 So, let us straight away go to tutorial problems, very simple, medium, little bit difficult. So, simple problem, it is a very practical problem. So, what is told in this problem is this that this is a joint, this is a very crude caricature for an artificial knee and artificial knee what we need is that we need a rotational stability. So, what we have done is that this M is representing the mass of the person or the mass of the person, the mass of the person, this is the leg, this is the knee joint. We are modeling that knee joint in a very simple manner as a torsional spring. Now what is torsional spring that if this beta is the relative angle between this and this clearly. So, this one link, this second link, this exterior angle is beta, then the internal torque will be given by K times beta. Like a normal spring, the internal force is given by Kx, the internal torque in the spring is given by K times beta where beta is this angle. And what we are asked to find out is determine the minimum value of K for which we will ensure the stability of a knee joint for beta is equal to 0. Okay, means when it is exactly straight, a tiny perturbation, okay, for example a tiny perturbation should come back again to ensure that stability what should be the minimum value of K, okay, for that. Very simple actually, it will take just like less than 10 minutes, less than 5 minutes to solve this problem. Only thing you have to realize is what is the potential energy? And by analogy, if the spring force is K times x, then the potential energy in the spring is half Kx square, use the same analogy and find out what will be the potential energy in that spring joint if the moment is K times beta. Same analogy and then you can write down the potential energy from the spring, potential energy from the mass and then ultimately do the differentiation second derivative and figure out what is the criteria for this to be stable. Very simple problem, okay, everybody got it? This one, most of you would have gotten it, this one is not difficult at all. So what do we know? If we just need to write down the potential energy from gravity which is nothing but if this is beta, this is beta by 2, beta by 2. So 2mgl cos beta by 2 with respect to this datum is the potential energy. What is the energy in this spring? Half K beta square. So total potential energy is Vg plus V, dv by d theta, you will see that beta is a solution. d2v by d beta square, you will get this simple answer. Put in beta equal to 0, you will see that minus half mgl plus K should be greater than 0, okay, at beta equal to 0. Which means that K should be greater than mgl by 2. So K should be more than that for this assembly to be stable, at point B is equal to 0, beta equal to 0, okay, fine, next problem, okay. So problem 2, it is like going to a theater and sneaking a video camera when the DVD is already out, okay. So an exploration device which unfolds from a body A, okay. Let us look at the second problem. So exploration device which unfolds from this body A of a spacecraft, okay, it's some rigid body, which is completely stable, okay, at least for all practical purposes that we are looking into. This mechanism, it is exploring apparently surface of Mars according to the book, or not Mars Moon, okay, maybe Mars will be later. So consists of a spring loaded pantograph with detector head B, okay. So this is the detector head B, it is desired to select a spring that will limit the vertical contact force P to 100 Newton in the position for which theta equal to 120 degrees, okay. So when theta equal to 120 degrees, we want the maximum value here to be equal to 100 Newton. So what we are given is that if the mass of the arms and head is negligible, okay, then what is the corresponding stiffness? And one thing which I did not say is that when theta is equal to 30 degrees, this spring is unstretched, okay. This I think I forgot to type it here, yeah. When theta is equal to 30 degrees is when the spring is unstretched, okay. Now with this information, okay, just note that theta not equal to 30, this spring is unstretched. What we want is that, what is the spring constant such that, this doesn't exceed 100 Newton, okay. Problem definition is straightforward. The moment you can write down the free energy and take dv by d theta equal to 0, you'll immediately realize what the equation is and from that you can find out what is the minimum value of k, very straightforward problem. Just only thing is to write the potential energy appropriately and not do any mistakes while doing the differentiation put in proper numbers. It's a very straightforward problem. Again note, I forgot to type it there, okay. Please note that when theta is equal to 30 degrees is when the spring will be completely unstretched. So we have to take that into account while writing down the potential energy for the spring as a function of theta. So did anybody get the answer already? Let us briefly discuss what is the procedure for the problem. What we need to know is the elastic potential energy and the potential energy due to this force. Now if I take this as the reference level, okay. This bar as the reference level. Then what is the potential energy by this force? B, 2B, 3B, 4B, 5B, okay, how much? So, no, 2, 8, so 2, 3, 2, 3, 4, 5, 6, 7, okay. 7B into sin theta by 2, so from this datum, if you take a particle from here up to here work against this force, what will you do? You will do negative work because the force is upwards. The potential energy is plus B into 7B cos theta by 2, okay. So just add all those numbers and you will see that each arm the length is 300 millimeters. So 7 times 300 millimeter is 2.1 meter. So this is P into 2.1 into cos of theta by 2, okay. Next, what is the potential energy in the spring? What is the length of the spring? When theta, okay, when theta is equal to 120 degrees, okay, what is this length? It will be 2B, no, we should not write 60 immediately. Why? Because we need to differentiate with respect to theta and then put 60 afterwards. We cannot put 60 immediately. So as a function of theta, that value will be 2B sin theta by 2. But what is the initial length? Initial length is when theta is equal to 30 degrees. So the 2B is equal to 0.6. So potential energy half K, 0.6 sin theta by 2 minus sin 30 by 2 square. This is the extension of the spring. This is the stiffness, half K extension square is the potential energy of the spring. Is that point clear? And we should not put theta equal to 60 degrees, 120 degrees to begin with. Why? Because we need to find the equilibrium position first, okay. And what is the equilibrium like for a given K, how will that P be balanced? And then we put theta equal to 120 degrees. So we put dV by d theta equal to 0. Now we get this simple equation. Put theta equal to 120 degrees. What you will get is a simple equation like this for K. And