 All right, so those who are online, please type in your names quickly so that we can start. Okay, we'll wait for a couple of minutes. So this is the second session on rigid body motion problem solving. We had one, I think, couple of weeks back. So this time I have taken questions from other books. They are not your past year J questions. All right, so I guess we can start now. Let others join meanwhile. All right, so you can start solving these two questions. Welcome, everyone. Oh, Apurva, you came finally to the class. Quickly answer these two questions. Okay, question number one is basically rotation about fixed access, right? So when a rigid body rotates about the fixed access, kinetic energy for that rigid body is given as half moment of inertia about the fixed access into omega square, right? So we have a mass M with radius R, okay? And then the cylinder is rotating about the axis with constant speed V. Now this is a little ambiguous here because not all the particles will have same speed V, okay? So we will assume that the rim has this velocity. As in you consider this as a cylinder, then it is rotating about its axis which is passing to the center and the rim has a velocity V, okay? So if rim has a velocity V, omega becomes V by R. That is the reason why it will become half MR square by two into omega which is V by R. So this is V square by R square. So it becomes one by four MV square, all right? Second one, second one is rolling. So you have to pay attention to the words. Here it talks about rotating and this is rolling, okay? So there is a linear motion as well as rotation in question number two, okay? Question number two, see many of you are saying, all right? So kinetic energy, if you do not find a fixed axis can be written as half M into VCM square plus half ICM into omega square, fine? So this we can write, you know, now constant speed V is given. So we'll assume that this speed is for the center of mass, okay? So this is half MV square plus half. Now it's a disc, okay? Of mass M and radius R. So ICM becomes MR square by two, all right? And since it is rolling, the point of contact should be at rest because the surface on which it is rolling is at rest, okay? So if VCM is in this way, then omega R is backside. So when V is equal to omega R, then this point is instantaneous, the at rest, okay? So that is the reason why omega is V by R. So this will become V by R whole square, all right? So we can say this is equal to half MV square plus one by four MV square, fine? So that's the reason why the answer is three by four MV square. All right? Any doubt, guys? We'll move to the next one. Okay, third is A, others? Okay. So we have a tube of length L filled completely with the incompressible liquid, okay? So let's say we have this tube of length L completely filled with incompressible liquid of mass M. Inside there is a liquid closed at both the ends. The tube is then rotated in a horizontal plane about one of its end. So like this, it is rotated. Okay, let's say angular velocity is omega. We need to find out the force exerted by the liquid at the other end. So here, what is the force exerted? We need to find out, okay? Now to do that, you can just look at the, let us say a small mass at a distance R from the axis. This length is supposed DR, okay? So at a distance R, this small mass, let's say DM should have an acceleration towards the center, which is omega square R, fine? So the net force from this side, okay? There will be two forces, right? There will be F1 and there will be F2, okay? But F2 is F2 minus F1 will give you as in F2 minus F1, which is actually small amount of force. The difference in the force should be equal to mass, which is DM into omega square R, right? So DF will become equal to, now DM is what? DM is area of cross section into density, into DR, omega square R, right? So when you integrate this, DF, it goes from zero to till wherever you want to integrate till here, okay? And this R will go from zero to L, all right? So you will get here A, rho and omega square, they are constant, they come out of the integral and it becomes RDR integral, which is R square by two. So when you put limits, it becomes L square divided by two, like this, okay? And we know that rho A into L is mass. So it becomes M omega square L divided by two, fine? So that is the reason why option A is correct over here. Okay, fourth many of you have answered. Let's see, a gramophone record of mass M and radius R is rotating at an angular velocity omega. A coin of mass small m is gently placed on the record at a distance R by two from the center. So we have gramophone, which is rotating with angular velocity omega. Now we are putting on, this mass is M, radius is let's say R, everything I'm writing here. And then we are putting a coin of mass small m at a distance of R by two from the center. So from here, the center to this coin, this distance is R by two, fine? So we are putting a small mass, fine? Now, the new angular velocity system we need to find out. Now the assumption is this small m will move with the capital M finally, all right? So if that happens, then of course, about this axis, I can conserve angular momentum, right? Because there is no external torque. Although there is a fourth between small m and capital M, but if I consider them to be part of my system, as in small m plus capital M is my total system, then the force becomes internal. So there is no external torque, it's just internal force. So internal torques will get canceled between small m and capital M. Fine, so let's conserve angular momentum about this axis. So initial angular momentum is i omega, which is MR square by two into omega, right? This should be able to final angular momentum. Now, in order to write final angular momentum, I should write here the new moment of inertia, which is MR square by two plus small m into its distance from the center square, because moment of inertia changes, isn't it? A new mass is coming here. This is omega dash or new angular velocity, right? So when you simplify this, this and then R square also goes. So you will get omega one to be