 Let us start let us start the something else we will now we are now at the far end of the chapter we are talking about the energy consideration now okay so basically we are discussing here how to write mechanical energy for the rigid body okay so for writing kinetic energy or writing potential energy is very easy for for a point mass or for a mass which is just going from one place to the other place without rotating if a bigger mass like let's say this glass is just goes like this without rotating then writing its kinetic energy straight forward half mv square okay but while moving suppose it starts rotating and it is moving forward then every point on this glass will have different velocities so which half mv square will you write or writing half mv square is correct there is some issue with that so that is what we are going to discuss how to write kinetic energy for a rigid body and how to write potential energy for a rigid body even potential energy for a rigid body is not straight forward for example this glass is there at a height h the bottom most point is at a height h from the floor should I write mgh or should I take the center should I take the top most portion which height should I consider to write the potential energy that is what we are going to derive here and then once we know how to write down the kinetic energy of a rigid body and how to write down the potential energy of a rigid body then what we can do once we know which concept we can use once we know how to write kinetic energy and potential energy we can use work energy theorem okay all right so let us first see how to write kinetic energy for a rigid body kinetic energy for a rigid body now if case number one case number one only only translation only translation without rotation as in only translation means without rotation only but I specified it then all the points all the points on rigid body same velocity okay so kinetic energy can be written as simply half capital M into V square how does it come wait you can assume that the entire rigid body is made up of small small point masses so you can say M1 into V square plus kinetic energy of second mass M2 into V square like this you will get if all the velocities are equal so half of M1 plus M2 and so on into V square so you can simply write it as half M into V square okay so here we have done it in very simpler manner right you can consider the entire rigid body for example this is the rigid body and all the points one two three four all the points they are all moving the same velocity so you can just take one by one add their kinetic energies you will get half into total mass into V square like that this is the case number one right but this is not how things will be this is the case which we have been using till now but now we need to account for the rotation of a rigid body also so case number two is about the rotation about the fixed axis case number two write down the rotation about the fixed axis for example fan now you can see that center of mass is not going anywhere center of mass is actually at rest but still it has a kinetic energy all the points they have some speed so there has to be kinetic energy for the fan like scenario so let's see how to write that so suppose this is a rigid body which is rotating about the fixed axis this is the fixed axis let's say so you can see that all the points are moving in a circle because it is a fixed axis right so if the angular velocity is given to us as omega you have this point mass m1 at a distance of r1 okay then what is the kinetic energy for mass m1 how to write it as half m1 into velocity of mass m1 square what is the velocity of m1 everyone omega into r1 omega r1 right so this you can write it as half m1 omega square r1 square so this will be half m1 r1 square omega square fine so the total kinetic energy will be like this you have to add all the kinetic energies right so half of m1 r1 square omega square plus half of m2 r2 square omega square and so on all the points this will be half you can take omega square common it will be m1 r1 square plus m2 i2 square so on into omega square what is the bracket thing what's the bracket moment of inertia right about what about what about the fixed axis so if if there is a rigid body if there is a rigid body which is rotating about the fixed axis about the fixed axis then its entire kinetic energy can be written as half i about the fixed axis into omega square you don't need to worry about anything else all right clear right now you you can see there's a pattern in the equations about the fixed axis torque equal to i alpha about the fixed axis the angular momentum is i omega about the fixed axis kinetic energy is half if omega square now tell me one thing will the kinetic energy change depending on about which axis you're finding torque changes right angular momentum angular momentum and torque depend on axis but but do you think energy depending on axis does it sound proper kinetic energy does not depend on axis are you getting it so even though you are finding i mean you when you write kinetic energy is half i of omega square you should not say that you have found kinetic energy about the fixed axis getting it you are finding the kinetic energy of the rigid body by using an equation in which if is there okay so if there is a rigid body which is rotating about the fixed axis you can write kinetic energy by using this formula clear so this is case number two case number three is rotation plus translation there is no fixed axis can you guess what should be the kinetic energy we have seen how you have written the angular momentum so similarly can you guess kinetic energy so with respect to the center of mass it will be i cm into omega square plus half the center of masses this