 First of all to start with what is potential energy? If we just quickly visit the Wikipedia you will see the definition is there. So, I will just read line by line potential energy is associated with a set of forces that act on a body in a way that depends only on the body's position in space. So, that is very important that potential energy of a force that is acting on a body it depends only on the body's position in the space. Now, remember that if the work of forces of this type acting on a body that moves from a start to an end position is defined only by these two position and does not depend on the trajectory of the body between the two then there is a function known as a potential that can be evaluated at the two positions to determine this work. Now, there is a very clear statement happening here that means if when we are calculating the work done by the forces that we are only considering during the calculation the end position and the start position. So, that means trajectory of the body when it is moved from a point let us say a to a point b the work done will only depend on these two positions and it is not going to be dependent on the trajectory then we can say that for these kind of forces which will allow you to calculate the work in this way we can define a potential and that potential we are going to call the potential energy. So, for example very simple example would be in our purpose look at the gravitational force on a body. So, for example if I have the weight of the body and if I am moving the weight let us say from a point a to the point b I can go in any manner from point 1 to point b, but the work done that will be calculated by moving the body from the a to the b will only depends on the vertical height that it travels it is not going to depend on the path for the trajectory and the reason would be very simple because the direction of the force is not going to change the direction of the weight is not going to change. Similarly, we can think of springs. So, for example if I really you know pull a string then we can see that the spring force that is acting on the body we call this a elastic force therefore, the potential energy would be the elastic potential energy and that will again depend on the starting position and the final position only the or in other words it will depend on the stretching of the spring. And we all know that elastic potential energy that is that spring problem that I just discussed coming from the elastic force of the spring that is given by the spring on the body that is really used to you know address a very challenging problem of that of an arc. So, it is seen here that you have this bow and arrow problem ultimately string is going to act as an spring therefore, it will store the elastic energy and when it is released the you know arrow is going to be released then the energy is going to be converted to the kinetic energy. The other example would be we all know that if I lift a ball and I am going to drop the ball then therefore, the ball at a certain height will have its potential energy and when I am releasing the ball when it is trying to hit the ground the entire potential energy will be converted to the kinetic energy. So, that is the energy conservation theorem again we are going to use that later on. But ultimately therefore, what is happening when we are going to you know define the potential energy of a force it will only be determined by its position nothing else ok. So, just recall the work of a force and that we have already done. So, suppose I have a particle A and a force is being applied such that displacement is given by d r ok. Then we can say that the work done during this small displacement d r of that particle is simply given by f dot d r. Now, what is f dot d r remember work done is a scalar quantity and f dot d r is nothing, but we can write it in terms of f d s cosine alpha, where d s is the magnitude of vector d r ok. So, what it means essentially that if I take the force and the displacement along the same direction right. So, work done will be basically defined by force multiplied by displacement along its own direction. Now, that will be positive work done if the displacement is in the other direction then it will be negative work done. So, we can we know all of this definition the other issue is that let us say I have the body weight body weight is downward and if I move the body in the perpendicular direction then the work done will be 0. So, that means, in that situation we have alpha equals to pi over 2 and therefore, the weight is not going to do any work if the body is moved horizontally ok. Similarly, we also know how to calculate work of a couple. So, that this is a rigid body and at point a and b you can see a couple is acting that means, f and negative f and due to the action of this couple let us say I have a line a b joining this couple that has been taken a final position a prime b double prime. Now, you can decompose the problem into two parts one is the translational part that means, the line has been displaced to a prime and b prime a prime b prime is parallel to a b. So, that is a rigid translation and then you have a rigid rotation that means, a prime b prime is now given by a prime b double prime. So, what happens in the process if you try to adopt the vector mechanics as we have done in the previous you know slide we have gone through this. So, what is the work done work done is going to be now this force is going to work negative f dot dr 