equal to two m divided by two m plus small m times omega. Fine, so that is why option A is correct. All right, any doubt, guys? Please type in yes or no. Solve this one, question number five. Anybody got the answer? Okay, so angular momentum about a point, you know it is written as like this, RCM cross with m into VCM plus ICM omega, fine? Now this RCM cross MVCM is basically perpendicular distance of the center of mass's velocity. Okay, it's like, you know, R cross force. Like how you find torque? You find the perpendicular distance of the line of force from the axis or from the point. Similarly here, the line of velocity of the center of mass is this. So I'm trying to find angular momentum about, oh, this is the perpendicular distance, okay? So perpendicular distance is R into m and VCM. Okay, now one thing you need to understand that this term has which direction? R cross MVCM, the sense of direction is this way, fine? So it is clockwise, okay? And similarly, i times omega is also clockwise. So you just add algebraically these two, but it may not be happening like this every time. It may be that the object is spinning and you know, moving backwards. So they might be sliding over this point, but then if that happens, then you need to subtract, not add. So always make sure whenever you add two vectors, check the directions. Plus, this is a disk, right? MR square by two into omega, right? Now VCM is omega into R. It becomes MR square omega plus MR square by two omega. All right? That is why you get three by two MR square omega. So option C is correct. No doubts, right? I'll move to the next one. All right, so you guys are getting D as the answer, D for Delhi. So we have a thin wire of length L uniform mass density rho, okay? So length is L. So mass will become rho into area of cross section into L. Right? This will help us to get the mass and then length should be also equal to two pi R. So you'll get R will be equal to L by two pi. Why we are doing all this? Because ultimately the formula for moment of inertia is M into R square, right? Neither M or R is given, but everything else, right? And we can use these two to get the answer. Okay? So I think the correct answer, assuming you guys are not making any silly errors. Okay, we'll move to the next question. All right, seventh is B because perpendicular distance, perpendicular distance is not changing. So when you're moving parallel to x-axis, it's perpendicular distance remains unchanged. So D into Mv will remain its angular momentum. What about the eighth one? Elastic collision takes care of the linear velocity. Okay, so that's the condition for that comes from the, you know, that velocity of approach should be able to velocity separation. Although omega will come into the play, but there is no direct correlation with the angular velocity. All of you getting it wrong? Question number eight. Okay, see what happens here is, let's say this is first sphere and then there is an identical sphere, right? Now when they collide, when they collide this sphere is anyway moving with angular velocity omega. Let's say omega A, okay? No, sorry, omega. It was moving with omega, okay? Then it hits this sphere also. Now when it hits this sphere, this sphere has omega equal to zero, okay? And V is equal to zero, all right? Now this sphere, let's see what is happening. This sphere will experience a force from the first sphere, which passes through the center. Then it will experience MG force, okay? And then it will experience a normal force. Any of the forces will experience. There's no friction between this and that. So there is no friction over here. Similarly, there's no friction from the surface also, okay? So if you see about the center of mass, torque is zero. And if torque is zero, alpha should be zero. Torque about center of mass is zero. So alpha should be zero. And if alpha is zero, omega remains zero only, okay? Similarly, this sphere will also not experience any torque, the first one. So it will maintain its constant angular velocity. So omega A will be equal to omega, okay? Is it clear? If there is any doubt, please type in quickly while I'm moving to the next one. Okay, this we have done. Yes, Atmesh, ninth, what is the answer for ninth? So solid metallic sphere of radius R having moment of inertia I. So I should be equal to M R square into two by five, right? So this you might be knowing. So it is melted and recast into solid disc of radius small R of uniform thickness. So the density will be same and the volume should be same. Okay, because mass has to remain the same. That is the reason why four by three pi R cube should be equal to thickness into pi R square. Fine, so pi goes away. R square is gone. Thickness should be equal to, oh, sorry. This has different radius. The radius is small R square. So pi and pi goes away. So you will have an equation for R cube by three is equal to R square into T. Okay, so you have these two equations. Fine. And then it says that moment of inertia of the disc about an axis passing through its edge and perpendicular to its plane is I. Now, how much is that? About the center of mass moment of inertia is MR square by two plus MR square. This is the moment of inertia of the disc about the line that passes through the edge and perpendicular to the plane. Okay, so this is small R square. So three by two small MR square, this has to be equal to I. Okay, so since mass is same, I will just equate these two, all right? So I think there is no need of equation two. So you can just equate two by five MR square is equal to three by two small MR square. So when you do that, you'll get option A to be correct. Okay, 10th option D is correct. So I'll give you a small hint of those who have not done question number 10. So a small object or informed velocity rolls up in curved surface with initial velocity V. It reaches maximum height this. We need to find what is this object? So height is given in terms of velocity. Now here simply you have to apply work energy theorem. Okay, that should be the first thing that should come in your head. So work done, put it as zero, okay? Potential