 Let us move on to our new topic and the final topic of the day which is minimum potential energy. So what we have done in the morning is we had discussed the principle of virtual work. Now virtual work also is an energy method when we say energy method what it means is that we do not deal with simply the force but we deal with work of the force. Now what we had seen earlier that suppose we take this particle and take it along this trajectory from O up to A2. Now what we know is that when we take this particle in an infinite symbol a very small displacement ds then the work done at that position will simply be given by f dot dr which is f ds cos alpha we had seen that in the morning. But now if I want to find out that what is the total work that is done on this particle to move it from O up to A2 then what we have to do is we have to do the integral of this quantity f cos alpha dx ok. This f may or may not depend on ds so that is why it is in the integral and you do this full quantity and what we say is that similarly for a couple du is equal to md theta. So if you take a particle from state 1 to state 2 then the total work done will be m times theta 2 minus theta 1. So now in the morning session we were talking about infinite symbol work and when I say infinite symbol work it just means that you give a small displacement what is the work done and now what we are talking about is you take this particle along a path or you give a body of finite rotation and you are asked that if you are giving a body of finite rotation with a torque or moment m what is the total work that is done on the body. Similarly if you take a body or take a particle move it from point 1 to point 2 by applying a force air what is the effective work done. So this is the effective work done and this is the effective work done for moment this is effective work done for force ok. Now to take simple examples these are elementary examples which we know from our high school. Suppose we have a trajectory where we take this mass and we take it from point 1 up to point 2 ok what we know is that that we are doing some work against gravity. So gravity on this particle acts downwards if I take the particle along this trajectory ok then from here to here ok a small motion of delta y what we will do we will do a work of minus dy what is this minus dy minus wdy is the work done by gravity because gravity is acting downwards this particle is moved upwards the work done will be minus wdy and this point that I want to emphasize that this is the work done by the gravity ok minus wdy. So what you say is that if I take this particle from 1 to 2 what is the total work done by gravity in this overall moment and the total work done will be simply this integral of wdy w is constant for the body and the total work done will be minus w delta y where delta y is simply y2 minus y1. So what this mean is that to take this body from 1 to 2 gravity does a work of minus wdy for obvious reason that gravity acts downward the particle has gone upward so gravity did a work which was negative. Let us look at the second example ok so you have a particle or a body which is attached to a spring which is in this undeformed configuration the stiffness of the spring is k now if you displace this spring ok by some distance x ok the force acting in the spring ok will be kx which will act in this direction. So the question we ask is that what is the work that this spring does if you pull it from this point ok which is coordinate which is x1 up to this stretched point which has a coordinate of x2. Now what we realize in this case is that unlike in gravity where the force was always constant in this case the resisting force F which is coming from the spring is not always constant but is proportional to the stretching which is kx and a work done will be kx times dx minus sin y because dx is in the direction opposite to the force exerted by the spring. So what is happening here is that the internal spring is causing a resistance to the outward motion and in turn is causing a is doing a work given by minus kx dx if you want to find out what is the total work done by the spring if you stretch this spring from x1 to x2 you just do this integral and what you will see is that the total work done by spring is kx1 square minus kx2 square divided by 2 so you have this factor of half coming into picture and this is x1 square x2 square ok simple example. Now what we do is that what we saw in the last slide or in the last couple of slides ok is what is the work done by a force we saw that we take a body we move it from below we take it upwards what is the work done by the gravity in the spring we take a spring spring at rest keep stretching it what is the work done by the spring in the stretch from x1 to x2 we saw that it is equal to half kx1 square minus half kx2 square ok. So there is a minus here because the work was done ok against the direction in which the force either from the gravity or from the spring was acting now we define this quantity which we call as a potential energy of the body with respect to any force. So we say that the work done from taking from taking the particle from 1 to 2 we define it is given by the potential energy of the particle at point 1 minus the potential energy of the particle at point 2 ok. So what do we get here is that that since u12 is negative you will see that this quantity potential energy at 1 of the particle will be less than the potential energy of the particle at point 2. So what we will see is that that to take this particle from 1 to 2 the work done against gravity was equal to minus w into delta y but by definition we are saying that by doing this work there is certain amount of potential energy stored in the particle which is negative of the work that was done by gravity on the particle which is simply equal to in this case w times delta y. So if you take the particle take it down and move up the potential energy now that is stored in the particle with respect to this datum ok with this point is simply w delta y. Now somebody will ask ok that w delta y is the potential energy stored in the particle why are we taking that to be equal to positive why this convention that the work was coming to be negative why are we saying that we leave the particle up its potential energy increases. This is from the simple observation ok that from the conservation of energy we see that if you take the particle up and drop it it will come down and gain some kinetic energy at the bottom from where we had left. So what does that mean that by taking the particle up the body had capacity to do some extra work or it has the capacity to generate energy and that energy now from potential it gets converted to kinetic energy. So we do work which is negative against the gravity but that gets stored or that is reflected as the potential energy of the particle. Similarly the potential energy stored in the spring will be exactly opposite ok the work done will be equal to half kx1 square minus half kx2 square and the potential energy that is stored in the spring will simply be equal to half kx square with respect to certain datum ok we will come to this point a little bit later ok. So what we again define here is that when differential work against the force for example either gravity or spring was done the work done ok is equal to minus the potential energy stored even for a spring for example we keep on pulling on a spring ok there is a body that is tied at the end ok spring is doing negative work but that negative work done by the spring is being stored as the positive potential energy of the spring. So when we release the mass we see that there is a rebound what is that rebound it is the potential energy stored in that spring that is now acting on the body and it is converting it into kinetic energy ok. So is the convention that what is the work done by force and what is the corresponding potential energy is that point clear fine ok that is a clear point. So what we do is that the convention