 So, we will now start the tutorial. So, for a disk what is the mass moment of inertia about the center? m r square over 2, I am just giving that I am not deriving it, but you can see the I think there is a yeah. So, if you look at the mass moment of inertia about the center about the z axis it is rotating. So, what is the mass moment of inertia? I have m r square over 2. So, just note this down and then we can solve this problem. By the way anyone could tell me that whether we have to really take into account gravity effect or not in this problem. So, we do not have to you can try to solve it by first considering the static equilibrium, because as we can see that there will be a static equilibrium configuration when this spring is going to stretch and this is actually coming down. We can start the problem from there, but ultimately if we really look at the oscillation from the static equilibrium configuration your equation of motion will not have any G on it ok. So, gravity will not come into the picture if I consider the static equilibrium configuration, oscillation from the static equilibrium. So, that configuration you can choose just along the line a b. So, line a b we can just say that ok it is my static equilibrium configuration ok and just give it a small rotation theta. So, it is a two line problem basically yes why not we have a spring no no no it is rotating you know the it is rotating along with the disc this bar is rotating along with the disc no that is attached to the disc that is firmly attached to the disc. So, that is a good point the bar is firmly attached to the disc and it is rotating otherwise it will collapse immediately if there is a thing. Yes the whole idea is to draw the free body diagram very quickly and solve that is all just draw the free body diagram get the equation of motion. So, time period of vibration omega n anyone got omega or capital T anyone no I have numbers no just use the numbers because to how much please no yes no it answer is 12.6 did anyone get that it is not that difficult you just there is a this force you find out no with the rotation is r theta. So, rotation is theta then we have the displacement r theta right and then that k r theta is the force and that moment of that is k r square theta and then from this side from the rod we are getting right ml square over 3 that will be added with the m r square over 2 I at least I am now expecting one answer because we have quiz also after 12 anyone got 12 12.6 is the answer yes. So, we will move on to the solution. So, here you know what we have done we have taken the I B directly here. So, that will have two components one is the m l square over 3 coming from the bar right because this bar if I look at the inertia force about these rotational inertia force then we are going to have m l square over 3 right and from the disc itself we have m r square over 2 ok and from the spring we have basically k r square. So, this is the equation of motion that we are going to get and omega n is going to be this oh that spring you see here what is the displacement we are getting here. So, if I move the system no. So, this is r theta in small displacement. So, spring force is k r theta we take the moment about a. So, that is k r square theta k r theta multiplied by r clear. So, remember in the mass here you can also show the mass you know at the lumped mass here with a translation if translatory force and the inertia force that can also be done or ultimate expect will come here that is m l square over 3 ok. So, in that problem. So, ultimately we have the time period of oscillation that is 0.496 maximum velocity at n day remember how we calculate that we know the solution for theta theta 0 is given what is theta 0 18 divide by 900. So, that is my theta 0. Sir. Yes. Can you go to the previous slide. Sir in this slide from D L M Byrd's principle in this we had written, but in this mass of rod is not you know taken care. No, I will just I think I will what I will do I will do this here right here because what is shown here is directly taking the I b there. So, we can actually do this see first we make a lumped mass here right and then we show that it is being rotated here. So, this should be l over 2 theta ok. So, that will be m l over 2 theta double dot right along with that I will have an inertia force to equilibrate this. So, direction will be this ok. So, this is the mass center is rotating. So, that value I know m l square over 12 theta double dot ok. Then remember I can draw this thing you know in a different way it was shown directly there. So, here I am getting the moment of inertia of the disc itself right. So, disc is rotating about its center. So, here I can show that this will be m capital R square by 2 theta double dot ok and then I have k r theta. So, that is the complete free body diagram we can draw. Now, remember these two forces that translator inertia force and the mass moment of inertia. So, these two if you are trying to take the moment about this ultimately become what? This will become equivalent system if you look at right that is going to be m l square by 3 ok. So, it will be ultimately m l square over 3 if you take a moment about this here and this is anyway going to add to that ok. So, ultimately we are going to get m l square 3 that will be coming from the bar about this point plus m r square over 2 ok. So, effectively I can show like this. So, ultimately what I have what I had shown there that I have a I b ok and that I b was equals to if you go step by step that is actually m l by 2 theta double dot. So, this is theta double dot I b that multiplied by l over 2 plus m l square over 12 theta double dot right plus m r square over 2 theta double. So, ultimately what we get is m l square over 3 theta double dot plus m r square over 2 is that clear now. So, we can simply show like this also, but it is always better to you know go to this mode because what we are doing is from the beginning this is the right approach. So, the approach is very simple as such we are just looking at you can connect this very easily with the virtual work that all I am doing here is that I am giving it a displacement right which is small comparable ok that both the cases we have we have you know we kept doing this one that we have to find the displacement at various point in virtual work you are just doing