 All right, so we we have started this chapter last class walk by energy Okay, write this down So what we did last class was that we have just introduced the concept of work Okay, and then we have introduced kinetic energy. Okay, and Towards the letter part we have got a relation between the work done on a mass Okay, and how much kinetic energy is changing. Okay, so we got a very simplistic expression that total work done Total work done on a single mass is equal to Change in kinetic energy. So K2 minus K1 is the work done. Fine. It's a very simple straightforward relation So this is for a single mass fine now if there are multiple masses Okay, suppose there are two or three masses Then the work done if you consider two or three masses together fine So you find out the work done on all the masses Okay, so work done on all the masses will be equal to change in kinetic energy Let's suppose there are two masses So we'll take final kind energy of both the masses and subtract it with initial kinetic energy For both the masses Okay, so we can not only use Work energy theorem for a single mass But we can use this work energy theorem for the Multiple masses also fine. So this is the total work done on both the masses All right So you'll see that work done by the internal force as in the force between one and two masses It will come out to be zero fine So ultimately what you will end up with is a work done by the external forces with respect to one and two both objects together All right. So this is what we have learned towards the end of the session Okay, so we'll start the session with a numerical. Okay, so that you properly recollect whatever we have done last Okay You guys remember the work done by the spring The work done by the spring force. We have derived it to be half K x1 square minus half K X2 square fine So I'm writing it here because we are going to use this Again and again fine where x1 is initial deviation from its natural length and x2 is the final Deviation from the spring's natural length. All right. So x1 and x2 can be extension or can be compression. It doesn't matter fine All right, let us take up this question all of you so I'm starting with you know simplistic of All the questions This is a spring constant K There is a mass There is a mass of mass m that is coming towards the spring Okay The mass is coming towards a spring with velocity v0. All right You need to use work energy theorem to find out the maximum compression in the spring. What is the? max compression on the spring all of you Find out quickly Welcome everyone. Please solve this question. Okay, but I've got an answer others What is the condition for maximum compression when the compression will be maximum? See this mass is moving forward with velocity when it hits the spring its velocity will decrease Right, so its velocity will keep on decreasing But then it is moving forward and as it moves forward spring Becomes, you know Spring get compressed more and more. All right. So when this mass momentarily come to rest Okay, that is the position where the spring will encounter the maximum compression All right. So after the mass m comes to rest after that point onwards It will start going backwards, right? So we need to use this work energy theorem Whatever we have derived in the last class between this position where the mass has velocity v0 and Where the mass comes to rest? So between these two point I have to use work energy theorem now Who is doing work on the mass? There are forces like mg Okay, mg is a force Normal reaction is a force. There is no friction fine mg and normal reaction they are perpendicular to displacement. So mg and normal reaction do not do any work But the spring is doing work on small m when the spring get compressed it applies a force And we know the work done by the spring formula is this So total work done is nothing but work done by the spring and this is equal to Half k x1 square. Now. What is x1? Initial division from its natural length initially Initially the spring in isn't in its natural length. So x1 is 0, right? Finally, let's say Spring has a compression of x. So this is the work done by the spring Okay, this comes out to be work done by the spring. This should be equal to change in the kinetic energy Final kinetic energy is 0 minus initial kinetic energy, which is half m v square Getting it. So using this relation You will get velocity Sorry, we need to find x, right? So you'll get the value of x to be equal to v0 Under root m by k it will come Fine any doubt with respect to this question Hello, yes, Krishna or Shitosh Any doubts with respect to this question? Okay Fine. So let's modify this question a little bit. Let's say that Let's say that the coefficient of friction between the surface and the mass is mu Okay, so there is a friction between mass m and the surface now, which is mu getting it All right. So mu is a coefficient of friction and it starts acting as in the the surface is rough From the point where the spring starts. So before the spring starts the surface is smooth Getting it. So after this when the spring starts Then onwards the surface is rough and coefficient of friction is mu Now you have to find out the maximum compression in the spring. Now, can you find out what is the compression in the spring? Surface is no longer smooth Now you will see that Not only work done by the spring will be there, but work done by the friction will also feature Getting it. So there will be two work done now earlier Friction was not there and normal and mg. We're not doing any work Okay, so only work done by spring used to come Now the friction is also there. So work done by friction will also come in this particular case anyone of you Got the expression So okay, not present today. I think many others At least five or six people who are Not present. Okay, Bharath sent me an answer Okay, Bharath X has to be positive. So although you get a quadratic equation you need to reject The roots of the equation where X is negative Getting it. So mathematics do not care about a physical scenario But when you're studying physics, you should use the mathematics Alright, and once you get the answer in to check whether it makes physical sense or not fine So X being negative doesn't make any sense So work done by the spring will be equal to again minus of half K X square only Fine now this X is not equal to that X. Okay, X is a variable and Work done by friction will be what? The friction will act back side backwards Fine and since it is sliding friction force will be equal to mu times n Okay, and the value of n is equal to mg Fine. So the work done by friction will be minus of Mu times n which is which is a mg Multiply for the distance it has moved So into X fine. So total work done on this block will become how much? minus of half K X square minus Mu mg X This is the total work done and this is equal to change in its kinetic energy Final cutting edge is zero minus initial cutting edge, which is half m into v square. Okay, so Any doubt with respect to this question, I hope many of you got this equation correct All of you got it any doubts Okay, I'm assuming all of you understood this particular question in case you have any doubts feel free to Message on the chat box or you can message me personally or whatsapp whatever it is. Okay, so Good, we'll take up a few more questions Because see mechanics is more of problem practice fine So we cannot just discuss theory and that's it So it is 90% problem solving and only 10% is the theory part anyways draw this diagram with me This is a This length is given as L. Okay, and this height is given as H fine, so particle slides along a track with elevated ends and The flat central part as shown in the figure so B to C is a flat central part Okay, the flat central part has a length of L is equal to three meters Fine the curve portion of the track are frictionless. So there are no friction on these curve parts. So a to B and C to D They are smooth fine, and for the flat part from B to C There is a friction with Mu is equal to point two getting it Now the particle is released from a height of This H value is given as 1.5 meter Okay, and Also ignore any bumps it can encounter Not near the transition between the Let's say a flat portion and the curve portion. So it's smoothly transition. So there is no loss of energy as such Okay, so you need to find out till what height it will reach on the other curved Surface Let's say it reaches here. You need to find out. What is this height? How much is this H? Letting it need to find the value of H All of you please try this out See whatever is not given you can assume. Okay, you can say okay masses M If it is not given probably the answer is independent of that, but then if you want just take it as M Okay, people are getting answers That's that's wrong Sukrit Bharat Ashutosh. That's not correct Sorry, sorry. I was looking at the other questions answer. Anyways, so Okay, let's see. Let's try solving this So between what and what points I can use this work energy theorem I'm going to use this work energy theorem between two points. Okay Work done is equal to change in kinetic energy. All right. So these two point. I'll directly take a and D Getting it. So point D is the second point and point a is the first point fine, so Work done can be split into two parts work done by the gravity and Work done by the friction, isn't it? Now what is the work done by the gravity between point one and two? So work done by the gravity can also be split into two parts work done by gravity when it goes from a to b Then work done by gravity when the object goes from C to D Getting it now. There is no work done By the gravity when the object is moving between B and C because gravity is work acting downwards and object is moving forward So it is 90 degree fine. So between B and C. There is no work done by the gravity So I'm just taking a to B and C to D fine So work done by the gravity between a to B is mg into H Fine because displacement along the direction of force Which is vertically down is H and mg is down fine So mg multiplied by displacement along the direction of force, which is H So work done by the gravity from a to B is mg H All right, and when the object goes from C to D then the movement of object is upwards Okay, it is against the direction of gravity. So you need to write it as minus of mg H Fine, so this is the work done by the gravity and then we need to take into account work done by the friction Fine, so that is equal to frictional force, which is Mu mg Into L and there will be a minus sign coming in here because friction is acting backwards But the displacement of object is forward. So minus sign will come Fine, so total work done will be equal to Minus of Mu mg L plus mg H Minus mg capital H. This will be equal to change in the kinetic energy and how much is the change in kinetic energy? Initial kinetic energy is 0 because velocity is 0 and Final is also 0 because at point D Velocity is 0 it comes to rest. Okay, so 0 minus 0 So from here you can get the value of H. You can see that M get cancelled out Fine so H You will get here as G times H minus Mu GL Okay, so Divided by G also sorry So you will get H minus Mu L Fine, so H is what H is 1.5 Mu is 0.2 and L is 3 so it comes out to be 0.9 meters Fine, so like this you have to do this particular question Okay, so, you know there is a tendency that you will break up the entire motion into two or three Component like you can say that okay for he'll apply working with theorem between A and B Then you apply between B and C and then between C and D. It doesn't