1 and this force will do the work f dot dr 1 plus dr 2. Now, what we can clearly see that these two terms cancels out that is negative f dot dr 1 plus f dot dr 1 that means, during the rigid translation pure translation there is no work done the only work done that is happening is during the rigid rotation because, we know that dr 2 is nothing, but r multiplied by d theta from the arc rule. Now, again what we know is that f multiplied by r that is the moment right that is the couple moment that we talked about in the very beginning therefore, we ultimately have m d theta. Now, in engineering problem they are for we will often have this situation either we have to you know calculate the work done by a concentrated force or by a couple. So, we have to keep that in mind. Now, remember these are represented in terms of small displacement. Now, how about if we try to calculate this for a finite displacement? Let us assume that again as I said before that we will be mostly focusing on the work due to a weight right that is we are going to define the potential of the weight as well as we are going to define the potential of the spring force right. So, we are going to attempt mostly these two problems as it appears in many engineering applications. So, in this case when we are trying to define the work of a weight let say that body you know is moving from a 1 to a 2 or body is taken from a 1 to a 2 and this is the trajectory. Question is what is the work done by the weight of the body? So, we can see from the definition of the work of a force that we can simply say that the small work that is being done once the body is moved from a small you know trajectory let us say such that vertical height is d y we can simply say d u equals to negative w d y. So, that is the small work done. Now, we will be interested in finding out how much work it is done when the body is gone from a 1 to a 2. So, therefore, we see the total work done by the weight is equals to w y 1 negative w y 2 or it is negative of w delta y. So, what we have studied here that work done by the body will actually be negative when the body is going upward direction. Now, why this happen? Why it is only determined by the two positions? Because as we know that weight cannot change its direction it is always going to be vertical. Therefore, when we are calculating the work done by this weight we can discretize this trajectory in small small vertical height and some of the this small small height is going to be the difference of the two heights that the body has been moved. So, it is going to be y 2 minus y 1 always. Similarly, if you consider the work done by the spring force, remember a body is attached to the spring and I am going to pull the spring. So, therefore, what would be the direction of the spring force? So, this is the force exerted by the spring on the body the direction is always going to be on the negative direction. So, stretching is in the positive direction force is going to be on the body in the negative direction. So, again if you think of it that I want to stretch the spring from position x 1 to position x 2 then what is going to happen? Now, again I am going to choose a very small instant x and let us say I am stretching the spring by dx. So, at an instant of stretching x I am just taking the stress to be dx. So, therefore, what is the force in a work done? Work done is simply going to be negative f dx. So, at any position of spring x I have the work done negative f dx if I stretch the spring a little amount dx. Now, what is f? We know that f is going to be equals to k times x that is that is the spring force when the spring is stretched by an amount x from is undeformed configuration. So, ultimately again we see that the work done as I stretch the spring from x 1 location x 1 to the x 2 is given by half k x 1 square negative half k x 2 square. Now, remember in this case what is happening? If you plot the x versus f we can clearly see that the force is defined by k times x. Therefore, the work done is simply going to be the area under this load curve. So, whatever area you get under this force displacement relationship that is going to be the work done. So, once we have studied the work done now we can try to see how do I define the potential energy there. So, by definition now what we see that we know that work of a weight is simply given by w y 1 negative w 2. So, remember work done remains negative and this work is independent of the path followed and depends only on what w multiplied by y that means the position of the weight is what defines the work done. So, we have already said that therefore, we can define the potential or potential energy of the gravitational force. So, potential energy of the body with respect to the gravitational force can now be defined as w multiplied by y. Now, remember if it is so then we can say that work done is equals to potential energy at position 1 negative potential energy at position 2. Now, what happens? Therefore, we can see that actually potential energy is increasing in this process right potential energy is increased whereas, work done remains negative. Similarly, if you consider the spring force example then we can again define a potential energy of the body with respect to the elastic force of the spring F and that will be simply we can say half k x square ok. So, therefore, we again have the change in potential energy if we look at it right. So, potential energy is again increased and work done is negative. Remember that