 do this this came in 2006 so is it D D for donkey yes sir yes D is correct all are we getting D yes sir a thin circular ring of mass m radius r is rotating about its axis with concert angle of omega two objects are attacked gently on the opposite ends of the diameter of the ring the ring now rotates with angular velocity omega dash equals to what now you know even if you are a little bit confused about the concept you don't have any other option other than using conservation of angular momentum okay so that should just pop up in your head that conservation of angular momentum I can use here about the center of mass so your system is these two objects and the ring so if you consider these two objects and the ring as a system then the forces between them becomes internal force so you can't say that that is external talk okay so you can conserve angular momentum when you do I1 omega 1 is equal to I2 omega 2 I1 is mr square but I2 is not mr square I2 is mr square plus capital mr square plus another capital mr square because moment of inertia when you place two objects changes suppose you know you have one object here another object there okay so moment of inertia of these two object is mr square and capital mr square both capital mr square so two capital mr square is a moment of inertia of these two objects itself plus moment of inertia of the ring which is mr square so this is new moment of inertia into new angular momentum sorry new angular velocity let's say omega 1 this is equal to initial angular momentum this into omega so you get omega 1 from this particular equation okay so option D will come out to be correct here any doubt on this anyone please type in say alright anyone what do you answer sir I'm getting B B for Bombay yes okay many are getting option D to be correct that must be correct only 20 yes B is correct B for Bombay alright so here I'll just give hint with respect to this question okay so let's say you have a bigger ring like this this is radius 2r and another ring this is disk correct this is disk of radius 2r then a disk of radius r is removed so that this circumference is touches so this disk is removed okay now the center of mass of the new disk is alpha by r from the center of the bigger disk the value of alpha is what basically you know the question is about finding center of mass of this remaining structure so this remaining structure is symmetrical about this horizontal line yes or no so center of mass should lie horizontally somewhere okay so let's say that center of mass coordinate is x here alright so we can also say that if I add the smaller disk on this particular structure then I will get the bigger disk okay if I add this smaller disk on this one I will get a bigger disk of radius 2r okay so without changing the coordinate axis which remains same I can locate the center of mass coordinate for this which is what r by 2 comma 0 okay the center of mass will be here somewhere and its center of mass is 0 comma 0 and this one is unknown now you can say that this bigger disk is made up of this and this smaller disk so the center of mass coordinate for this bigger one should be equal to let's say mass of that is m1 into x plus mass of that is let's say m2 then m2 into r by 2 divided by m1 plus m2 this is center of mass location of this bigger one which is what 0 okay so when you get that to 0 you'll get the value of x which is the center of mass of this structure okay now you get the well you get a relation between m1 and m2 you need not know the value of m1 m2 because you know you you can cancel out the constants from numerator denominator or anyway it is equated to 0 so the mass is proportional to the surface area because the mass is uniformly distributed over the area so just write constant times area of the structure as mass and that's how you get the value of x and then you can answer this question okay if there is any doubt on this please message or say it I'll move to next one so won't it be m2 into r m2 into r see the rate the radius of oh yeah sorry this is our distance of this point is our the center is our for the smaller one so you're correct this is our only okay I've taken diameter as R so this point x coordinate is our only so m2 into R okay good then there any other doubt I'll move to next one this one try to solve this this is I think a simple one don't get you know bottom by just looking at number of marks and subjective question it can be solved easily this is a last question for the day you can easily you know you can take your time get it correct okay and you have five to ten minutes to solve this question sir the answer almost three no there are two uniform thin rods each of masses 0.01 0.02 rigidly joined the combination pivoted at lighter end I can it can rotate about point p a small object moving hits the lower end sticks to it fine so now when the when the small object hits the hits the rod okay can you conserve the linear momentum no can you conserve linear momentum when the block hits when the small mass hits the rod yes or no can I conserve linear momentum others those around yes sir can I conserve linear momentum just before and after you cannot conserve linear momentum just before and after collision because it is hinged at point p okay so at from the point p there will be external force getting generated okay because of which you cannot conserve linear momentum okay and that external force at point p could be impulsive also large amount of force could be generated at point p right so you cannot conserve linear momentum okay then in this collision what you should conserve just before and after see coley during collision they only in elastic collision there is conservation of energy that happens and here it is not elastic collision it sticks to it so you cannot conserve energy just before and after so what you are left with conservation of angular momentum nothing else there has to be some equation that links what is happening before with what will happen later on okay so you are going to conserve angular momentum you can see that even though there is a force from point p but because of that force the torque with respect to point p or any axis passing through point p will be zero yes or no because of perpendicular distance of that force which is acting at point p from the point p itself is zero so the torque due to that hinge force will be equal to zero with respect to the axis passing through p all of you agree yes so since yes since that is axis where the net external torque is zero even though there is net external force but net torque about axis passing through p is zero so you can conserve angular momentum about that axis you cannot conserve angular momentum about any other axis because there is a prerequisite to conserve angular momentum you can only conserve it about an axis about which net external torque is zero and that axis