, you write the angular momentum of all of it separately. For example, this mass, see VCM is zero. This mass is moving with omega into two A velocity. That mass is moving with whatever is angular velocity into A, right? Okay, the center of mass remains same, is it? It is two M is at a distance of A and M is at a distance of two A. Center of mass remains same, but then suppose it is not same. Suppose the center of mass doesn't remain same. Then what you have to do is this. M into velocity after the collision into two A plus two M into velocity after the collision into A plus ICM, which is eight M. How much a distance two A, two A, four, five, six. Eight M, six A square by 12 times omega. Is it clear to all of you, this equation? How I wrote the angular momentum of three things separately. Mass into its velocity after the collision into perpendicular distance. And then for the rod separately, you could have written, because center of mass remains unchanged, you could have written two M into A square plus M into two A square plus eight M, six A square by 12 times omega. That exactly what it comes if you take omega common in these three expression. Clear to all of you, type in Okay, someone is asking even if center of mass changes what is the issue in writing the angular momentum like this? See angular momentum about the fixed axis is I F into omega. Angular momentum of the center of mass is ICM into omega. And angular momentum about any other point other than the center of mass and the fixed axis is our perpendicular interval. So this is our perpendicular into M into VCM plus ICM omega. So then you have to use this expression and you don't want to use this. You want to keep things simple, isn't it? That is why I mean, I have taken this separately but it turns out luckily this question the center of mass doesn't change, do this. You can use impulse equation anywhere. Impulse equation is more generic than the torque equation, more generic than the conservation of angular momentum. You can use it anywhere. It is in a crude manner. The most basic thing is impulse because of impulse only everything is happening. If you're comfortable with the impulse equation. Do this, find out the final answer, say Dan how does it matter? Equations are written find it out. What is the previous answer? By the way, anybody found out V by six A those who want to write it V by six A was the end of velocity. Okay, Aditya is getting something else. But that's what I'm telling you, solve it your own. Okay, solve this question your own now, all of you. You're not here to create notes. This is not a session where you are creating notes for yourself, all right? So I have written equation, that's it. You should be able to do it now. Okay, Ritu and Siddhas got two different answers. What will be the final velocity? Immediately after collision, the velocity will become what? When it comes to rest, zero. No, total momentum will be zero. Velocity will be exchanged. Now left is fear will go in the left sphere which was going on the right hand side will go on the left and right is fear which is going on the left hand side will move on the right. Okay, velocities will get exchanged. V naught, V naught, okay. But what about the angle of velocity? Will angle of velocity change the direction? Yes or no? Angle of velocity will remain like this only, okay? Angle of velocity will be like this. Angle of velocity will be like, okay, fine. I was going this way, suddenly what happened? So this is the angle of velocity. What is the value of angle of velocity? V naught by R only, right? And because of the symmetry of situation, whatever happens to the one sphere, same thing happens to the other sphere also, okay? Direction of friction at the bottom most point is what? Which direction it is? You can clearly tell because the bottom most point is going on the left hand side. So friction is this way. So acceleration is F by M. Same thing you have to do here, but you have to take care of the directions, all right? So V velocity is V naught minus F by M into T. This is the velocity V and the torque. Friction into R is equal to two by five MR square alpha. Alpha is five F by two MR. So omega is equal to omega naught. Now, alpha is opposite to the omega or not. Angular acceleration is opposite direction or in the same direction as omega. It is opposite. So minus will come. Minus of alpha T. Omega naught plus alpha T. Now, how do you relate V and omega? Tell me. When pure rolling starts after a long time, how do you relate? Everyone? No. The direction of omega and direction of velocity will remain like this even if pure rolling starts or it will reversed. We reversed, right? Here, we are going to reverse. We reversed, right? Here, which direction you are considering positive? For angular velocity, which direction you are considering positive? Clockwise or anti-clockwise? Clockwise. Here, left-hand side, right? Suppose the sphere starts pure rolling on moving on the left-hand side. So this is what will happen? The omega will reverse finally or not? Direction of omega should change. It has to do pure rolling. So that is why V should be equal to minus of omega into r. Clear to all of you? Okay. Do this. Get the value of T and then you substitute here to get the answer for the V. Can anyone get it quickly? Solve it quick and tell me what is the answer you are getting? 