matter Okay, you can directly take initial point and the final point if you're able to calculate the total work done. It does not matter Getting it. So This is how you have to solve this particular question in case you have any doubts, please message immediately Else I'm moving to the next question any doubts Niranjan says no doubts on behalf of everyone others also please You should not feel shy just communicate. Okay. Anyways, we'll take up Couple of more questions All of you draw this Fine, so two blocks are connected by the string as shown in the figure. Okay They are released from the rest Show that after they have moved a distance L Or After they have moved the distance of L. What is the velocity of M1 and M2? Okay, where coefficient of friction between The horizontal surface and M1 is mu Okay, who is massless and frictionless the string is inextensible and massless All right. So get the value of the velocity of Of M1 and velocity of M2 when both the block move by a distance of L They'll both move the equal distance right M1 if it moves by L M2 also moves by L Now you have to apply work as a theorem on both blocks together M1 and M2 velocity of M1 and M2 But at the answer you sent me it's difficult to understand what you've written Sorry Can I send it again? Those who get the answer you can WhatsApp me just take a pic of whatever you're doing and send it across It's very easy Okay, Sukirt has sent one answer which is independent of coefficient of friction looks like You're saying it doesn't depend on coefficient of friction. Is it? Yes. No, it is correct. Okay. I'm getting very Different answers Okay, I got multiple WhatsApp messages. Anyways, so I'll try to solve it. I Hope everybody of everybody have tried this out Okay, so see at times you will get the answer By many different means but then you need to understand the kind of intricacies that are involved in a particular question Okay, so focus on how it is solved So I am going to apply this work energy theorem on both the masses together fine so kinetic energy For M1 final counting a G for M1 final counting a G for M2 minus Initial counting a G for M1 plus initial counting a G for M2 This is a total work done fine and This work done is a work done on M1 and M2 together fine. So the work done Will be some of the work done on M1 and M2 so who all are doing work Friction on M1 is doing work. Isn't it? So if this block moves a distance of L So how much the friction will do the work? minus of mu M1 G Into L because friction is acting opposite of the displacement So minus mu M1 G into L is a work done by the friction on M1 Fine and who else is doing work? You may say that tension which is acting on M1 is doing work All right, so let's include that for some for a while plus tension force Into L even tension is doing work here, right? Now this is the work done this total work done on M1 gravity is not doing any work on M1 because gravity is acting down Which is perpendicular to displacement and normal reaction also same case Okay, now let's add up the work done on M2. So if M1 moves forward M2 will go down By the same amount let's say L Fine, so if it goes down that is a direction of gravity. So gravity will do the work on M2. So that is M2 G into L Right and also there is a tension acting T Okay, now tension is acting upward and displacement is downward. So minus T into L will come Fine, so you'll see that the work done by tension get cancelled off Okay So whenever you find the work done in a system, you have to only consider the external forces Which is in this case friction and the gravity Fine, so this is the total work done. This is equal to change in kinetic energy So final kinetic energy for M1 is half M1 into V square Okay, plus for M2 it is half M2 into V square Right Because velocity of both the masses will be same. They're connected with an inextensible string This is the final kind of energy of the system minus initial kind of energy of the system is zero for both the masses Fine, just you equate these two and you'll get the answer for velocity Okay, so I hope you have got this Let's move to next question Okay, what is the final answer is what Sukirt has sent is correct just multiply M1 with Mu Then the Sukirt answer is correct The M1 and the numerator if you multiply with Mu then the Sukirt answer is correct All right, so let us take up one final question before we start the next concept Okay So You have a hemisphere Okay, draw a hemisphere this is Hemisphere the radius of the hemisphere is R Fine radius is R Okay, the hemisphere is smooth All right, you are keeping a very small mass over here Okay Very small mass is keep you're keeping there and then you are slightly nudging it you're just Slightly touched it and it starts going down Getting it. So what will happen? It will move down Along the sphere All right, and at some angle it will leave the sphere fine So it may not move all the time along the sphere only So there is a point where it leaves the hemisphere fine So let's say that that angle is theta with the vertical fine You need to find what is the angle theta at which this mass M Leaves the hemisphere. Okay, try to do this Okay, you need to find the angle theta at which the mass M will leave the hemisphere Barat got one of the answers others Those who agree with Barat can say that okay Barat's answer is correct Yeah, I'll solve it Everybody everybody tried this Okay, what is the condition? when the mass Just tell me the condition when it will leave the hemisphere No Barat Yes, all the contact forces goes to zero because it is losing the contact right So since it is losing the contact so contact forces Will become equal to zero fine. So What are the contact forces here? There is only one contact force that is normal reaction out here So with normal reaction when it becomes zero That's the point when it leaves the hemisphere Okay, so all the contact forces even if there is a friction even friction force will become zero at that moment fine Okay, now if suppose it has a velocity V at that moment, let's say it has a velocity V at that moment Okay, then since it is moving in a circle before leaving the hemisphere It will have an acceleration of V square by R To this center. So this one will be the acceleration. So if you draw the free-body diagram of This mass you will have mg like this and Then you will have a centripetal acceleration V square by R Getting it and this angle will also be theta fine So if you draw this mg force, this angle will be equal to that angle All right. So the force along the direction of radius is what? mg cos theta Fine this force should be equal to mass time acceleration. So MV square by R All right, so I am just ignoring the normal reaction normal reaction goes to zero When it leaves the surface fine So this equation should be true at the point when it leaves the surface Okay, if it is not leaving the surface then what will be the equation? mg cos theta Minus normal reaction. This is the net force This net force will become MV square by R But at this moment normal force is becoming zero. So mg cos theta is MV square by R All right. So this is the condition for which It will leave the hemisphere now what I'll do. I'll Write work energy theorem between point one and point two Okay, so work done by all the forces will be equal to change in the kinetic energy What is mg cos theta beta mg cos theta is the component of mg along the direction of acceleration Component of mg along the direction of acceleration is mg cos theta Fine as you those let me answer if anybody asks me So mg is acting downwards its component along this direction is mg cos theta Anyways, so Work done by all the forces will be equal to change in kinetic energy, which is half MV square half MV square is the final kinetic energy right and initial kinetic energy is zero Fine so this is equation one and that is equation number two All right now I need to find the work done who all are doing work is normal reaction doing any work when it is sliding down Normal reaction is perpendicular to the displacement all the way Okay displacement is tangential normal reaction is perpendicular to that. So work done by normal reaction is zero The only force that is doing work here is work done by gravity Fine this is equal to mg multiplied by this displacement, which is along the direction of gravity Okay, let's call it as y so y will become how much? This total length is R fine and this much is R cos theta fine You can see that there's a right angle triangle over here So this length from here to here this length is our cos theta So y is equal to r minus r cos theta. So one minus cos theta times r Fine. So work done by gravity is Mgr 1 minus cos theta Fine. So all you have to do is just substitute there Mgr 1 minus cos theta is the work done by gravity This is equal to half MV square. All right and how much should be half MV square? So if you look at here You know half MV square should be equal to Mgr cos theta by 2 Okay, I'm just utilizing this equation So I would like to write MV square in terms of Mgr because Mgr will get cancelled out from left-hand side okay This is equal to Mgr by 2 Into cos theta fine. So now you just have to solve So cos theta will come out to be 2 by 3 fine, so theta is cos inverse 2 by 3 Fine any doubt with respect to this question, please type in Okay, how I get why first of all you need to understand why is what why is a Displacement of this mass when it goes there Along the direction of mg because in order to find the work done by gravity I need to find how much it has displaced along the direction of gravity fine So you can see that there is a right-angle triangle Like this Okay from here to here this length This length is r Okay, this angle is theta and since this is 90 degree this length is R cos theta Getting it. So I know from here to there topmost point the distance is r and from there to here the distance is r cos theta Fine. So the y is equal to r minus r cos theta, right? So that's how you get the value of r So that's how you get the value of y, okay Okay, just have to solve this guy is you'll get cos inverse 2 by 3, okay So cos theta will come out to be 2 minus 2 cos theta So 2 cos theta you get it that side You'll get 3 cos theta is equal to 2 and cos theta is equal to 2 by 3 How you got the value of work done by mg see work done by mg is Force mg multiplied by displacement along the direction of force which is r into 1 minus cos theta Okay, so we have got the value of y just like that Fine, so I'll just make this figure a little bit more clearer so I'll draw it here So mass goes along an arc Like this fine. So this is our and that is also our and this is theta Fine, so if I drop a perpendicular from here, this is 90 degree this length Will be what our cos theta this length, okay? And I need to find that length which is why right the total length is So why is equal to r minus r cos theta? That y is the displacement along the direction of gravity. So like this you get this particular Work done. Okay. Okay. Shall we move to next? Let's move to next concept now We are talking about Conservative and non-conservative forces Right down as a by the way that work energy theorem that we have derived Which is work done is equal to k2 minus k1. Okay, you can apply this theorem At a frame of reference where the reference frame is At rest Okay, or the reference frame can move with the constant velocity also Fine, so if reference frame moves with a constant velocity, then you observe relative velocity So the measurement of kiting energy will change if you are observing the object with respect to a