 What do you have to do here with concept you have to use? D for Delhi. Yes, correct work magic theorem you have to use and D is the correct answer here so work done is equal to u2 plus k2 minus u1 plus k1 okay so w is 0 all right and let's say this horizontal level is zero potential energy small object why it is given small object it rolls of the curve so this is small object so you can ignore this radius compared to the ramp it is taking otherwise the potential if there is initial potential G which is MGR okay so you can ignore MGR potential G is center of masses above the surface of by a distance of r so you can say that approximately initial potential energy is 0 initial kinetic energy you can write it as half M into VCM square plus half ICM into omega square okay and then its potential energy that you can write it as MGH where H is given as 3 V square by 4 G so this is 3 V square by 4 G okay and K2 okay K2 is 0 it comes to rest at the highest point yes or no so also omega is equal to VCM by r since there is a pair rolling that is going on fine so when you substitute all of this you will get moment of inertia about center of mass to be equal to MR square by 2 and that is true that is true for the disk only fine this is this came in year 2007 should I move to next any doubts on this please quickly ask let me find out some of the latest ones anyone got the answer this doesn't look to be from a rigid body motion but then we should not hesitate to take any other chapter's question also since now is towards a fag end of your preparation okay kushal saying D Sundar is saying D others no one is getting answered other than these two so once second one second is over okay so let's see first of all what is the velocity what should be the velocity for it to reach B how much that should be equal to reach B we have done this in work energy chapter under root 5 GL okay so the velocity at velocity should be equal to root 5 GL for it to complete the vertical circle okay now we are giving the speed which is half the angle theta at which the speed of Bob is half to that of a okay you need to find out where is a point where the velocity becomes if this is v1 the velocity should become under root 5 GL by 2 so where is that point okay so simply you have to use work energy theorem work done is u2 plus k2 minus u1 plus k1 okay so w is 0 you can say that this line represents zero gravitational potential energy that red line okay so u2 is what mg into this distance which is equal to what this distance is equal to r minus r cos theta because this distance is r cos theta and total distance from here is r so r minus r cos theta so this is equal to mg r 1 minus cos theta is u2 plus k2 is what half m into v2 square v2 is this all right minus u1 u1 is 0 to begin with plus half m into v1 square where v1 is this so from here you get some value of cos theta okay for here you'll get some value of cos theta and then so if I see you might have got under root 5 GL by assuming that it at point B the velocity becomes 0 okay so when it is moving a vertical circle and it has to reach B then velocity at point B cannot be 0 okay if you assume velocity to be 0 there then you get initial velocity to be under root 4 GL okay but if it has to move in a circle then you know the velocity should be such at point B the tension become just 0 over there otherwise you know what will happen is this after a point it will start moving in a parabola okay and then it will move something like this and may fall like that so even this point is also at a distance of 2 r only okay but this is not point B here velocity is 0 because the string became slack but if to ensure that it goes to the point B okay then it should move in a circle go back and revisit that vertical circular motion that we have done in class 11th okay the tension just becomes 0 over there okay so that's how you get under root 5 GL and just revisit that once here is the next question this came in J mains 2013 okay Sundar is saying see I am not saying a others I'm here saying a okay now which concept should we use here see ultimately when ultimately when the object starts rolling you know when the object starts rolling then we center of mass should become equal to omega r because let's say there is this hoop okay so if it has to perform pure rolling okay then this point of contact should be at the rest okay so with respect to center of mass suppose the angular velocity is like this okay so with respect to center of mass the back side the velocity is omega into r okay and since it is moving forward with VCM velocity the total loss you will be omega r and VCM combined so net net velocity will be VCM minus omega r okay when this become equal to 0 then it means that there is no slipping okay so this is a no-slipping criteria where VCM should become equal to omega into r okay now you can solve this question in different ways okay one of the obvious ways is just use that you know basic as in there are forces like MG okay this is MG force and then there is a normal reaction from the surface like this and then there will be a friction force let us say that friction force is FR okay and till the point this object starts pure rolling it is sliding friction or kinetic friction so the friction value remains constant okay after it starts rolling then it becomes static friction okay and also I'm just randomly telling you these facts that you know few of the things like if the object is pure it is doing pure rolling the work done by the friction you should take always to be 0 because that is static friction and the point of contact is at rest okay so never take work done by friction if object is pure rolling but here it is not pure rolling okay alright so the total force is what total force on this hoop is FR only okay so accretion is what friction force divided by mass alright so velocity after time t okay is u plus 80 initial velocity is 0 plus FR t divided by m okay see basically what I'm doing here is I am assuming that it has been given angular velocity like this okay so it is this point is slipping right now okay so velocity along this direction horizontally that will be created by this force will be