 Okay, let us see. We'll take up few questions to start with. You can start solving these questions. These are filling the blanks. They used to come long, long back. Hello, everyone. Please start solving these questions which are there in front of your screen. All right. How many of you got the first one? Can start solving the first one by drawing a simple figure representing the situation? Okay. You have a cube. Okay. Fine. And then you have a force applied normal to one of the faces at a point that is at a height 3A by 4. So you have a force acting like this. Okay. You have this force F, which is at a distance of 3A by 4. Fine. You need to find minimum value of F so that the cube begins to tip about the edge. All right. Now, when the cube starts to tip over from this edge, all right, then it will just lose the contact from everywhere else. So normal reaction will be applied through this point only. Getting it. So this will be the normal reaction applied from here. Usually we take it from the center. Okay. But that there is no reason why it should be from the center in this situation. Okay. Because here the point of contact when it is about to topple is this only and normal reaction is applied from the contact point. All right. And then you have a gravitational force, which is what MG hitting it. So this is a situation when it is about to topple. Getting it. So if it is not about to topple, then normal reaction will be applied this way sideways. But then if it is about to topple, this will be the situation. Okay. And I know that MG will be acting from center of mass. So this distance has to be a by 2. Fine. Now I can apply torque balance about this point. So about this point, it is about to rotate. Okay. So about this point, the torque is about to be greater than zero. This acts like axis of rotation. All right. So if I write torque due to force, that will be F into 3 a by 4. Right. And which direction the torque will be, it will be trying to rotate in this way. And MG is trying to rotate in opposite way. So that is when you add it up, these torques, they have to, you need to reduce MG's torque from F torques. So MG into a by 2, this should be equal to zero. This is a limiting condition. If this is just greater than zero, then torque will not be zero, then this will be equal to I alpha. So if alpha is there, it is rotating. Okay. So like this, you can see that you will get F equals to 2 MG by 3. Fine. What about the second one? Now, whenever you see such situations, immediately conservation of angular momentum should come in your mind. Okay. So if you consider the beats and this rod to be one system, then there is no external force or external torque about this axis. Even MG, when your bead will move, one will be moving this side, other will be moving that side. So torque due to the gravity will get cancelled out about this axis because it is symmetrically placed here and there. So torque due to MG is also not there. So torque about this axis is zero. Hence, I can conserve the angular momentum about this axis. All right. Now try to do it quickly. Six omega by ML. No. You're saying, Kushal, are you saying that it is independent of small m? It is independent of mass? Okay. So when I will conserve angular momentum, so conservation of angular momentum. All right. Then I will have to use this expression. I1 omega 1 is equal to I2 omega 2. Fine. So basically, angular velocity at the second point will be equal to I1 by I2 times omega 1, which is what omega not over here. Okay. Now I1 is what I1 is moment of inertia of entire system. Fine. Now system consists of two beads of small m which are located here near the axis first. So if a mass is located near the axis, then moment of inertia for that mass about the axis is zero. Okay. So hence, only the moment of inertia will rod will come up over here. Moment of inertia of the bead will be zero. But when the bead is at a distance of L by 2, okay, then you will have moment of inertia for the rod as well as the moment of inertia for the bead, which is 2 times m L by 2 whole square. Fine. So this is what you have to substitute here. I1 and I2, you'll get value of omega 2. Okay. Any doubts with respect to this question? Please type in yes or no. Those who have answered, they don't need to say anything. Okay. We'll move to the next one. Do this quickly. This is third and this is fourth. Okay. So those who have got the third question, move to the fourth one. I'll solve the third one meanwhile. So a stone of mass m tied to the end of the string is moving in a horizontal circle. So if this is the horizontal circle, okay, then it is moving around this circle. Let's say radius is R. The length of the string is gradually reduced. Okay. Keeping the angular momentum of the stone about the center of the circle constant. So about this, the angular momentum is constant. Okay. So if angular velocity is omega, let's say at any moment angular velocity is omega, then the angular momentum of this mass is what? Suppose the mass is m, angular velocity is omega, radius is R. Angular momentum is what? Angular momentum of a point mass is mv into R. Okay. Actually it is r cross p. Okay. But here you'll see that R is perpendicular to p. So r cross p will simply become mvR. Okay. Now v is what? Omega into R. Okay. So I will get mR square into omega. So this is the angular momentum which should remain constant. Okay. And if I do the force balance, okay, if I do the force balance, let's say I assume here, let's say some, if something is moving in a circle, you know, if something is moving in a circle, then this force T should be equal to the centrifugal force, which is m omega square R. Right. So T should be equal to m omega square R. Fine. And we know that angular momentum is constant. Okay. So I can substitute the value of omega in terms of angular momentum. Right. So omega will come out to be equal to L divided by mR square. Okay. Now why I'm substituting omega in terms of L? Because right now here you have two variables. Okay. You can't say that T is proportional to R ratio power 1 because omega depends on R also. Fine. So you need to get an entire expression in terms of a single variable. That is the reason why. So when you substitute omega over here, you will get tension to be equal to mL square divided by m square R to the power minus three. Okay. Hence, now L square and m are constant. So T is proportional to R ratio power minus three. So n is equal to minus three. Okay. Now start solving the fourth one. Vaishnavi, are you watching over your mobile phone? Don't do that. Watch it on your laptop. Okay. Shweta is asking something. I for a circle. Shweta, I for the disk is mR square by two. Okay. It is not a disk. Okay. It's just a point mass which is rotating. So point mass's moment of inertia is mR square about a point which is at a distance of R. Okay. Guys, don't watch this on your mobile phone. The font size will become very less. If I increase the font size, then the space for me to write will reduce. Okay. This is not meant to watch on mobile phone. Okay. So