one important thing here is to remember that in this case y is always measured from a fixed datum ok. So, once the y is measured from a fixed datum depending on where is your datum right the potential energy the sign can change. For example, if my body is above this datum which I have chosen then in this case potential energy is positive. In many problem it may appear that you can choose the datum to be a fixed reference you know point and if the body is actually below that reference point then potential energy of that body will be negative. So, definition of potential energy is simply w multiplied by y, but important thing to note it comes with a sign ok. Similarly, for the spring force however we need not to worry about it. So, you know because you have a square term here. So, therefore, the stretching square is always going to be the potential energy. So, in this case the sign is not coming into play, but for the work you know for the potential energy of a body sign will play an important role. So, in summary what we have studied that work done is equals to the negative of change in potential energy and what is important for us to understand that work done is negative then potential energy is increasing if work done is positive then potential energy in decreasing. So, these are the you know this is the point that we have to always keep in mind. One more thing to note that forces for which the work can be calculated from a change in potential energy will be called conservative forces just by definition. That means, if I can calculate the work from the change in potential energy I will call this conservative forces. And remember work done by this conservative forces do not depend on the path followed. So, as we have seen in case of a body weight that means, gravitational force as well as in case of a spring force work done by these forces is independent on the path followed. However, if you look at really friction force we have studied that remember friction force it will always depend on the trajectory because it is it will keep on changing the direction depending on how the trajectory is, how the body is moving from one point to the other point. So, therefore, the friction force will be a non-conservative force and I would not be able to define the potential energy for that. So, another important point is to always remember that when we are coming to you know solve some of the engineering problems it is very important to choose an appropriate datum and refer the potential energy consistently with respect to that datum. So, just keep that in mind. So, these are some simple illustration now you can think of what would be our objective in this case. Our objective is to first find out what is the potential energy of a rigid body system. And then what we want to achieve remember what we want to achieve is that what is the equilibrium configuration and whether that equilibrium configuration is stable or unstable. So, therefore, why potential energy is so important because it is not only giving you the equilibrium configuration, but also it is able to tell that whether that equilibrium is a stable equilibrium or unstable equilibrium or a neutron equilibrium. So, therefore, you can see in this problem of a spring and mass let us say spring is unstretched at this point. So, v you know equals to 0 was there this is the you know datum we have chosen. So, that is the datum. Now, once you hang the mass therefore, it is going to be stretched that stretching is let us say x. So, what we have the potential energy of the spring force that is given by half k x square. And the potential energy of the weight that is mg that is going to be now since I have chosen the unstretched configuration of the spring as a datum. So, therefore, what we have the potential of the weight is equals to negative mg x. So, what is shown here that look we have the potential energy of the spring that spring potential energy is going to be like this. So, that is half k x square whereas, potential energy of the weight that is going to be coming down and therefore, if you sum these two algebraically. So, that is the net potential energy ok. So, net potential energy now once I you know draw this kind of figure graph we can clearly identify what would be the stable you know equilibrium configuration. So, firstly we should be able to tell that where is my equilibrium configuration equilibrium configuration will be determined where the potential energy is either minimum or maximum. That means the slope of this potential energy graph has to be 0. So, wherever I have slope of this potential energy graph is equals to 0 that function is 0 there I will have the equilibrium. Now question is whether it is going to be stable equilibrium or unstable equilibrium I am going to come to that next. So, as I have already said in words we can now think of one important point that how do I let us say I have the potential energy function right. I have defined the potential energy of a rigid body or rigid body system in terms of some position which is the variable to me. Now, can I calculate the equilibrium configuration how do I do that now we have already learnt the principle of virtual work. What is the principle of virtual work? If a rigid body or system of rigid bodies they are in equilibrium so the point was the body was already in equilibrium and I give it a infinitesimally small displacement right. So, that infinitesimally small displacement was virtual displacement then what we have said we