you can figure out that if K is 1664 Newton per meter, okay. Then or any K which is less than this value, then you will see that P will not be more than 100 Newtons. Is the procedure clear? So what we are getting by putting dV by d theta equal to 0, what we are getting is for any given P, okay, when will the system be equilibrium or to put in this way, what is the value of K that will bring this system in equilibrium for P equal to 100 Newtons when theta is equal to 120 degrees. And from that we can backtrack what is the value of K, you will get a simple linear equation and K will turn out to be 1664 Newton per meter or 1.66 kilo Newton per meter, fine? Fine? Okay? Last question for the day, okay, before we begin the quiz, what we have here is a very simple problem if you know the right knack of doing this problem, otherwise it can be extremely messy. So what we have is we have a rectangular plate, the rectangular plate dimensions are given, this is a, a, all are given. At all the four points a, b, c, d, four corners of the rectangular plate, we put four vertical springs, the spring constant of each of the springs is K. Now what we are doing is we apply two forces P and minus P along the centerline of this plate and what we are asking is for given values of K and a, what can be the maximum value of P or what is the range of value of P for which the assembly is in stable equilibrium at this position is the question and with respect to what like to make the matter simple what I have further written about which the equilibrium position is stable with respect to small rotation about the center of the plate, which means I have also given you what is the degree of freedom that you require for this problem, that we want to check the stability of this system with respect to a small rotation about the center. And we can use small theta approximation and simplify this problem greatly, okay? So please attempt the problem, is the question clear? You can also verify from writing the potential energy that theta equal to zero, that the spring in this position, sorry the plate in this position is in perfect equilibrium. What you want is that if you give a tiny perturbation in the form of rotation, will this assembly what values of P will this assembly be in stable for this K and these dimensions, clear? You use small theta approximation just to want to ensure that the potential energy should be quadratic at least and then figure out what is stability. First you ask yourself that what will make the system unstable for a tiny rotation, what can happen? You will see that when the system has a tiny rotation, these two P's will try to provide a torque and the other the two forces will try to provide a torque, which will try to rotate the system whereas the springs will try to provide a counter torque and what you want is that what should be the corresponding forces and case or for given case and is what should be the corresponding values of P's such that the counter torque will always balance the system and bring it back to its original position, which is theta equal to zero. So that is the logic and with that logic use the rotation as the degree of freedom small rotation, write down the potential energy and check the stiffness. Stiffness is check the second derivative to find out what should be the maximum value of P that can keep the system in equilibrium for this K and this A. Is the problem clear? Think about it, very simple problem if you know what is a simple equation to write. Very similar to what we had done a few moments ago with respect to those cross springs and a platform, very simple, very similar. So let me briefly discuss this problem before we start our quiz. It is not too difficult a problem, only thing you have to realize is this that the rotation we are providing is a small rotation. Just note that if we perturb this by rotation about the center, small perturbation what is a top spring will get compressed, this spring will get stretched, this will get stretched, this will get compressed. How much will be the stretch? This is A, A by 2, so A by 2 theta, A by 2 theta all of them. So two stretching, two compressions. Now second thing which you want to know that what is the potential energy contribution from this and this. What you will immediately realize is that this force after rotation will come inwards. What is this distance? A by 2 cos theta which you can also write as A by 2 1 minus theta square by 2 if required. So put that in. So P A by 2 cos theta into 2 because there are two forces will be the potential from the force. Four times half K A by 2 theta square which is the stretch is the total potential energy dv by d theta equal to 0. You will see that this equation becomes 0 at theta equal to 0. To check stability what do we need d to v by d theta square should be greater than 0. Just check what you will get, you will get simple equation at theta equal to 0 given by minus P A plus K A square which has to be greater than 0 which implies that P should be less than K A. So that is the criteria you need for the system to be stable in this conformation with respect to rotation about the center. Now I had given that with respect to rotation just for simplifying the problem but you will see that the only instability that can come in the system is with respect to that degree of freedom. Everything has a translation in the horizontal or in the upward direction is completely stable. There is no issue with that at all. That was the only instability can come from the rotation and I have already mentioned that just worry about that degree of freedom. Any questions about this? So see reasonably clear. So what we had done is that this could be far more complicated for finite theta but because we want to check only stability about small rotation, just take small rotation but ensure that your potential energy is at least quadratic in theta so that we can get d 2 v by d theta square. Now what happens exactly at P is equal to K A? You cannot know because it says that d 2 v by d theta square equal to 0 then we have to do higher order analysis. So what we have to do is that cos theta you cannot write it simply as 1 minus theta square by 2. So you have to write it to further degree plus theta to the power 4 by 4 factorial if required but I am not asking you to do that but if you require to find out the stability at P is equal to K similarly the extension in the spring was A by 2 theta. You have to add extra component higher order term which will be cubic and so on and then you write the potential energy you will see that you can do the further analysis that what happens when you have to take further derivatives of v with respect to theta that we do not bother. But only thing we can say for sure is if P is less than K your assembly is stable. So with that we conclude our session on minimum potential energy.