let me solve the second question as well here. All right. So a uniform disk of mass m and radius R is rolling up the inclined plane. Okay. So a typical situation in physics where you have an inclined plane. Okay. And then you have a disk. Okay. You have a disk of mass m and radius R that is rolling up the incline which makes 30 degree with the horizontal. So this is 30 degrees. Okay. The coefficient of static and kinetic frictions are equal to mu. All right. The only forces acting are gravitational and friction. Then the magnitude of friction force and the direction is what? Okay. Let's see. So we can say that normal force from the surface will be like this. Okay. So this is the normal force. Okay. That is the normal force. Then you have component of gravitational force here. This is mg sin theta. Okay. And then you also have a component in this direction mg cos theta. Fine. And let us assume that the friction direction is, you know, up the incline. I may be wrong. If I am wrong, when I solve the equations, I will get friction force to be negative if this direction is not correct. Okay. Now this is written that it is rolling up the incline. So when it rolls up the incline, of course, the sense of rotation will be like this. Fine. Now all these forces are represented. Now we can solve this particular equation. And okay. So we have along the incline plane. Okay. If I write along the incline plane, I will get friction force minus mg sin theta is equal to m into a center of mass. Okay. Let us assume that a center of mass is upwards. Fine. And I also have torque acting on it because of friction. Right. Now you can see that because of friction, if I write torque about center of mass to be equal to i alpha. Okay. If I do that, then you will see that the friction due to torque is in opposite direction. Are you getting it? So we have assumed alpha in this direction, but torque due to friction is in opposite direction. The sense of rotation because of friction is in opposite direction. Getting it? Hence, when I write the torque equation, I'll have to write f r into r is minus of i into alpha. Now this is a disc. So m r square by 2 into alpha. Fine. So this is my first equation and this is my second equation. Fine. Now here there is an assumption that it is rolling without slipping. Okay. If it is rolling without slipping, this point, the point of contact should remain at rest. Fine. So according to the figure, the point of contact will be, you know, moving up with the center of mass with acceleration ACM and it will go down because of the rotation with alpha into r. Fine. So the total acceleration of the point of contact should be ACM minus alpha into r. The point of contact should be at rest because inclined plane is at rest. This should be equal to 0. Fine. Many a time student directly write ACM is equal to alpha into r. That is also correct if the surface is stationary. But if surface is not stationary, then you have to analyze the situation further. Okay. So all you have to do is now solve these equations and you'll get the answer. Fine. Any doubt guys with respect to these? Type in yes or no. Okay. I'll move to the next one. Whenever you see a question that we have taken up already, immediately point it out. I'll replace that one. I do these two questions. Okay. I'll start solving the first one. Those who have got please move to the next one. So we have a rod which is supported by two parallel knife edges A and B. Okay. Let's say this is A and this is B. Okay. These arrows represents normal reaction from the knife edges. This is N1 and this is N2. Okay. Center of mass of the rod is at a distance X from A. So this is A. This is B. So let us say this is a center of mass. Okay. So the entire weight will act through that. Fine. So this is where the weight is acting. So in this chapter, you know, you just can't ignore from which point the force is acting. Okay. You need to not only track the direction of force, but also from where it is applied. So this distance is X. This is X. Okay. And that distance will be D minus X. Right. Because the distance between A and B is D. Okay. So now I need to find out what the normal reaction at A and normal reaction at B. If I do the force balance, I will immediately get N1 plus N2 to be equal to W. Fine. So whatever answers you are getting, N1 plus N2 should be equal to W. Normal reaction at A plus normal reaction at B should be adding up to give you W. Okay. Now, if I use a torque equation, let's say about point A. If I write torque equation, it should be 0 because this rod is not rotating. So angular acceleration should also be equal to 0. Okay. So that is the reason why if I write torque equation about A and equate that to 0, what I'll get here is X into W. W is trying to rotate this way. So if that direction is taken as positive, but normal reaction is trying to rotate in opposite direction. So X into W minus N2 into D. Okay. This should be equal to 0 because the distance of N2 from A is D. So from here, you'll get N2 is equal to XW by D. All right. So all you have to do here afterwards is substitute the value over here. You'll get normal reaction at A also, which is N1. This is normal reaction at B. Okay. Okay. Let's read the question. A symmetrical lamina of mass M consists of a square shape with a semicircular section over the edge of the square as shown. The side of the square is 2A. The moment of inertia of lamina about an axis passing through the center of mass and perpendicular to plane is 1.6 mA square. The moment of inertia about AB is asked. Okay. Now here, you will first of all notice that it is lamina. Okay. Since this is lamina, I can assume that this is a two-dimensional figure and I can use perpendicular axis theorem. Fine. So what I say here is that, let's say moment of inertia, which is given, is IO. Okay. This should be equal to IX plus IY. Now, why I am doing this? I am doing this because I have to find out moment of inertia about this, which is not passing through center of mass. Okay. So if somehow I get moment of inertia about this dotted line, then all I have to do is shift this axis from center of mass to this and I use parallel axis theorem. But I don't know moment of inertia about this axis. So in order to find moment of inertia about this one, let's say about IX, I need to use perpendicular axis theorem. Okay. I'll say that moment of inertia about the axis coming out of the plane passing through O is equal to IX plus IY. IY is this. This is IY. Okay. And due to symmetry, due to symmetry, I have IX equals to IY. All right. Hence, two times of IX will be equal to I naught, which is what? 1.6 MA square. All right. So I will get IX, which is 0.8 MA square. Now, what is so special about IX? One special thing about IX is that it is passing through the center of mass. So I can use parallel axis theorem. So I about AB is equal to IX plus M into D square. Now, what is D? D is this distance. Okay. Which is what? From here to here, it is A. And then again, this is a semicircle, which should have radius starting from