have said that work done virtual work done by the forces moving through the virtual displacement is equals to 0. That means idea was if the body was actually in equilibrium then if I allow the virtual work principle then it says that work done by the active forces equals to 0. So, now you think of it that we have already established the fact that work done should be negative of change in potential energy. So, what is shown here that work done delta u is equals to negative of delta v. So, delta v is the change in the potential energy of the body as you displace the system by an infinitesimally small amount let us say. Therefore, now what is my delta v dv by d theta multiplied by delta theta if the potential energy is a function of theta where theta will determine the position of each and every force that is acting on the rigid body. So, therefore, now since virtual work should be equals to 0 and delta theta is arbitrary in nature therefore, what we have dv d theta equals to 0 that will determine the condition of equilibrium. So, virtual work it has assumed that if the body is equilibrium virtual work is 0. So, we have simply taken that and we have said therefore, dv d theta equals to 0. So, that is the condition of equilibrium. So, now that is going to play an very important role and we are only going to keep our discussions on the single degree of freedom system that means, position of all the forces are going to be dependent on only single variable that variable could either be x or theta depending on the problem as we are working. For example, in the previous slide I have shown you a spring with a mass the position of the forces are determined by x. In this case now look at this problem right here it will be dependent on theta. So, how this problem looks like you have two bars A, C and CB. Now what is happening in this case the unstretched length of the spring is dA. So, spring was unstressed when it is dA that means, that point B right here should be at A therefore, both the bars are going to be absolutely vertical. Now what we have done let us say both the bars were at the vertical and then you have placed a weight w. Now my question is if I place the weight w then you know what would be the equilibrium configuration suddenly as I place the w as I put the weight simply A, C and CB is going to take a shape like this which will be determined by a position theta. Question is for which theta it is under equilibrium. So, if you now again try to understand what are the you know what is the potential energy of the system in the displaced configuration as I said that first we have to fix a datum. Remember potential energy is not you know not an absolute quantity it is a relative quantity measured from a datum. Now datum can be anywhere ok. So, ideally speaking is that we can choose datum anywhere, but what is my variable? Variable is theta. So, if you have a constant term in this you know expression and when we do the derivative of that that constant term is not going to play any role ok because ultimately everything is a function of theta. So, if I choose my datum somewhere you know here and if I just have an additional height h for each and every forces they will simply disappear ok once I do the d v over d theta equals to 0. So, therefore let us look at let us choose the datum at point A. So, I will choose my datum at point A in this line a b. So, let us look at what is the potential energy of the spring? Spring is stretched by amount x b. So, the potential energy is half k x b square ok and the potential energy of the weight right here which is acting at point C that is simply going to be w multiplied by y c which is the height. Now you have to express that in terms of theta we can do it very simple way. So, what is x b? x b equals to 2 L sin theta and what is y c? That is L cosine theta right. So, my potential energy function is absolutely defined in terms of theta. Now I do d v by d theta equals to 0 right. So, what we get? We will get that this will be equals to 0 that will give me two positions of equilibrium. One is theta equals to 0 and the other one is cosine theta equals to w divided by 4 k L. So, these are the two equilibrium configurations possible. Now I have to examine then which one is going to be give me stable and which one is going to give me unstable ok. So, how the stability of the problem comes into picture? So, if I go to the next slide. So, then you can see very nice representation is here. What it is? Let us assume I hang a bar. So, let us say this bar a b has a weight that weight is passing through is mass center ok and I am trying to understand that what is the potential energy of this system and let us understand this way that whether this system is stable at its own configuration or not. That means whether this position is stable position or not. Then what we do? We basically disturb the system by an amount theta. If it is stable right then what will happen then it has to again come back to this configuration that is the whole idea. Now we will get all the links that we talked about in this problem. So, suppose I plot theta versus v then how it would look like? So, at theta equals to 0 right it has some potential energy if you keep one datum position. So, let us say you know some value is there. Now as you go up then what happens? Your potential energy will gradually increase right. As I rotate the bar potential energy increases. Remember work done will be negative. Why work done is negative? Because w is downward