Frame that is moving with constant velocity Fine, not only kiting energy will change, but the work then will also change because then you'll see a different displacement Getting it. So that is slightly more involved when you solve questions You will encounter some scenarios where if you change the frame then the question gets solved easily Fine, so we'll take up questions like that changing frame of reference to find the kiting energy Later part of when we revise or when we revisit this chapter will take up those scenarios. Okay All right, but then your frame of reference cannot accelerate if it accelerate There'll be pseudophores and pseudophores also does the work Fine, so pseudophores is another concept that I will have to take Separately that I have not yet discussed Fine, so I'm just telling you so that you know if you are like if you want to move ahead You know, don't hesitate. You can ask me individually also in case you you want to Okay, Bharat is asking something. Yeah, I just discussed that All right, so we are talking about conservative and non-conservative forces now the forces can be, you know broadly classified into conservative and non-conservative forces depending on how you calculate or How the work done by the forces will behave, okay So if write down if Work done by a force independent of frame independent of of the path taken and only depends on initial and final points Okay, then The force is conservative Fine, so you soon understand why we are defining conservative and non-conservative force right now just Take it as a definition Okay, now suppose this is point one. Okay, and that is point two Fine now You are going from point one and two suppose you are going like this and Then you are reaching point two like this Getting it. So it doesn't matter what path you take if initial and final points. They are same Then the work done by the force Does the same work, I mean it The work done by the force will be same in both the path Let's say this is path one and this is path two Okay, although the displacement is independent of path. Okay, displacement will be what from point one and two Displacement will be this only this is a displacement Okay, and if it is a constant force fine. So if the force is constant anyways You know, it will be independent of the path because displacement you have to take the dot product with the force Fine, so displacement is independent of path and force is a constant force So f dot displacement will give you the work done fine But I'm talking about in a generic sense even if the force can change its magnitude and direction both In that case also there are few forces Okay for which the work done by the force is independent of the path It only depends on initial and final points Okay, so in that case That force is the conservative force fine but in a case where the work done by a force depends on the path taken by the Particle or an mass or a mass then That force is not a conservative force Fine, now how you test whether a force is conservative or not you test whether a force is conservative or not by just finding the work done Just find out work done by a force if an object Moves in a loop What do you think the work done by conservative force will be if an object moves in a loop? Let's say object moves in a circle Okay, like this so in one full circle how much will be the work done by the conservative force? It will be zero because initial and final points are same Okay, so work done by the conservative force in a closed loop will always be zero getting it Fine now I'll just take one example of non-conservative force so that you can understand the difference Take an example of this This is an Enclosure, it's a spherical enclosure. It is a hollow sphere. You're taking a mass from here Okay, you're taking a mass. All right, you are pushing it All right from both sides from top and the bottom and this object Moves like this goes to the top when it goes to the top you're pushing it like this okay, and You're making sure it completes one full circle It starts from one and comes to the same point so initial and final point both are the same Fine, so is the work done by friction zero? Work done by friction is it zero? It is not zero right friction will do some work Friction will do some work All right, so friction is a force which is you can say an example of non-conservative force Fine so it depends on what path you have taken When you calculate the work done by the friction, so you'll see that all the while when it is Sliding up that is sliding up friction is always in opposite direction of displacement So you should add up the small small work done by the friction All the small work done by the friction will be a negative quantity So all the negative quantity cannot add up and equal to zero Fine, so there will be a net work done by the friction which is negative Because all the small small work done by the friction. They are negative. So you just add them up. It will never become zero Fine, so work done by the friction is not zero when it moves in a closed loop So friction is an example of non-conservative force Okay, now take an example of gravity What is a gravity force mg? mg is the gravity force and it is always acting vertically down Okay, so what is the work done by the gravity force work done by mg force? What it is right now? How much is a displacement along the direction of gravity? It will be zero in case your connection is failed. Please come live and watch it Okay, you can revisit what you have left later on Okay, Niranjan is saying minus mg into 2r Niranjan if you are talking about point one and let's say this is point three So if the object goes from one the bottom most point To the top most point then The work done by the gravity is minus mg into 2r Because a displacement along the dirt it is now against the direction of gravity actually This is the displacement which is 2r and gravity is acting downwards Fine So mg is downward and displacement is 2r against it so minus mg into 2r when the object goes from 1 to 3 Fine, but if I'm saying that object comes back The displacement becomes zero it goes up by 2r and comes back by 2r so displacement is zero Okay, so work done by the gravity is Zero Fine, so gravity could be an example of And conservative force fine So we'll take up a couple of more examples and we'll ascertain that The gravity is the conservative force and then we'll analyze it further Okay guys do this You have an inclined plane. This is angle theta Okay This is the height edge And now I am removing this incline and making it a wavy path like this Fine without touching the initial and final point initial and final point remains the same Okay This is height edge Now you're keeping a mass here fine So what is the work done by the gravity in both the cases? In both cases what is the work done by the gravity? That's not correct miranjan Oh, that's correct actually Yeah, it is mgh gravity force is acting down and displacement is also down so The work done by the gravity is mgh in both the cases Fine, so you see that the path is changed, but then still you're getting the same work done by the gravity all right, so since The work done by the conservative force do not depend on the path. It only depends on Where it is right now And where it is finally Okay, so it depends on only two points. So it is like a state function Okay, and since it is uh, it depends on only Where it is and where it will go All right Hence we can define something which is a function of only What is the location of the object? Fine, so we need not calculate the work done by gravity By looking at the path Fine, so we can cater to the work done by the gravity in a different way All right, or let us say work done by any conservative force we can Find out quickly if you know its initial state and the final state fine Do you guys recall any other physical variable that you have encountered till now in physics? That depends only on the state initial and final can you message it quickly? Which variable which physical parameter you have learned that is only the function of initial and final states Exactly internal energy right so we have learned Correct so we have learned this physical parameter Delta u which is just equal to n cv delta t in case of an ideal gas Fine All right, it only matters. What is the change in the temperature? Okay, so change in internal energy is a state function now. Similarly, you can Define the You know a parameter which only depends on initial and final points Okay, and I think you guys know what I am talking about here. I am talking about potential energy here write down Now the way potential energy is defined okay, it is Potential energy is the energy write down it is the energy the mass has due to its due to its position shape and size Okay, so How it is located where it is located? What is its shape and what is its size? okay, so it It is the energy that is possessed by an object because of these factors Fine, so if if it is constrained as in if you take a let us say If you take a mass if you take it at a height h, okay You're keeping it at that height. All right. You're keeping it that height So what I'm saying is just by the mere fact that it is at some height Okay, it is at some level. It has some energy Okay, just the fact itself that it is there it has some energy All right, and if you remove all the constraints or all the other you know surrounding things that Is you know making it possess that much potential energy Then ideally every object wants to just give up all the energy it has all the potential energy it has and Comes to the state where it has the minimum possible energy fine This is how every object in the universe behaves right so natural tendency of every object is right down natural tendency is to lose all potential energy or to minimize minimize the potential energy so by the way potential energy is represented by a letter capital u so that's something which everybody has accepted Okay, how does shape affect the potential energy? So I think the spring is the best example of potential energy due to the shape You just compress it its shape changes right so if its shape is changed Right i'm saying that it has energy And how do you know it has energy you just release it And if you have a mass connected with the spring it will transfer its potential energy and create the kinetic energy for the mass fine All right So every object has a natural tendency to minimize the potential energy Okay, so whatever potential energy an object has given a chance It will just give up all the potential energy it has and that will manifest in the form of kinetic energy fine Now I can take up multiple examples right so if I take a mass m Like this in air Okay, if I release it It will gain kinetic energy It'll come down and it'll gain kinetic energy Right, so if I assume that there is nothing called potential energy that exists in the universe Then how come energy gets created? Energy got created out of nowhere That is against the universal law of energy conservation fine So there must be some sort of energy that is getting converted into the kinetic energy Okay Fine and that energy We are defining as potential energy All right, so And enough of the introduction to the potential energy I'll directly get into uh the definition mathematical definition of the potential energy now I think you have You know You already know what is potential energy. You might have used it in your lower grades. All right fine So the way potential energy is defined is this You say that change in potential energy Change in potential energy Is