is the formula please write it down okay so even if even if as a fixed axis then also both formulas of kinetic energy gives same value so you can use anything but if there is no fixed axis you don't have any other choice than to use this thing all right all of you clear how to write kinetic energy everyone so this is the kinetic energy let us see how to write gravitational potential energy for a rigid body okay this chapter will get over i think this chapter has taken three plus two five weeks to complete so i hope those who have done it systematically you can appreciate that your end your entire physics got revised whatever you have done in kinematics laws of motion work by energy everything got revised in a single chapter again isn't it so suppose you have a rigid body like this you have to write the gravitational potential energy this orange line is zero potential is your reference gravitational potential which is ug equal to zero all right now how to write it again i'll divide into multiple point masses let's say there is a point mass over here m1 which is at a distance of y1 this is at a distance of y1 m2 is there which is at a distance of y2 and so on so i'll just write down the their individual potential energies and add them up to get the total potential energy total potential energy at the total mass however is capital m2 so total potential energy is m1 g y1 plus m2 g y2 and so on like this right i've taken one one point masses and add up their potential energies so add all the point masses one by one this will be equal to m1 y1 plus m2 y2 and so on into g what is the assumption when i write like this everyone tell me what is the assumption here can you tell the assumption g is constant at all points okay that is the assumption correct so you can you can add and multiply summation of all the masses m m1 plus m2 plus m3 plus m4 that you have multiplied and you have divided also what that will do this divided by summation of m what is this bracket term what is this if this orange line would have been x axis then what this would have been bracket term y center of mass right y center of mass so basically potential energy is summation of m into y center of mass into g summation of m is what total mass y center of mass you can write like this height of center of mass g so potential energy gravitational potential g is mg h center of mass okay this is how you write the gravitational potential energy all right so now we have learned how to write the potential energy for the rigid body and how to write the kiting into the rigid body no matter whether it is rotating or not rotating about the fixed axis or not about the fixed axis we can write the kinetic energy similarly we can write the potential energy so like what you said we will be utilizing this definition of kinetic energy and potential energy to put in work magic theorem which is w is equal to u2 plus k2 do you remember this work magic theorem all of you this we have done i guess 40 to 50 question on work magic theorem better you do not forget rest of your life okay so just one small thing before the break if if there is a pure rolling rolling without sliding then at point of contact then at point of contact there is no relative movement hence work done by friction will be how much will be zero so you know when pure rolling is happening don't start finding the work done by the friction because that is zero okay so if if pure rolling is there other force is doing work w is zero w zero working by friction is zero so then u2 plus k2 is equal to u1 plus k1 so we can conserve mechanical energy potential g plus kinetic energy is mechanical energy and this is conservation of mechanical energy conserve mechanical energy and you don't need to actually know this you can just use whatever is written at the top of the page you can use that also i'm just telling you for simplicity sake right so going forward most of the cases there will not be any dissipative forces only friction will be there friction won't do work so we'll be using u2 plus k2 equal to u1 plus k1 most of the time and over here potential energy is mg h of center of mass kinetic energy you know three cases how to write the kinetic energy right so with uh by the way even though work done by friction is zero is friction applying any force or torque or even force in torque is also zero everyone is the force and torque that also zero or that is not zero it applies force that creates acceleration it applies torque that creates angular acceleration all that are there okay but the work done is not there it can i mean it is a little weird right it is weird in a way that this force can create acceleration this force will cause the object to accelerate this force will cause the object to rotate but this force is not doing any work okay so a little weird it is but that's how it is okay so after the break we are going to take up questions on energy make sure i mean this is the last thing for the chapter make sure you end the chapter well and this is an important thing okay so we'll meet after the break at 6 20 p.m fine so then continue let's take few basic questions then we'll proceed for now for example you have a rod like this this rod can freely rotate about this axis okay it is right now horizontal like this and you're holding it then you release it what will happen it will go down like that okay to rotate and go down okay so you need to find out what will be its angular velocity when it becomes vertical like this find out it started from here mass m length l it can rotate about this axis and you need to find what is omega when it becomes vertical omega is how much there is no friction and all so you can use this right conservation of mechanical energy just substitute the values get the answer all