is that the work done on a body is negative of change in the potential energy of the body now the point is this it is very important to choose an appropriate datum and we always refer to the potential energy consistently with respect to that datum. For example if I want to find out what is the potential energy of the particle at this point now it depends that what is our datum if I say that this is my datum that this is my reference level then by degree you say that the potential energy is 0 because you can define whatever potential energy you want to. So we say that potential energy is 0 here and with respect to this the potential energy at point 2 will be equal to simply W times y2 with respect to this datum the potential energy of the particle at this position is simply equal to W times y1. If you decide that I want to take the datum or the reference level at this point then we know that with respect to this reference the potential energy at this point is 0 and with respect to this level the potential energy at this point is simply W times delta y. This point will become more clear as we proceed further and solve a few problems and these kind of forces it will become very clear for example if you teach structural dynamics later on or teach some courses in solid mechanics where we have to discuss about energy the forces for which work can be calculated from a change in potential energy are called as conservative forces. For example damping and friction are non conservative forces because they cannot define a potential with respect to which you say that the force can be obtained or but whenever we can define a force in terms of gradient of some kind of potential energy we say that those forces are conservative and we don't need to go into details if you're interested you can ask me afterwards but just for clarity I'm just for completeness I'm emphasizing this point here. A simple illustration now. An illustration is this that suppose we have a spring which is at rest here that the spring where it is unstretched is at this location to the end of the spring we put a mass and what we are asking is that what is the total potential energy of the system is a simple question we ask. Now this coordinate X okay what will happen here is that when you apply some put some mass here initially the spring was rest at this level let us say that this was our datum level that at this datum the potential energy of this entire system was 0. Now when you apply a mass and the mass gets lowered by a distance of X there is a stretch of X in the spring so the potential energy in the spring now is half Kx square but what is the potential of the energies of the mass we saw a couple of slides ago that the with respect to this datum the potential energy is positive w y but here with respect to this datum you are actually going down so the potential energy will be minus mg into X you see that point okay why because if you go from here to here work done by gravity will be positive equal to mg X so the potential will be minus mg X so the total potential energy of this system is half Kx square minus mg X now you ask me okay we got this potential energy okay so clap clap now what okay the idea is that this is a very simple problem but we can write down that as a function of coordinate X how is X defined rest position any distance X below that is the way we have defined the coordinate X so this is the overall potential energy now can you tell me how can we put that to any reasonable use now that X can even become positive if desired but okay we got this potential energy now can you find any good use for this potential energy can we learn something by knowing that this is the potential energy of the system now in this case for example there is no question about any force force is gravity it is getting stored in the potential it is getting stored in the spring so gravity is doing somewhere getting stored as a potential energy of the spring but the point is that can we know something about when the system will be in equilibrium from knowing this energy minimum potential energy so the way we use this okay and it can be shown okay there is a detailed derivation in beer and Johnston Mary M. Kray if you want to have a look at it you can have a look but it can be shown that if you can write down the potential energy of a system in terms of some displacement parameter in this case which is X then the X where the potential energy is minimum or let me put it this way where the potential energy is extremum what is extremum in calculus language is if you draw the potential energy as a function of X as we have drawn there then the slope of that curve wherever it becomes zero is called as an extremum special condition is is a minima and you will see that in this problem this blue curve is the spring energy this magenta curve is the potential energy minus mg x coming from the gravity and a combination of this will give you a curve which look like this and this point where the slope of this effective energy curve is zero it is at that point where you will see that the body will come to equilibrium now how do we get that point if we take du by dx equal to zero what do we get kx minus mg equal to zero so x equal to mg by k you said good that's what we expect from Newton's law that mg is the way that should balance the force coming from the spring kx equal to mg so but why this simple example what we could see is that Newton's law okay which says that all the forces should be balanced for a particle okay is perfectly valid for this conservative system where we can define some kind of potential energy there is no dissipation in the system and you say that the minima or the extreme of this potential energy is the point where the system will come to some kind of equilibrium okay so this in short is a principle of minimum potential energy that if you can write down the overall potential energy of the system in terms of any of its kinematic coordinates be it rotation be it displacement be it a combination of this and if we can extremize the energy with respect to those coordinates then the solving those resulting equation of resulting equations of equilibrium whatever values we get are the positions are the coordinates where your system will be in some kind of equilibrium now you will ask me okay system is in some kind of equilibrium okay good but what is the benefit we did virtual work principle we did simple Newton's laws or movement balance for movement balance force balance what is the big deal about this what you will see is that there are certain category of problem and there is one problem which I will like really demonstrate that with in which cases principle of potential energy is extremely useful okay that it is very very difficult in many complicated problems to write down the to write down all the free body diagrams especially if all these problems are like for example there are many springs involved okay where there can be huge amount of deformations in the system till it reaches some equilibrium configurations so especially in those problems and many of those problems we are going to solve this principle of minimum potential energy becomes very good second type of problems where it becomes very good is I had just discussed a few moments ago that we just need to make that d u by dx or d u by d theta whatever the kinematic coordinate is should be equal to 0 but it is only an extrema okay Newton's law will only give you where d u by dx equal to 0 which means for example wherever the the overall force acting on the system or overall torque acting on any free body diagrams or part of the full free body diagram is 0 but what it doesn't tell you that if you displace the system from that position a tiny perturbation then will the system remain stable or not stable and that kind of considerations can be easily be done by using potential energy approach that where we can also say that not only if the system is in equilibrium but also we can say if that equilibrium is