negative of the work done by the conservative force Our focus on the class don't uh Focus on answering others query. I am there for it Okay So change in potential energy is defined as negative of the work done by conservative force And work done by any of the force Is what integral of force With the displacement Fine, so this is the definition of the change in potential energy Getting it now this definition is true. No matter what potential energy we are talking about We may be talking about the spring potential energy Or we may be talking about the gravitational potential energy or let's say we'll talk about electrostatic potential energy What are potential energy you may talk about the definition of the potential energy is this that change in potential energy will be equal to negative of the work done by That conservative force for which you are defining the potential energy Getting it fine So if potential energy is there If there exists potential energy There has to be A force corresponding to it Isn't it That's how potential energy is defined if there is a potential energy. There has to be a force corresponding to it All right, so the way potential is defined is this now same thing. I can write in a differential form I can say that du is equal to minus f dot dr fine Now what is f dot dr? You can write du is equal to minus of magnitude of f Into magnitude of dr Into cos of theta Getting it dr cos theta is a displacement along the direction of force Let's say displacement along the direction of force is dl Getting it No, but a du is not a vector It's a dot product between force and the displacement. Isn't it so dot product will give you a scalar only all the energies No matter what it is all the energies they are scalars All right So dl is what dl is a displacement in the direction of force Fine, so I can say that du is minus of f into dl So from here I can get f is equal to du by dl Fine, this is a special derivative. What you're doing. You're taking derivative along the direction of force only Fine, so we call this as partial derivative What is partial derivative? You will learn later on that When you take up the partial derivative, suppose you're taking derivative with respect to y Okay, suppose u is let's say x square plus y Okay, now force along y direction is minus of du by dy Okay, I am differentiating with respect to The direction Along which I need to find the force So when you differentiate or partial differentiate with respect to y you take x to be constant The fy will come out to be minus 1 newton's And similarly fx will come out to be minus of du by dx That will be equal to minus 2x So fx is this and fy is that so total force will come out to be minus of 2x icab minus jgab Fine, so we'll take a few questions on the you know On the partial derivative. I mean there is nothing more to it It's just a matter of a couple of questions and then you will start feeling comfortable with it So what we have done till now is that we have Got a relation between Change in into change in the potential energy And the force for which you are finding the potential energy fine Now let us try to define potential energy Due to the gravity and potential energy due to the spring because these are the two forces That you will be dealing again and again when it comes to you know solving a mechanics question Fine, so let us try to quickly find out the potential energies for the spring and for the gravity So first we will try to define write down potential energy Due to gravity All of you might be aware of the formula already that potential energy due to the gravity is mgh Okay, we'll get into more details about it how it comes out and what is the implication of this particular formula Okay, so we'll take a simple, you know example. Let's say You have a ground like this If a ground like this, okay now you take a mass This is mass m. Okay. So what you're doing is you are lifting this mass m from this point To a point which is at a height h This is height h Okay, we are trying to find out The formula for potential energy for the gravity All right, so We will start with the definition of potential energy potential energy is equal to negative of the work done By the force for which we are finding the potential energy and we are finding potential energy due to the gravity force Right and gravity force magnitude is mg Okay, so work done By the gravity force is how much all of you How much is work done by the gravity work done by the gravity is minus of mg h right Because you go from point number one to point number two the displacement is h and that is against the direction of gravity So work done by gravity is minus mg h Okay, so according to the definition, this will be equal to the change in potential energy This should be equal to delta u Okay, so delta u is what u2 minus u1 Sorry, uh, this Let me erase this. I missed a sign over here Okay, so according to the definition Negative of the work done by the gravity is change in the potential energy fine So negative of the work done by the gravity which is equal to mg h Fine, this is equal to change in potential energy. Which is what u2 minus u1 Fine, so I can say that u2 minus u1 is equal to mg h Okay Now I am interested in finding the absolute potential energy I want to know what is the potential energy at point two, but what I am getting is difference in the potential energy All right, I'm getting difference in potential energy, but I want to know what is the potential energy Okay, so one thing you guys please understand is this There is no concept of absolute potential energy Fine, so you will never know what is the zero potential energy. Okay, hence In order to find the potential energy you just you can randomly assume anything to be zero Okay, now Now I'll explain you Uh, how the answer will not change because of that first. Please write down That we need to assume a location Where potential energy becomes equal to zero Fine, then only we can talk about absolute potential energy fine Although it is not absolute, but then you know, it will help you to visualize better. That's it so You can assume that u1 is zero. So if you assume u1 is zero then u2 will come out to be mg h Are you getting it? Then u2 will come out to be mg h So potential energy we have defined at a location which is at a height h from where The height h from a location where you have assumed Uh, the gravitation potential energy to be zero Okay, so it need not be zero, you know The height need not be zero. This could be the case like this in that that it is On a table which itself Has a height of small h Are you getting it? It it could be on a table that is at a height of small h from the ground and this ground could be The second floor of the building Okay, so You can assume any horizontal line to be zero potential energy Getting it so every question Going forward understand I'm talking about with respect to problem solving so every numerical or every question If you have to find the gravitational potential energy fine, we need to first assume a horizontal line that represents gravitation potential energy to be zero Getting it. So once you assume a horizontal line In a particular question Then after that you can write down the potential energy relative to that horizontal line So if it is at a height, uh H above Then potential energy will be equal to mg h And if it is at a depth h below if it is h below of that line the potential energy will be Minus mg h fine. So potential energy could be positive if that is above the horizontal line And potential energy could be negative if that point is below that horizontal line Okay, so like that in every question you can have a new horizontal line and you can say that that is my zero gravitation potential energy All right Now once you assume a horizontal line for which you say that gravitation potential energy is zero, you cannot change that line in between the question Are you getting it? So for entire question that becomes the zero gravitation potential energy line fine so for Different different question you can have a different horizontal line Up to you what is convenient to solve the question you can assume any horizontal line to be zero potential energy But once you assume a zero gravitation potential energy line, which is horizontal You cannot change it midway of the question Okay, so I hope I am very clear about this in case you have any doubt Please type in you have any doubts type in yes or no So understand one thing very very clearly H is The distance which is vertical And in a particular question if you are not sure what is vertical It just find out what direction the accession due to gravity is So I'm talking about the distance along the line of accession due to gravity Okay, and all of you please do this question quickly You have suppose a string At the end of the string there is a mass m The length of the string is l Okay, you are pulling this mass With some force and that because of that Let's say the force is f it always remains horizontal fine So this bob Moves a distance It moves by an angle of theta I show in the figure okay the length is l you need to find the Change in gravitation potential energy how much it is Between this location, let's say this is location one and this is location two between these two locations How much is the gravitation potential energy that has changed? A way h1 and h2 are not given See you can assume this line to be Zero gravitation potential energy fine. So once you assume that this line represents gravitation potential energy to be zero Then the potential energy of point two will be what? Suppose this is the height h Potential energy of two will become equal to mg into h Fine and what is h? It is very similar to the previous one where h will come out to be l minus l cos theta Okay, so this is a right angle triangle from here to here This length is l cos theta Okay, so u2 will come out to be mg l One minus cos theta Fine So u2 minus u1 will come out to be this only because u1 you have assumed to be zero Okay, now if you assume, let's say this horizontal line to be zero Will delta u change will delta u which is u2 minus u1 Will it change or will it remain the same? I'm asking you if I'm asking you if You are changing the Line for which gravitation potential energy is zero with the change in potential energy Very or it remains the same It remains the same, right? It's like this, you know That you take a length You take a scale, all right, so if this is 10 centimeter and this is 40 centimeter You have a reading it start with 10 centimeter and ends at 40 centimeter This length which is difference between 40 and 10 will remain 30 Okay, it doesn't matter you can take This as zero if this is zero this could be probably 30 and this could be minus 10 Then 30 minus of sorry If you take that to be zero this could probably become 20 fine So 20 minus of minus 10 still it's 30 So length Which is a difference between the two uh readings. It remains the same fine. So you can assume any Gravitational potential energy line to be zero the change in potential energy remains the same Fine, so that's the best part about it. You can uh, you know choose gravitation potential energy to be zero at any location And you will be able to find same change in potential energy Okay, all right, so I will uh now quickly talk about the implication of the potential energy The last uh concept we have learned that work done is equal to change in kinetic energy Right, this is the verb energy theorem Now you can modify this theorem by saying that work done can be work done by conservative force Plus work done by non