 can you see this problem see ultimately you know we are trying to design an artificial leg and there is a rotational spring so that is an additional part coming into play what does this rotational spring will do it will take that torque that is m that is k times beta so if the spring constant is k now so that is rotational spring constant multiplied by theta theta is in this case is beta as shown here so therefore what is the energy of the rotational spring spring energy half k beta square same as the linear spring linear spring it is half k x square but here it is half k beta square okay so you can think like this as if the stretching is equals to k time k multiplied by beta okay so that is the torque torque multiplied by theta and then you have to add an half because it is going from one you know position to the other okay so half k beta square that will be the work done so strain energy will be half k beta square so just keep that in mind and then the problem is very simple okay I am getting the answers actually coming up yes so answer would be mgl divided by 2 that is the k k should be greater than mgl divided by 2 and many of the remote sensors are posting the solution now okay so we have displaced the solution now so look at the solution carefully so firstly we calculate the potential energy of the mass that is calculated as you all see here right and then we have the potential energy of the rotational spring as I said it will be half k beta beta square so that is an additional information we get so this is the you know rotational spring so only thing is half k x square x is replaced by theta here so theta is beta so now you see the total potential energy of the system and then we can do the dv d beta equals to 0 right so in term indeed beta equals to 0 is indeed an equilibrium configuration okay now question is when it is going to be stable so for that we are going to get d2 v d beta 2 right so that should be greater than 0 if okay so at beta equals to 0 we substitute beta equals to 0 that should be greater than 0 so we can say k should be greater than mgl over 2 okay so now next let us move on to the next problem so uniform rectangular plate of negligible mass so we are again not considering the weight of this body itself and it is attached by 4 springs okay now remember why we are studying this problem so we have p and negative p we want to find out for a small rotation whether it will maintain and stable equilibrium configuration or not so that small rotation is given in terms of theta with respect to the center we are going giving it a small rotation now remember why this problem is important when it rotates about the center that p and negative p will create a couple so it is going to create a disturbing torque about the center can you see that it is going to create a disturbing torque and the springs will try to restore the body in its original configuration so spring force is going to produce the restoring torque about the center now as long as this disturbing torque is less than the restoring torque then the body will be stable otherwise unstable okay so ultimately we want to find out through the potential energy but remember this problem can also be solved if you just think this way that you have a disturbing torque and you have restoring torque coming from the spring forces and as long as disturbing torque is less than the restoring torque we can solve this problem as well so that will be based on the force and moment concept that we have developed alright so we just want to analyze at this point through the potential energy concept so think of potential energy concept and just give me the answer so p should be in terms of k and a so what is the condition prevails between p and k and the geometry okay is now lot of remote center is giving the answer main thing here is that we should brainstorm how to get the potential energy function so give it a small rotation theta and try to see how do I establish the potential energy function that is all rest is simple math so p should be less than k as long as p is less than k we have the stable equilibrium while display the solution again and since we have already done the virtual work remember how to get the small displacement when you are rotating the body at individual points at different points how to get the you know displacement that is already clear to us so in this problem as you rotate the body what are the small displacement that is going to come here this spring will have a 2 by theta so all of this displacement in the vertical direction the spring is going to be stretched by a by 2 theta in some cases it will be stretching in some cases it will be compressive so we have to be careful on that okay the next thing is that so here you can see basically half k a by 2 theta square so all the spring the displaced by a by 2 theta just keep that in mind so that is one thing and the second thing is that what is the potential energy of the force so this force right in both case you have two forces one is this one so you can calculate the work done and from the work done you can simply put a negative sign also so ultimately in potential energy if you look at it p a by 2 cosine theta so a by 2 cosine theta is the distance of the force from the mass center okay so a by 2 cosine theta will be distance of the force from the mass center that multiplied by p would be the potential energy of this force and that is multiplied by 2 because there are two forces that is acting okay so that is the total potential energy and then the rest is very simple okay so the last problem that again a challenging problem we are going to consider now so the idea is to design the you know an adjustable lamp which has a parallelogram linkage so this member is parallel to this one so it always it will always remain parallelogram okay and you can adjust its height but remember there is a spring so ultimately everything is controlled by the spring what is given that the unstretched length of the spring is b over 2 okay and then you see determine the necessary spring stiffness k for equilibrium at a given angle theta with the vertical so first thing is to find out the potential energy of this system remember there is a mass of this that mass is in terms of m and this mass is composed of mass of lamp as well as mass of fixture now we are not concerned about where that mass is located you really do not have to find it and that is the whole issue of this problem you can take the mass somewhere and as I was describing earlier that ultimately remember you will just have a vertical height and you can see clearly what will happen that as you rotate the body there is a fixed distance let us say from this point where the lamp is connected to this pin there is a fixed vertical height let us say okay so for the mass center what will happen there will be always a fixed vertical height that will be l cosine theta so l cosine theta plus some h so l cosine theta plus h is always going to be the height that we should consider when we are considering the potential energy of the force that force is mg so potential energy of the weight will always be l cosine theta plus h some h okay is that clear so that is the main concept here now you can do this very easily so only trick here involves is that I do have a mass center somewhere I do not know where it is exactly but since the problem demands that I connect that mass center in terms of theta so what I will do I will always take l cosine theta plus some height h okay and that h will never change that h is constant no matter where the entire assembly goes okay so if you buy that then we should be able to solve it so I will just ask you to do this you know little bit brainstorming at maybe take it as a exercise problem and homework problem so I will show you the solution it is everything is there okay so please look at the solution to this problem and kind of rethink it when we go back so solution is displayed now just think very very carefully so as I was saying that you can consider the mass center somewhere but the distance is always going to be fixed about this pin here so that is h so therefore you have mg l cosine theta plus h remember the spring length at the configuration so on at theta is given by 2b sin theta 2 from the geometry so the stretching is 2b sin theta 2 negative b by 2 okay that is the stretching x so 2b sin theta 2 is now the distance that is the final length negative original length was b over 2 okay so therefore we have the stretching now the potential energy is completely set up so we have spring force springs energy plus the gravitational energy so those are 2 things coming into play now we have to do dv the theta equals to 0 for equilibrium now what it will immediately give it will give you a solution for k and solution for theta is also obtained see what is happening sin theta if you do it as 2 sin theta 2 cosine we can clearly see that I have one solution here another solution will come from this right so there are two solutions coming into play so from this if I set it equals to 0 I get a value of k now what is important to understand this k will go to infinite at some value of theta what is that theta that theta is actually equals to what is given in the problem that sin inverse 1 over 4 right so that was given in the problem okay you can see that k will go to infinity so what does that mean that means that you need a very stiff spring right to maintain this configuration at theta equals to that angle that angle will come up as 29 degree that theta equals to you know sin inverse 1 by 4 multiplied by half so almost close to 30 degree I need a very stiff spring remember theta cannot be less than you know 30 degree if it is so then k should be negative so we do not have any solution so first I have that theta equals to 29 degree but for that I need a very stiff k now you can think of the ultimately for any mass okay I need that stiffness k tends to infinity for theta equals to 30 degree so it is completely controlled by the geometry of the problem then you see for more theta now you have to understand that for more theta what is happening and then we go to the d2 v d theta to substitute the you know theta expression down there okay and then we can say that whole range will give me the value of k so k will be 4 mg l 3 divided by 3 b square so k equals to infinite for theta equals to 30 degree and k equals to this which is now a function of mass at theta equals to 180 degree so in this range it is going to give you a stable equilibrium configuration