indeed a stable equilibrium unstable equilibrium or a neutral equilibrium so that is the reason why it is really good to use to learn minimum potential energy approach another point is for example in the morning I had said that there are many advanced techniques in solid mechanics in structural mechanics in finite element method or approximate methods where Newton's law using Newton's law does not really help us okay we need to resort to some kind of energy principles in order to obtain say forces or moments in statistical indeterminate systems or if you want to solve a problem approximately get an approximate solution of a problem then Newton's law don't help you but minimum potential energy principle as well as virtual work principle okay we can use them to our advantage in order to get approximate solutions of various problems okay those are topics of those are advanced topics okay which which is not that not the subject of this course okay so far so good that this is the minimum potential energy so wherever you can write down u in terms of some kinematic coordinate okay now this kinematic coordinate did not be a small coordinate this x need not be small x can be finite okay x can be finite and then we can write down that what is the overall potential in the system minima not minima the extremum of this where du by dx become 0 is the point of equilibrium that is the principle of minimum potential energy now let us do this simple example okay what we have here is this that for the structure what we know is that the unstretched length of the spring is da which means what is this system this this spring is fixed here is attached to the roller at point b this there is a hinge here to which you have two rods okay pinned at a pinned at c and then correspondingly these two rod meet at c and what we are asked in this problem is if we tie a weight w at point c then what will be the resulting configuration okay or what will be the resulting angle if we define this angle as theta then what is the resulting angle theta at equilibrium is what we are asked to solve in this problem the simplest thing to do is let us write down the potential energy for this system what we know is that that when theta is equal to 0 rod ac will be vertical rod bc will be vertical this length will be 0 so that is when the spring is unstretched so what does that mean that this length AB is nothing but the stretch x in the spring so the potential energy of the spring is nothing but half K into AB square but what we know is that AB is nothing but 2l sin theta this is one part is l sin theta other part is l sin theta so it is 2l sin theta further what we realize is this that if we say that from with respect to point a what is the vertical coordinate of w what is that equal to it will be equal to l cos theta minus the length of the string l cos theta will be the coordinate of point c vertical coordinate of point c so with respect to this datum l cos theta is the vertical coordinate of point c minus the length of the spring this this string not spring string is w but that string is a constant length when we differentiate it will not matter so we say that that constant will left out so essentially we are saying that v is equal to half K 2l sin theta square plus w l cos theta so far so good so we could write down in the previous problem which we had shown the kinematic coordinate was x in this case it is theta which is the kinematic coordinate and what we want to know is at what theta or at what thetas will the system be in equilibrium you can solve this problem using Newton's law it will be more painful this will be much easier you try it using moment balance force balance not extremely painful but somewhat this is much more transparent in this case now you tell me is this is this point clear so far yes somebody has a question yes please please w l cos theta I think this is not right why it is not right can you tell me the displacement is l cos theta l cos theta will be the displacement what is l cos theta you tell me if if you take this bottom line as the datum as your datum plane okay with respect to that what is the height of w look at that if you replace that that w k what is the height with respect to the base plane this is you tell me it will be w cos theta minus the length of that string yes or minus the length of the string is a constant I can put that length but ultimately when I differentiate with respect to theta that length will go away this will go out in del u it will be there in w l cos theta minus length okay but then let me do this thing whatever length of the string is I will say that I will take a datum l below so then it will also go away because you have the full freedom to choose what is the base with respect to which you can define your potential energy okay so you just take the datum l below so that minus l plus l now because you're already energy is l plus w l cos theta minus the top l then it is just this so potential energy there is no absolute value that's what I'm saying there will be always some constant which will depend on where your datum or the base plane is is the point clear okay is are there any other question okay it was a very valid point but note that there is no absolute value of potential energy there will always be some constant which will depend on where you take your datum yes please sir it is let us assume it is initially at C what there is no string there is no string okay just the weight is let it be at C yes initially it will be when this theta is 0 it will be at the height l yes and when the it is stressed it will be at height it will go little bit below so if you want to then so change in height will be l minus l cos theta let's do it okay let's do it now let us take what you are saying is that I will keep datum on the top is what you are saying that you take top as the reference level yes but now the body is going down okay with respect to top it is going down so potential energy will be minus mg l times 1 minus cos theta yes so that there will be a constant of minus mg l but but minus mg l into minus cos theta again plus mg l cos theta so you will get this plus another negative constant but that negative constant is just because there are fact that you are kept datum upwards okay so it doesn't matter so you can choose one datum and be consistent whatever you do your only thing that you will get is an additional constant here which is of no consequence because ultimately your potential is shift up or down that doesn't change the feature is that clear point so you can do that also you will get a same answer any other question yeah okay so let's move on then what do we know that this is the overall potential energy of the system which is a combination of V e or the is the elastic potential energy and V g which is the gravitational potential energy and now to find out which is the extrema okay let us not say minimal let us say it's an extrema we said d V by d theta equal to 0 you will get an equation of this form now clearly what do we know one solution will be theta equal to 0 okay now is theta equal to 0 does it make sense from our from our Newtonian thinking or from our force or moment balance thinking theta equal to 0 means both of them are perfectly vertical does that make intuitive sense to us yes or no yes because everything is vertical no force in the spring it's like you keep it is like keeping something hinged about it okay a rod hinged about a bottom point and we are just like having a weight which is passing through the bottom point okay so no force no torque you draw any free body diagram you will see that the system is fine it's in equilibrium okay so theta is equal to 0 actually makes sense to us only thing what we realize is that that that is not a stable equilibrium okay what we will realize when we go for stability analysis now second solution will be when 4kl cos theta minus w is equal to 0 now that solution will give you what cos theta is equal to w by 4kl and you know that w by 4kl only that