of you do this which horizontal line you want to take as zero potential energy top bottom most which one bottom most top okay i think more most of you have taken no it is half half actually doesn't matter but if you take the bottom most like this then potential energy will be there initially as well as finally financial won't get changed so bottom most is not a appropriate choice you may you choose in such a manner that at least one of the potential energy becomes zero so you can take this top one as zero potential energy that way i guess it will be better right so u1 is how much u1 is what what should i write u1 as zero k1 is what zero u2 is what what is u2 all of you tell me what is u2 you need to find out center of mass height right center of mass is here mg l by 2 or minus mg l by 2 there will be a minus sign right it is down the center of mass has gone down so negative change in potential energy is negative potential energy has decreased so minus of g l by 2 how should i write kinetic energy in terms of omega what will the kinetic energy plus half is there a fixed axis if there's a fixed axis i can directly write kinetic energy as half i about the fixed axis into omega square how much is the fixed axis moment of inertia l square by three ml square by three into omega square clear is this clear to everyone whatever i have just done this is like the beginning of things to come omega is root over three g l no three g by l all of you clear type in okay simply up to just i mean don't think too much with respect to applying energy conservation you know how to write kinetic energy you know how do i put energy just write the equation and feed the values okay next um suppose you have a slope is theta you have a sphere at the top this is the sphere at the top whose center is at a distance of h this is the edge okay now it starts rolling and comes down you need to find out what is the velocity of center of mass with which it moves forward vcm is what of course apart from vcm it will have angular velocity because i am telling you that it rolls without slipping at no point there is a slipping it is pure rolling everywhere pure rolling get the value of vcm and omega mass m and radius r is also given to you find out can pure rolling start without friction is friction required or not if there is no friction can there be pure rolling started friction is required friction is required for pure rolling because only friction can create the torque initial angular velocity is zero so if there should be angular velocity developing in this there has to be some torque only friction can create the torque over there every other force is passing through the center of mass right torque equal to icm into alpha right so alpha will create omega so friction is required pure rolling if it starts initial angular velocity is zero in order to create omega torque is required that is where friction is required because friction only can create torque okay do it but friction is not doing any work so you can use here u1 plus k1 equal to u2 plus k2 now do it substitute the values get the answer archib got something others should I wait or start solving this ground you can take zero gravitational potential g ug is zero okay we have also got something u1 is what initial potential g g h continue is how much k1 is zero u2 is what mgr its center of mass is at a distance of r right k2 is what how to write k2 is there a fixed axis there is a fixed axis or not actually there is instantaneous fixed axis okay so the this point is at rest there is a fixed axis but okay leave it you don't need to write regarding that with respect to that you can use the other formula half mvcm square into half i omega square that is also fine so half icm omega square plus half m into vcm square is this equation clear to everyone the one which I have written just now all of you whatever I have done is this clear now this is rolling on a fixed surface pure rolling on fixed surface so what can I say if it is pure rolling on fixed surface what is the relationship between omega and velocity alpha and acceleration what is the relation a is equal to alpha r right and v is equal to omega r if it is pure rolling on a fixed surface okay so I can use that and moment of inertia about a center of mass for a solid sphere how much it is 2 by 5 mr square so icm is 2 by 5 mr square omega is vcm by r so v square by r square is omega omega square plus half m into v square this is equal to mg h minus r so you can see that m get cancelled away m need not be given to you okay so r square is also gone so 7 by 5 so 7 by 10 7 by 10 v square is equal to g tends h minus r so from here velocity is root over 10 g h minus r by 7 all of you clear any doubts quickly type in if it is clear type in that it is clear it is very simple you you just substitute the values of potential energy and kinetic energy straight forward right now these are the straight forward school level or ct level questions let us see you know slightly uh this thing so that your once you do the difficult ones then your school level will be very simple for you okay so let's go ahead and do something else also so next question is this suppose um first we'll do that there is a disk it's not very difficult also okay don't worry about all those this is m radius r this disk can rotate about this axis this axis passing through the plane of the disk okay this axis is at a distance of r by 2 from the center this distance is r by 2 it can rotate about the disk sorry about the axis it's like this assume that this is a disk about this axis it can rotate so how to rotate it can rotate like that okay so after it rotates one complete this thing it will become like this after having one 180 degree rotation if I merge it like this it is like