conservative force Okay, plus work done by let's say other forces Fine, and you can say this is equal to change in kinetic energy k2 minus k1 getting it all of you So I am modifying the work edge theorem only so work done by conservative force We have just discussed work done by conservative force is negative of change in potential energy Plus work done by conservative force Plus other forces work done will be equal to change in kinetic energy Fine, so if you modify this you will get Okay, what is wo? Wo is the work done by other forces like you're not sure whether it is conservative or non conservative So you just uh, you know take it as other forces work done by conservative Sorry, this is non conservative Because work done by conservative force becomes change in potential energy. So this is non conservative force So work done by non conservative force Plus work done by other forces is equal to change in kinetic energy plus change in potential energy Okay, so if I club non conservative forces and other forces together So we can say that total work done By all the other forces be it non conservative or any other force Will be equal to change in kinetic energy, which is k2 minus k1 Plus change in potential energy, which is u2 minus u1 fine. So this entire thing gets modified into k2 plus u2 minus k1 plus u1 Fine, so this is another variation of work energy theorem here. We are writing the You know, we are writing in terms of potential energy also Till now we never uh factored in the potential energy We always used to take the work done by all the forces Fine. Now what we are doing we are converting work done by conservative force Okay, in the form of potential energy fine. So if you take potentiality of the gravity Then do not find out the work done by the gravity now Are you getting it? So if you are taking Potentiality of any force then you should not Find out the work done by that force and put it on the left hand side fine So these are the other forces Other than for which you have considered the potential energy Okay, any doubts till now, please text No doubts okay so We have modified our work energy theorem And now we have written like this guys, this is the equation which You have to use in every question of this chapter fine So make sure you are familiar with this And uh you you are able to use it freely, okay U2 Is the final potential energy k2 is a final kinetic energy U1 is initial potential energy and k1 is initial kinetic energy fine this potential energy and kinetic energy sum That together they're called as mechanical energy All right fine and if only conservative forces Are applied If only conservative forces are applied on the system or mass or Non-conservative forces Don't do any work fine then Work done by other forces will become zero This is equal to u2 plus k2 Minus u1 plus k1 Fine, so you will get u2 plus k2 Is equal to u1 plus k1. What is u2 plus k2? This is final mechanical energy This is final mechanical energy sum of kinetic energy and potential energy And this is equal to initial mechanical energy Fine, so mechanical energy will be conserved if there are only conservative force acting on a system Or let's say other forces are there, but they are not doing any work Then the mechanical energy will be conserved Okay now we have uh talked about uh the The gravitation potential energy all right. Let's talk about quickly the spring potential energy also because even this Work done by the spring Does not depend on the path. Isn't it? We have already found out that work done by the spring is what Work done by the spring is Half k x1 square minus half k x2 square right So it doesn't matter how x1 has come and from x1 to x2 how you have gone it only depends on x1 and x2 that is initial and final position Okay, and hence you can define the potential energy for the spring force also Okay, so according to the definition change in potential energy is equal to negative of the work done by the spring Okay, so negative of the work done by the spring will give you half k x2 square minus half k x1 square All right Okay, so this is the negative of the work done by the spring and this should be equal to change in potential energy So this is let's say u2 minus u1 Okay Now let's say if x1 is zero What does it mean? Initially The spring in its natural length Initially the spring is in its Natural length Fine, so if x1 is zero This term will become zero. So what do you get u2 minus u1? You will get as half k x2 square Fine now u1 is what u1 is potential energy of the spring When extension or compression in the spring is zero Fine, so if you assume that in such scenario if potential g When x is zero if you assume that to be zero Okay, again, we are assuming here's something to be zero Okay, so if you assume that spring has zero potential energy when it is in its natural length Then the potential energy Of the spring will become half k x square fine, so I can write down a generic Thing potential energy is half k x square for a spring Okay, and since it is a unique point where the spring Is in natural length Then we are assuming potential g to be You know zero and since it is present in all the spring all the spring will have a natural length, right? So we can all universally accept that okay. Let us call You know potential g of a spring to be zero when it is in its natural length fine So if we don't have to again and again find out or again and again have to assume When the potential energy of the spring will become zero So everyone has accepted that they will Say that when the spring in is in its natural length will all assume That potential energy of the spring is zero fine now the x x can be compression and x could be extension x can be anything Fine, but it doesn't matter The potential energy will be equal to half k x square the spring potential energy Will always be greater than zero it can never be less than zero fine This is something unique about the spring potential energy, which was not the case with Gravitational potential energy because gravitation potential energy can be negative Getting it so spring potential energy is half k x square and hence The spring potential energy will always be greater than zero Okay, all of you clear about this any doubt that k is a spring constant Okay, that's great Now let me tell you one thing where most of the students falter See potential energy due to spring at times is tricky You need to look at Initial compression or extension in the spring Okay, this is something which Many students ignore fine. So spring can be already compressed. So spring can have initial potential energy itself Getting it for example, if I have a spring like this and on top of it if there is a mass Mass M. So spring is compressed. So it already has Up potential energy initially Okay, so there are a lot of such scenarios. I have taken the most Simple example to you know, make you aware that potential energy need not be Zero initially all the time Fine, so be cautious with respect to this see now let us Further look at our modified version of work energy theorem work done by the other forces What is the other forces other forces are the forces for which you have not considered the potential energy Is equal to this Okay Now this potential energy It can be potential due to gravity And potential due to spring both can be present. Okay So make sure you take care of both gravitation potential energy plus spring potential energy. All right, and Usually you will have friction as the Force which is other than the force for which you are considering the Potential energy. So this could be Due to the friction Could be Fine, so Any doubts till now anything Message me any doubts no doubts Okay, so we are all good for Taking few questions Let's take a question now guys going forward. I am assuming that you guys will use this particular relation Although you can solve the same question By just using total work done is going to change in kind of energy. You don't need this particular relation But then since we have derived it Okay And when it comes to spring using this relation is easier than using Work done is equal to k2 minus k1. So try to use this particular relation get familiar to this because You know at times this will give you a quick answer All right, so uh, all of you draw this figure There is a mass m Like this spring with spring constant k is there Okay, so mass m is released Uh from the rest from a height h As shown in the figure It slides and compresses the spring of stiffness k. You have to find the maximum compression in the spring Okay, find out the maximum compression in the spring quickly root of 2 mgh by k See whenever you use work energy theorem you are using it between the two points Point number one and point number two. You can pick any of the two points Fine, so I'll take the two points where the point number one is this and point number two is where the spring get compressed maximum right so spring will uh get compressed till The velocity of mass m goes to zero fine So at the maximum compression the velocity of mass m will become zero Fine, so let's try to use this particular relation only so work done Is what zero Gravity is doing the work, but that work you are considering it as a potential energy Fine, and you're saying that this line this line could be the logical Point the line for which you are assuming gravitation potentiality to be zero fine. So You one is what initial potential energy Will be equal to mgh fine k one is Zero it was at rest Okay, then you two is what? Finally the block comes to the level horizontal level where uh, I mean that is the level where you assume the gravitation potentiality to be zero so this is zero This is gravitation potentiality, but then there is a spring potentiality at point two Initially there was no spring potential energy now There is a spring also so spring will gain a potential energy half kx square And we know that k2 is zero So u2 has two potentiality spring and gravity So if you substitute it you get zero is equal to u2 Which will come out to be half kx square Uh, gravitation potentiality is zero. So this plus k2 is zero minus u1 Which is mgh Plus k1 which is zero Fine, so from here you'll get the value of maximum compression in the spring which will come out to be 2 mgh by k Okay Let's take up next question. You guys have any doubt on this particular question. Anyone of you have any doubts? Sir, why is the work zero? See who can do the work? Normal force doesn't do any work in this case because normal force is always Uh perpendicular to a displacement it acts perpendicular to surface fine gravity can do the work but That work you are already factoring in in a form of potential energy. So you can't factor Factor in the work done by the gravity twice Fine, so there is no other force. So only gravity is there and When it compresses the spring all the spring also applies a force But then even the spring force you are considering it as a potential energy Fine, so no other force is left That is why work done Will be equal to zero because this is the work done not by all the forces, but by the other forces For which you have not considered the potential energy see now, let me caution you Just give me one second. I lost the chat So there's a small caution here Now what will happen is that you will develop your own version of conservation of energy Okay, so you will be like If friction would have been there then we would have considered it right. Yes, you will consider the work done by friction the friction is there Okay, so the caution is this You should write it down because this is something