quantity is less than or equal to 1 will there be a second solution or else there is no second solution to this problem point clear okay because if you have w by 4kl k is extremely small okay or l is extremely small okay then this quantity can become more than one there is no second solution the only possible solution equilibrium is the vertical solution where theta is equal to 0 and you saw you try doing this problem with simple Newton's method is not impossible but it will not come out to be as clean as we have shown here any doubt about this is the entire procedure if you can just explain this expression we just with the science please this yeah last last line previous line previous line previous line to that yeah this one yes okay so half k this is positive that I am not getting really this positive okay this potential at w is weight what we have done is this that we have taken datum at the bottom if you take datum at the bottom then think about it if you take a mass from below take it up to distance yc what is the work done by gravity minus minus w yc and so the potential energy as we have discussed is negative of the potential energy that's why the plus so if you are in doubt what you do is that forget about everything else just that force or that mass you say if I take some particle where the force is acting go all the way up to yc what is the work done that work done come minus is what you add there okay so if you have any doubt like for example if it is a force rather than having a weight if it were a force you say that if this force were here and I take that party there is some particle along with the force you just take it upwards work done is how much minus that force into distance so the potential energy is plus f into that distance with respect to that datum if I have it here the datum is at the top okay what happens you take this work is done with respect to this so plus w into that distance so potential will have minus w into that distance and you will see that no matter what datum you take if you follow this convention properly you will get the same v up to a constant okay is this point clear any questions any doubts okay these are very important points that for example because sign convention only simple rule is that forget about anything else when you want to write down the potential energy of a force just take that go wherever you want to find out work done potential is minus of that simple now let us discuss that the stability the concept which we do not discuss strictly in all this to equilibrium into the equilibrium or even in virtual work we do not do any discussion about stability of equilibrium but what we say see now is this let us look at this very very simple example let us take an inverted rod which is pinned at a it's a one degree of freedom system that if you rotate the rod by theta the entire position of the rod is automatically known so it's a one degree of freedom problem now what we say is that if there is gravity acting downwards if you plot what is the energy of the body of this rod as a function of theta you will see that at this position when theta equal to zero the energy is a minima what is this absolute value will depend on what is a datum that you are using okay that absolute value depends on like if your datum is this or this or somewhere in middle but what you will see is that the minima will come at theta equal to zero and the energy always increase now think about it if for example we have an inverted cup like this we keep a tiny marble somewhere what will it do it will try to go and settle at the bottom okay so what it says is that at wherever there is an energy minima the system that that that marble or a small particle is that degree of freedom it will try and go and settle at the bottom forget about the oscillations okay we don't worry about the oscillations but it will try to settle at the bottom now the question is this suppose you have a particle settled at the bottom you want to know that it's in equilibrium why do you know it's in equilibrium first of all the energy so slope is zero for the energy even thinking from Newtonian point of view just look here that the force acting is gravity okay which is passing through the hinge so this entire assembly this this rod is in equilibrium at theta equal to zero but we want to know that if that equilibrium is a stable equilibrium or it's an unstable equilibrium now how do you know that we put up support for example we have a minima like this we have a small marble stuck at the bottom we put up what happens if it's a minima like this the marble will try to come back again right it will try to come back again so whenever a system when put up from its equilibrium position tries to get back to that position we say that that position is a stable equilibrium okay it's a stable equilibrium and what is the mathematical definition for this the mathematical definition is that that if this is the potential energy v this is theta then what does that mean that if you look here slope is negative slowly becoming positive positive positive increasing what does that mean that a slope is increasing and slope is increasing means that d times the d by d theta of the slope should be greater than zero which in other words says that d2 v by d theta square which is the curvature should be positive so curvature at that point being positive okay is a sign that it says stable equilibrium for small perturbation the equilibrium position is that point clear that when the curvature at that point for small perturbations is positive then that equilibrium position is a stable equilibrium position straight forward any question any doubt about that yes please oh it should be negative sorry this should be negative okay my error yes I will come to that okay I will come to that at least small large is again relative I will come to that okay in the next slide I will come to that it's a very good point because during equilibrium we are telling the differential change will not change the energy not change the energy but what is that but to the now we have to think out to the first order to the first order to the first order means what if you if you have done Taylor series expansion what you will see is that that expansion of a function about some point will be equal to v0 plus dv by dx dx plus half d2 v by dx square energy minima just tells you that to the first order dv by dx equal to 0 means the energy is stationary or energy does not change to the first order but this what it is telling you is that to the second order the energy is actually increasing if you put up so that's the order means for example if you do a Taylor series expansion for v that you will see that if let think about it like this that locally it looks flat but there is an additional correction which is given by this so the first correction is dv by d theta equal to 0 the second correction is that this is dv by d theta square into delta theta square physically a small change what it means is this okay let me do it this way if you remember okay let us do this thing here if you recall what we had done in the morning okay that this is the rod AB suppose we give a small rotation to this rod what happens is this a small rotation which is given by delta theta what we know is that the horizontal displacement okay is how much really speaking if this length is L then really speaking delta H is equal to L sin theta okay sin delta theta you agree with me that is really the true answer true expression but what we say is that if delta theta is small what is sin theta approximately theta so we say that this is equal to approximately sin delta theta and when did sorry approximately equal to L delta theta and when delta theta becomes smaller and smaller this becomes more and more accurate now what is the vertical displacement of this point it is equal to L 1 minus cos theta or cos of delta theta you agree with me if I do it if I rewrite this expression what does this become 2L sin square delta theta by 2 now