is an axis passing through p okay now what is the initial angular momentum of the part of the system total is there any angular momentum of the rod initially no no so angular momentum of the particle plus angular momentum rod initially should be equal to angular momentum of particle plus angular momentum of rod finally so angular momentum of rod is zero but what is the angular momentum of particle perpendicular distance into its momentum right so let's say velocity is v so mass times velocity into its perpendicular distance which is what this much yes or no this distance is how much let's say this distance is 2l each has length l which is 0.6 meter so m v into 2l is there any doubt on this no this is an initial angular momentum this should be equal to final angular momentum now see the thing is when the particle hits the rod it becomes the part of the rod itself okay so I can say that these two together can be treated as a single mass a single rigid body that is equal to moment of inertia about the fixed axis into omega okay and what is the moment of inertia about the fixed axis if you look at this the thinner rod the thinner rod has mass let us say m1 that is about p its moment of inertia is m1 l square by 3 okay plus let's take the thicker rod thicker rod I need to use parallel axis theorem I will translate this center of mass axis from here okay this is going into the board okay from here till here the moment of inertia about this axis which is this one which is center of mass axis we know that it is equal to let's say m2 l square by 12 okay plus this is center of mass axis plus m2 into l plus l by 2 whole square this into omega okay sorry this is the moment of inertia I'm pointing fine this is the moment of inertia of these two rods plus there's a particle also whose mass is m let us say so m into 2 l square so this is the total moment of inertia after the collision about that axis fine so from here you'll get the value of omega fine all of you understanding what I'm doing here yes those who are on youtube please type in yes or no are you getting it is there any doubt okay okay all right so I have just got the angular velocity immediately after the collision okay now what should I do I have the situation that it is just able to become horizontal this entire system there is this thin rod and then there is a thick rod this rod along with the mass it becomes horizontal fine initially it was like this fine so what should I do between this point and that point I have it's angular velocity what should I do working on it working as a theorem so simply I'll use w is equal to see this question is subjective question okay so this kind of question may not be asked in objective question as objective question because this will consume a lot of time to solve but then a small part of this question can be asked they can just ask you what is the angular velocity after the collision comes in like that so but then you'll get to learn many concepts from the single question that is why we're doing it k1 plus u1 okay so w is 0 and k2 is also 0 it just able to become horizontal as you guys see many a times our focus is to get the answer and you know don't focus on the answer and in that process at times I see that you know you tend to become so focused about telling the right answer that you you resort to other means also so those who are not getting the answers correct don't get intimidated by first of all that many people are able to answer the questions you are not able to you I mean it is perfectly fine okay so just keep a moral high and keep working hard your hard work will not go waste that you can take in writing from me okay so and also those guys I mean don't try to you know focus only on the answer look at the understanding of it how we are analyzing the particular scenario and all that because this exact same question will not come but the learning will remain with you so the final potential energy you can see that one way of finding potential energy is to find out the location of center of mass of this entire rod plus mass system and find out how much height it has gone up to okay that is one way of writing potential energy the other way is just this just you can split this entire thing into three parts okay for example potential energy of the thin rod let's say tr then potential energy of the thick rod for that is also tr so we can say th r just for I mean just it's a thick rod okay the second potential energy or we can just write t1 and t u1 and u2 u dash and u double dash and this is potential energy for the mass okay how much is u dash u dash is mass of this thin rod which is m1 into g and how much is center of mass has gone up by this distance which is what l by 2 so m1 g l by 2 u double dash will be equal to m2 g you can see its center of mass was here which is what distance this from here to here l plus l by 2 okay so this is 3l by 2 so mg into 3l by 2 okay and potential energy for that mass which was moving is mg into 2l because it has gone from here up till here so this is how much it has moved up fine so you can split the potential energy and write no need of finding center of mass okay the value of k1 is what half i about the fixed axis into omega square okay and u1 is a initial potential energy so you can see that this horizontal line which passes from the below this horizontal line represents the zero potential energy line okay so with respect to this line you can write down all the potential energies you can write down potential energy for this see the potential energy initially for this rod is not zero guys okay so I think I missed a point here so see the thing is I have assumed this as zero potential energy line okay so with respect to this final potential energy for this position is m1 g 2l m2 g 2l and mg 2l because all of them are at a height of 2l from the zero potential line okay initially their potential energy will be m1 into g into this height okay which is 3l by 2 okay plus m2 into g into l by 2 because this center of mass is this much height and this small m is at zero potential energy so that is not coming in picture so when you solve this equation then you'll get the answer getting it so this one was little involved question I urge you to again revisit this question try to solve it yourself you will learn many many you know small small concepts from this single question itself okay so that's it for today I hope you have learned many things today and don't forget to like the session okay before leaving and let me know if if there is something bothering you or you want me to revise any other concept why don't you suggest a concept immediately after this class you can WhatsApp me what next you want to you want me to revise or some concern you have with respect to any topic here and there okay so that's it for today and we'll see you next time thank you sir thank you sir