here to there, which is again A. So this distance will be equal to 2A. Okay. So I will get 0.8 MA square plus capital M into 2A whole square. So from here, I will get 4.8 MA square. Okay. Fine. Now, again, see some of you or at least half of you might not be getting these questions right in the first go because you are doing it for the first time. Okay. Maybe people who are answering have already seen these questions. So don't get intimidated by all that. Okay. You will get surprised when you see everybody's marks at the end of the day. Right. So don't get intimidated by intimidated by anyone. All right. So let us move to next question. See, these are true and false questions. Okay. So they also used to come much before you were born. They were from 1980s. Okay. Do these questions. Why it is false? Tell me the reason also. Right. So you know that this equation torque is equal to I alpha. This equation you can write about fixed axis. Fine. So if you write this about A, then you have to use moment of inertia about A. So this is first equation. And if you write that about B, then you have to use moment of inertia about B. Okay. And clearly moment of inertia about B is different from moment of inertia about A. When you look at about A, the more mass is further away. When you look at from B, lesser mass is away from B axis. So we know that IA is more than IB. Okay. That is the reason why if you apply the same force, okay, then the torque is same. Okay. This is let us say D. Then this is also D. Okay. In both the cases, you are applying the same torques. Angular equations will be different. Since IA is more than IB, alpha A is less than alpha B. Fine. So like this, you have to do the seventh question. What about the eighth one? Eight is false. A uniform disc, circular disc and mass and radius are rotating in a horizontal plane about an axis passing through its center perpendicular to the plane with angularity omega. Another disc of same dimension, but mass m by 4 is placed gently on the first disc coaxially. So again, you know, whenever such kind of situation come up, you should immediately try to see that if conservation of angular momentum holds good or not. Okay. So if you consider both the disc, okay, as one system, then whatever is a torque between them, that is internal one. Okay. So if you conserve angular momentum, you should have I1 omega 1 equals to I2 omega 2. Fine. So omega 2 will come out to be equal to I1 by I2 omega. Fine. Omega 1 is omega over here. Now, I1 is MR square by 2. It's a disc. Now, and the disc is placed gently on the first disc. Now, the assumption is the second disc is not rotating initially. Okay. So I2 is the sum of momentum inertia for both the disc. That is MR square by 2 plus m by 4. Dimension is same. So m by 4 is, sorry, m is m by 4. Okay. So MR square by 2. This is I2. So when you substitute it, you'll get a different answer. So hence, even 8th one is false. This is also false. Both the statements are false. Okay. In case of any doubts, please type in quickly. I'll move to the next one. You can help each other on the chat. If somebody is asking a question, you can reply to that same. Nickel m is m by 4 for the second disc. Read the question carefully. Okay. So MR square by 2 is a formula. Now, if m is m by 4, it becomes MR square by 8. This is 9 and this is 10. Try solving these two questions. What about the 9th one? Saimir, please explain how you're getting it. Okay, Saimir, can you explain how you got that both of them will reach together quickly, waiting for you. Saimir, that's, I'm not asking you whether their velocities are same or not. How you got, what is your approach? Just brief it immediately. How you got their velocities will be same. Anyways, let's do this. A ring of mass. So mass of ring is 0.3 kgs. Okay. The radius of the ring is 0.1 meters and you have a solid cylinder whose mass is 0.4 kg and the same radius. Okay, the radius is still the same. No Saimir, that's not how. Okay, so the kinetic energy is same in both the cases. Okay, so kinetic energy in case of ring, this is what? Half i omega square. Fine. So here i is MR square. All right. So I will get this as MR square omega MR square. Yeah, this is kinetic energy of the ring. Now kinetic energy for the cylinder, kinetic energy for the cylinder is half, this is ring. Okay. This is half IR, half IC omega square. So here you will get half, this is half M of the cylinder. This is of the ring. Okay. This is M of the cylinder, R square by 2 into omega square. Okay. Now this is just rotational kinetic energy. Okay. This is not translational. This is rotational kinetic energy. Okay. Now translational kinetic energy for the ring is half M into VCM square. Right. Now here we are assuming that it is pure rolling. So that is some assumption which is not clearly stated over here. So that makes this question very ambiguous. So I did expected anyone to solve this without, I mean, anyways. So this is half MR, VCM is omega square R square. Fine. This is kinetic energy of the ring. This is translational kinetic energy. Okay. Now translational kinetic energy for the cylinder is similarly half M of cylinder omega square into R square. This is translational. Okay. Now when you add translational kinetic energy of the ring with translational kinetic energy, sorry, translational kinetic energy of the ring with rotational kinetic energy of the ring, you will get total kinetic energy. Okay. Similarly, when you add the cylinder's rotational kinetic energy with cylinder's translational kinetic energy, you will get cylinder's total kinetic energy. Okay. You will see that both their kinetic energies when you equate, you will get same omega, same velocities. Okay. And since they both are moving with same velocity of center of mass or same angular velocity, when they are released on the ground, it means that they will move forward with the same, you know, same velocity and hence they will reach together. Okay. Try to do the second one as in the question number 10. Yes, Atmej, that makes this question little ambiguous. I can understand your concern. Okay. Now the thing is, it is given that as soon as it is placed on the surface, they begin to roll. Fine. And the friction is not there. Okay. If friction is not there and they are rolling, it must have initial velocity as well as initial angular velocity. Otherwise, it will just spin. It will not roll and move forward. Getting it? Because friction is not there. So friction cannot create torque for it to start rolling. It must have initially itself acquired some velocity and angular velocity and then it is rolling even though there is no friction. Fine. So that makes this question very ambiguous and I was not expecting any one of you to solve this. Do the next one. Very ambiguous is the wrong word. It makes this question very tricky. Rolling even though there is no friction because initially itself it has velocity and angular velocity. Okay. Anybody got the 10th question? Okay. Nikhil will explain how we got it true. So two particles mass 1 kg, let's say m1 