which Destroys the chances to solve the question Do not create your version Of work energy theorem that you should never ever do If you start doing this, you will be always confused Okay, because I have seen students, uh, you know making version of the work energy theorem themselves like they'll be like, okay Laws in potential energy will be equal to gain in kinetic energy or loss in kinetic energy will be equal to gain in potential energy Or some random stuff Okay, so do not put your brain in uh, you know for creating a formula Always fall back to this particular relation Before every question first write down this and just one by one find out the values u2 k2 u1 k1 and work done by other forces All of you clear about it. Do not create your version of work energy theorem You may get some questions right. Okay, you may get it right. I'm not saying you'll not get it right But you'll confuse yourself and you'll never be confident Okay, so let us move to next question In case you have any doubts feel free to message you can message your whatsapp You can message on the chat whatever you may feel comfortable And I will not take your name another best part about the online class is that you can you know Secretly ask a doubt You can be anonymous So if you are feeling uh You know conscious when you ask a doubt I mean you can freely text me your doubts I was very conscious of asking doubts in a class. So I always used to keep quiet and later on when uh Everybody used to go. I used to ask anyways So a block of mass m Okay, there is a block of mass m that strikes a light pan fitted with a vertical spring So there's a small pan Like this It is a light pan. Okay it Strikes a light pan fitted with a vertical spring after falling through a distance of h So it falls a distance of h then strikes this pan Okay, if the stiffness of the spring is k find the maximum compression in the spring Get the value of maximum compression in the spring Maximum compression in the springs, okay What is the maximum compression? No Niranjan that's not correct To the one who has sent me the whatsapp message. It was not correct You can just take a pick of whatever answer you get and send me across Should I do it? Those who are sending me Answers over whatsapp Please understand. I cannot chat with you. I can only look at what you are sending. Okay So don't ask me questions on whatsapp that I mean don't try to chat with me or you can post your doubts that is fine That I can address Here guys see This is the probably The initial phases of you learning the application of work energy theorem, right? So start the question by writing the You know formula itself. So always write this down And then start solving it, right? And when you write potential energy of the gravity, you need to understand what point you are taking zero potential energy All right No till now nobody has sent me the correct answer No Good many of you are sending me a message One of you send me an imaginary answer As in root over some negative quantity That will be not correct. Definitely Hmm. Who is this? Sukheer that's correct Let's see how to solve this particular question. Maybe Barad you are doing some silly error probably the conceptually It may be correct, but then does it matter? So do it slowly and get it right the first time Okay till now only Sukheer has sent me the correct answer fine. Let me now solve this particular question I want everybody to participate see you can clearly see that I have got at least 12 answers from people who have whatsapp me and Only one was correct. So don't hesitate to answer You can be wrong right now. You can see that only one person was right and everybody else was wrong. So don't hesitate Okay, let's try to solve this now Let's assume that when it hits the spring the spring get compressed By a distance of x. So this is the maximum compression. Let's say x is a maximum compression fine So I'll take my second point There only This is my point number one and point number two is the point where the spring get compressed maximum All right. So what I'll do I'll draw a horizontal line And I'll say that the line that passes through the maximum compression point of the spring Represents gravitation potential g to be zero Fine. So what does it mean? u2 automatically becomes zero And u1 will be what? u1 will be equal to mg H plus x I'm talking about gravitational potential energy fine Many of you might not have taken x While finding the gravitation potential energy okay The u2g and u1g now there is a spring potential g finally u2 s is also there which is half kx square Okay, so what is left now? k1 Is zero right Initially the object was at rest and finally also it is zero now the one assumption. We are making throughout Can you guess what is the assumption? We are making here When we are applying this over here The assumption we are making is that no energy is lost Due to collision fine If energy get lost due to collision you cannot take point number one and point number two when the spring get compressed because in between there's a collision okay so There is no other force that is doing work other than gravity and spring for which you have already considered the potential energy right, so you get zero is equal to u2 Which is a spring potential energy that is half kx square plus k2 which is zero minus U1 now there is no spring potential energy initially, but there was gravitational potential energy that is mg H plus x fine Plus kinetic energy zero So you get a quadratic equation just solve this and you look at the answer Okay, the answer should be this x will be equal to this is what you should get Fine, so I hope it is clear in case you have any doubt. Please message immediately Any doubts So let us move to next question That's time for something right. Let us take up the next question It refers to a figure, so Let me draw the figure first All of you please draw it with me This small m This is the last question. Then we'll take up another concept. So this is a scenario. Okay So two blocks of mass small m and capital m They are placed like this. This is the ground okay They're connected by a light spring of stiffness k They kept on a smooth horizontal surface as shown what should be the initial compression of the spring What you're doing is you're compressing this small m Okay, what should be the initial compression of the spring so that system will be about to break off from the surface So when you compress it right so suppose you compress it up to this level Okay, and then you release it So when you release it Okay, what happens is block will Try to move up because of spring force Okay, and when it moves up spring may get extended also fine So you need to find out how much should be this compression in the spring initial compression x naught so that capital m can break off from the surface so capital m You need to find x naught what is x naught for which capital m breaks off from the surface attempted Breaking of the surface means that capital m leaves the surface Capital m jumps off Should I do it What is the condition for capital m to leave the surface? What is the condition all of you? Normal force to be zero Okay, now if if it is placed just like that normal force will never be equal to zero because there is a gravitational force acting on it Which is mg and since it is not accelerating There has to be a normal force Which is equal to mg Fine, but what if there is a spring force upwards Which is let's say k x one And then if you balance out the forces red k x one plus n minus mg Then this is equal to zero You'll get condition when normal force Become zero so normal force becomes zero when extension in the spring becomes equal to capital mg by k This should happen then only it will leave the surface Okay, let me call it as x2 Initially it was x one Fine Now try to do this Which chapter from book one is going on in school? Is somebody sent me the answer No, that is not correct You need not know the spring length Okay, because that doesn't matter spring constant is given so Deviation from the length of the spring is what determines Okay So we are at par with whatever is going on in school. I have received another answer See there will be initial compression already in the spring You need to find out further how much you need to compress it Okay, now let me do this I'll do this now Okay, so let us say that When it is kept at rest the spring has some compression Right, let us say x0 was an initial compression Which is mg by k Okay, this much compression will be already there because small m is a m is sitting at the top of the spring Okay, now let us say that you have further compressed it By a distance of x1 Okay, and then released it So what will happen now is that from? a compression Of x0 plus x1 it should go to extension of Then only capital m will be able to jump off Okay, so this is our initial point This is our initial point and this is our final point. So between these two points I am going to apply work energy theorem fine. Let me take up another color pen I'm using only yellow and white How is this? This is nice Okay, so right Fine, so I'm going to first write down work energy theorem work done Is u2 plus k2 minus u1 plus k1 Now you'll see that Only gravity and spring forces are there and both of them you are considering potential energy So this left hand side becomes zero Okay, so w is zero u2 is what the final This thing final potential energy. So u2 will have Gravitational potential energy. Let's say u2 g and it will have a spring potential u2 s Fine, so we need to first assume. What is a zero gravitational potential g? So we can assume that The level where it is initially compressed That is The gravitation potential g to be zero fine, so Yeah, I can say that initial gravitation potential is zero fine, so final gravitation potential g will be what? mg into x0 plus x1 Plus x2 because x1 is compression. So from this much Down it should go that much up. It should get extended the spring. So total Height from this zero potential g line will be x0 plus x1 plus x2 Okay, I hope it is clear in case not please message Final spring potential g will be half k times x2 square Because x2 is the extension in the spring now All right And then I need to quickly find out initial gravitation potential g which is zero because that's where you have assumed the gravitation potential g Because zero and initial spring potential g is half k times what? Half k into what square? x0 plus x1 square guys Okay, u2 g see I have assumed this line to be Zero gravitation potential g. This is a line Initially the block was in and right now block is compressed okay so This is initial Position now finally what should happen finally the spring should get extended by x2 right now It is compressed compressed by how much x0 plus x1 so it should move up by x0 plus x1 And then further go up by x2 So if you if it moves up by x0 plus x1 it will go to the natural length and then it should go to x2 further up Okay, so that's how you get gravitation potential g as half x0 plus x1 plus x2 Sorry mg into x0 plus x1 plus x2 that is gravitation potential g and spring potential g is nothing but half k into Whatever extension or compression square Is it clear now? So this is u1s Okay, and w is zero so just substitute This is u2 sum of gravitation potential g finally Plus spring potential g and this is u1 sum of gravitation and spring potential g And anyways Since we are finding the minimum compression Initial kinding edge will be zero and final also i'm taking it to be zero Is it clear all of you? It was not a very straightforward question, but i