energy is MGH, K2 will be zero at the maximum height. Fine, U1 will be zero and initial kinetic energy. You can write it as half M into V square plus half I center of mass into omega square. Okay, and we know that omega should be equal to V by R. It is pure rolling. Fine, so when you substitute all of this, you will get ICM is equal to MR square by two. And that happens for the disc only, okay? All right, we'll move to the next one. So here you see that there's a hinge. So while the bullet is striking the rod, I mean huge amount of force get developed at the hinge. So you cannot conserve linear momentum because the external force from the hinge, okay? And that is impulsive. But whatever force it is, it has to pass through the hinge only. It says hinge is creating the force. And that's the reason why about the hinge, the external torque because of that force will be zero. So that is the axis about which you can conserve the angular momentum, okay? Now when you conserve angular momentum, you have to take both bullet and rod together because there's a force between bullet and rod also. So if you take just rod, there's external force because of the bullet and that is creating torque. But then if you take both of them together, the internal force between bullet and the rod, I mean, whatever is a force that becomes internal force and torque becomes internal torque. So angular momentum of bullet plus angular momentum of rod initially should be equal to angular momentum of bullet plus angular momentum of rod finally. So angular momentum of bullet about this axis is M into V into L, okay? Plus angular momentum of rod initially is zero. This should be equal to, now when the bullet get embedded into rod, it becomes a single mass, fine? So now when the rigid bodies, you can say that rigid body is a single mass, the angular momentum about an axis is just moment of inertia about that axis into omega if the axis is fixed. So about this axis, we need to find the moment of, sorry, angular momentum and this is a fixed axis also. So moment of inertia about that is M into L square plus rod's moment of inertia is ML square by three into omega, right? Hence you can find it like this, three MV divided by three M plus capital M on by L, okay? So option C is correct, fine? Any doubt, anything you want to discuss with respect to this, do this? Okay, I think we have done many such type of questions, okay? So what we have to do is, we have to assume that this plus a small disk will give you a compound, okay? Will give you a complete disk, okay? So moment of inertia about this axis, let's say this is I naught and this is I one, then I naught plus I one about this axis should give you the moment of inertia due to the full disk, okay? The moment of inertia due to the full disk is MR square by two, this should be equal to I naught, which we have to find out plus moment of inertia of this small disk about this point, right? Now this small disk has what mass? Total mass of the complete disk is M. So M divided by pi into R square is the mass per unit area. This into pi into R by three whole square if you do, you will get M by nine, okay? So the mass is M by nine, alright? So you will get M by nine. Now R is also different. So this is R by three whole squared MR square by two plus M into distance square, which is two R by three whole square, fine? So from here you get the value of I naught, alright? Let's move to the next one, 13th one, two. What do you mean, Atmesh? One object will have only one center of mass. But if you divide this object into two parts, let us say total mass is M, then this will become M by two and this will become M by two, the horizontal portion and the vertical portion. So if you say that this entire object is made up of two objects, then center of mass of the lower object is here and center of mass of the upper object is there, fine? But then center of mass of these two combined is only one, alright? Now you can easily find out here the center of mass, the X coordinate of center of mass is M by two. Now this length is from here till here, this length is one by four because from here to here it is one by two because total is one. So M by two into one by four plus M by two into zero divided by total mass M, okay? So I will get one by eight and similarly Y center of mass I will get one by eight, okay? So distance will be under root of X center of mass plus, sorry, under root X center of mass square plus under root Y center of mass. So here there will be a center of mass somewhere. In terms of centimeter also you can do it. So 100 by eight you can say 12.5, this is also 12.5, okay? So like that you can do root two into 12.5, okay? Now do the fourteenth. Only one answer for fourteenth. So if velocity is V for a mass which is rotating in a circle, then angular momentum for that mass about the center is what R into MV or MVR is the angular momentum of a point mass, right? It's perpendicular distance from this axis is nothing but the radius only. Let's call it as R. So it should be MV into R, okay? Now if angular velocity is same, okay? Then initially let's say L1 is the angular momentum. Initially angular momentum is MV1 into R. Now R is let us say R1 over here. So V1 is what? Omega into R1. Now why I'm writing in terms of omega? Because omega is keeping constant. So MR1 square into omega is L1. So L2 will be equal to MV2 into R2, okay? So this will come out to be M into... Now omega is constant. So V2 is omega into R2. This becomes MR2 square omega, okay? So L1 by L2 is equal to R1 by R2 the whole square, okay? So R1 is L and now the string is hard. So R2 is L by 2, okay? So it becomes 4, L1 by L2. So L2 is L1 by 4. So that is why option A is correct, right? No doubts, right? I'll move to the next one. Okay. You are able to see my screen? All right. Any answer? Aerial velocity is area swept per unit time. DA by dt. Now we know that area is... You can just look at the movement. This distance is around, let's say, V into dt, okay? And this is R. So area is half per unit distance, let's say, R into V into dt. This is small area, okay? So DA by dt is V into R