the work done right. In this case I am just saying take a moment you can also solve this using virtual work as we have seen any question on this clear. So, next problem therefore, just a little variation the problem statement is clear there is no friction about O. So, that is any case all the time ignored and we just have to displace the system that means we have to rotate a system by small amount you can give it in any direction you want it is a single degree of freedom system or is it a single degree of freedom system or not it will be because ultimately this mass is going to come down depending on how much I rotate. So, they are connected interlinked ok I have it in little different form, but you are correct yes. So, as such it is a very simple problem and remember you can always exercise this whether the gravity is going to you know affect it or not. So, it is a very good exercise if it is if you really have to take the gravity into account or not ok. So, I will again I will just ignore the gravity effect part, but we can show that you know how the individual things looks like. So, actually if you can consider these as a static problem also first to begin with. So, what happens due to this mass there will be rotation theta s t right. So, what is that theta s t can we calculate that. So, that can be calculated very easily you may or may not want to take two free bodies, but we do so because there are indeed two bodies. So, we can detach them. So, just this spring is going to take care the mass here ok that spring force. So, ultimately what will happen this is my static equilibrium consideration if we do then we take the moment about O right. So, that is static you know displacement or rather static rotation can be defined as this. So, make sure to substitute T s t equals to M g. So, that is my static configuration I have then we give additional rotation. So, now system is moving from the static equilibrium configuration right by an amount theta then again you redo the problem. Everything has to be you know recalculated when you go to static equilibrium configuration then I give it a small oscillation just take a look how the free bodies looks like what are the forces in the free bodies. So, that is my UAM for the block remember what is happening. Now, we are in dynamic right due to the dynamic effect this mass will now experience the inertia force translator inertia force that is equals to M r theta double dot remember since I have detached it. So, there is an increment in the therefore, tension in that cable right. So, T s t plus t. So, that T is the new tension right because we know there is inertia force already. So, T has to you know change it cannot be T s t ok. So, as I am saying you may want to detach it or you do not detach it it is up to you ok you may not want to detach that mass at all. So, ultimately you see I do this exercise just go through it very simple. So, for this mass small mass M g I have this equation of motion established ok and then you do the equation of motion for the dicks right. So, that will be some of moment about its point O equals to 0 ultimately you can prove that gravity will not play any role ok. So, there will be a part cancel this T s t M g this equation is in turn going to be carry forward and it will cancel. So, ultimate equation will become this right and some of you have got this right. Now, probably you have done. So, without taking the gravity effect that is fine because indeed I can prove that gravity will not play any role in this problem ok. So, just look at it you know these are exercise we give to understand that whether you know gravity effect is coming into play or not we can also do this without considering the gravity effect that is what all of you have done. If from the very beginning I do not say M g is there you see I have dropped the M g here I do not have M g now that is dropped I can solve it. So, I have shown you now both approaches one is to you know just look at the static deflection then from there we start the motion and there are problems where it may affect there are problems where it may not affect and it will be affected gravity will affect it as long as you have a inverted pendulum or a pendulum type problem ok that is pinned about point A somewhere you know vertical bar going like this. But in all of these cases we are not going to get any kind of gravity effect is that approach clear again you may see this one as I said since there are two free bodies therefore, I have detached it, but it is not necessary you can simply keep just say M r theta double dot and do not take into account t that can also be done. So, then you just say M r theta double dot is directly acting on this body in the direction of the cable ok and you take the moment of that thing about any question on this clear yes. Which one? See if you rotate it by a theta clockwise right clockwise one is one is compressed yes. So, if compressed then how it gives the force back to this that way right then other one is in the tension yes now that what it does that actually pulls that pulls. So, ultimately you have compression as well as you have the pull. So, both are in the same direction ok any question done? Now, last problem is going to be interesting. So, you want to determine the P that will cause the natural frequency of the rod approach to 0 for smallest whether very very nice problem again it is a its concerns actually stability of the system ok. So, you will be able to find a P where t tends to infinity time period tends to infinity or omega n equals to 0. Now, we can show statically also using the potential energy approach that if the P is less than that then it will never be stable ok at that theta equals to 0. This problem is just two lines problem and I would expect that more of you get involved and just give me the answer get the equation of motion first and set omega equals to 0 or t tends to infinity whatever way. See, you have to make sure that in this problem gravity will enter ok say inverted pendulum ok. So, you see that. So, gravity has to be taken into account from the very beginning you cannot ignore gravity and we have solved the similar problem remember there was a bar with two springs and attached mass was there. So, gravity was that there ok. Yes. It is a very large value we are going to get. What was that you said kilo Newton