is 1 kg and m2 is 3 kgs. Okay. They move towards each other under their mutual force of attraction. No other force is there when the relative velocity of approach of the two particles is 2 m per second. So m1 and m2, let's say this is v1 and this is v2. What is given is velocity of approach v1 plus v2 is 2 m per second. Okay. The center of mass has the velocity of vcm has the velocity of 0.5 m per second. Okay. When the relative velocity becomes 3 m per second. Now after some time v1 dash plus v2 dash becomes 3 m per second. The center of mass velocity is 0.75. Okay. Of course that is not true. It is false. Okay. Because there is no external force, no external force on the system. Okay. Since there is no external force on the system, velocity of center of mass has to be constant. There cannot be any acceleration of center of mass. Okay. So velocity of center of mass will remain 0.5 m per second only. Let's read out guys on these questions. Please type in quickly. Yes or no? Okay. So I'll move to next question. So in case you see a question which is repetitive, okay, which we have done earlier, immediately point it out. No, v1 is a velocity of m1, v2 is a velocity of m2. Okay. Do these questions. Okay. Don't delete your messages. Why you have to retract your message? It's not your Facebook or WhatsApp status. No need to delete anything even though you are wrong. Okay. Do these two questions. This is let's say 11th. This is 12th. Single option correct. Okay. Okay. Yeah. Even I was thinking 11th. We have done. But now that it is already couple of minutes past, we'll keep it as it is. Okay. So I'll quickly solve the 11th one. A thin circular ring of mass m and radius are rotating about its axis with constant angular velocity omega. Two objects, each of mass m gently on the opposite ends of diameter of the ring. The wheel now rotates with the angular velocity of m1. Now, immediately, I mean, now you must have been able to understand that you will be applying conservation of angular momentum here. Okay. So simply I1 omega 1 is equal to I2 omega 2. You have to apply. Okay. So I1 is mr square and I2 is mr square plus I'll write here. I1 is actually capital M into r square. I2 is capital M r square which is ring plus two masses angular, sorry, two masses moment of inertia, mr square. Fine. Now, since they are placed dramatically opposite to the ring, their external torques because of gravity on both of them will get cancelled out. So no torque with respect to that axis. Okay. So you can write this and get the answer. 12th one. What is the answer for the 12th? Pratik, how you did it? So basically work done is change in kinetic energy. Okay. So you have two point masses that are fixed at the ends of the rod of 1.4 meter and negligible mass. The rod is said rotating about an axis perpendicular to its length with a uniform angular speed. The point on the rod through which the axis should pass in order that work done for the rotation of the rod is minimum. Now, whatever is the work done should be equal to the kinetic energy. Okay. So if I say that the rod is rotating, let's say about a point this, let's say point P. Okay. And you have two masses at two ends. This is first mass and this is second mass. So this mass is point 3 kg and this one is point 7 kg. Fine. So let us say this is no Pratik, that's not how you should do it. Anyways, you can say that this is x. Okay. If that is x, P is the axis of rotation. Then this distance is what? This distance will be 1.4 minus x. Fine. So work done is equal to change in kinetic energy that work energy theorem. Right. So this is equal to half i omega square. So it is, I'm not considering change in potential energy because of gravitation. Okay. Because this is immediately acquired kinetic energy. So there is no change in height for the center of mass of these two point masses. So I will not consider that. Okay. Now I is what I is a moment of inertia. Rod is massless. The moment of inertia about point P is what point 3 times x square plus point 7 times 1.4 minus x whole square. Right. MR square is a moment of inertia for a point mass. Okay. So work done should be equal to half point 3x square plus point 7 1.4 minus 1.4 minus x whole square. Okay. This is i into omega square. Yes, you can say that. Now basically you need to minimize moment of inertia so that the work done becomes minimum. All right. So you can do it like this. You can differentiate work done with respect to x and equate that to 0. All right. You'll get the value of x. Okay. And you also know that moment of inertia for any system is least about an axis that passes through the center of mass. So you can just find the center of mass of these two point masses wherever it is lying. So that is where you know you should be rotating this system. Okay. So I'll move to the next question. I mean you can start solving the question if you are able to see it. You don't have to wait for me. Others. Anybody got the answer? 13th is B. Two different answers. Sundar, can you explain why it is B? Okay. See, mass M is moving with constant velocity along a line parallel to the x-axis. But it is not saying that it is moving on the x-axis. Fine. It is moving parallel to x-axis. So let us say that this is my x-axis. This line is my x-axis. So mass is moving parallel to x-axis. So let's say this is the mass M moving with velocity V. Okay. And if I say that this is my origin. Okay. This is let's say origin. So we know that the formula for the angular momentum is what r cross M into V. Right. Now r cross M V is what? The magnitude of angular momentum will come out to be r into M into V into sin of theta. Okay. Now r sin theta is basically perpendicular distance from the line of velocity. Right. We have done this many a times. That into M V. Now perpendicular distance of line of velocity is what? So this is the line of velocity. Okay. And if I drop a perpendicular on it, this length is a perpendicular distance which is r sin theta. Okay. You can also see that this is r vector. Okay. If this is r. Okay. This is the angle between r and V. Let's say this is theta and this is r. This automatically becomes r sin theta. Okay. Now no matter what is the angle theta and what is r, the perpendicular distance remains the same. Right. That is the reason why magnitude of the angular momentum is a constant over here. So option B is correct. And you can, you know, apply some logic like this also that if it is zero, it has to be constant also. Right. Like if you get zero as a answer, it has to be B also correct. So but then this is only single option correct. So like that you can write at answer B. But that's little, you know, it's like tricking the question that is not recommended. Why should we option B? See kinetic energy can be same. Okay. Because kinetic energy not only depends on omega, it depends on V also. So if V and omega both can change so that some of it remains constant. Okay. But the key reason over here is that it's a smooth sphere. A is a smooth sphere moving on a frictionless horizontal plane. Okay. It is hitting