this it will become like that do you understand it is like that the disk is like this you slightly push it it will turn like that okay all right so you need to find out what will be the angular velocity when it flips 180 degree like that initial angular velocity is 0 you just nerfed it and it goes down and gain some angular velocity what is that omega is what okay everybody understood the question if you have not please ask I'll clear it okay I've got something others the formula is same same thing you have to use here also okay that way there is no difference all right okay should I solve or should I wait you have to use that thing only u1 plus k1 equal to u2 plus k2 there is no change to that you can take this as 0 gravitational potential energy u g is 0 how much is u1 everyone what is u1 g r by 2 isn't it k1 is what 0 u2 is what what is u2 minus of mg r by 2 what is k2 is there a fixed axis for this rigid body is there a fixed axis all of you there's a fixed axis so how can I write cantigan g how can I write cantigan g I moment of nature about a fixed axis into omega square simple right so half I about the fixed axis omega square is equal to mg r fine now moment of inertia about fixed axis is how much if can you find out about this white line how much is the moment of inertia anybody else calculated the moment of inertia how much is the moment of inertia about this axis that passes through the center of mass but the diameter how much it is do you know that about a natural axis moment of inertia of the disc is mr square by 2 you know that right now I want you to use parallel axis theorem right so but then parallel axis theorem requires one icm and this is what I want I but but this is not parallel to I so I need to find axis that passes through the center of mass and parallel to this I so I will use perpendicular axis theorem to find out the moment of inertia about this horizontal blue axis okay so this will be equal to Ix plus Iy both are equal due to symmetry so two times of icm I can say so icm moment of inertia about the diameter is mr square by 4 getting it so moment of inertia about this axis how much mr square by 4 plus this is icm plus m into the distance r by 2 whole square mr square by 4 plus mr square by 4 which is mr square by 2 all of you is it clear till now type in till now is it clear so substitute this value of the moment of inertia about the fixed axis over there you'll get half of this mgr omega is two times root of g by r anyone has any doubt let's proceed further why can't we use parallel axis parallel axis we have used right parallel axis we have used which is icm plus m into d square this is icm this is m into d square but mr square by 2 is not the parallel axis to I I need to find axis passing through the center of mass and parallel to this which is this blue line it is not this line this line is not parallel to that it is perpendicular understood so I have to first find out what is that blue lines moment of inertia okay so there are some questions in which there is a collision happening okay just like in the in the work by energy chapter you had you need to use conservation of momentum then conservation of energy similarly there are a few questions here in which you have to use conservation of angular momentum and conservation of energy both okay so for example this one you keep it simple suppose there is a rod which can rotate about this fixed axis okay of mass m and length l mass m length l okay there is a small mass m coming like this with velocity u just before hitting the rod its velocity u and after hitting it comes to rest after hitting after hitting let's say it sticks okay after hitting it sticks to the rod that will be interesting sticks to the rod you need to find out the maximum angle rod will swing the rod will swing sorry you have to find what is this angle theta the mass is stuck here can I conserve energy can I conserve energy directly from initial to the final point conserve mechanical energy before collision till the end can I do that why those are saying no why what is the reason direct Rohan got it yes good so there is no because energy is energy is lost during collision so you cannot so what you have to do first you have to go from before collision to after collision and after the rod attains omega from that point onwards you can conserve energy okay so how do you get omega of the rod can I conserve linear momentum can I conserve linear momentum yes or no during the collision is there an external force in horizontal direction is there an external force in horizontal direction during the collision what do you think Rohan Aniruddh Irritu Kranau mega Kiran Gayatri what do you think can I conserve the linear momentum horizontal direction you cannot conserve it because there is reaction force from the hinge okay there is this external force which is impulsive in nature so how you get omega what you can do what you can do which concept we will use right conservation of angular momentum about which axis when you say I will conserve angular momentum you have to also decide which axis about the hinge axis about the hinge axis we can conserve angular momentum we can conserve angular momentum for rod plus mass okay so initial angular momentum for the mass m is what how much it is what is the perpendicular distance l right so m u into l m u l this is the angular momentum of the mass angular momentum the rod is zero this should be equal to what moment of the rod plus mass afterwards which is how much will the mass becomes part of the rod will the point mass after collision becomes the part of the rod right so entire thing is rotating about the fixed axis so I can say this is equal to moment of inertia for the fixed axis into omega because about that axis angular