by 2, right? So angular momentum we need to find. Angular momentum is nothing but M V into R, okay? So M V into R is 2 times M into DA by dt, okay? Now here A is not taken as area. A is itself is DA by dt. So I have my A here is area. This question's A is DA by dt, okay? So that is why let me put it as A1. So DA1 by dt is A. That's why option B is correct. What about 16th? Others? So let's say this is the disk of mass M2, okay? And the radius is R. This is R, okay? And then there is a uniform rod, uniform rod of length L and mass M1 riveted at the center. So this is the scenario. Mass is M1 and its length is L. We need to find center of mass of the combination from the center of the disk and also along the rod. So along the rod where is the center of the mass, okay? So of course center of mass will lie along the rod only because that's like axis of symmetry or line of symmetry, okay? So in order to find where it is, if consider this to be your y-axis and this to be your x-axis, fine? So x center of mass will be basically M1 into L by 2 plus M2 into 0 divided by M1 plus M2, okay? So you will get option 2 to be correct, option B, okay? I'll move to the next one. In case any doubts, quickly type in. If radii of gyration is K, moment of inertia is M into K square. So that's the definition of radius of gyration. Anyone? 16th. Sorry, 17th. So angular velocity of the body changes from omega 1 to omega 2, okay? Now torque is not applied. So angular momentum should be constant, right? So L1 should be equal to L2. So that is the reason why I1 into omega 1 should be equal to I2 into omega 2, okay? So I1 by I2 should be equal to omega 2 by omega 1, right? And I1 by I2 can be written as M into radius of gyration for the first one square divided by M into I2 can be written as M into radius of gyration for second square, okay? So K1 by K2 is under root omega 2 by omega 1. So that's why option C is correct. From here, K1 by K2 is this. What about 18th? See, it is asking ratio of corresponding radius of gyration. So first one is omega 1, second is omega 2. So I have assumed that they are asking K1 by K2, okay? So you can claim that there is a little ambiguity in the question. But then if such ambiguity is there, you go by putting the first object first in the numerator. So I just went by how it is written. But ideally they should have written that what is radius of gyration of the first object divided by radius of gyration of the second object. 18th, none moment of inertia about the axis that comes out of this plane. Let's say that is I0. This should be equal to I1 plus I2. Any doubt here? This will be equal to 2 times of I1 because I1 is equal to I2 due to symmetry, okay? And I0 similarly will be also equal to 2 times of I3 because I3 plus I4 is I0 perpendicular axis theorem, right? So I get I1 equal to I0 by 2 and I3 is equal to I0 by 2. So I1 plus I3 will be equal to I0, option C. So moment of inertia about any of the accesses I1, I2, I3 or I4, all of them about all of them moment of inertia is same. So you can take any two and some of that will be equal to moment of inertia which is perpendicular to the plane. Any doubt on these two questions? Quickly type in. I'll move to the next one. Quickly do these two questions. Okay. Good that you have tried the easier one first. That's how you have to do an exam also. Angular moment of particle moving in a circle orbit with a constant speed remains conserved about the center of the circle, okay? Because there will be, I mean, if a particle is moving in a circle, okay? So if a particle is moving in a circle, then there will be a force that will be towards the center of the circle. That's like centripetal force that is required. Otherwise it will not move in a circle. And that's the reason why the torque due to this force will be zero about the center. So hence since that is the access, it has to be passing through the center of the circle, okay? Nineteenth, cylinder rolls up the inclined plane reaches a some height and then rolls down the directions of the frictional force acting on the cylinder. So when it is rolling up, the friction will be down. So and while it is coming down, its friction will be up, okay? Because you know when it is, first of all, let's talk about coming down. That is easier to deal with right now. So when it is coming down, then it is getting accelerated because of the gravity. So it this point tend to skid forward. So that is why the friction force that is coming down is upward, okay? So up while descending and when it is ascending, so let's say this cylinder is moving up, okay? So it has an angular velocity, alright? And of course its angular velocity should decrease as it moves up, its velocity decreases. So angular velocity should also decrease. Now in order for angular velocity to decrease, the friction should create a torque in opposite direction. So that is the reason why again while going up also the friction force will be upwards. So both ways it is upward only. That's why option B is correct for nineteenth, okay? So these kind of questions you can easily go wrong, okay? So even though you will have this temptation that, you know, it is not a numerical, I can just think and answer it. So please avoid that temptation and in case you have any doubt or any hesitation, so just leave it. Why aren't we assuming pure rolling while descending? It is doing pure rolling only, Atmesh. It is doing that only, okay? But then let's look at angular velocity, omega. Omega should increase when it is coming down, right? So there has to be a torque which should increase this angular velocity in this direction, right? And who can provide that torque? Only friction. That is why the friction torque should be in this direction. So hence the friction is in upward direction while it is coming down. And while it is going up, omega should decrease. But angular velocity is now in this direction. So that's why friction again should be in that direction. So both ways it is upwards. Is it clear now? Any other doubts? Let's move to the next. 21 should be A, right? Because there is no external force to the system, okay? So