thought of taking it up. Is it clear? Kindly message yes or no if you have any doubts Any doubts? What break Niranjan? Are you not enjoying it? We'll see Ashutosh. We'll see all right. So let us uh I'll leave a little early Niranjan So let's not have a break Cut me Okay, so many times uh, you will encounter a scenario in which you know, uh Chain is there Okay But when it is a chain How you will take care of That particular scenario. It's it's a special case. Uh, how to deal with A problem in which the chains are there. Okay, what is chain? Chain we are considering a chain which has Distributed mass Along the length Your mass is distributed fine. So let's assume that you have this chain The mass can be uniformly distributed or non uniformly distributed. It depends on the scenario So let's say this chain has mass m and it has a length of l Okay, so Suppose this is the ground For which you are saying that gravitation potential g is Zero, okay By the way, the work energy theorem that we have derived is not only valid for the For the small masses. It is valid for the bigger masses also Fine like in the case of chain It is valid here as well But the problem which you will be facing here is to write potential g and kinetic energy for the chain Okay, because this entire chain is not located at a particular Location at at a particular point, right? So since it is not located at a particular point When you write mg h as potential g You will not be sure what h to take whether to take h from here There or there we are not sure about it, right? So that is the reason why we have we have taken the special case of chain here Okay So my question here is Find out the total gravitation potential energy of the chain mg is what you need to derive it Okay, no but we don't know what is center of mass Yes, you can assume the chain is just touching the ground the hint is Take a small mass Let's take this mass is dm and This Is dy this dy Is at a height of y okay, so Yes, mg h at no, but what m mass of what total chain no So potential g of dm is what see dm is a point mass. It's a very very small mass It has a fixed location. So you can write down the potential g of dm As du which is equal to dm G into y Okay, now the total potential g will be nothing but the integral of this now the problem with integral of y dm Or dm y is that both m and y Both are the variables Okay, so you need to write one in terms of others then only you can integrate Right, so the width of this dm is dy so total length l has a mass of m So dy will have what mass m by l which is mass per unit length Into dy Okay, so this is equal to dm So just substitute it here. You'll get g times m by l y dy now, uh, you can integrate you and mg l Integral of y dy will come and this you can integrate from zero to l that's how the y coordinate is changing So what you'll get here is that mg l by two This is what you'll get as potential g so in case You have uniformly charged Sorry, not charged uniformly distributed mass In the chain Okay, then all you have to do is to find out where The center of the chain is Okay Where the center of the chain is and find out what is the height of the center of the chain? Okay, and once you find out the location of the center of the chain, you just multiply that with m into g so m into g into height of the center Will give you the potential energy of the chain Fine, so we will learn uh the distributed mass in greater detail When we talk about the rigid body motion later on but right now you can use it as a thumb rule If the chain is uniformly distributed mass Then the potential energy is just mg into h center of mass Getting it any doubts Any doubt to write the potential energy of the chain? Kindly message All you have to do is to find out where the center of the chain is Then entire chain will behave as if it is a point mass Located at where the center of the chain is located Here we have also integration part So this is how we have integrated We have got dm g dy. Okay. We cannot integrate it right now because it has two variables Fine, so we first write dm in terms of dy so that there is only one variable y and then I can integrate It's changing y from 0 to l. Okay, so I'll get mg l by 2 It's fine. So you can also write this as mg into l by 2 and what is l by 2? l by 2 is the height of the center of the chain Fine, so the thumb rule is Thumb rule is write down if Chain is straight then mg Into height of center of the chain Will give you the potential energy of the chain Fine, you're assuming to be straight. Okay Any doubts? Yes or no, please message They go for length l Total mass is m Okay, so for length dy What is the mass? Parade length into dy Okay, that's how you have to apply getting it all right, so let us take up a Small numerical on the chain itself and see how we can apply what we have done till now is That we have defined the potential energy of the chain Okay, and that too for a special case where the chain is straight and if chain is not straight Then we have to follow the integration procedure to find the total uh potential energy of the chain Okay, for example, if chain is placed on a hemisphere If chain is placed on a hemisphere like this Okay, then you cannot just find out the center of the chain and just you know write potential energy accordingly What you have to do now is First you have to assume some zero potential energy Then take a small Length of the chain find out its potential energy and then integrate it throughout the length Okay, so that you learn when you take up questions like these. Okay, so let's not Learn it as a theory. So once you try out these questions message me then I can help you with this Okay, meanwhile, let us take a question Suppose you have a table like this on the table you have a chain placed with L by four length hanging This is l by four total length of the chain is l fine mass of the chain is m only Okay This is a smooth table. It's a smooth Both Fine Now what will happen the chain is will slowly slide down Okay, you need to find the velocity of the chain When the chain completely come out of the table find out Velocity of the chain when it Comes out completely anyone So if you see with respect to kinetic energy The chain is a straight forward case because entire change Chain if it is sliding will be moving with the same velocity So you can directly write down half m into v squared to be kinetic energy of the chain Where v is the velocity of the entire chain So with respect to kinetic energy There is no ambiguity. I have seen it others Let us assume this to be zero potential energy This is zero gravitation potential energy over here. Okay. So if you break the chain into two parts fine 3l by 4 part has no potential energy u is zero for 3l by 4 and for l by 4 potential energy is what initially minus of m by 4 Which is the mass of the hanging part g into center of the hanging part l by 8 Fine. So initial potential energy will be equal to The potential g of this part which is lying on the table Which is zero plus potential to the hanging part hanging parts mass is m by 4 Okay, and its center is lying below A distance of l by 8 from the Level where you have assumed to be zero potential energy Okay, so u1 will be equal to m gl By 32 This is u1 Okay, k1 is what inertia kinetic energy is zero Okay, u2 is what? u2 is when the entire chain comes down the table That is minus of m g l by 2 if entire chain comes down the table Then the center of mass will move down to a distance of l by 2 Fine and k2 you can assume it to be half m into p square All right, and w is anyway zero. There is no other force other than gravity Which is doing work for which you have considered the potential energy So you can say zero is equal to u2 plus k2 That is minus mgl by 2 Plus half m into v square Then minus u1 plus k1 mgl by 32 Okay, we just solve this equation you'll get the velocity Okay, so like this there will be uh, you know, I can understand there is a discomfort in taking up the problem related to chain So this you can only master when you, you know Try to look for the questions which involves the chain Are you getting it? so if you have any books Just look at the question that have That involves the chain in it. Okay, don't just count the number of questions every day number of questions what you have solved You should also be on a lookout of some of the unique type of questions fine, so try to Get hold of at least five or six questions related to chain and solve it yourself. Okay, and in case you have any doubts Feel free to get in touch Okay So we'll take a small break now. We'll take a couple of minutes break right now. It is 11 Right now It is 11 So we'll learn 50 We will meet at 12 Okay, is it fine? Okay So guys, let me tell you uh, we have done enough So that you can crack any question from this particular chapter now Whatever we are learning are different different scenarios like if the chain is there what we'll do if this is there what we'll do right so But when with with respect to the the concept, okay, we have learned everything So let us take up another scenario. This is a very important scenario This is called motion in a vertical circle All right write down all right now Do you have few examples of motion in a vertical circle? First of all, what is this vertical circle vertical circle is a circle Which is moving In a plane that is vertical Okay We define vertical with respect to accession due to gravity fine. So if this plane contains Accession due to gravity g then this plane is vertical Fine So naturally if there is an object that moves in a vertical circle Its potential g will change Isn't it If you say that this is the height Or this is the level where gravitation potential g is zero Then the height of the object is changing in a vertical circle Fine. So if it is here Let's say this is quarter of a circle. So this is R Yes, okay, that's correct So at this point if this is potential energy is zero So if this is a potential energy zero At this location potential energy will become equal to m g r Fine So you can see that when an object moves in a vertical circle Then the potential energy of the object keeps on changing Fine And if if the object moves in a horizontal circle fine So if it moves in a hundred circles since height is not changing the potential energy of the object is not changing Fine. So that makes the vertical circle motion a special one because here the potential energy also changes Okay So we can I know think of many examples Uh Quickly write down the typical examples. I'll talk about of course. There will be many others as well example one example one could be The string and mass Like this. So if you give this mass that's a mass m A velocity like this what will happen this mass Will move in a vertical circle Okay Guys don't discuss among yourself focus This is one of the example So it's a pendulum sort of thing fine So pendulum actually oscillates and here you just go in a vertical circle like this Okay, so this is example one There can be many other example. I'll quickly take few Example two could be you know a circular track Like this In which an object goes like that Fine. So this track may not be a complete circle fine, but then it is a part of a circle So when this mass moves It'll move in a circle like this Okay, then third example Could be a bead in A circular wire There could be a bead like this that can move in a circular wire fine, so Example four could be that there is a tubular passage Okay in which The mass moves Getting it. So all these are typical examples of All these are typical examples of the Motion in vertical circle. Okay