 So, let us try to look at the, you know, tutorials. Now, tutorials are displayed in the video. So, AB is a 7 kg slender load, right. It is riveted to 5 kg uniform disc. So, there is a proper fixity here. It is not hinge. It cannot be hinge because otherwise this bar is going to collapse. So, there is a fixity at this point, ok. Now, a belt attaches the rim of the disc to a spring that holds the rod at rest in the position shown. Now, what has been asked? If the end A of the rod is moved 18 millimeter down and released, determine the period of vibration and the maximum velocity of end A. Is the question clear? So, again the AB you can assume to be fixed with the disc at B. Remember, there is a mass of this bar and that is very, very important. So, the problem needs to be solved by taking the mass of the bar into account. So, there is going to be a static equilibrium configuration. First of all, we have to understand the effect of the gravity, ok. So, you can solve the problem. Can I really bypass the gravity effect and solve the problem? And do I get the same answer as that of saying that there is a static equilibrium configuration and I solve the problem from that, ok. So, think about it and just get into the answer. So, answer is the period of vibration and the maximum velocity of end A. So, you can actually assume the static equilibrium configuration as a original configuration because in this case, the gravity will not play any role. 1013, center 1013 has put in an answer. I do not see lot of, you know, centers are actually giving the answers out. As I said, do not worry about the gravity and all that. Just ignore the gravity. Just give it a small rotation theta. Just rotate the disc by a small amount theta. Therefore, bar will be rotated by theta and you do the way we are doing the problem. Someone said velocity is 9.6 meter per second. That is not correct. So, answer came correctly. I do not think it should take that long time actually because what is not given here? We know the, you know, we have already studied what is the mass moment of inertia of a disc about its center. We know the mass moment of inertia of a bar about its center. So, inertia forces can be easily considered. Yes, so answers are coming. So, time period of vibration will be roughly 0.5. Time period of vibration roughly would be 0.5 second and the velocity at the tip, the maximum velocity at the end A will be 0.228 or point roughly, let us say, 2-3 meter per second. Yes, so some of the centers are now giving the correct answers. What I will explain on this thing needs to be really solved in 5 to 10 minutes time. So, let us just try to go through the solution. It is a very simple problem. There is nothing. So, see how I have taken the forces on the free body diagram. As I said, in this problem gravity will not affect anything. So, you can assume the original configuration as the static equilibrium configuration and look at the way the forces are shown on the entire body. So, spring is isolated. So, you have the spring force here. Remember, it is given a counter clockwise rotation. So, based on that I have the inertia forces m capital L by 2 theta double dot. Let us say this is total length is L and this is the inertia force rotational inertia that is m L square by 2 L theta double dot. These two are acting at the mass center of this bar. Now, how about the disc? Disc will have also inertia force which is shown like the way it is. So, that is small m R square over 2, small m R square over 2. Now, look at the spring force. Remember, the spring will be stretched by how much? R theta. Therefore, the spring force will be K R theta. The spring force will be K times R times theta. If you take the moment, you will get K R square theta. So, ultimately equation of motion is derived here. That is the circular natural frequency. So, you can convert that to time period and I got a lot of answers. So, 0.5 second roughly that is the time period, correct. And then remember what will happen? Now, you can convert this problem. You know, you want to calculate the maximum velocity here, ok. So, we have the equation of motion in terms of theta. So, first you figure out what is the theta 0 initial condition, right. Initial condition will be 18 divided by L, is not that? 18 divided by L that is your theta 0 at t equals to 0, right. So, therefore, if you look at the angular velocity theta dot that is theta 0 omega n and therefore, the translational velocity, right which is in question that will be X dot L theta dot. X dot should be L theta dot. Is that clear now? Remember, there is only one degree of freedom. You can either choose theta to be degree of freedom or maybe you know this one at the tip that is translational motion at the tip can be taken as degrees of freedom. But the translational motion at the tip is related to the theta that motion is related to the theta. So, it is a single degree of freedom and as I was telling that it is always going to be a generalized single degree of freedom system, ok. Now, given that let us move on to the next tutorial problem very quickly. These problems as we see has to be solved in you know 5 to 10 minutes time. So, again in this problem you can show that gravity will not enter into the equation of motion. So, therefore, what we can do? We can simply ignore the gravity effect from the very beginning. Then problem will be much simpler. So, you need to express this answer in terms of the algebraic quantities involved. So, we need to derive the expression of omega n. Omega n should be derived in terms of stiffness k, the mass of the disc and the small mass that is connected by the rope, ok, small m. So, capital M, small m and k and of course, R will come into picture. Yes, so one answer came up correct absolutely correct answer from center 1114. So, again as I said it is all about the free body diagrams there is not much you know we should be talking about as long as we can do the free body diagram. So, give it a small rotation theta try to get the free body diagram correctly, ok. So, let us go to the solution. Just take a look now you know if you do this using even let us say you try to find out the static equilibrium configuration. You know you can try to find out static equilibrium configuration and then from the static equilibrium configuration you start you know give the vibration. So, give a small oscillation theta you will see ultimately that gravity will not enter the equation of motion. So, therefore, what we can do and this part I am not going to explain I mean that you know doing how to really look at the static equilibrium configuration and then we displace it by theta or rotate it by theta that is already there in the solution it will be on the module. But right now just think of if I really understand that gravity is never going to come into the picture if I measure it from static equilibrium configuration then I can simply assume that original configuration