velocity center of mass should be unchanged. So initially if it is zero, it should remain zero. That's why option A. What about 20 second? So one impulse is given and the 10 kg block started moving with a velocity of 14 meter per second. What will be the initial velocity immediately after 10 gets impulse? What will be the velocity of the 4 kg block? How much it will be? Immediately after the impulse, the velocity of 4 kg is? It is zero, okay? Because 4 will take time to change its velocity because spring is applying force on it, which will slowly grow and then it will have acceleration from there velocity will come, okay? So velocity of center of mass will be equal to 10 into 14 plus 4 into zero divided by 10 plus 4, which is 10 meter per second. Okay, we'll move to the next one. This one? Phone, anyone? Okay, let me do this. So we have a pulley, pulley of radius 2 meters, okay? And moment of inertia 10, okay? So it is getting pulled tangentially from one of the ends, the force of 20t minus 5t square. Okay, this is the force. So here we need to find out time when the direction of motion is reversed or when omega becomes negative, alright? So we know that torque about center of mass should be equal to i about center of mass into alpha, okay? So alpha will be coming out to be, what is torque? Torque is radius into force because this force has a perpendicular distance of radius only, right? And there is no other force that can create the torque over here, right? So torque is 2 into 20t minus 5t square, okay? This is a torque divided by moment of inertia. Now moment of inertia is 10, okay? So here you will get this as equal to 2t minus t square by 2, alright? So this is alpha. Now alpha is what d omega by dt. So this is equal to 2t minus t square by 2. So d omega will be equal to 2t dt minus half t square dt, alright? So the initial angular velocity is what? Okay, the initial angular velocity will be 0 because only this force is causing the rotation. So let's say from 0 to omega, so 0 to t, okay? 0 to t, right? So you will get, what are you saying Saimir? So omega is equal to t square minus t cube by 6, fine? So this is the omega as a function of time and just we need to find out how much time it takes to reverse the omega direction, okay? So basically if you equate this to 0, you will get that 1 minus t by 6. And equate to 0, you will get t equal to 6 seconds, okay? Alpha is torque about center of mass is force into radius 2 divided by 10, oh-ho. So you understand, right? There is a calculation error. Of course you guys are not paying attention right now. I mean, only Saimir got it. It's 4t minus t square, okay? So from here then everything changes. So just do that, do the math properly, you'll get the answer, okay? Fine, we'll move to the next question. 24 is B. All you have to do is just, you know, add up the torques. So torque by the forces about the center is asked, okay? So of course the torque because of the 6th Newton will be 0, okay? And when you take component of 8 Newton force, there will be this component which is 8 cos 30 for which torque is 0. So only component will be, sorry, only torque will be because of this component of force which is 8 sin 30, okay? Which is 4, okay? So only 4 Newton force will have the torque. Now you can see that 9 Newton has torque this direction, 4 Newton has in this way. And this 8 sin 30 has the torque in this way, okay? So torque due to 9 Newton is 9 into 0.2 minus 4 into 0.2 plus 4 into 0.2, okay? You'll get 1.8 Newton meter, okay? So that's why option B is correct over here, okay? So now we'll take a small break. Right now it is 6.10. We will meet around 6.20. So this is our break time. All right, so let us start these two. It's a hollow sphere but then it has a thickness. Till r radius, there is nothing. Between r and 2r, there is material. That has mass and... How will you go about this? Now you can consider this... I mean the method of solving this problem is similar to the earlier ones which has some cavity, right? So moment of inertia for the remaining one will be equal to moment of inertia due to the full sphere, okay? This is due to the full minus moment of inertia, the sphere for radius r, okay? So this is... Full sphere has a radius of let's say 2r, okay? Now what is given here is the mass of this sphere. It's not the mass of 2r or r, okay? So mass is corresponding to volume. So density will be equal to mass divided by 4 by 3 pi into 2r cube minus r cube. This is density, okay? So rho will come out to be m divided by 4 by 3 pi r cube into 1 by 7, okay? And that is the reason why the mass of the 2r radius sphere will be rho into 4 by 3 pi r cube. So r is 2r there, so it becomes 8 by 7m, okay? And mass of r radius will be m by 7, okay? Hence moment of inertia becomes 2 by 5m of 2r is 8 by 7, 8m by 7, 2 by 5m r square. So this is 2r whole square, 2 by 5. Now m is m by 7, m by 7 into r square, okay? So this will be equal to 2m r square by 5 into 7 is common, both, so 35 into 8 into 4, 32 minus 1, okay? So you will get 62 by 35m r square, right? So that's why option number D is correct over here. So I guess many of you didn't get it right, so you have to be careful, okay? Mass is mass of this, so mass of 2r radius and mass of r radius, they are different. Similarity is for the density, not for the mass, if the material is same. All right, now do the question number 26th. So again here, you know, these are like standard kinds of the questions where you conserve the angular momentum, okay? So I1 omega 1 will be equal to I2 omega 2, okay? So omega 2 will be equal to I1 by I2 times omega 1, fine? Now when the moment of inertia becomes 75% of the earlier one, then I1 by I2 will become what? 1 divided by 0.75, this into omega 1. Now 0.75 is 3 by 4, right? So omega 2 will be equal to 4 by 3 times omega 1, okay? So this you can write it as omega 1 plus omega 1 by 3, so there is increase of 33.3%. That's why option A, okay? Any doubts with respect to these two questions? Type in quickly, yes or no? These two questions, what is the answer for 27th? Okay, others, right? So it's a massless chord and