 వాపివాల్పాసితిలుటెలే. తాదిలునిస్యాస్త్తైస్స్లియాన్నినిచేటినార్సండందిన్ అంటి మంనుస్వా కిసి఍ధావై మరేనుచేలిటాందికేటి. And the reason to do so is that you have seen that whatever topic we have discussed that was really at a great depth. And if we want to deliver this again back to the students like this way dynamics will take one more full semester and for you another four days. So what we thought of doing it basically in most engineering applications we actually encountered a simple vibration problem. In fact, different disciplines like mechanical, aerospace, civil, they have their own kind of tailor made courses they have done on dynamics so that they only give those input to the students. In our case we thought that in all disciplines actually what is necessary is to establish the equations of motion for specially you know problems involving vibrations. So we are going to keep it very simple for example we see that you know machine foundations we can use the spring mass damper system and we can study how the equation of motions are going to be solved. So likewise in civil applications we see you know we have to always resort to the equations of motion for variety of problems. And so on so forth. So ideally what we are going to do today is we are going to study planar motion of a rigid body. And we are simply going to you know find out what is the equations of motion how to derive those equations of motion. Now as we know that for a rigid body if we look at this way that it is subject to external force system now they are dynamic in nature. And therefore we can actually go to the Newton's second law and say that the resultant force will be equals to mass times acceleration and the resultant moment. That will be equals to mass moment of inertia multiplied by angular acceleration ok. Now what happens in this case in the very first day we have actually studied the equivalent force system. So what I have presented at the very you know top left corner right here is the fact we can say that this system and this system is equivalent in nature. In an way such that these equations hold. So we are saying now these are actually effective forces on the body. And body is being accelerated in this direction and undergoing an angular acceleration also about its mass center ok. Now when you think this problem like this way although it can be solved this kind of problems. But we want to keep basically very uniform pattern throughout. That means what we want to do since we have already learnt the statics. We want to really convert this problem to a static problem. Now how do I do it? So we basically take the D'Alembert's principle. Now what is this I will just explain it in a very quick manner. It is basically that we want to introduce the inertia force onto the system. But how that is going to happen? But what we are really watching here is that let us say I talked about a particle. And that particle is subject to a resultant force which is sum over force here. And therefore we say that the particle is accelerating in this direction. And we can apply the Newton's second law sum of force equals to mass times acceleration very nicely. As long as we are concerned about a fixed reference system that fixed reference system is XY. So we say now particle at A the absolute acceleration is whatever you know the acceleration at this point A. And we say sum of force equals to mass times acceleration. Now what happens if I really move this reference system and attach this to the particle itself. That means we are saying now that the observer itself is accelerating with an acceleration A. So my reference system is being moved. So that means now I have to really attach this negative MA such that net force should be equals to 0. So that is the basis of D'Alembert's principle that we are going to work. So ultimately now we are going to talk about two different systems. So this system is actually our inertial frame system so fixed frame system. Whereas this one that is attached to the particle is actually non inertial reference frame system. So that's the basis of D'Alembert's principle. Likewise if we look at the rigid body motion of a let's say a planner system. Then we can say that okay whenever we are looking at this body we are simply going to add a mass times acceleration on its CG. That will be my translational force, inertia force. And I should also have a rotational inertia force. Now these two forces are going to always oppose the motion. Yes, J is the mass moment of inertia about its mass center. You can call it I also usually you see I so that is J. So we refer to J. So it's a general description of the problem. Now we are really going to make it very very simple. So now what we have learned that okay I have a rigid body it's a planner rigid body. And what I am going to study I am going to study simple spring mass systems. Therefore what we will do we will simply try to draw the free body diagram of that system. And we will put the inertia forces into the free body. And simply take the static equilibrium okay. So that will go very very fast and we can finish this session also as early as we can okay. Now let us assume that we have a mass and we have also spring attached to that mass. And this mass is on a frictionless surface, smooth frictionless surface. And what we are essentially doing is that we are giving it a small oscillation. So let us say if I apply a force Pt we can either talk about a forced vibration or we can either talk about a free vibration both ways. Let us say what I have displaced is here forced vibration. So I give it a Pt. Now question is as per the Newton's second law. How do I write the Newton's second law? P minus Kx equals to mass times acceleration. So P minus Kx is the resultant force along x direction. And that must be equals to mass times acceleration. So