3V naught by 7. Correct. This is what everyone is getting. I hope it is clear to all of you. Type in everyone. F will be mu times normal reaction till the pure rolling starts. Yes. As soon as pure rolling starts, friction is not mu times normal reaction. Someone is asking, why did they mention after a sufficiently long time? Was it to consider after time T after time T after collision? You will get it. Basically, once the pure rolling starts, they want that situation. Okay. Because pure rolling will be happening for some time at least. They should have written clearly though. But I just guessed it. Do this. Read the question and first tell me the condition for that to happen. Okay. Others, what is the condition? The condition is the structure rotates opposite direction. The torque generates. Have you observed how many of you have taken ride on a boat? Not the ship boat. Smaller boats. Have you observed that the boat shakes like this whenever it moves forward? It's not that the boat doesn't oscillate. Boat moves but still it doesn't topple over ever. Even the bigger boats also where 10 to 12 people are there. Sometimes the boat has ground floor and the first floor also. Even there also, however much the oscillation happens more and more like say the water is moving. Then also the boat doesn't topple over. The reason is that if the boat rotates like this then a torque gets developed in the opposite direction. So immediately when it goes this way quickly it comes like that because of the opposing torque. So it tries to bring it to that position. Now why it happens? Let us quickly do that here. See, suppose this rotates little bit. I may not be draw the perfect figure here. Suppose it rotates this way. This is the center of the sphere. This is the ground. So the normal reaction from the ground will it pass through the center of the sphere or not? Everyone, will it pass through the center of the sphere or not? This is the point. The ground is a tangent. Ground is tangent to the hemisphere and normal will pass through the center. All of you type in, is this clear? Now if center of mass is here if center of mass is here then the gravity force will act like that. So will it rotate more this way or it will go back? What do you think will happen? More it will rotate. So this is unstable. This is unstable. But if center of mass is somewhere here let's say assume it will be there then tell me what happens? About this point will it try to go back or it will rotate more? It will try to go back. So this is a very important criteria of designing boats. The buoyancy force is there. The gravity force is there. So like that they design it. Here maximum value of L so that it remains in equilibrium. So if the center of mass comes over here then the gravity force will act here only. At max you can have it over there for stable equilibrium. At max your center of mass can be on this line. It should not be on the right hand side of the line. Getting it. So let us assume that center of mass is at the center of this hemisphere over there. Not center of the hemisphere center of the sphere for which the hemisphere is a part of. So that is the condition. Center of mass lying over there for both the structure. Can you tell me location of center of mass for hemisphere? What it is? We derived it. For hemisphere. 3R by 8 or R by 2? Two different answers people are giving. R by 2? 3R by 8? Okay. Two people have said 3R by 8? It is a shell. Okay. Shell is R by 2. Solid hemisphere is 3R by 8. Now do it. Get the answer. Archer got something. Others? Center of mass of entire structure should lie here. See because of symmetry of the problem center of mass must lie on this line. Right? So the only thing when it rotates what will happen is that the normal will go like this. Whenever is the normal it will always go through the center. So center of mass has to be on this line only and make sure that it doesn't become unstable. It has to be here at least. So maximum value of L so that it remains in stable equilibrium. Mass of the hemisphere is proportional to the surface area. Sigma into 2 pi R square the mass of this cylinder that is also hollow. Yeah. Sigma into 2 pi R L. So let us say my Y axis is this, X axis is that. So Y center of mass which is R should be equal to sigma 2 pi R square this is M1 from here to there the center of mass distance is R by 2 plus sigma 2 pi R L center of mass is here which is R plus L by 2 Anybody did like this what you get answer as another like I have not use that simpler way is taking O as the center of the axis that would have been lot simpler I think okay. Yeah. If the center of the coordinate axis is this then 0 should be equal to sigma 2 pi R square into minus of R by 2 plus sigma 2 pi R L into L by 2 right this will be simpler. Are you getting option to write do this suppose we don't get enough time to do orientation then we can do little bit of it next class as well but I think we will get time okay anyone fine so let us do this shall we do it now hmm uniform mass M radius R point mass M