to be the static equilibrium configuration. So, as if I will just ignore the gravity effect from the very beginning. So, now let us look at what we have done I have given it a clockwise rotation theta. So, disc has been rotated about the hinge by theta. Now, let us look at the free body diagram remember why it is a single degree of freedom system because this mass m small mass m right that you know the translation of this mass is dictated or determined by the rotation theta as we all know. So, what you see here you see the free body diagram of the disc and the free body diagram of this small mass. So, remember what is happening therefore, there is a tension in the cable. Now, that tension is simply going to you know take care of the inertia force of this mass you look at this inertia force right here that is small m capital R theta double dot why capital R theta double dot because the translation of this mass m is simply going to be capital R theta right. Therefore, we have the acceleration to be capital R theta double dot. So, this free body is clear. So, now you can see t equals to negative m r theta dot that is the equation of motion for this block t is going to go to this ok. So, now you take a moment and when we take the moment equals to 0 right. So, t is already taken care of right. So, that is also coming here. So, ultimately what is the spring forces k theta r k theta r because one spring will be under compression the right spring under compression the left spring under tension, but the direction of the force on the body should be on the same direction ok. So, that is what it is now I have also included i theta double dot right. So, just take a moment about 0.0 answer will come out very easily ok. So, now let us get into the next problem. The next problem is again very interesting one and as I said it involves also buckling issue right that is the unstable system. So, we are going to look at this problem we have a vertical rod with a load p on the top, but it has weight also 60 Newton. Now with that it is maintained in a vertical position by 2 identical springs the spring is k and k determine the vertical force p that will cause the natural frequency of the rod approach to 0 for small oscillations. Again you want to solve it using dynamic approach fine, but as I said you can also use static approach as if you have solved this problem before using potential energy approach because what is the condition I really want to determine the p that will cause the natural frequency of the approach to 0. That means, time period is infinity that means what would be the p for which the system remains unstable statically if you think of it because at that condition omega n equals to 0 right. So, remember in that way you can solve also potential energy approach and if you want to solve dynamic wave fine just go ahead with the theta rotation give it a theta small rotation set up the equation of motion get the time period set it to infinity and solve for the force p. I would really urge you to solve this using potential energy approach also those who have solved using dynamic approach fine, but static approach can also be considered principle of minimum potential energy as we have discussed and you will see how nicely things are connected. So, it is a good practice and it delivers the concept what is actually unstable equilibrium what is again stable equilibrium at what condition I am going to get into the unstable situation. Yes the answers coming are correct I think 34 kilo Newton will be the answer and you can check that using potential energy also. So, all I am saying just you know this gives students immediate feedback how to draw the free body diagram see that is all once they know how to draw the free body diagram there is nothing. So, let us go to the solution quickly. So, give it a small rotation theta first we should draw the displacement diagram what are the displacement that is required and then we should think of what are the forces. So, the forces are shown you have the vertical load p now in the mass center of this rod I should have the translator inertia force rotational inertia force and the rotational inertia force and then the spring force remember 0.2 k theta you see here. So, one spring will be in compression now there was again question I do not see why you know some of you are unable to find out the direction of the force. So, probably for the direction of the force just think of it what happens to the spring first see the spring that was on the right side that is compressed. So, spring compression means on the spring it is compressive. So, it should be directed rightward therefore, as per Newton's third law on the spring it is directed rightward. So, on the body it should be leftward similarly the spring on the top will be under tension. Tension means therefore, on the spring if you draw the spring free body what will happen tension means it will be on the positive x direction let us say therefore, on the body it should be on the negative x direction. So, therefore, although one you know spring is under compression another is in tension, but idea would be that on the body the direction of the forces will be same. So, draw the spring you know this thing first spring free body first and then simply adopt the Newton's third law to transfer the force to the body is that clear now. Remember this term right here m L square over 3 that is really you know these two are added. So, when you take the moment m L square by 3 that is really coming from when I take the moment of the this force and the this force about A. These two inertia forces if I take the moment about A will be combined to m L square over 3 which is also natural process that is your I A. So, ultimately using this you know D'Alembert's principle technique I can get the omega n set it to 0 and I should get the solution for capital P that is 33 roughly 34 kilo Newton. As I said you could do the potential energy approach it will be a good you know homework exercise actually because again potential energy will be revisited just think of it. I am trying to determine whether theta equals to 0 is an stable equilibrium or unstable equilibrium at what condition it will be unstable right. So, I set up the potential energy expression this is what we have done also set dv d theta equals to 0 and then you can see theta equals to 0 is an equilibrium condition. Now, I have to find out at what condition that is stable. So, d2v d theta should come into play substitute theta equals to 0. So, P should be less than this for stability. So, for unstable P should be greater than equals to this and it should give you the same answer. So, P less than this is always stable condition. So, if you give P greater than this what will happen it will simply go to infinite mode of vibration that means it is never going to come back to its original configuration very nice problem you know to look at from both angles.