equal to this yes or no any doubt on this and friction into R is a torque about the center of mass no other force creates a torque about center of mass only mg and normal reaction are the other forces and for those forces about center of mass the torque is 0 so FR into R is the torque due to friction this is equal to moment of inertia which is mr square by 2 into alpha i alpha okay so fridge sorry the angular velocity alpha will become equal to 2 FR divided by m into R alright this is alpha so the angular velocity will be equal to omega not which is an initial angular velocity minus alpha into t see the angular acceleration due to the friction is opposite direction of angular velocity angular velocity like this but alpha is in opposite direction so that is why omega not minus alpha into t okay so this is omega after time t and this is velocity after time t okay so if this has to start the pure rolling then we should be equal to omega into R so basically FR t by m should be equal to omega into R or that is equal to omega not r minus 2 FR t by m okay so from here you get t alright from this equation you get the value of t and then one word to find out you need to find out the velocity right so then you substitute that value of t over here you get the value of velocity okay so this is one way to solve this is like the more conventional way and the more obvious way how you can approach this question other ways to use conservation of angular momentum okay so you can conserve momentum about this axis alright so let me show you how you can do that so initial angular momentum when it is just sliding sorry when it is just rotating is ICM into omega okay about this point because it is not going anywhere so VCM is 0 initially so ICM into omega is the initial angular momentum this will be equal to the final angular momentum which is r cross mv so m into VCM into R plus you get ICM into omega dash and omega dash is equal to VCM into R okay so you can write down omega dash as VCM divided by R so from here you get the value of VCM okay so like this also you can get the value of velocity by just using conservation of angular momentum right any doubt on this particular question those who have solved some questions on this chapter might be knowing that these are like routine ones I think you will encounter such question during your practice also there are couple of them like this in HCRMA as well okay do this one anybody is able to get the answer no it is not D others okay Puriq is getting A that is correct size getting each and every answer correct huh good I hope you repeat this performance in test also okay so let's see how we can solve this particular question hmm so we have a solid sphere okay from the solid sphere of mass m and it is our cube of maximum possible volume is cut okay so cube of maximum possible volume is cut so if you look from the front it will look like a square to you okay and hence you'll be able to calculate the side length how much you got the side length as hmm if you look from diagonal to diagonal this this point to that point how much is this distance if let's say the the side length of the cube is a how much is this distance this diagonal so we root 3a root 3a correct so it's not a two dimensional thing all right so this is not root 2a this is root 3a getting it so it's a three dimensional figure this is a cube it's not a square actually it looks like a square when you look from the front but this is one point one end of the diagonal to the other end of the diagonal it's a major diagonal of the cube okay so this root 3a should be equal to 2 times r fine so the side length of the cube will be 2r by root 3 okay now we know that moment of inertia of a cube do you know it is m a square by 6 about the center of mass yes right now m is what the mass of the cube all right this is not same as the mass which is given over here this is different mass this is what the spheres mass so let's say this is m dash so you need to find m dash in terms of m you already got the value of a in terms of r so now try to find m dash in terms of m how will you find out you can assume that the mass is uniformly distributed over the volume okay so the ratio of mass m by m dash is same as ratio of the volume is this something which is reasonable yes right so v is the volume of the sphere which is what 4 by 3 pi r cube okay divided by volume the cube which is a cube only all right so just substitute the values of you can substitute value of a you'll get m by m dash and from here you get m dash in terms of m and then substitute here okay and you'll get option a to be correct all right so you can see that here more than physics it is about how you think or how you break down a problem that is what is tested here it's not about reading the theory or reading the concept again and again it's at times about opening up your mind and thinking freely okay so you have to develop that okay so that will get developed only when you get lots and lots of tests when you take the test with time constraint you will first of all get to know there are many many varieties of the question and you will know how to you know think then and there and get the answer okay initially you'll not be able to solve many questions correctly like this but slowly and slowly when you are getting accustomed to that kind of rigor you will be able to get even these kind of questions correct easily but then these questions cannot be you know you will not be able to solve this by just you know revising the same thing again and again and trying to finish up something I mean just for the for for your meant you know for for your own satisfaction like so just to satisfy yourself or just just to feel good that okay I have completed something and moving ahead you know all that doesn't matter you know so whenever there's a test there's a mock test take it it doesn't matter whether you finish the syllabus or not because anyway you're not aiming for 100% marks nobody gets 100% marks there rarely it happens okay if 100% marks are not too impossible you are aiming for 60% marks then what is the logic of finishing the syllabus and then writing a mock test doesn't apply