and we are moving up right ok because it is rotated so displacement is upward weight is downward. So, work done is negated with indicate that potential energy has to increase. Therefore, what we see if we really plot the potential energy function it is simply going to look like this from a negative theta to a positive theta. So, therefore, we can clearly see that equilibrium configuration is given by what? Theta equals to 0 because the slope of this potential energy function is going to be 0 right here ok. Now question is whether that equilibrium configuration is stable or not. As I said idea is simple think like this I take a ball. If this ball here and let us try to put the you know a ball on this surface right. If I put somewhere here then it is going to come back to its original configuration. So, that means it is a stable equilibrium configuration. So, in mathematically we say this is the minimum potential energy where d 2 v d theta 2 equals to 0. What does that mean? That means curvature is going to be concave ok. So, the curvature of this system is greater than 0. So, if that is happen then the system is in stable equilibrium configuration ok. Similarly, look at the next one. Now I have simply make it a inverted pendulum ok. So, I have just inverted the problem. Now remember if I really look at the potential energy function here how it will look like? So, with respect to the theta it is also going to be a convex shape. So, it is going to get an convex shape ok. So, now theta equals to 0 is the only equilibrium configuration here. Now question is whether that is an stable configuration or unstable configuration. We can clearly see that if I just simply try to deviate that bar a little bit rotate the bar a little bit it is not going to come back to its original configuration. So, therefore it is a highly unstable system. You think of a ball you keep a ball and try to you know push the ball little bit it is simply going to go down the curve. Therefore, what is happening in this case? If I take the higher order derivative so d 2 v d theta 2 that means the curvature of this is less than 0. So, in other word whenever I see the potential energy is taking a convex shape if the shape is convex then I will say that equilibrium configuration is unstable configuration. So, curvature is less than 0. Now how about this? So, in this problem what happens that hinge or pin is put exactly on the weight. Therefore, as the body rotates about the center then we see that potential energy is not changing at all. So, it is everywhere it is flat. So, that means I can tell that it is a neutral equilibrium. So, everywhere I am going to get this equilibrium. Now if that situation arises in reality then I have to examine higher order derivatives. That means I have to now look at what is happening with d q v d theta q and so on. If they are greater than 0 then I can say now this will be always defined by neutral equilibrium. So, we must examine in this case higher order derivatives and how that is coming into picture. Although you know in many problems we really do not get into this one neutral equilibrium most of the problems will be governed by these two that is stable equilibrium or unstable equilibrium. So, we can clearly see that how to decide you can also take a graphical you know input. So, let us say I have first of all let us say I have the potential energy function I can simply plot it graphically and if I plot it graphically then you can see what are my equilibrium configurations possible. One would be right here that is the minima, maxima, minima and so on so forth. So, these are all possible equilibrium configuration wherever I have slope of the potential energy function equals to 0. The question is which is my stable and which is my unstable that has to be determined by the curvature of it as I said if d 2 u d theta 2 greater than 0 that will be stable if d 2 u d theta 2 less than 0 unstable. So, this will be stable this will be unstable and so on. Now, this will be neutral. So, we have to go to the higher order derivative to examine this particular you know problem in higher orders you know things go like in a similar way. So, if this situation happen then it is a unstable system in two dimension if this is the situation then it is a stable system again you have the saddle point coming into play. So, that is going to be the neutral equilibrium. In two degree of freedom system we know that equilibrium is going to be. So, now you have theta 1 and theta 2 are two variables. So, this is going to be the equilibrium you know configuration. So, I am going to get the equilibrium solution based on this and then the higher order if we look at the stable solution will be determined by these three conditions. If these three conditions are satisfied then we are going to get the stable equilibrium configuration. So, we are not going to really look at in this course the two degree of freedom system we are simply going to restrict ourselves into the one degree of freedom system type of problems. So, what I will do I will quickly finish this topics with examples as I said the concept is very easy all will try to do how to calculate the potential energy and once we know the potential energy of that rigid body system then the rest is really max. We just take the derivative of this potential energy function equate it to 0 gives me the equilibrium then we go to the next higher second derivative and that will determine whether it is a stable equilibrium or unstable