circumference rolling without slipping station of point O normal reaction the disc I have to find out. So free by diagram nothing else normal reaction up MG down and let's say friction is backwards no other force is center of mass accelerating vertical direction alpha is A naught by R or not it is pure rolling where is center of mass by the way where is center of mass center of mass is here in the middle R by 2 distance away right so all of you do you agree that it will have a tangential acceleration in downward direction which is alpha into R by 2 this way the expression is A naught do you all agree type in vertical direction if you write down the forces MG minus normal reaction is equal to not MG 2 MG there is MG from here also point mass MG also this so 2 MG down minus normalization I am just looking at forces I am not looking from where it is applied vertical direction 2 MG minus N is the force that should be equal to 2 times into alpha into R by 2 okay so 2 MG minus normal reaction should be equal to 2 M alpha is A naught by R A naught by 2 2 and 2 is gone so normal reaction is MG minus M into A naught so it is less than 2 MG clear to all of you type in we will come back to these questions if time permits later on but anyways solve this one at least one numeric type simple question many of you got the answer already others any other answer alright okay see these are the kind of question you should never get it wrong in the exam so never hurry up I can see that some of you getting the wrong answer and when you read the question it appears it's a very simple and straightforward question right don't get it wrong for your own sake moment of inertia of the uniform semi-circular wire about an axis passing to the center of mass perpendicular to the plane this is the if let's say semi-circular wire is of mass M circular wire will be of what mass 2M so moment of inertia would be 2MR square for full circle and for half of it it will be half of that so for this moment of inertia is MR square but that is the moment of inertia without this axis what they are asking is moment of inertia passing to the center of mass this moment of inertia they want do you know what is this distance center of mass to this points distance is what 2R by pi that is 2R by pi so according to the parallel axis theorem ICM plus M into D square is equal to I so ICM plus M into 2R by pi whole square this is equal to MR square actually which you have got okay so anybody did like this 1 minus 4 by 10 okay 10 is in the denominator so 6 by 10 so X is 6 I hope this is clear to everyone Archer understood are you run clear okay someone is asking sir can we find the moment of inertia of any axis through the center of the semi-ring in the same way I didn't get what are you asking I'll unmute you speak sir can we find the moment of inertia through the semi-circular ring through any such axis it doesn't need to be any symmetric axis but through any such axis what such axis means what like any moment of inertia of the full circle and when you remove half of it symmetrically moment of inertia becomes half and you realize that that is not the center of mass center of mass has shifted down so use that only yes sir but can we do it if that axis of that we are considering it was any axis that let's say it was on the plane of the ring at some angle here no there and passing through the centers so it's on the plane the axis is on the plane oh okay huh that suppose you know the for full circle moment of inertia about that axis is MR square by 2 so for half of it will be half same thing you can do yes can do that way okay why not you should ask why not right is there any flaw in this solution like that you should critically analyze okay but then yes you can do it alright let's move ahead how many questions we have here never ending anyways we know that rigid body dynamics is never ending chapter this looks interesting so we will solve this question before starting the gravitation okay do this whatever it is like if we are getting centimeter just tell me the speed forget about what is the value of x tell me the speed directly with the units don't worry about this thing but they should have specified speed is in centimeter per second or meter per second okay so that got it anything or anybody else okay nobody else is it lendy alright so let us do it the speed of the block when it descended by 2 centimeter spring was unstretched so I hope all of you understand we will be using conservation of mechanical energy here this is 2 centimeter it has descended u2 plus k2 is equal to u1 plus k1 potential if we have two types one is gravitation potential other one is spring potential you have to take both so let's say the initial gravitation potential energy hold on initial gravitation potential energy about this red line is let's say 0 that is our reference line so u2 is minus of 10 g is also 10 let's say mgh 0.02 this is the final gravitation potential energy how much spring will be extended will it be the same thing right so 0.02 whole square so this is the final potential energy final potential energy is this now if rather than getting extended if spring get compressed then potential energy remain this only