another identical sphere. So even though it is rotating with different angular velocity, when it hits another identical sphere, there is a slipping on this point. Right. And since there is a slipping on this point, if there's a friction, it will apply a friction and because of that some torque might be applied to create an angular velocity. Okay. But then since it is a smooth sphere, it cannot apply any tangential force on the second sphere. Only force will be normal reaction that anyway passes through the center. So normal reaction do not create any torque. So because of that, the second sphere will not rotate. Okay. This is identical sphere not only in terms of the shape and size, but also it also has same angular speed and angular velocity. So their angular speeds will not get changed even though they might be rotating in opposite directions like that. Fine. So omega A will be equal to omega B since there is no friction between them. Since there is no friction between the sphere, there is no friction to change the angular velocity. So angular velocity will not change. Whatever is the other force that is passing through the center of mass. So there is no torque because of other forces with respect to center of mass. Here we are talking about angular velocity, not velocities. Okay. Let's move to next one. This we have done. These two questions. Okay. So if you guys know how to solve 15th, please proceed to the 16th question. I wanted to discuss 15th, so that is why let's keep it. Okay. Anybody got any of these two? No. 15th is C. See, the angular momentum about a point is given as this formula. R cross MVCM plus ICM omega. Okay. Now R is what? R is the vector that connects the point from which you need to find the angular momentum to the center of mass. Okay. So this is R. R cross VCM. Now VCM vector is like this horizontal. When you do R cross VCM, you basically get perpendicular distance into M into V. Okay. Pappinical distance is radius only. So R into M into VCM plus this is a disk. So ICM is MR square by 2 into omega. Okay. Now this is rolling. There is no slipping. So this will come out to be MR square omega plus MR square by 2 omega, which will be 3 by 2 MR square omega. So that's why option C is correct over here. Okay. What about the 16th question? Vishal is getting A. Okay. Should I do it? Okay. If I ignore the gravity, it is written that gravity is ignored. Okay. And this rod is rotating with angular acceleration alpha about A. Okay. About A it is rotating. So if bead is, let's say bead is at a distance of, okay. Initially it has a distance of L. Okay. Let's say, let's say the bead is at a distance of L and it is about to slide. Okay. Then what will happen? You'll see that when you draw free by a diagram of this bead only. Okay. You will have a centrifugal force, which is M omega square L. Okay. And you will have a frictional force, FR. Okay. So frictional force will be actually, if this bead is stationary, then frictional force should be equal to M omega square L. Right. And the maximum value of friction force is mu times N. So when this will become equal to mu times N, afterward it will start slipping. Okay. Now the situation is that it has angular acceleration alpha. So omega is increasing with time. Once this reaches, after this point, omega will further increase and it will start slipping. Okay. Now the problem here is to find the normal reaction. Okay. Now how will we get the normal reaction? We know that the rod is rotating with angular acceleration alpha. Okay. Suppose it is going into the screen. So the acceleration of this bead is what? Alpha into L. Okay. This is tangential acceleration. Getting it? Fine. So normal reaction, if it is acting on the bead, should be equal to mass into tangential acceleration. See normal reaction cannot be radial. Okay. Normal reaction cannot be along the rod. Normal reaction has to be tangential. Normal reaction that is creating the tangential acceleration on the bead. Fine. So this is equal to M into, now tangential acceleration is alpha into L. Fine. Alpha into R is the acceleration. So alpha into L is the tangential acceleration over here. Okay. So M omega square L, when this become equal to M alpha L, then it will start slipping. M alpha L is a normal reaction, this into mu, sorry. Okay. So mass get cancelled off. L also gone. So omega comes out to be under root of mu times alpha. Okay. Now omega should be this. Initially, it is set to with constant angular acceleration alpha. So initially the angular velocity is zero. Initially its angular velocity should be this. So what should be the time? We know that this formula omega is equal to omega naught plus alpha into T. So T will be omega divided by alpha. Okay. That will come out to be mu under root mu by alpha. Okay. So like this you can solve this particular question. Okay. Any doubts with respect to this question? Please type in yes or no. Next question. So you can quickly tell me the answer. We will move to the next one. See moment of inertia is clearly changing. These masses, they are moving away from the axis. Fine. So moment of inertia is changing. So C cannot be correct. Here D also cannot be correct. Okay. Now what about total angular momentum? Now angular momentum is constant because there is no torque. No external torque about this axis. Fine. So since there is no external torque about this axis, angular momentum is also conserved. Okay. So even A is gone. So it's only B. Okay. All right. So let's go to the next one. You guys are so conscious about answering something wrong. Immediately retract your message. Someone will feel about me. Don't feel ashamed of answering something wrong. Do these two questions. In C axis, you can add and subtract moment of inertia. Okay. So had it been a full circle, as in the full disk, as it been the full disk, its moment of inertia would have been what? Its moment of inertia would have been, first of all the mass is of the sector. This mass is m. Okay. So had it been the full disk, the mass would have been 4m. Okay. It's one fourth of the disk. So moment of inertia would have been 4m r square by 2. Okay. Had it been full disk? Okay. Now it is just quarter disk. A by 4 is the answer. That is m r square by 2. So option A. So it's a trick question for those who are in a hurry. So you will quickly get minus one marks. Nineteenth one. Nineteenth, what is the answer? Again, it's about inclined plane over here. Okay. So it is rolling up the inclined plane. Fine. So of course, this should be the sense of rotation. This should be the, this is the sense of rotation and acceleration will be down. All right. This is the accession of center of mass. Now, you'll see that this point, okay, when it is rolling down, it is the entire, I mean, there is no support force. You're not pushing it up. You're going up on its own. So this point is trying to skid downwards. Okay. Although this point is at rest, okay, because it is a pair rolling. So it is instantaneously at rest, but its tendency is to go down, to slip down. That is the reason why