what when sin delta theta is small can't I rewrite it approximately as L by 2 delta theta square okay now note one thing so what we are saying is that that delta H to the leading order is L delta theta and delta y to the leading order is L delta theta square but what we say is that that because delta theta is small I can say that this quantity is much larger compared to this quantity because when delta to the small delta theta square become even smaller so we say that to the first order the only dominant displacement to the first order why first order because delta theta to the power 1 that is first order so to the first order we say that my displacement is only horizontal and then we say that to the second order the displacement of the vertical point is delta theta square there are further quantity because cos sin square delta theta is not just equal to delta theta square by 4 there are many other quantities in terms of delta if you do the Taylor's expansion you will see but to the leading order okay which is 2 I say that this is my displacement now similarly if for example you have a potential energy okay where the minimize happening at x naught so what you say is that that if I give a tiny perturbation of delta x then this can be written as v at x naught plus dv by dx at x naught into delta x plus half d to v by dx square delta x square plus terms which are delta x cube delta x 4 and so on what we know that at equilibrium this quantity becomes 0 so to the first order we say that the system potential energy stationary but to the second order there is an extra correction and what this quantity is telling you that if you perturb about this if this quantity is positive then your energy is actually going to increase in a second order sense okay so think about it okay so this is mathematical so that's what it means so it means that to the leading order and the geometrical meaning is like this that to the linear order the energy stationary to the second order okay is when you want to do the energy expansion and this you will see that when so to the leading to the first order energy stationary but when you go to higher order you will see that it is a curvature that will decide that is the energy increasing or not increasing so d to v by d theta square greater than 0 means that a tiny perturbation is going to increase the energy which means that it will want to come back so it's stable equilibrium and the second quantity okay I made a mistake here it should be d to v by d theta square less than 0 which means that the shape of the minima or the extrema looks like this that a tiny perturbation from this will actually reduce your energy so there is no incentive for the for the system to come back here it will want to reduce its energy further and it will just roll down here for example if you take an in system like this you put a particle it may precariously be placed here but the moment you put up it will just roll down why because the energy is decreasing further and that's what the particle want to do it's fine and the last case here and that's the case which is shown here is the case of an inverted pendulum you see that for example if you want to balance a pain like this okay so it is so much more difficult why because even though if I exactly balance the pen okay the line of action and the reaction will exactly pass through each other and it is an equilibrium position a tiny perturbation will just make it fall down okay this exactly what is happening here at any tiny perturbation will let the system go away from equilibrium so it will not come back to that position and ultimately neutral equilibrium is this position that if you pin it at the center then the energy just does not change no matter what theta you put energy is constant and in that case you say that no matter where you move the system doesn't have a particular preference so is this is this classification overall clear to you here you are okay with this any questions any queries fine now let me give you a typical for any complex system okay a typical energy landscape may look like this so what you can have you have a saddle point what's a saddle point saddle point is a point where d2v by d theta square or is an inflection point saddle is strictly speaking in two dimensions that at this point the second derivative of v or u the energy with respect to theta will be equal to 0 it's a minima here what's a minima it's a stable point where the curvature or d2v by d2u by d theta square is positive you can have a maxima you can have a minima and when I said small perturbations the reason is that suppose your particle is in minima if I give it a tiny perturbation here it will roll back but if I give it a huge perturbation then it will just jump over the maximum and go here okay and this is a very important problem in so called optimization where the system can get trapped into local minima where you actually want to get a global minima okay is this point fair last slide last part the neutral equilibrium just think about it in this problem suppose the rod is uniform the center of gravity is at the center okay now if you pin the rod at the center just think about it you perturb it anywhere the center of gravity does not go anywhere doesn't change it's not a function of theta then as far as that parameter theta is concerned the energy is independent of theta which mean that any position you put the system has no preference one way or the other in Newtonian language if you take the free body diagram of this no matter what orientation you give system has no preference to change okay that's what it means and it also means that everything is flat the energy landscape is completely flat and what it will also mean in some very crude terms that if this inflection point is reasonably wide then it is another way of saying that the inflection point is a point of neutral equilibrium here actually I should not use the saddle point should point of inflection saddle point tool is for two dimensions okay so this point is where your second derivative of energy with respect to theta is actually 0 and what happens is that think about it like this in neutral equilibrium all derivatives of energy with respect to theta are 0 all so you can say that if more and more derivatives with respect to theta are 0 you say that the landscape is flatter and flatter there and then you can say that the saddle point or the inflection point can be taken to be a neutral equilibrium point if it is a reasonably flat area okay but it is not a hard and fossil there are some many complications let us not go into those details but saddle point is a point where locally it is flat so one side there is a maxima if you see here if I put a bit on one side the energy will increase so the particle would want to come here but I put a bit on the other side it will just roll down whereas in a minima you put a bit either here or there the particle would still want to come back and in a maxima you put a bit either way the particle still want to go out the difference between these this and this is that here one way it's stable other way it's not stable okay we do a few problems and then these things will become more clear it 2d for example we don't need to go into this 2d it will look like this so let us look at a sample problem so what we have here okay is a big a big pulley a big wheel okay which for example there has a small inner wheel which is rigidly attached okay so this is not a 2 degree of freedom system there is a small wheel the small pulley which is rigidly attached okay to this big wheel and the entire assembly is pinned about this point O okay so this can rotate like this about point O now on the top of this okay there is a mass a weight of which is given the radius of the outer pulley is given the radius of the inner pulley is given and what we are asked is that that a spring B is unstretched when theta equal to 0 which means that when a is vertically up here the spring a is unstretched any subsequent motion if you