or it will change it will remain same it does not matter whether spring is compressed or extended potential energy is half kx square only this is the final potential energy plus kinetic energy half 10 into velocity square plus half i omega square for this what is that cylinder right whatever it is we are assuming it is rotating half i omega square omega is u by r u by 0.1 whole square right so this is u2 this is k2 this is equal to u1 which is 0 plus k1 which is 0 so those who got the answer have you done it like this get the answer now quick what are you getting i think some of you might have ignored can i guess what kind of cylinder you did you ignored kinetic energy of this cylinder no then what do you i ignored work from the tension tension is internal force and i mean tension won't do the work actually whatever work tension does here get cancelled over there why does pulley rotate when there is no net torque who says there is no net torque first of all it should be written actually it is not written the question it should be written clearly that the cylinder at top rotates without or at least should be written that the rope doesn't slip over the cylinder that should be written okay it rotates because the tension on both sides they are different they are not same t1 and t2 are not same so there is a torque what do you get answer here as everyone u is what 3 by 10 meter per second correct okay so let us start the gravitation now let us start the gravitation now everyone in gravitation what all things are there in gravitation we have we have the first is universal law of gravitation which tells us the force between the two point masses okay then we have Kepler laws 3 Kepler laws you remember that or not law of areas, law of periods and and what else law of areas, law of periods and law of orbits okay but we ignore law of orbits in this chapter at least for j main we ignore the fact that thus satellites or planets are moving in a elliptical path we say it is a circular okay so we are studying a circular motion fine now gravitation we have already accounted for till now in work by energy we had gravitation force and we have the gravitation potential energy all that we had but in this chapter we are studying the gravitational force again and gravitation potential energy again because when I say gravitation force is mg there are two things first I am assuming only earth is applying gravity force which is wrong even this bottle is applying gravity force on this object even these two have gravitational force so I am ignoring that I am only assuming that earth is applying gravity force second thing which I was assuming was that earth is flat which is a reasonable assumption if you are near to the surface of the earth but as soon as you go very far away or let's say you are taking the gravitational force between moon and the earth you can't assume earth to be flat third thing was we assume that gravity is constant throughout all of these assumption goes for a toss if we talk about the larger distances so that is why there is a separate chapter in fact this chapter is so big that people spend their entire lifetime studying only gravitation Einstein was one of them okay then what was that other person named the string theory Stephen Hawkins he also did only the gravity study so gravity is the fundamental force and if you understand gravity people say that you have understood the entire universe also in a way but anyways our task is to get the marks in this chapter so even though the universal law of gravitation f is equal to g m1 m2 by r square is there with us but then this is rendered useless if we talk about the force between a point mass and let's say a rod you take rs sorry r is this distance this distance that distance what you take you can't take any of them because the gravitational law is valid only for the point masses so you define something called as gravitational field which is force per unit mass which is gm by r square gravitational field for a point mass so what you do you find out the gravitational field of this entire rod at this point and then multiplied by the mass to get the force between rod and the point mass so that is the idea so that is why gravitational field is defined now gravitational field into a ring that is something which comes quite often so let's say there is a ring at the centre everyone gravitational field is a vector quantity like all of you know right it is 0 at the centre it is 0 because the force per unit mass at the centre is 0 if you tensor this mass this point mass it will apply force this way so like that force per unit mass get cancelled away in all directions but if you take like this kind of scenario along the axis if you have to find out the gravitational field mass m radius r over here which direction the gravitational field will be can you tell me is like this like that circle is on my palm and this is the axis about the natural axis basically along this natural axis only this way how much it is if that distance is quickly find out what it is let me know once you are done see in case I mean I can understand there might be some things which you are seeing for the first time okay you should also understand