when it goes up, the friction force is upward direction. Fine. Now when suppose it's coming down. Rotation is like this. Acceleration is again like that only. Now also, this point, the point of contact is trying to skid forward. So again, the force of friction is upward only. So option B is correct over here. Okay. Now you can solve the same question by, you know, by, by assuming some angular velocity, angular acceleration. And you'll, if you solve this entire question, you'll get the correct direction. Okay. But then that will be like wasting a lot of time. So intuitively, you have to solve these kinds of questions. Okay. So that is why here option B is correct. Okay. Let's move to the next one. While it is descending Ramcharan, the point of contact has a tendency to slip down. Because the gravity force is pulling it down and that is the case while it is going up also. Both the cases, gravity is pulling the disc down and there is no extra force which is pushing it up. So tendency of the point of contact is to slip down. Do this one and even got the 20th. Why are you saying D Sondarya? Why D? Okay. Suppose this is the path of the tortoise. The red line is the path of tortoise. And this is the center. Okay. Tortoise is moving along a chord. Okay. With a constant velocity. So tortoise is here right now. Then it is reaching there. Here, again, since the entire platform is rotating about this axis, you can observe the angular momentum of that axis. Again, the net torque is 0. Now you can say that the tortoise will experience some gravitational force. So hence there is a torque due to the gravity on the tortoise. But the direction of the torque is perpendicular to the axis. So r cross v is perpendicular to the axis. So that is the reason r cross mg is perpendicular to the axis. So that is the reason why I talked to the tortoise mass is not along the axis, which is perpendicular to the axis. Hence, we can conserve angular momentum about the axis. So along the axis, there is no external torque. Okay. Now when you conserve the angular momentum, then we have I 1 omega 1 is equal to I 2 omega 2. So omega 2 should be I 1 by I 2. Times omega 1. Right. And what is I 2? I 2 is moment of inertia for the platform plus moment of inertia of the tortoise with respect to center. Okay. Now you can see that when the tortoise is going from A to B. All right. Suppose it is at A. The moment of inertia of tortoise is m into r square, where r is this distance. Okay. Now when it reaches here, its distance from the center decreases. So at the center, when tortoise is there, the moment of inertia of tortoise with respect to center is least. Least at the center of the chord. Fine. And then it further increases. All right. So whatever is the case from here to here will be the case from there to point B. Okay. And it will achieve minimum. Moment of inertia will achieve minimum. So I 2 will be minimum at the center. Hence omega 2 will pass through the maximum. Fine. So at the center, omega should be maximum. So this is not true and that is also not true. Okay. Now it is between B and D. Okay. Now it is easy to get confused between B and D. But if you write the formula for moment of inertia for tortoise with respect to center, you can see that let us say tortoise has moved along the chord a distance of let us say V into T. V into T is a distance. Okay. And this is the center. So if this is r, moment of inertia of tortoise is m into r square. Right. So I can write r, this r in terms of VT and this distance. Okay. That distance is fixed. Let us say that is h. Okay. And this distance is what? This distance is r minus VT. Okay. So the mr square will become m into h square plus r minus VT whole square. Okay. This is moment of inertia of tortoise. So this plus moment of inertia of platform which is constant throughout will go in the denominator. All right. So you can see that the relation is having quadratic equation. T square term is coming. So it's definitely not linear one. So D is not correct. And we'll go with option B. Okay. So all of you clear about it. Any doubt with respect to this question? I don't know. Let's move to next question. These two. Which question was earlier? See impulse is change in momentum. It is integral f dot dt also. Anyone? Okay. So here I think some of you might not be comfortable with the impulse thing. So the angular impulse, write down the angular impulse is equal to change in angular momentum. Fine. Now you can use this formula about center of mass. Okay. Now about the center of mass, the angular impulse is MV which is linear impulse into L by 2. Okay. This is equal to change in angular momentum. Initially, there was no angular momentum. Finally, let us say it has angular momentum and it has gain angular velocity omega. Okay. Like that you can find out the angular velocity omega. So for that, first you need to get moment of inertia which is what? M into L by 2 whole square. So you will get ML square by 2. I'll just substitute the values. So from here, omega will come out to be V by L. Okay. Now also write down linear impulse formula to find velocity of center of mass. But then nobody is asking here. So impulse is change in momentum. So from here, you will get M into V is total mass of the system into velocity of center of mass. Okay. So velocity of center of mass will come out to be equal to V by 2. All right. A is correct. What about 20 second? Perform circular motion about which point angular momentum of the particle remains conserved. Okay. Now I think this is a straight forward question. So that is center of the circle. Okay. It is MVR the angular momentum of this particle which moves in a circle is MV into R and it remains constant because the reason why it is moving first of all in a circle is definitely because of some force towards the center. Maybe it is tied to a string. Okay. If it is tied to a string, then string is applying a force on it. Okay. But that force is radial in nature because of that about the center whatever is a central force that is creating the circular motion the torque due to that center force will be zero with respect to center. Okay. Hence angular momentum will be conserved about the axis passes through the center. All right. Do these questions for 21st Sondaria we have used impulse equation angular impulses change in angular momentum. See Sondaria impulses okay. This is impulse. Now if you write the angular impulse which is R cross with J this is equal to R cross F into DT okay. R cross F is what torque okay. So integral of torque DT is the rate of change of angular momentum. DT is gone. So integral of DL is change in angular momentum. Fine. So impulse multiplied by perpendicular distance will give you change in angular momentum. So like that this is what we have used in last question. 