go in this direction what will happen the spring will get compressed if you turn clockwise spring will have an extension and what we are asked to find out that what are the positions of equilibrium and do we know that if those positions are stable or unstable positions of equilibrium is the question so let us write down what is the potential energy the idea is that that the potential energy now has two components one is the potential energy of the spring other is the potential energy of this mass coming from gravity so it is Ve is for spring elastic Vg is for gravity for this top mass and what we want to know is equilibrium will come when dv by d theta equal to 0 that will give us some values of theta where the system is in equilibrium and then what we want to do is that for those values of theta what is the value of d2v by d theta square is it greater than 0 or is it less than 0 okay so let us down what is the overall potential energy now let us say that a spring gets extended by distance s so the so the potential energy of the spring will be half ks square okay and potential energy of g will be mg y okay how do you decide this y let us take the point O okay let us take this point this point as the datum or the reference point and we will measure our potential energy with respect to this base point now note one thing that if you rotate this by an amount delta theta by by an amount theta what will be extension of the spring just b times theta b is the inner radius so b times this theta will be simply the extension of this spring is b or a sorry a a times theta a is the inner radius will be the extension of the spring with respect to this datum what is the height of this point it is nothing but b cos theta so what we know is that a total potential energy is half k a theta square is the inner point plus mg which is the weight times b cos theta what is b cos theta is nothing but the height of this point a from this center point O you agree with me so far yes no no it says change of the height with respect to what with respect to a datum we take the center O as the reference point what what sorry can you repeat so the mass will be there yeah but that's what I have been telling like so last half an hour ago I told that to one person there you can do that also let us take top point as the reference level yeah definitely okay yeah then what is the potential energy with respect to that reference level how much does the mass come down by yeah mg b 1 minus cos theta yeah okay but now the particles come down what is the direction of gravity down if you go from top little bit down what is the work done positive what is the potential energy minus so potential energy minus mg times b 1 minus cos theta if you expand it what will you get plus mg b cos theta minus mg b that minus mg b is just a constant which is coming because you are putting a datum at a distance b which is higher just right so it just depends what is your reference level you can take a reference layer at a top right everything accordingly the answer you will get will exactly be the same up to some constant which doesn't depend on theta now instead what do I do for convenience let me take point O as my reference level with respect to that reference level if I take this mass up by distance of b cos theta how much is the work done minus mg b cos theta what is the potential energy plus mg b cos theta just try it think about it try it okay you will be convinced if you just write down few equations okay you will be convinced immediately any question it's just a matter of choosing an appropriate reference point or a datum point there is no absolute value of potential energy ever it just depends on what is the base level that you are choosing any question other question now so if there are no question is is everybody on board with respect to this equation okay that is the potential energy now it's a one degree of freedom system only variable is theta minimum potential energy theorem tells us that dv by d theta is equal to 0 for the system to be in equilibrium so what will you get dv by d theta is equal to k a square minus mg b sin theta so we have to solve an equation of this form that sin theta is equal to k square by mg b times theta substitute all the values the equation that you have to solve is equal to sin theta is equal to 0.8699 theta just put in all the numbers now note one thing one solution to this equation is theta equal to 0 if you put theta equal to 0 this equation is satisfied does that make sense theta equal to 0 means point a exactly at the top according to a Newton's law what does that mean when this point is exactly at the top the spring was unstretched so there is no torque acting on this point is there any torque by gravity provided by is gravity providing any torque with respect to this point no the system is in equilibrium we are fine but now the second point will come from solving this non-linear equation you can solve this graphically you can solve by a Newton Raphson method or simple fix point iteration whatever way if you solve this you will get that theta is equal to 0.902 radians or 51.7 degrees okay so theta is some value here now does that make sense if theta becomes larger what is happening this spring gets extended if the spring get extended it it will exert a tensile force what torque it will exert about point B anticlockwise torque what is the torque exerted by gravity okay when you take moment about this point clockwise so those two clockwise anticlockwise torque will now balance in the moment equilibrium language and you will see that even if you do it that way you will see that the equation you will get is exactly the same and a particular value of theta you will get will be this okay now the question is that even without doing a single calculation what do you think theta is equal to 0 position is it a equilibrium position or do you think it is not an equilibrium position theta equal to 0 it is unstable equilibrium because we just saw that you tiny perturbation okay you will immediately see that the restoring torque will not be enough to bring it back to its original position and so the system will be an unstable equilibrium and if you do energy arguments you will see that d2v by d theta square is simply ks square minus mgv cos theta okay now if you put all the numbers okay whatever numbers we have at theta is equal to 0 if you put all the numbers you will see that d2v by d theta square is equal to minus 3.83 which means that equilibrium is unstable but note one thing that without doing the calculation you cannot say that a system is an unstable equilibrium look at this equation if you put theta equal to 0 then the value of d2v by d theta square will be equal to ks square minus mgv so if k is very large then you will see that even theta equal to 0 position is actually an equilibrium position any question any point yes yes please yes back slide okay in a theta a theta yes we take a theta when theta is very no no no now here theta we are getting 51 point no no no no that is not true because what will happen is that that because everything is connected to each other like it is like so it is a circumference because in an in listen to me in an infinite symbol say you give it a tiny rotation what you will see is that b times delta theta is the extension of the spring now at that position give another tiny rotation further extra is what b times a times delta theta more tiny rotation a times delta theta so if you keep summing that up that's all approximate method no no no they are not approximated it's exact okay it's exact for example there is no other way otherwise how will that spring length be conserved you can take the circumference of the small pulley for to find the s to find the s but circumference of the pulley unless and until you are rotating that that circumference of no use only when you rotate that pulley complete pulley only after rotation is that string this part of the this particular spring it is getting sucked into that pulley