that I cannot stop everyone and start teaching a concept this is not a class where I am teaching a concept fine so if you are seeing something for the first time you have to learn that concept okay to learn that concept you have to watch the entire classroom recording for that you get in touch with me I will share all the recordings with you whatever is your requirement let's say this angle is theta so if you consider a point mass over here dm mass let's say this dm will have gravitational field this way okay that's a gravitational field since you have learned the electrostatics so when you learn gravitational field in grade 11 and after learning electric field you can correlate very well whatever is happening in the gravitation same thing happens in the electrostatics also electric field is like gravitation field exact same thing dE is g dm divided by that distance is square that distance is square is a square plus r square okay this dE now if you consider a point mass over here that will also have the field this way and if you see the symmetrically symmetry here along this direction the field will get cancelled away anytime you take one point mass let's say if you take it here diametrically opposite there will be one more point okay so they will cancel out perpendicular to the axis so only field remaining is the horizontal so why to consider vertical component we'll just consider the horizontal one which is g dm a square by r square into cos of theta cos of theta is a divided by root over a square plus r square now since all the components are along the same direction you can integrate them dm becomes m so the field is g a m gam a square plus r square this is what you get so now if you put the point mass over here of mass m the force between ring and this point mass m would be this field multiplied by this m okay so I just did the derivation actually I should not have done the derivation just directly written the expression variation in value of g g is nothing but force per unit mass just look at the name it is acceleration due to gravity so it is like if only gravity force is acting what would be the acceleration if only gravity force is acting net force is gravity force so acceleration will be forced by mass so that is how you find out g on the surface it is gm by radius of the planet square when you go above let's say from center if you are at a certain distance of h from the surface so re plus h whole square is g now in our textbook they have given us some approximation right approximately g is this 1 minus gm by re square what is that to take re 2h by re right do you remember this so this is g0 1 minus 2h by re don't use this this is an approximation in numericals never use it it assumes that h is very less compared to re use this if you are at a height h from the surface when you are going down below then your g will vary like this 1 minus d by re this is not approximation height variation is the approximation and you know g can change with the rotation of earth also even that sometime you have to take into account for example here get the answer we have to take a break after this question we will take done anyone ok I can see some of you got the answer this is our mother earth about the pole the earth rotates so let's say earth rotates like this you are here ok so you know you will realize that there is a pseudo force due to rotation that is acting on you which is m omega square into radius of the earth so this is also due to earth only right this is due to earth only so the total force is not just gravity force gravity force minus this force is the total force due to earth the total force is mg minus m omega square re this should be equal to mass times g effective at the equator so g effective should become 0 which is g minus omega square re so omega is root of g by re this is omega ok now you can find out the length of the day so time period would be 2 pi by omega so 2 pi re by g this many seconds ok so I think I have done some calculation couple of years before and time of rotation comes out to be about one hour something so if earth completes one day in one hour then this will happen ok now potential energy you know the potential energy between the two point masses is minus of g I mean don't read it right now whatever I am writing focus there g m1 m2 by r between the two point masses so till now whenever we wrote gravitational potential energy again I am repeating we only considered that the potential energy between earth and some other object we ignored the potential energy between the two objects so now we are considering potential energy can be between any two masses so between any two point masses this is the potential energy minus of g m1 m2 by r don't write mgh in this chapter so but why minus this here earlier it was plus mgh because it was plus mgh because you assume something as zero and you are comparing potential energy because of that now what you are doing is you are assuming zero potential energy when the distance between the two masses is infinite so when distance is infinite potential energy is maximum which is zero so all other potential energy will be less than zero which is why negative and yes you can use work