there is a mass capital M, sorry, small m attached to it. This is small m and it's a ring of mass capital M, alright? Radius is r. Now I need to find out the angular speed of the ring when the mass m has fallen through a height of h, okay? Now here, the first thing that should come in your mind is conservation of energy or conservation of mechanical energy, okay? So we are going to use work energy theorem over here. W is equal to U2 plus K2 minus U1 plus K1, okay? W is 0, let's say U2 is 0. So when mass comes over here, the potential energy of mass is 0. Let's say this is 0 and then we are, our system is capital M and small m both. So I have to look at both when I write kinetic energy and potential energy, okay? So I'll assume the potential energy of the, you know, this disk remains at the same height. Let's say it's disk, this is a disk which remains at capital H only. So when you subtract the potential energy from initial and final, so this potential energy will not feature. So no point writing that. So we'll just ignore that and kinetic energy of the disk is, sorry, ring is half into moment of inertia of the ring, which is MR square I into omega square, okay? Minus U1 is MGH plus K1 is 0, okay? So from here you will get the answer. Oh, sorry, there is kinetic energy for the small m also that we have to write, okay? I missed that. This is kinetic energy of the ring plus kinetic energy of the small m, small m, okay? This is half M into V square where V is omega into R because this point is not slipping. So the velocity of this point on the ring as well as on this string should be same and velocity of the string is velocity of the mass. So V should be equal to omega R, right? So from here we'll get the answer. Any doubts? Well, you'll not get D, I guess. You will get C, isn't it? Yeah, you'll get C. Atmos, did you get? How will you get C? Okay, 28th. All of you please try it, 28th. Okay, so are these questions difficult? Are you guys finding it difficult? I don't see many of you able to do this. Do you find it difficult? Yeah, I would be shocked if you say that these are difficult ones. We have done much difficult than these in when we used to do past year J questions last session. There everybody was very active. Okay, so this is server disk of radius R which is free to oscillate about this axis. Okay? The disk is turned through an angle of 60 degrees. So if I track the center of mass, this is 60 degrees. We need to find angular velocity when it reaches the equilibrium position. Now equilibrium position is what? The equilibrium position is the position where net force is 0 and net torque is also 0. Okay? So when the disk becomes vertical, then you can see that net torque about this axis becomes 0. All the force will pass through the axis, even mg. But right now, mg is not passing through the axis about the fixed axis. There is mg force and it has a perpendicular distance. So there is not equilibrium position 60 degrees. Okay? So let's try to conserve energy between this position and that position. Okay? So we'll use the same age old formula. W is equal to U2 plus K2 minus U1 plus K1. So W is 0. U2 is, let's assume that this is the level of, this is a 0 potential energy, let's say. Okay? So U2 is 0. K2 is half i into omega square where i is about the fixed axis. Okay? Minus U1 is what? U1 is mg into h height of center of mass from here. Okay? So that is equal to mg r 1 minus cos of 60 degrees. Right? So this is 60 degrees. This distance is r cos 60. So that is why I'm writing it like this. Plus 0. Now moment of inertia is, you can just translate this axis about the center of mass. The moment of inertia is mr square by 2. So just translate it to the fixed axis. So this plus mr square. So you will get this to be equal to 3 by 2 mr square. This is i. So you substitute that, you'll get the answer. Cos of 60 is half. So you've done like this only, right? And you're getting C as the answer. Let me tell you the final answer. Answer is B. B for Bombay. This is what you'll get. Tell me one thing. Don't you think I can skip the steps and get the answer quickly? I can solve the same question in just one step. Okay? So if I am, you know, you're not sitting in exams. So don't skip the steps. All right? So I am not skipping the steps. And I have done such type of questions, hundreds of them. Don't skip it. You can skip in the exam. Don't do it right now. 29, 30th. Solve these two. Okay. Anybody else getting the answer? Okay. This rod is hinged about this point. Mass M and length L. It is hinged on the floor. So this line will represent the floor. Will represent the floor. Okay. Now this rod will slightly pushed. And then this rod will just rotate and fall on the ground. It will come like this. Okay. We need to find angular loss of the rod when it's NB strikes the floor. This is A that is B. Simply conservation of energy you have to use. Work energy theorem. W is equal to U2 plus K2 minus U1 plus K1. Right? Now W is 0. U2 let's say it is 0. Ground I am taking is 0 potential energy. Plus K2 is what? K2 since there is a fixed axis I can directly write moment of inertia about the fixed axis into omega square. Right? Minus U1 is what? M into G into G into L by 2. Plus K1 is 0. Okay. Now moment of inertia of the rod about the end is ML square by 3. This into omega square minus MG into L by 2. Right? We have M gone. We have 2 gone. So omega will come out to be under root of 3G by L. That's our option C is correct. 