it is just saying that for example if you are like flying a kite you rotate that thing how much ever you rotate by omega times the r of that is equal to how much length that came in so it is simple conservation of length that whatever length like for example if I'm flying a kite okay you are like the spindle you are turning it in if you turn it by theta the radius of the spindle times theta is the amount of string that coming so it is simply the conservation of length that is at play here and one thing that we are assuming is that that the length okay the individual length of that that free portion that b times whatever is not changing is what we are assuming that it is not stretchable if that was stretchable then that what you are saying is fine that the amount of length that goes in we need not be exactly equal to whatever stretching this stretch is happening here but if you say there is a conservation of length okay that there is no stretch of that top portion then there is no other choice however much you turn the spindle by that much length will be sucked in and when that much length is sucked in that point this point here okay this point there okay that will keep getting pulled up by that much distance and that is the extension in the spring okay we don't have choice about that at all and even calculus was it will work to be the same way you just add infinite similar rotations infinite similarity is true you keep adding the rotations you will see that everything adds up any other question yes yeah fine let us move on so again I just want to emphasize that there is no limit on what the value of theta is for this assumption this is a perfect assumption this is not an assumption this is a perfect quantity here only assumption is that this particular portion is non stretchable that is the assumption we have okay that this is not because if you say that if I use this portion the portion which is getting pulled in here if that is also stretchable then you don't know because then some small portion may get stretched in but that's an additional degree of freedom that you are introducing in this problem okay do I move on yes okay now this is only because of these numbers what we see is that at theta equal to 0 this d2v by d theta square is now negative so there is no restoring torque for a tiny perturbation about theta equal to 0 and this assembly we say is unstable if d2v by d theta square becomes positive then this is greater than 0 this entire assembly is in stable equilibrium at theta equal to 51.7 so what we have to do is this second derivative we have to put in value of theta put in value of 51.7 okay and then you have to check what is happening but note here that this a theta this theta is in radiance this cannot be in degrees the moment we say that s equal to a theta then the implicitly we are using theta in radiance okay yes please after stretching if we leave this force what will happen to the what will be the position of the mass if after stretching what I don't understand does stretching if I just lose it lose it means stretching condition it is what will be the position my point is what will be the position it is theta as because there is a stretching condition there is a force stretching condition on the bottom spring right yeah yeah now if I lose this what will be the position of the mass so where are you losing the mass suppose do you have some theta yes already yes is it in equilibrium are you keeping the theta at 0 or 51.7 where I where what is your theta to begin with yeah say about 51.7 okay if you have theta equal to 51.7 now I am losing now if you just lose it system will be in that position only but if you perturb it if you perturb it up then it will go to its original position yes if you perturb it down it will go even further but then to really know where it will go to what you have to do is you have to assume that theta can be even larger okay but what will happen is that most likely is when you go when you drop that perturbation so you are saying is that if I perturb it about 51.7 degree what will happen where will it go okay where will it go it will return it will return there sir it should return there it will not return it is an unstable what is the stable equilibrium it is returning there sorry sorry sorry it is a stable equilibrium so at tiny perturbation you will get back to 51.7 okay but about 0 if you perturb it about 0 it will go to that equilibrium position is does that answer your question that if you do it in the opposite side it will go in the opposite side it will go to the opposite because the torque there is even more yeah here we are just assuming theta to be greater than 0 strictly speaking we should also look at theta which are negative you have any other question you have any further question okay fine we one thing I want to clarify here is that we have not taken theta to be negative here okay strictly speaking we should also take theta equal to negative and then see that are there any other solutions in negative theta side also any question but my task to you is this okay if you go back now you can do one simple thing you have the function energy v as a function of theta why do not you just plot that energy for these parameters for a range of theta and then see what happens you will see that this point will be a maxima this will be a minima and just check that if they are graphically do you see any other maxima or minima okay just check any question let us move on so what we have in this problem okay very simple problem is that the spring is unstretched when theta is equal to 30 degrees at any position of the pendulum the spring remains horizontal why does the spring remain horizontal because it is connected to a roller whatever you do if the if the spring gets tilted the roller will move up and keep the spring always in horizontal position now what we are asked is that if the spring constant k is equal to 15 Newton per meter at what position will the system be in equilibrium so when theta is equal to 0 this spring is unstretched and we want to know that at what position will the system be in equilibrium straightforward how do we write down the potential energy of the system if you take the top point as a datum for this mass what will be the potential energy due to gravity this is 5 plus 5 10 10 cos theta into into 100 and minus because why with respect to this okay the potential energy is negative so minus 100 into 10 cos theta what will be the potential energy stored in this k as a function of theta okay so 5 sin theta minus 5 sin 30 square into half okay so this is 5 so what is this 5 sin theta is this length but original unstretched length was how much 5 times sin 30 so the total extension is 5 sin theta minus 5 sin 30 half into k square that is done now only thing you have to plot is you have to you have to find out what is the minima you will see that dv by d theta looks like this stability is not asked but you can still do if you wish to okay okay it is perfectly fine you can solve this you can solve this equation exactly can you tell me what does the final equation look like thousand sin theta plus 50 sin theta okay 15 to 5 250 cos theta sin theta so okay so most likely this equation would not be solvable by simple methods okay you will have to solve a non-linear equation for that and if you solve this okay by Newton-Rosson method or any other method you will see that the solution will come out to be 15.8 degrees and at 15.8 degrees you can find out the second derivative it is a very very simple problem d2v by d theta square plug in theta equal to 15 degrees and you will see that the value of d2v by d theta square is 2196 is greater than 0 so the equilibrium position will be theta approximately equal to 15 degrees and at that position the system will be in stable equilibrium okay really straight forward nothing nothing about this any question sir considering the first year course we kind of go like energy method in details or linear or non-linear analysis can we assume