energy theorem here also don't hesitate to do that conservation of momentum conservation of angular momentum torque equation whatever we have learnt everything you can use wherever it is required you can use anything whatever we have learnt just that potential energy you should be careful you should not write mgh because mgh is an approximation for gravitational potential energy now many a times many a times you know we are not dealing with point masses I mean most of the time actually we are not dealing with the point masses we are dealing with the bigger masses but the formula for the potential energy is for the point masses so we need to define just like we define the gravitational field for the bigger masses which is force parent mass we define gravitational potential potential energy parent mass so if you tell me that potential gravitational potential of this object over here is v then if I put a mass over there potential energy between mass and this object would be m into v like that so potential for a point masses minus of gm by r by the way I keep on saying point mass for a uniform is spherical shell outside the shell the spherical shell behave like a point mass located at the center okay so whatever we had we are writing for the point masses value for the sphere also outside the sphere so for example for ring of mass m radius r at this distance potential would be what it's a scalar quantity just like electrostatic potential it is scalar quantity so you can see that all the masses are at some fixed distance this is a under root of r square plus a square is a distance so potential is minus of gm divided by root over are you able to relate whatever you have learned earlier type in are you able to recollect from your memories sweet memories of gravitation are you able to remember that type in if those sweet memories are not coming to you make sure you revise the gravitation okay gravitation is a very simple chapter in grade 11 you should even ignore the difficult topics but you can never ignore the simpler topics if you combine 11th and 12th I can tell you 60 to 70% of physics is straightforward but the biggest mistake students do is that they spend 70% of time doing 30% of the stuff and 70% of the syllabus which is so simple they don't spend much time over there so you need to score marks so focus on the simpler things first master it and then go to the difficult ones can relate with electrostatics yes you can do that also okay only one difference is that in electrostatics forces can be attractive and repulsive both but in gravitation everything is attractive this we just did look at this we have done all of these derivations you remember this we haven't done the cone actually but we did all of it remember or not okay I can see where if you are saying yes if you don't remember then what we can do what I can do it is you have to revise I can't revise on your behalf isn't it so spend some time now I can tell you this is where the people the people who are less prepared will make mistake this solid is fear solid is fear if there is a solid is fear inside at a distance small r from the center the potential is if you remember it and if it comes others will get it wrong and you will get it right inside it is this okay so when r is equal to 0 the potential of the solid is fear at the center is minus of 3 gm by 2 r this is where most of the students will make mistake and here also I know that I am telling you this 60 students are there out of 60 only 10 to 15 of you are focusing right now and when it comes to the exam you will again say that the center potential is 0 or something else okay someone is asking why field is infinite due to the cone it comes out like that if there is a point charge if you try to define field on the point charge itself it is not defined actually infinite is the wrong word not defined do this after this we will take 2-3 minutes break okay so many of you getting the answer here ring mass m radius r is placed on the yz plane the center at the origin just do that axis is on the x axis particle is released from here x equal to 2 r speed when it reaches the center this mass m radius r okay what is the first thing that comes in your mind what you have to do working as a theorem and in this chapter most of the time external force is absent okay like there is earth there is moon or stars or satellites are there it will never be that there is an external force which is pulling the earth towards the this thing so work done is most of the time 0 so that is the reason why you can directly write u2 plus k2 is equal to u1 plus k1 in this chapter specifically but not always I mean I am saying 99.9% this will happen so final potency when the object reaches here is mass into potential over there kinetic energy half m into can you tell me what is the assumption here while I am writing the equation this that is equal to m into v1 plus 0 what is the assumption correct the assumption is that the ring is fixed it should be specified here okay ring is not moving if you allow ring also to move then what you have to do to solve the problem conservation of linear momentum and energy you have to conserve the linear momentum then let's say final