23rd B you are getting B for Bombay let's see for the 23rd we have a circular plate rotating about the vertical axis passing through the center with angular velocity omega naught moment of inertia of system doubles. So conservation of angular momentum okay. So omega finally which should be able to omega naught by 2. All right. So kinetic energy initially is half I into omega not square. This is initial moment of inertia is half moment of inertia is not double 2 I naught into omega naught by 2 whole square okay. So we will get half of half I naught omega naught square. So if initial kinetic energy is K the final kinetic energy becomes K by 2. So that's why option number B is correct 23rd 24th options are not visible clearly. I will upload it again. Do this 24th Kaushali saying B B for Bombay others what are you getting see whenever you have to find out the total velocity of a point on a rigid body you split it into two parts okay. First you find velocity with respect to center then you add the velocity of center of mass on it okay. So like that you can do it none of you see this is C the center this is point P okay. Now this is rolling without slipping with angular velocity omega. So first of all velocity of center of mass becomes equal to r omega okay. Now with respect to center this point is moving up omega into D where D is the distance of point P from C okay. And this point P is then moving forward with velocity of center of mass VCM which is nothing but r into omega okay. So total velocity of point P will become what under root of omega D square plus omega into r square omega into r is VCM okay. So you will get omega times under root of D square plus r square this is the magnitude of velocity of point P what about Q now Q is there now you can see that for point Q the velocity with respect to center is tangential right because this point is moving in a circle with respect to center so this value is omega into D and then you have VCM like this fine. So this angle is acute it is an acute angle over there right. Wait it's sorry here there is a small error the sense of rotation if VCM is like this that. Okay so with respect to center then even this is not upwards that is coming down okay but the magnitude will be unchanged because still it is 90 degree here you will get like this this is omega into D so total velocity is vector sum of this and that okay. So this angle is obtuse okay so let's call it Q so VQ is under root of VCM square which is omega into R square plus V center of mass sorry plus omega into D whole square plus two times of omega D into VCM into cos of angle between them now you can see that this is an obtuse angle so cos of that obtuse angle is less than zero negative so we can see that the magnitude of P okay magnitude of P is greater than magnitude of velocity of Q alright so P is greater than Q so B is satisfying that and P is VC Q is this definitely this is not satisfying okay now what about VC now VC is just omega into R okay this is VC okay now you can see here that VP is the highest VP is more than VC as well as VQ so that is you know clear over here so since VP is more than both the velocities even D is not correct okay and VP is definitely not equal to VC VC is omega into R and VP is omega into root over D square plus R square so even C is not correct see we the sense of rotation will come depending on which direction you assume it to be moving forward so if you say that the disk is moving forward like this the disk has to rotate like this if you say that disk is going backwards then you can say okay it will rotate like that fine since it is rolling without slipping VCM direction and omega direction related yes okay that is correct okay should we move to next question any doubt guys typing quickly yes or no any doubt this we have done this done if we have done any of these two please quickly tell me I will replace yes all these two questions anyone got the first one that is 25th RQ should be equal to pi R square into T disk is like a cylinder now moment of inertia of the disk about an axis passing to the edge and perpendicular to the plane now disk moment of inertia center of through the center of mass is M into R square by 2 okay see the mass will not change okay when you invert the sphere into disk mass will remain M only so like that so about the edge moment of inertia will be I center of mass plus M into R square parallel axis theorem so this will come out to be 3 by 2 M R square fine so this is also equal to the moment of inertia of the sphere 2 by 5 M R square this should be equal to 3 by 2 M into small R square okay so here you will get option B to be correct okay so this information is not used here so unnecessarily it was given anyways question number 26th anyone okay should be saying A okay so there is a drawing which has been drawn with the ink of uniform line thickness okay so we need to find the Y coordinate of the center of mass the X coordinate of center of mass is clearly 0 because Y axis is the axis of symmetry over here okay so Y coordinate of center of mass this is M3 M4 M5 so M1 into Y1 center of mass plus M2 into Y2 center of mass Y3 center of mass this divided by M1 plus M2 plus M3 plus M4 so this formula we have to use straight away okay M1 is what M1 is the bigger ring the outer ring which is 6 M mass okay its center of mass Y coordinate is 0 so this into 0 plus M2 is the horizontal line so whose mass is M and its Y coordinate is minus A so M into minus A center of mass of the horizontal line plus M3 is the vertical line so M into vertical line center of mass is 0 so M into 0 plus you have okay there is M5 also plus M5 Y 5 center of mass plus M5 okay so these are the 3 masses then M4 is the smaller ring which the left inner circle whose Y coordinate is A so I can write both of them together that is 2M into A this divided by sum of all the masses so that is what 6M plus M plus M plus M okay so denominator will have 6, 7, 8, 9, 10 denominator will have 10 and numerator will have this is 0 anyways this is gone 2M A minus M A so this is M A divided by M also will come in denominator so this will be A by 10 okay so 26th option A is correct fine so like this you have to do this particular question alright fine let us move to the next one this one okay Sanjana is saying A the hint here is that the rate of change of angular momentum will give you the torque okay and here we are talking about the entire torque total torque if we are able to find rate of change of angular momentum that is a torque okay should I do it Sanjana is saying C others so the insect is starting from point O and it is moving along the rod when velocity V is with respect to the rod okay it reaches the end of the rod angular speed of the system remains constant we need to find the magnitude of torque about O the magnitude of torque about O is the rate of change of angular momentum about O okay now omega is constant so omega comes out of this derivative so this is Di by dt okay now what is I I is the moment of inertia for the rod which is M L square by 12 plus let us say insects mass is small m small m into its distance from the axis square let us say the insect is at a distance of r now r should be equal to V into t isn't it it is moving with uniform velocity so M into r square is M into Vt square okay so Di by dt Di by