30th what is the answer? Moment of inertia of a thin rod of mass M and length L about an axis passing to the point at a distance L by 4. Now this is straight forward right? L by 4 from one of the ends. So from center what will be the distance? From center also the distance will be L by 4. Okay. So moment of inertia will be equal to ML square by 12 which is ICM plus M into L by 4 whole square. Fine. So you will get here ML square by 12 plus ML square by 16. Okay. So you will have option A to be correct. Any doubt on these 2 questions? Type in yes or no? Why can't we do ML square by 3? See Atmesh moment of inertia is ICM plus M into D square. If the parallel axis theorem when you apply this is moment of inertia about the axis that passes through the center of mass. You can't take ML square by 3 which is the moment of inertia passing through the end of the rod. That's the parallel axis theorem. Okay. And also remember that there will be infinite axis that passes through the center of mass. Okay. So you need to find moment of inertia. ICM is the moment of inertia about the axis that is parallel to I as well as passes through the center of mass. Okay. All right. These are next 2 questions. Okay. 31st people have solved 2 guys. Atmesh and Kondi. A uniform rod of length L suspended from one end such that it is free to rotate about an axis passing through that end and perpendicular to the length. What minimum speed must be imparted so that rod completes one full revolution. Okay. So here suppose this is the hinge. So this is the rod. Okay. This is the rod. Now with uniform rod of mass, uniform rod of length L, what minimum speed must be imparted to the lower end. So what should be the speed over here so that it completes the full circle. Now the critical point is when it becomes vertical. What is the condition for it to complete the full circle? At least that its velocity should be zero at the top most point. Okay. It is not a string so there is no question of the string becoming slack. Okay. So just you need to take care that at the top most point velocity just becomes zero. So even if slightest of the velocity is there it will further rotate and come down complete the full circle. Okay. So we will apply conservation of mechanical energy. Okay. Just give me one second. All right. Did anyone get question number 32? Okay. The height of a solid cylinder is four times its radius. Okay. So it's a solid cylinder. This is H and this is let's say radius. Okay. So height is four times the radius. It is kept vertically at t equal to zero on a belt which is moving horizontal direction with velocity 2.45t square. So velocity is changing with time 2.45 into t square. If the cylinder does not slip it will topple over at time t equals to what? Now if cylinder doesn't slip then the only force which is acting on the cylinder in forward direction is friction only. There is no other force horizontally. Right. Now that friction force must be equal to mass into acceleration. Right. So if velocity is 2.45t square acceleration will be dv by dt that will be equal to 4.9t square. Fine. So the frictional force will be equal to m into a. So 4.9m into t square is the force of friction. Okay. Now if it is about to get toppled. Okay. If it is about to get toppled normal reaction will shift towards the edge. Okay. Because everything else will be lifted off slightly. So normal reaction is shifted towards the edge. Okay. And if you do the force balance then normal reaction should be equal to mg. I will I also make a lot of celeros. So normal reaction is equal to m into g. Okay. Now if it is about to get toppled then torque about center of mass which should be equal to i alpha should be just becoming 0. Okay. If it becomes slightly more than 0 then it will topple. Okay. So net torque about center of mass is actually friction which is 4.9m into t this friction into h by 2 this torque is rotating anticlockwise. Okay. Oh wait. It will topple in other way right. So normal reaction will shift in other direction. So normal reaction will go to this edge. This is the normal reaction. So if it is like it will be like this and this minus torque due to normal reaction will be mg into r. Okay. This will become equal to 0. All right. Now m will get cancelled off. Okay. So you will get what is h? h is equal to 4r. So this and this if I cancel numerator will have 4. This will get 2 and 4.9 will get cancelled. So here you will get 2. Okay. So t is equal to 1 second you will get. Option A is correct. All right. We will move to the next one. These two questions. You can attempt 34th question first in case you are struggling with 33rd. Okay. So let's take 34th one first. Now we know that moment of inertia is minimum if it passes through the center of mass. So that will just remove BC and AB because other two options have axis that is parallel to AB and passing through the center of mass. Okay. And parallel to BC passing through the center of mass. So it is between C and D. Okay. The difference between C and D is that for EG the mass distribution is further away from the axis. Right. So ultimately moment of inertia is summation of MIRI square only. So if more mass is away from the center of sorry more mass is away from the axis the moment of inertia will be automatically more right. And if you look at HF the mass distribution is closer to the axis here. All right. Hence option C is correct over here. Now let's talk about 33rd. Here it is a question where a solid sphere volume is increasing. So delta V by V is given as 0.0 or let's just talk about percentage term 0.5 percent. All right. So basically when you look at this you need to get a relation between delta V by V and change in moment of inertia. Okay. Now before you do that first you need to understand that it is clearly written anyways that no instant torque acts. So you should know that angular momentum is a constant. So L which is I into omega is a constant. Okay. So you can take log I mean this is a standard procedure in case you have not done this. So there is always I mean you should learn this it is useful. So when you differentiate this log of L is constant. So this is 0. It becomes Di by I plus D omega by I. Okay. Now if delta's are very less I can say that delta I by I is equal to delta omega by this is omega by omega in terms of magnitudes. Right. So of course if one increases other will decrease because some of these two should be 0. Okay. And it is written what that volume increases. Now if volume increases moment of inertia will increase because mass is going further away from the axis. So if moment of inertia increases omega will decrease. Right. So this and this they are out of question it is between B and D. So let's see which one of them is correct. Now the problem is the volume