Just give me one second As a meanwhile, you can please Try out this particular question I'll take two minutes You have a mass. Okay. This mass is Uh, you know down Vertically down like this hanging on a length l You have given this mass A velocity Which is equal to under root of 5 gl You need to find out its velocity when the mass becomes horizontal Okay Do it. I'll be back in two So this is position one and that is position two All we have to do here is to apply work energy theorem. So Is the tension force doing any work? Tension force is not doing any work because tension is always perpendicular to the direction of motion Okay, so this perpendicular to the motion Hence work done by tension is zero and the only force that is doing work here is gravity and for that You have considered the potential energy. So w will come out to be zero This is equal to u2 If you assume this line to be zero gravitation potential g So u2 will be equal to mg into r k2 will be equal to half mv square minus u1 will be zero Plus k1 will be equal to half m into this velocity square Okay, let's Call it as v1. So this is half mv1 square. So when you substitute it, you'll get the answer Okay fine, so what I was talking about here is That in a vertical circle, ultimately it is what a circular motion whether it is Whether it is horizontal circle or a vertical circle Ultimately, it is a circular motion only and if it is a circular motion Then at every moment, there will be a centripetal acceleration of magnitude v square by r fine so at every moment It will have an acceleration towards the center, which is v square by r fine So it must have a force that will create this much acceleration Fine. So towards the center, there will be a force called centripetal force that will be acting I think we have already learned this in laws of motion chapter I'm just reiterating that make sure you are Taking care of this fact that it is moving in a circle And since it is moving in a circle, there will be a centripetal acceleration of magnitude v square by r towards the center Fine. And since there is a centripetal acceleration towards the center, there has to be a force towards the center Getting it Right Now keeping all this knowledge, let's try to Solve this particular question See this is not a part of theory. Okay. I'm just taking a scenario and Trying to Tell you how you can Tackle this scenario. Suppose this is Mass m It is hanging. Okay length of the thread is l. All right Now if you give it a small velocity, what will happen if you give it a very small velocity it will Move like this slightly this way Okay, then it will come back That way fine, so it will oscillate Left and right Getting it So if you give it a very small velocity, it will just oscillate Okay, now if I give it slightly more velocity It will Be reaching horizontal level and it can swing one horizontal level to the other horizontal level Okay, now if I give it slightly more velocity It will reach there Okay Now it will reach there and suppose Tension becomes zero here Then what will happen if this tension In the string at that location becomes zero then what will happen at that moment if tension is zero What will happen to the motion of the object? Will it move in a circular motion? so if t is zero There is Absence of necessary centripetal force For it to move in a circle Fine, so if tension becomes zero at that location Then it will not move in a circle fine. So it will be as if an object is thrown With velocity v So it will become a projectile So it will move in a projectile way In and it will not be a circular motion They're getting it so from here to there From initial point to the point where tension becomes zero It moves in a circle and then as soon as tension becomes zero it becomes a projectile motion It no longer moves in a circle fine So the necessary condition for this object To move in a circle is what tension should not become equal to zero Fine. Now tell me where is the location where the value of tension Most likely becomes zero Most likely where the tension of this mass Or of this string Can become zero highest point or top most point So at top most point Tension is Most likely to become zero Fine So if T is not zero at top most point Then can I say that T will never be becoming zero T will not be zero Anywhere is this statement correct? If tension is not zero at the highest point, then can I say that tension Will not become zero anywhere So if tension is not zero anywhere, it means that the string is not slack anywhere String is always top And if string is top, there is no motion along the radial direction. It is always tangential direction So I'll say that if tension is not zero anywhere, then The object will complete Full circle Fine if tension is not zero then the object will be able to complete the full circle The mass will be Just completing write down The full circle if tension Tending towards zero At the top most point Getting it So this mass If tension tends towards zero because tension will keep on decreasing decreasing And it will become minimum at the top most point. So if minimum tension itself is not zero Tension will never become zero Fine So the limiting condition for which it will just complete the full circle is the tension at the top most point Is zero Okay Now listen to the question any doubts till now kindly message. Yes or no you have any doubts kindly message Okay, this is the last question for today Then uh, whatever is left over will take another class to finish this chapter. All of you should get the answer. Okay The question is like this You have mass m and length l Okay So you need to find out what will be the minimum velocity What should be the minimum velocity you so that this mass completes the full Circle or full vertical circle try to do it Clear about the question, right? I'll write here. You have to find minimum velocity So that the mass completes full circle Not correct Niranjan. Don't be in a hurry Okay, answering quickly doesn't impress me. You know that I was joking See, I know the uh, the The thing is that if tension is just becoming zero at the top most point Then it will be able to complete the circle. Fine. So let's try to balance out the forces here So let's assume that at the top most point its velocity is v Okay, so Down there will be a gravity force. Let's say mg is that force and there will be a tension force like this fine And there will be a centimeter acceleration of magnitude v square by r which is l Radius is l. Okay. So if I write down equation there, I'll get t plus mg Is equal to mass time acceleration Fine, so if tension is just becoming zero at the top most point, then what will I get? I'll get mg is equal to m v square by l fine, so m v square I will get this as equal to mg l fine, so let's Uh, I'll quickly write this as v square equals to gl but This is what this is not the initial velocity. This is a final velocity. Okay, so if final velocity square is equal to gl okay, then Tension will become zero at the top most point. This is a minimum required final velocity that this mass should have Okay, so minimum required final velocity will give you minimum required initial velocity Okay, now I am going to write down work energy theorem between this point and that point So w is equal to In case you have any doubt, please keep messaging u 2 plus k 2 minus u 1 plus k 1 Now is there any force other than gravity that is doing work here? You can see tension is a force but tension is always perpendicular to the motion of the object Motion of the object is tangent to the circle and tension is normal to the circle fine, so work done will be zero by the tension Okay, and gravity is doing the work, but for gravity you are anyway considering potential energy So work done will be zero u 2 is what there is only gravitation potential energy. There is no spring potential energy So if I assume this level to be zero potential energy Then u 2 will be equal to what u 2 will be equal to mg into 2l because the Height of this object second point is 2l now l and another l Okay, so mg l plus k 2 k 2 is what half m into v square minus u 1 is initial potential energy, which is zero plus initial kinetic energy half m u square Okay, so all you have to do now is to substitute the values you will get mg 12 plus half m into v square v square is gl finds this minus half m u square will be equal to zero So u will come out to be under root of 5 gl So this is the minimum required velocity for The mass at the lower most point to complete the full circle Okay, when it is tied with a string, okay This is the condition which you should remember because at times you can use it directly in problem solving Okay, so that's it for today and there There are a few topics still left in this chapter and a lot of problem practice We need to do In this chapter to master this Okay, so at least 200 to 300 questions you have to solve your own to master this topic Okay, and let me tell you that is not something which i'm asking And nobody else is doing everybody who is serious Is practicing lots and lots of questions. Okay, so the homework Nobody has asked me any doubts from the previous homework Okay See guys, you know, I I can only tell you from my experience. Okay, and I don't want you to make Errors because you will not be able to learn from your errors and implement it to correct yourself because You'll get only one chance, right if you make error in that chance itself It the you'll not get a second chance Okay, so make sure you do what I'm saying All right Follow exactly what I say nothing less nothing more and you'll be through And i'm not asking something which is like beyond As in nobody else is doing everybody is working equally hard those who are serious Anyways You need to solve entire hcv on Work power energy Okay, this is for j mains only if you're serious about j mains Solve entire hcv on work for energy by next week Okay, and if you are aiming for j advanced then I'll be sending a worksheet you message me Then i'll be sending you Once you're done with this okay j advance is not like you should not do this and then you start doing the advanced level questions No, that's not how you prepare for j advance. You have to Prepare from the basic level up Okay, so once you're done with this message me I'll send you a worksheet of around 100 to 200 question for j advanced Okay j advanced is this Plus a worksheet that I'll be sending you once you ping me that you're done with this All right Okay, and those who are preparing for kvpy Those who are preparing for kvpy tomorrow will have a class 4 30 p.m 27 30 p.m Okay, I'm going to talk about uh, you know system of particles Okay, this is the start of rigid body motion and that we we are anyway doing it After work by energy chapter. Okay, so you don't I mean you're not missing anything if you're not attending the kvpy lecture Okay, so this will be anyway doing towards a later part of our syllabus. Okay, because kvpy is happening in November So I have to hurry up with respect to this particular topic. Fine. So those who are attempting kvpy only for those fine Please attend this class 4 30 p.m to 7 30 p.m tomorrow Okay, so that's it for today. I hope you have learned many things today go back and Straight away jump on to the problem solving and finish off the homework as soon as possible and I would love to hear At least some of you finishing the homework in next two days itself. Okay Thank you very much Yes, Ashutosh. That's in the jc levels But that we are anyway doing later on so don't worry that if you miss Yes, it is online session. Thank you very much guys. Bye. Bye