dt will give you 2M V square t okay so torque is equal to Di by dt which is 2M V square into omega into t alright so hence it is a linear dependence on time linearly the torque should vary that is why option B is correct and as soon as the insect reaches the end and stops Di by dt will become equal to 0 because angular sorry the moment of inertia doesn't change after that hence torque becomes 0 after t equal to capital T so you can see only option B is satisfying that any doubt guys on this question anything okay now we will take up questions which have more than one option correct okay so multiple options may be correct these two I think 29th we have already done solved 28th what is the answer for 28th okay anybody got 28th this one yes it is 0 see because initially both are at rest okay so VCM initially is 0 and since there is no external force VCM should remain constant and that constant has to be 0 because initially itself VCM was 0 okay so as no external force on the system VCM should be a constant and since initially it is 0 it remains 0 these two do both of them I think we have already done this I think it is okay it is okay I think we have already done then we have to finish it for one month okay guys so 30th C A and B for 30th two different answers for 30th alright let's see how we can do this question others what happened only I can see couple of them participating okay so at the highest point it is a projectile motion right so at the highest point maximum height which is h its velocity will be along only in x direction right so this is v cos 45 okay we need to find angular momentum about the point of position about this point okay so just extend this line of velocity okay and drop a perpendicular from the point where you need to find the angular momentum so this is h perpendicular distance so r perpendicular into mv is the angular momentum which is m into vx into h okay now vx is v cos 45 so m into v by root 2 1 by root 2 is cos 45 into h okay and h how can you find h h is u square what is the formula for h this okay so you will get velocity square sin square theta is 1 by 2 so v square by 4g okay so if you see first and second equation if you substitute the value of h over there you will get option b to be correct 31st moment of inertia of a uniform square plate about an axis ab that passes through its center and parallel to the two sides so you have a square so moment of inertia about this axis ab iab is given as i cd is a line in the plane of the plane that passes through the center and makes angle theta with the ab okay so you have let's say this line it makes angle theta with the ab you need to find moment of inertia about that line now okay anybody got this 31st okay can you explain why it is a why it has to be a it was just a guess okay you can use perpendicular system to get perpendicular okay fine so the moment of inertia about this white dot and line that is coming out of your screen okay let us say that is i z okay I can use parallel axis sorry perpendicular system and say that i z is equal to ix plus i y okay so if I use ab and let's say a dash and b dash these two as my x and y moment of inertia then this simply becomes two times ix is equal to i z alright so ix will come out to be z by two okay now it happens like that because ix which is iab and i a dash b dash they are symmetrical and hence they are equal that is the reason why two times ix will come out to be equal to iz okay now if I draw a perpendicular axis to CD to use perpendicular axis theorem what I will get here is this now can you answer me CD and c dash d dash are they similar axis yes or no can you answer me if i cd is equal to i c dash d dash are they symmetrical they are symmetrical okay hence now also two times of i cd will be equal to i z and i cd will be i z by two okay hence option a is correct okay so again this question is I mean meant to be solved pretty easily like this and in case your concepts are not clear you will get stuck okay alright let's take I think last few questions we can just take one more question now last question for the day so it has to be good start solving this question this is the last question for the day of yes it is more than one option correct see inner disk is rotating anticlockwise and the ring the outer ring rotates in clockwise so sense of rotations are not the same they are different this yellow line is for the ring okay so if the center is moving with v0 then v0 should be equal to omega into 3R okay this is the center of mass velocity and it has to be along x axis this is correct option a is correct all of you got a correct so in case you have already got option a correct and you are uncomfortable about the entire scenario you do not need to waste your time you can mark one of the correct option and you can move forward at least you will get the part marks but if you take a wrong answer you will get minus 2 in J advanced okay what about points p's velocities now other 3 options are with respect to p's velocity point p is making 30 degree with the horizontal so it is somewhere over here somewhere over here now you can observe point p point p with respect to with respect to center with what angular velocity point p is moving with what angular velocity point p is moving with respect to center it is on the disk right and disk is moving with angular velocity of omega by 2 okay so now if I just draw a line this is o and I can say that this is point p okay and this angle is 30 degrees okay so because of omega by 2 it will have a velocity in this direction perpendicular to this line that is omega by 2 into r okay not only that it will also have a velocity because o is moving and with what velocity o is moving o is moving forward with a velocity of 3r into omega getting it so along x axis along x axis the velocity of point p if you have to find out then this is 30 this one is 60 degrees okay so along x axis your velocity is 3r omega minus omega by 2 into r cos of 30 this is the velocity along x axis so I will say this is i cap okay so that will come out to be 3r omega minus omega r by so this is cos 60 sorry cos 60 so omega r by 4 so 3 minus 3 minus 1 by 4 into omega r i cap this will come out to be 11 by 4 okay so without even going you know thinking anything else I will just mark option b and that is it because the other two option along i cap they are telling different velocities so this will be 11 omega r by 4 i cap okay and this axis is z axis it's not y axis fine so along z axis along z axis the velocity is omega by 2r sin 60 so that will give me root 3 by 4 r omega k cap okay so since here we need to actually find out in a vector form what is the velocity hence we need to know where is x y axis and we will use components to find velocity along i cap vector and velocity along k cap vector and then just vectorally add them up to get the correct option okay any doubt guys anything you want to ask okay so that's it for today and I hope you have learned something today and you liked the session we will meet again with new chapter thank you very much