velocity of this and that is v1 and v2 so m into v1 minus capital m into v2 is equal to 0 this is one more equation that comes but we are ignoring actually okay keeping simple the potential of the ring at the center is g capital m divided by r no minus actually minus plus half m into u2 this is equal to mass potential over there if you consider that distance this distance is how much and 2r root 5 so minus of gm by root 5 r this is the equation all of you understand this type in okay let's take a small break because maybe you need one let's meet after 5 minutes alright am I audible fine so let us move this is the motion of planets and satellite like I told you we are assuming that the planets and satellites they are moving in a circular path and if anything moves in a circular path the first thing the first equation that you should write is that the force towards the center which usually is the gravitational force is equal to m v2 by r this is the first thing that you should write this equation and as a result of this what comes out this this is what I was writing here okay kiting edge is this and total edge is this have you seen this in modern physics when electron revolves around the nucleus something similar happens have you seen it right very similar to that so again I will write here that the potential energy is minus of gmm by r kinetic energy is gmm by 2r the total energy of the planet kiting edge plus potential edge which is minus of gmm by 2r so this is minus actually here total energy is negative it means that it is favorable for them to remain like that if they separate their total energy will become 0 which is more than whatever is the energy as of now so that is why they don't leave each other now these things for geostationary satellites these satellites they are relatively at rest when you are seeing from the earth their orbit must be circular they should rotate same way the earth is rotating it should be along the equatorial plane the rotation if earth is rotating about this axis so the satellite should also rotate like this only it can't rotate like this okay time period must be 24 hours it should rotate west to east just like earth is rotating let me find some objective questions here okay escape velocity is just like a made up concept here what is escape velocity it is the minimum velocity by which we throw so that the object doesn't come back doesn't come back as in it goes to infinity so initial kiting energy plus initial potential energy should be equal to the final potential energy which is zero and since it is minimum velocity the final kiting energy also we liquid that to zero and escape velocity comes out to be this okay radius of the planet is r we know that g is equal to capital G mass of earth divided by r square so the velocity comes out to be under root of 2gre now does it matter in which direction you are throwing the object this is the earth does it matter whether you are throwing it upwards in this direction like that or you can throw any direction escape velocity will be this only does it matter in which direction you are throwing it does not matter okay doesn't matter ultimately the object reaches infinity and the earth becomes like a point mass when it goes very very far kepler's law we already discussed solve these questions okay siddhas and aditya got something others siddharth got something alright so we have infinite masses each of mass 3kg plays at 1, 2, 4, 8 marks expression of 2kg at the origin will be what what we will do we will find out the gravitational field due to all of them okay it will be equal to minus of g 3kg divided by 1 square plus g 3kg divided by 2 square g 3kg divided by 4 square and so on right so this will be 3g 1 by 1 square plus 1 by 2 square what does this becomes is 2 raised to power 4 this is 2 raised to power 6 and so on this you can say 2 raised to power 0 it is a gp infinite gp so you can add it up this is a divided by 1 minus r this is your field so the force is this is how much field how much it comes out this is 4g field is 4g and field into mass this is the force 2 into 4g is the force this is equal to mass times acceleration so acceleration is 4g only field is the acceleration because field is what force per unit mass which is acceleration also if that is the only force acting do this okay siddharth and rithu got something which law will be using you will be using here tell me law of periods you will be using here or not kepler's law kepler did only for the planets but same thing is valid for the satellite also because the way the planets move around the sun same way satellite move around the planet so t square is proportional to r cube so t1 square by t2 square is equal to r1 r1 cube divided by r2 cube so geostationary satellite revolve around the orbit of radius small r so r1 is a small r angular velocity is doubled so t2 becomes t1 by 2 all of you agree so it will be 2 square over here r1 is r we need to find r2 alright so r2 is r divided by 2 to the power 2 by 3 all of you understand option c is it clear okay let us move do this rithu and siddhas got something hariaran got something radius got reduced to r by 2 and mass is constant radius is increased mass is constant