theta is a small and we will go for simple calculations what considering the first year course or first year so what we will do is this okay I will I will take your offer on making theta small and I am going to take that offer now okay so let us do this problem on that offer okay only thing is that if you take theta equal to small there is one small issue about just doing expansion about small theta because for stability analysis your potential energy should at least be quadratic in theta because if your potential energy is only linear in theta your second derivative will be 0 so we have to take some amount of expansion okay we cannot just take what we did for virtual work you will see that for example if you make delta theta to be very small and use what we did in virtual work no question of stability arise so your potential energy should be quadratic in theta and so long as we satisfy that and we will see a few few questions where we can do that expansion okay then the problem can be drastically simplified I totally agree with you that we need to do the small theta approximation but about what point and up to what order so let us work on that okay I like your offer okay and we will do it so the question we are asking is a very important question for the simple reason that if any of you would be teaching solid mechanics for example later on or if the students take solid mechanics okay in the in the second year then for example one of the problems they encounter will be buckling of systems and this is a very simple model to emphasize that what is the buckling instability and why and how it can happen so what we have here is we have a combination of two rods this is one rod AC second rod CD okay this rod AC is guided by a roller okay within these two set of guides at point C we have a pin connection now what we are doing here is that that this spring C okay there is a horizontal spring that is connected to point C and connected at another rigid support all the dimensions are given to us now what we are asked to do in this problem is determine the range of values of P for which equilibrium of the system is stable in the position so sure is what we are asked okay so even without going to the next slide can you tell me is that a one degree of freedom problem two degree or undetermined degree of freedom problem how many degree of freedom does the system have one degree if I can know what is the angle which this makes I automatically know everything which I need to know okay now what are the contributions that are coming to the potential energy what will contribute to the overall potential age of the system in this problem spring spring force will clearly contribute that is V of E what else energy of load energy of load will also come into picture okay so now suppose your system is now deformed in this confirmation okay that this is theta this is phi theta and phi are not independent there will be relation between them so what will be the total potential energy of this system if this is theta what is the extension of the spring now we are going to make use of small theta approximation here because what we are asked in this question is we are asked to find out that verify that this position is an equilibrium position and then determine the range of values for P for which the system is in equilibrium okay so this is our base position theta equal to 0 and phi equal to 0 now you tell me if some theta is given what is the extension of the spring a sin theta okay a sin theta is the extension in the spring so k by 2 okay a sin theta square in minus it does not matter a sin theta square will be the potential energy of the spring what is the potential energy of this mass now that is a critical question how do you define the potential energy of the force that is being applied at the top if you do with respect to bottom what does that mean what it means is that that from the bottom forget about the entire system if I take a particle and take that particle against this force from top to bottom how much work is done work done will be height minus of height into P because force is downwards work you are applying a upward displacement so the work done will be minus P into height so the potential energy will be how much plus P into the height now what is the height clearly it is a cos theta plus 3a cos phi okay so P a cos theta okay plus 3 cos phi now what is that theta by 2 it comes we will decide that later now you tell me what is the relation between phi and theta strictly speaking the relation should be that if theta is and phi's are large then 2a sin phi should be equal to a sin theta truly speaking but as the professor pointed out that okay all are small theta approximations so why don't we use a simple approximation that theta is small because we want to look at the stability about theta and phi equal to 0 position so we approximate sin theta is theta sin phi as phi so what do we know we know that 2 phi is equal to 2 theta so phi is equal to theta by 2 is what we have used there so now what do we have we have our potential energy written purely in terms of theta theta is small okay further so let us do this now further more okay let me finish this then I will get to another point now how do we decide okay that what is the equilibrium position if we do dv by d theta you will see that dv by d theta comes out to be this put in theta is equal to 0 here you will see that dv by theta equal to 0 is true when theta is equal to 0 which means that theta equal to 0 indeed is an equilibrium position now what we want to find out is the system stable or not what we do we do dv to v by d theta square we will get another expression put theta equal to 0 what we will see is that that at theta equal to 0 d2 v by d theta square will be equal to k square minus pa 1 plus 3 by 4 just solve this what you will know is that p should be less than 4 by 7 k for this entire assembly to be in stable equilibrium and this is a precursor to buckling that if we apply huge amount of load you will see that a system buckle that theta equal to 0 is not a stable position it will go into another position okay because of buckling another thing we know is that if k or stiffness is large then the system is stable if stiffness is small the system is not stable this is again same thing that happens in buckling the rod is stiff no buckling even for large load if the rod is weak or not in stiff inner then even for small load it can go into buckling mode there is a precursor to buckling problem now coming back to the question of small displacements we can also do another approximation what we can do is if we want to avoid all this theta cos and sin just replace sin theta with theta because what we want is our overall potential energy should be at least quadratic because we want to take stability so the potential energy should be at least quadratic so if you replace sin theta by theta this will just become a theta square and you will see that answer does not change whatever we do similarly there is this cos theta so what is the easiest resolution to cos theta 1 minus theta square by 2 okay you can do that also you can replace cos theta by 1 minus theta square by 2 and you are good you can then completely do away with sin sin cos but only thing is that your potential energy should be at least quadratic which in this case it will be if you replace sin theta with theta and cos with theta to be equal to 1 minus theta square by 2 okay and then you are good to go you do not need to use cos and sin everything will be simple after that because we are looking only at the stability this is good but if you want to find out the second position of equilibrium which is at finite theta then this approach will not work but if you are interested only in knowing if theta is equal to 0 is stable or not this is perfect much easier and much simpler to do and we will solve a few problems after the t break where we will actually not go into this complete sin and cos theta but use this simple approximation