momentum of everything now the mass is a part of the rod is i f into omega what is i f how much is i f m l square by 3 that's it what about small m small m won't affect anything will moment of inertia increase or not how much it will be m plus m l square by 3 and you're telling me m plus m l square by 3 is the moment of inertia it's like a rod of mass m plus m uniformly distributed is the mass uniformly distributed total m plus m no so you need to add moment of inertia's individually don't add the mass in the formula there has to be some logic so m n square by 3 plus small m into l square to omega so moment of inertia of the point mass plus the rod about that axis will be their individual moment of inertia is adding up everyone clear clear to everyone okay so this will give me omega okay as you omega is now known to you you don't need to calculate suppose omega is now known so i just made up the question so probably the expression may not be look very nice not look nice so now tell me what is the angle theta how much maximum you will swing and no no no integration conserve mechanical energy you can write u1 plus k1 equals to u2 plus k2 yes or no point number one is just after the collision point number one is just after collision point number two is when the maximum angle is there find out at the expression you don't need to calculate as such tell me once you're done don't do not try to find out cos theta and theta like that have you written the expression where is your zero potential energy what do you want to take it as they enjoy okay fine take the top a u1 is what u1 center of mass is this you need to understand point number one is when the small m is already attached with the rod the small m is attached and it has gained angular velocity just after the collision okay so tell me now for there will be two potential energy potential energy of the rod potential energy of the mass and getting it or you find out center of mass of both capital m and small n then you can take it as one mass but that is like unnecessary calculation so we can split them and add them individually potential energy for the rod plus potential energy for the small mass tell me how much it is potential energy of the rod do you all agree it is minus of mg l by 2 it is down mg l by 2 potential energy of the rod tell me is it clear plus other potential energy is also negative minus of small mg l this is the total initial potential energy okay if you consider this mass also k1 is what k1 is let's call this thing as if ml square by 3 plus a small ml square let's say if k1 is half of if into omega square all of you agree if is ml square by 3 plus a small ml square okay this is equal to u2 u2 is what now this center of the rod is here u2 is what how much is this distance this is the height you have to consider right how much it is this is l by 2 right this is theta that is l by 2 cos theta use little bit of trigonometry you will understand so it will be minus of mg l by 2 cos theta and for the small m what is the potential energy tell me minus of mg l cos theta k2 is what what is k2 k2 is 0 this is our equation number 2 so these are the equations I was looking for that's it you have to solve these two to get the value of theta everyone clear anyone has any doubt in this last red box last red box which part of the red box it's a big box I have taken this it depends on which horizontal line you take zero potential g I have taken this as zero potential g the top horizontal line so using that I have to write potential energy and counting energy all right now when I write potential energy the formula for potential energy is mg h center of mass okay now if you want to consider rod and mass small m as one single mass you have to find its center of mass then you can write capital M plus a small m g into h center of mass which will not be l by 2 okay so rather than finding the center of mass and then writing m plus mg at center of mass I have chosen to write their potential energies individually I've wrote for the rod I've written potential g minus mg l by 2 then I've written for the mass and added them and tell that this is the total potential energy of the rod and mass m plus the entire thing is rotating about the fixed axis rod and the point mass so moment of inertia of the entire rigid body rod and point mass is about the fixed axis if that into omega square half if omega square is the kinetic energy that is the kinetic energy plus potential g that I've equated to the final potential g of rod plus final potential g of the point mass plus the finite energy which is zero okay okay so now you are experts of this chapter okay anyways you can solve this quickly I'm more interested in the third part or let's do this you remember doing that question right wherein there was a rod um there was this rod horizontally line okay there was a mass small m this was going with velocity v okay this is mass capital M total length is l this velocity line of velocity is at a distance of l by 2 you remember that did we find out the uh okay immediately after collision immediately after collision and comes to rest we have done this right somewhere we have done it do you remember this doing here okay we have done this exact same question right so we have got the value of omega and velocity center of mass tell me what is the omega we have got what was the answer for omega your notes are you making it don't just sit without omega I'm asking omega this is the omega which we have found and velocity center of mass after collision how much we have got it as huh this was l by 4 how can it be l by 2 l by 2 will be at the other end vcm is what mv by capital M okay so you need to find out find out how much the center of mass will move forward one one thing you I I hope you