increases given but moment of inertia is 2 by 5 m r square. Okay. So now if volume increases by 0.5% mass will be constant. So delta I by I will be equal to 2 times of delta r by r. Right. Same procedure you have to follow. Mass will be constant and 2 by 5 is anyway constant. Delta I by I is 2 times delta r by r and also volume is 4 by 3 pi r cube. Okay. So delta V by V is basically 3 times delta r by r. Okay. So delta r by r is one-third of delta V by V. So delta I by I is two-third of 0.5%. Okay. So this is a 2 by 6 or 1 by 3. It is 1 by 3%. Okay. So that's why option B is correct. So we have done this type of thing. I think you might be remembering units and measurement chapter. It's a routine stuff. You can say you're finding error or deviation and similar procedure is followed here as well. Okay. In case of any doubts type in I'm moving to the next one 35 and 36. I think this is very similar to the previous one. So we will I'll just leave it for you to try out yourself and I'll give you a hint for 36. Okay. Original rotational kinetic energy is K. The new value will be what? Now kinetic energy is half I omega square, right? So this is 2 by 5 MR square into omega square. So just substitute the values of what is the change in omega and what is the change in the radius. You'll get the answer. The percentage change in radius in terms of percentage is equal to change in diameter. No, but anyway, that is not required. We have directly given the radius reduces to 1 by N. Okay. So just substitute here and there. You'll get the answer. Okay. Here we do I1 omega 1 is equal to I2 omega 2. So omega 2 will be equal to I1 by I2 into omega 1. Right? So I1 is proposal to radius square. So I2 becomes R by N. So this will be N square omega 1. Okay. So that's why it is D. And once you know the variation, this will be M into R is R by N. So this is R square by N square. Omega square is N to the power 4 into omega square. So it becomes N square. Okay. So the new value is N square. Okay. Fine. We'll move to the next one. So please ignore this. It refers to some question which we haven't done. Try doing this one. This is 35. No, this is 37. Anybody else 37th? So all we have to do is to find out individually the moment of inertia of these four rods about this axis and just add it up. And luckily DA is same as CB and DC is same as AB. So moment of inertia of CD is equal to moment of inertia of AB is equal to ML square by 12. Okay. So moment of inertia of AD is ICM which is 0 plus M into L square which is L square by 4 which is also equal to moment of inertia of BC. So total moment of inertia will be 2 times of ML square by 12 ICD plus IAB plus 2 times of ML square by 4. And when you use parallel axis theorem for AD this axis is parallel to that axis. So you need to use this axis and then translate there. Now this axis is passing through the rod. So all the masses lie on this axis only. So ICM is 0. Okay. But ICM about this axis that comes out of the board or your screen is ML square by 3. But about this it is 0. Fine. So the answer you will have 1 by 2 ML square 1 plus 1 by 3. So we will have 2 by 6 ML square by 3. ML square by 3 you will get right. ML square by 2 into 4 by 3 sorry 2 by 3. You get 2 by 3 ML square by right. So none of you get it. Guys be careful. These are like standard routine questions you cannot afford to miss it. And you are I mean missing is fine. You are getting it wrong. You get minus 1. I mean be 100% sure about these questions. Okay. Just simple calculation of moment of inertia. You need to be little bit careful and you will get the answer. This 38th no integration is required. Exactly. So you can treat it like a half of the square. Okay. So if you do that then you can just extend from A and then can do it like this. Okay. So it becomes a square thing. Okay. Now moment of inertia about this is what? For a square. We know that moment of inertia about this is what ML square by 12. It behaves like a thick rod. Isn't it? So this is about that axis. Now moment of inertia about this axis is also ML square by 12. Right. So moment of inertia about that axis that comes out of your screen. Okay. Which is IZ. If I use a Pandit's term it will be IX plus IY. Isn't it? So IX is equal to IY which is ML square by 12. So IZ will be equal to ML square by 6. Okay. Now why I am finding this moment of inertia? Because now you can translate this IZ into these two perpendicular axis. That are the diagonal axis. Okay. Let's say this is I1 and this is let's say I2. So this is also equal to I1 plus I2 because I1 and I2 they are perpendicular to each other. Isn't it? And because of the symmetry I1 is equal to I2. Right. So when I1 comes out to be equal to I2 that is equal to ML square by 12. Okay. Now in reality it is just a triangular plate. It's not a square. Okay. So about this axis for a complete square it is ML square by 12. Okay. Now for a rectangular what it should be? It should be equal to half of it. That should be equal to ML square by 12 into 1 by 2. Because of symmetry the upper portion and the lower portion will have same moment of inertia. Right. So now this half if you take a numerator it will be M by 2 square L square by 12. Now if you remember this M you have taken for the entire square. What is given is mass of the triangular plate. So this M by 2 let me call it as M dash. M dash by 2 is actually M. So that is why option A is correct. All right guys. So that's it for today. We have solved close to 40 questions in 3 hours. Right. So like that only you should judge your I mean how you are spending your time. So at times you sit with your phone. You don't realize that more than half the time is taken by your mobile phone. Okay. So you can count you know number of questions which you are trying out. Okay. And also make sure you take different varieties of questions because variety is another important parameter other than just quantity of questions you are trying to solve. Right. Okay. So that's it for today. We will have another session soon for problem solving. Okay. Bye-bye. I'll share this PDF with you guys and also I'll share the link again.