then what factor the density has to be changed so as to keep g the same what is g g is capital G m divided by radius of the planet square okay increase by factor of 2 keeping the mass constant okay mass is constant but density can be changed how come when we increase the radius of the earth and mass is constant I am keeping density fixed then by which factor the density has to be changed looks factor of density has to be changed to keep g the same if I keep if I change the density then how will I keep mass constant okay so it looks like this question is not proper what you guys did how you got the answer they are locking the density right you can't change it radius becomes 2 times mass is constant yeah I think let us ignore that mass is same let us ignore mass is same so just a small correction over there now mass can be written as density 4 by 3 pi r cube by r square this is small g so small g can be written as 4g rho pi r divided by 3 is small g so if r is becoming 2 times okay if r is becoming 2 times the density has to become half so that g is constant so that is option D clear to all of you they are asking about density and the radius so first get the expression of g in terms of density and radius right now expression of small g is in terms of mass and radius you want to do this solve this question alright anyone close so here we will be using u2 plus k2 is equal to u1 plus k1 okay so when they come and hit each other they will be moving symmetrically with each other so this is what will happen at the center they will meet so the final potential energy would be minus of g m1 m2 which is m square distance between any of the 2 is 2 r like this there are 3 pairs between 1 and 2 2 and 3 3 and 1 so 3 times of that is u2 k2 is also 3 times quantity of 1 of it this u1 is 3 times the 1 pairs potential energy which is this okay substitute u2 k2 u1 k1 is 0 what you get u as are you getting d as the answer when you do it like this I hope there is no silly error if there are no silly errors and okay many of you getting activity so that is fine aditya got something 2 spheres of mass m1 m2 distance d apart gravitational potential where the field is 0 so let's say at the distance of x from here the field is 0 field due to m1 with this way field due to m2 with that way so they will try to cancel out each other so g m1 divided by x square minus g m2 divided by d minus x whole square is equal to 0 so we are getting m1 divided by x square is equal to m2 divided by d minus x whole square take a square root root of m1 by x root of m2 divided by d minus x so we can solve for x also here d minus x so x is equal to root of m1 divided by root m1 it looks weird actually but that's how it is this is x now potential at this location is g m1 by x minus of g m1 by x minus of g m2 by d minus x is what it is ok so let us try to evaluate how do you get the answer quickly so take g outside and m1 divided by x when you do so d will come out so it will be root of m1 multiplied by root m1 plus root m2 plus I think it will come out like this only m1 plus root m2 so if you do the careful calculation you will get option d yes or no anybody got d yes what is the difference between this and that oh there is a factor of 2 there is no factor of 2 how come you got 2 in the denominator you cannot afford to make mistake on simple questions understand that understand let me see if I can get something ok do this whoever got something others is it very straight forward 2 bodies of mass this and that separated by this at a distance p by 11 km the intensity field is 0 oh I think yeah from the previous question itself very similar to the previous question you will get p is equal to 1 just try it yourself exactly on the same principle whatever we did previously let me say this ok let us do this also quick last question for today you can use this result differences omega square r at the pole radius of the revolution is 0 you are on the axis omega is 2 pi by time period time period is 24 hours 60 minutes 60 seconds that is square ok calculation is tough is it what is pi square closest to it pi square is like 10 so make it 10 4 into 10 6.4 into 10 to power 6 that divided by 2.4 into 6 into 6 the whole square 1 2 3 10 to power 6 which is gone 64 into 4 6 divided by how much is the denominator 36 into 24 is 26 6 12 6 7 4 6 864 it is 6.4 basically it is 6.4 square is how much calculation is bit on the higher side wabbo used calculator to get the answer ok anyways so since wabbo has told us the answer p is equal to 4 try it your own ok don't use calculator unnecessarily you will not get that in the exam ok and there will be some questions in which calculations will be on the higher side like this so they are not testing you on the concept they are testing you whether you can do some calculation under pressure ok everybody knows how to solve but only few will get the correct answer alright so that's it from my side we will meet next week and see what we have to do probably j main will get postponed to march or something not guaranteed though if it get postponed then we will see like we can have couple of more extra sessions bye for now yeah as cbsc sampler became those who are focusing that to it