might have noticed one thing no matter where this mass m hits this rigid body will always have a straight line center of mass velocity center of mass will always go in a straight line no matter where you hit you can try it after the class you can just put the pen on the floor which is almost like frictionless floor put it on the floor and just hit at one of the ends the pen you then just notice the center of mass the pen will rotate and move but the center of mass will always go in a straight line okay so that will go in a straight line with vcm you need to find out how much center of mass will move forward by the time rotates by 180 degrees okay so how much time will it take for it to rotate will it keep on rotating with constant angle of velocity after collision all of you will it keep on rotating with constant angle of velocity after rotating after mass after small and hits the rod there's no other torque torque is zero so whatever is the angle of velocity it'll keep on maintaining that only when torque is there change in angle of velocity will be there right so angle of loss will be constant so you can find out easily omega into t should be equal to pi so from there you'll get how much time it will take to rotate by pi so pi and l divided by 3 m v this is the time it'll take to rotate 180 degree now how much center of mass will move forward is center of mass with constant velocity forward with this only after collision is there any force on the rod there's no other force so the center of mass will also keep on moving with the constant velocity so the distance d will be equal to the velocity v into the time by 3 m v okay clear so you can see that it doesn't depend on the masses right so pi it doesn't depend on the initial velocity also so pi l by 3 it is independent of initial velocity how much distance it will move and it doesn't depend on the mass also okay so like this you can sometimes correlate what happens with the rotation with what happens in the translation now there are many you know there are many many fine fine fine things which you will encounter while problem solving so I cannot create infinite list of small small things and teach you everything okay so the theory part is long over I am trying to cover what all things can help you in problem solving but still there are many things you will learn by solving problem no one can teach you each and everything that you will encounter in questions this okay is it ending energy is mg l by 2 many of you saying what is the answer for the first one I mean these are simple questions guys don't think too much so we know that k 1 plus u 1 should be equal to k 2 plus u 2 okay so initial kind of energy is 0 initial potential energy also you can say 0 initial final kind of energy you have to find out final potential energy is minus of mg l by 2 so kind of energy is how much mg l by 2 only which is option 4 here it's straightforward okay second part solve second part first angular velocity what is the angular velocity everyone the cool got angular velocity others you need to calculate angular velocity how much it is angular velocity how much how can you get hen's reaction without finding angular velocity so question number three should be question number two and two should be three kind of energy is what half moment of nature body fixed axis which is how much m l square by 3 into omega square mg l by 2 so omega is root over 3g by l okay this is omega now find out the hinge reaction at point a now to the second question actually yeah anyways find out I didn't know the third question is about finding angular velocity draw the free body diagram then only solve it you're not experts even I don't skip steps the moment you skip it you get a wrong answer remember that every time and you have enough experience of getting negative which direction the hinge reaction will be at point a you can you can randomly say okay fine there'll be two directions the reaction can be this one and that one this reaction you can say r x this is r y you don't know which direction it is right what are the force will be there any other force on the rod all of you mg how can you forget mg right mg is with you mg any acceleration of the rod there is no other force right on the rod there is no other force but is there any acceleration center of mass do you agree it is moving in a circle is there an alpha someone is saying that there is an alpha is there an alpha for alpha there has to be a torque about a fixed axis is there a torque about fixed axis no there is no torque about a fixed axis right now there is no torque but when it was not vertical mg was creating torque because of which alpha was there if if you tell me if you tell me that rod is like this you know for example if if you draw rod like that then because of mg there is a torque about the center of mass sorry about the fixed axis so there is alpha because of mg at that moment but when rod becomes vertical the mg passes through the axis the fixed axis so even mg's torque is zero about the fixed axis so when it is vertical alpha is zero do you all understand alpha is zero i'm saying alpha is zero when rod is vertical all of you understand this so this implies that tangential acceleration is zero tangential acceleration is zero yes sir i was saying alpha is zero because there is no torque about the fixed axis mg was creating alpha but now even mg has zero torque so net torque will do i alpha about the fixed axis so alpha is zero so tangential acceleration is also zero is there a radial acceleration radial direction towards the center is there an acceleration all of you what do you think everyone how much it is if it is there the radial acceleration is how much okay v square by r in terms of omega can i write it v is equal