right and not 15 50 is not. Anyone just draw the free body diagram clearly what are the forces by the way 34 kilo Newton is the answer yes. So, just take a look at the free body diagram here quickly. Now here again you know I have taken here this as equivalent moment that is coming from the rod. So, this is m L square over m L square theta ok. So, ultimately we have this force right here there is another spring force both should be in the same directions. If I oscillate the system this way then remember there is a compressive force in this spring. So, that will act towards left this will also be tensile this will act towards left. So, there are two forces just calculate the displacement properly. Then we have the P that P will create a torque about this m g will create a torque about this and ultimately there is a distributed mass here. So, that will be m L square by 3 is this you can also lump it here no issue. If you lump it then there is a translational force and there is a rotation about the own center right. So, that can be done that way also, but see ultimately I am transferring the effect here already to here and that we have already learnt right m L square over 3 will always be the inertia force about the pin ok. Is that steps clear everyone ok. So, ultimately as you got the answer. So, roughly 34 kilo Newton 33970 Newton roughly 34 kilo Newton is the answer. Now, you can do a very you know good exercise here you know what I would ask you to do. Can you use potential energy approach and show that theta equals to 0 is a solution ok and P less than this will be the solution where it is stable. See as soon as P is greater than this it is going to be unstable even at theta equals to 0. So, ultimately start from potential energy forget about the dynamic ok. Suppose at omega equals to 0 we already said. So, now it is a static problem I have a P right. So, can we show that fall P less than this system is stable otherwise it is unstable is that clear using potential energy right that was done yesterday. How do you do the same approach? Now, there is no dynamic there is no inertia force all I am saying now the bar is straight right here and determine the stability that is all at theta equals to 0 configuration determine the stability of the bar whether it is stable or whether it is unstable period. Can we do that. So, what would be the potential energy of the system all of these are actually going to be you know there except for that you drop this one you can also do it in a finite rotation fine ok sin theta cosine theta can come to play, but small deflection you know your potential energy you just have to put together what is the potential energy. So, that will be half k delta square for the spring then you have the force right and you have the mg. So, all of these needs to be taken into account ok. So, I will show that quickly. So, there is no virtual work there is a mistake here. So, this is the we are just talking about potential energy. So, can we see that this is the expression for potential energy. If we can get that that is the energy you know expression then problem is solved then what we do dv d theta equals to 0. We can actually show that theta equals to 0 is indeed a equilibrium configuration right first and then we check for d 2 v d theta 2 greater than 0 right what is d 2 v d theta 2 greater than 0. So, the next step would be if you get that potential energy that is it. So, dv d theta we do right and ultimately see you have two cases here and we can show that theta equals to 0 is indeed an equilibrium condition right then question is whether it is stable or unstable. So, we substitute we do that d 2 v d theta 2 greater than 0 and for theta equals to 0 we get this expression back. So, this is the expression which says that p should be less than this. So, I have not incorporated all the algebraic quantities so far sorry it was done in terms of algebraic quantities. Now, let us put the numbers here. So, if I put the numbers you see the problem ok. So, p less than this it can only be stable see what happened in dynamic case with that p it has actually gone to infinite oscillation right. That means, it cannot come back as soon as see this is actually limiting condition that p equals to that is the limiting condition it cannot come back. So, what we have shown through potential energy that as long as p less than this it will be stable as soon as let us say p equals to 34 kilo Newton actually it is going to collapse it is unstable. So, you can link see what I have done today essentially I have also told you that virtual work why it is necessary I have ultimately told you that why potential energy is also coming into play to really do certain amount you know certain dynamic problems ok. So, everything is interlinked and as long as we feel that civil engineering we really do not consider those rectilinear motion curvilinear motions see ultimately we do not you know use the problems involve really lot of you know curvilinear rectilinear impulse momentum type of stuff what we really focus on the vibration for civil engineers. In fact, mechanical engineers they have lot of problem for vibrations aerospace lot of vibrations problem. So, we really in dynamics what we thought of doing in IIT and as you have seen that all the other things that we have done is very exhausted ok. And therefore, we just made a common platform of all the disciplines and we tried to say ok let us try to at least tell you know students that we can derive the equations of motion for spring mass system type of problems where you know which is applicable to many disciplines. And ultimately we are able to give you know some kind of flavor that ok we can link those problems with the you know stability, unstability. We can also do the virtual work instead of saying that moment equals to 0 can we use the virtual work to solve the equations of motion ok. Is that all? Any question? We will just do the you know but we can definitely discuss if you have different requirement we can try to see if that can be fit in into the main workshop for the dynamics ok. But for statics even we have to think whether we are going to be so exhaustive in the you know kind of in the main workshop. But we felt that you know whatever we do we do we do at a great depth I mean that is what most important than just doing it bits and pieces ok.