to what r is over here the radius of the center of mass is l by 2 isn't it this distance is l by 2 from here to here v how can i write v in terms of omega omega into l by 2 omega into l by 2 whole square divided by l by 2 this will be omega square l by 2 omega or you just remember in terms of omega it is omega square r in terms of velocity it is v square by r okay so now how should i write the equation what horizontal direction there is no acceleration there is no force do you all agree that rx is equal to zero reaction horizontally will be zero when the rod is vertical all of you agree with this rx is zero rx is zero find out r y what will be the reaction for what is the formula for r y what is the formula net force is r y minus mg this is equal to mass times acceleration which is omega square l by 2 net force is equal to mass and acceleration everyone understood this equation everyone understood see final answer i don't care i don't give a damn okay don't just start telling yelling the answer i am here to talk to you how to solve the question not to check what answer you're getting so do you all understand this everyone type it right now put the value of omega over here put the value of omega over here you will get 3 by 2 mg right 3 by 2 mg so from here you will get r y equal to 5 by 2 mg so this is how you solve this question all right so here you can see not only conservation of energy is utilized but also the force equation is used sometimes you have to use angular momentum energy torque each and everything you have to use sometimes in the question okay so do not constrain your thought process that first you look like okay fine what to use here which formula to use should i use this or that you need to use whatever is required in that question you have to think in a very open-minded manner and that will happen only when you struggle with lot of questions yourself fine so i guess we can take one more question and this list of question i thought i would do today but i created my own this thing there's something similar this is we have done something similar done something similar okay i think we have done many of those similar types let me give you one question again i'll just so you have a structure like this you have four rods no four or two two rods mass m and length l both the rods they are welded they are welded at the center okay guns are installed guns are installed at the ends fine this entire structure can rotate about the axis passing to the center like this okay the structure can rotate like that fine basically i'm saying that it's free to rotate like that fine so the guns simultaneously fires simultaneous fire bullet of mass small n and with velocity small v okay direction of the fire is as shown all right mass m is very very small compared to capital M so capital M remains unchanged even if bullet is released you need to find out what is the angular velocity immediately after bullet is fired do it what concept to use here all of you first tell me the concept collision no no can i conserve angular momentum no work material how can you use it the firing will create lot of energy loss or chemical energy got converted into mechanical energy so energy got mechanical energy got created you can't conserve it which concept to use here all of you let me ask you that way can i conserve the angular momentum about this axis that comes out from your screen right not momentum conservation of angular momentum momentum the forget you can't conservation of angular momentum you can do about the natural axis so initial angular momentum of everything is zero so zero plus zero now what is the angular momentum of all the bullets tell me bullets for the bullets plus angular momentum for the rods for bullets what will be you can see the three bullets are in one direction this one in this way angular momentum that is that way this is that way so three bullets are having clockwise angular momentum one has the anti clockwise so three times m into v into l by two minus m into v into l by two do you all understand this everyone understand that this is the angular momentum for the bullets cumulative after firing type in angular momentum for the rod is moment of inertia but the fixed axis which happens to be the axis about which you are conserving the angular momentum i f into omega this will give me omega i f is what tell me i f is how much ml square by 12 ml square by six there are two rods right ml square by 12 into two two rods are there so ml square by six is the moment of inertia all right so you can one more thing you can see here that if angular momentum of the bullet is clockwise the angular momentum of the structure will be anti clockwise then only they will add up to zero that is why omega will come out to be negative okay so angular momentum of rod is negative of the angular momentum of the bullets all right guys so i guess we have extended a bit five six minutes so that's it this chapter is over all of you are now free to solve any question from anywhere from this chapter okay so it's a huge chapter we have spent about a month and a half to complete this chapter you've done it in great detail right now everything is fresh make sure that you do the homework and assignment in time so that later on you don't have to spend time in this chapter fine so make sure that center modules are even bigger what do you mean by that kinshu so make sure you do the assignment yeah so you can take the center module for the exercises i'll put it as homework fine so make sure you practice a lot of questions your own then only you'll be comfortable with this chapter just sitting in classes is nothing okay that everyone does all right guys so we will meet next week with the new chapter thermal physics till then bye for now