body is actually accelerating along the same direction. But if I want to go to the D'Alembert's principle then a priority I will just say that there is a inertia force. And that inertia force is simply going to be mx double dot where x is the displacement and we put this double dot in order to reduce the time of writing notation. So d2x over dt2 that is my acceleration. So now I can write the D'Alembert's principle that sum of force along x equals to 0. And ultimately what we see in this way is of the same form as that of the Newton's second law. So both is going to give me the same expression but I am simply exercising this problem in two different ways. That force is a function of time. Force is a function of time. In this case everything is a function of time. My displacement, force, velocity, acceleration everything is a function of time because I am talking about a dynamic problem. What advantage the D'Alembert's principle will give me and we will immediately figure out because we are talking about a planar motion. Specially when we want to use the Newton's second law we always resolve the moment about the mass center. And then we say that resultant moment equals to I times α that is mass moment of inertia multiplied by angular acceleration. But in D'Alembert's principle if we adopt we will see that we do not have to do this thing at all. We can in fact take the moment about any point in the body. And we can say that again sum of moment about any point equals to 0. So that creates the basis of D'Alembert's principle and it simplifies our life. So let us go step by step. So now we have talked about a spring mass system where the gravity is not going to play an important role because the motion was horizontal so gravity was perpendicular to the reaction and the acceleration was in the horizontal direction. Now if I really hang this system now so what happens? First thing that we have to really understand that due to the gravity effect I am going to have a spring force so first I have to study the problem in terms of static equilibrium. So where is my equilibrium configuration? So once I do that I can clearly see that spring force is going to be equals to K times ΔSt now ΔSt is the static displacement of the mass that is equals to mg. So that is my static equation. What I have done? I have simply calculated what is the equilibrium configuration of that mass. Next what I am going to do? I am simply going to now pull this from the static equilibrium configuration let us say by an amount xm so xm is the magnitude of the displacement that I am giving and also I give let us say velocity V0 and then release the system. So I just pull give it a velocity also and release the system. So it is basically going to now oscillate about the static equilibrium configuration. So now I am studying the problem from the static equilibrium configuration. Now what happens in this case? We can clearly see if I look at the now motion of the body let us say I consider the motion of the body at a distance x from the static equilibrium configuration so I can clearly draw a free body diagram. What is my free body diagram? Remember now the spring force that is equals to K multiplied by the total displacement total displacement is ΔSt plus x. I also have the weight that is the gravity mg and mass times acceleration now remember this one is only here I am concerned about the displacement x because ΔSt is not a function of time. So mass times acceleration that is my inertia force that I put in the body in a opposite direction. Now I take again equilibrium of that so I say sum of force along x equals to 0 what I get is this equation right here. Remember I have already said that K ΔSt equals to mg therefore mg and K ΔSt they cancels each other so I am going to get mx double dot plus Kx equals to 0. So for this problem if I measure the vibration from the static equilibrium configuration that means I am assuming body is oscillating from the static equilibrium configuration gravity does not influence the result gravity does not come into play at all so there is no g in the equation of motion so I will now rewrite the equation of motion that x double dot plus K over mx equals to 0 and I will just define a parameter which is ωn square now we call it natural circular frequency of vibration this is called I will come next but ultimately remember ωn square equals to K over m then I can simply solve this differential equation the general solution will be what is shown there you want sin ωnt plus c2 cosine ωnt right ok so now on this what I am going to apply I am going to apply my initial conditions right because my initial condition was that when t equals to 0 x equals to xm right and velocity was equals to v0 ok so x dot was equals to v0 at t equals to t0 ok so once I do that then we can actually put the initial condition and I can solve the problem based on the initial displacement ok and initial velocity so x0 and v0 there was a misprint here so this should be c1 ok because at t equals to 0 and this should be x0 by the way ok so there is a small misprint there now how do I when I plot this so I am simply going to plot it so therefore I can look at its motion so this is the displacement versus time here now remember what is happening in this case how do I actually tell describe this motion why that's angular you know circular natural frequencies coming into play so think about this that at a given time t my solution is x that means the displacement of the spring is actually OP right so what is happening the motion along the x axis of the spring is governed by a particle that is actually moving in a circumference of a circle ok so in other words any position of this spring can be defined as if a particle is moving along this circle and we can clearly see that this is my maximum amplitude that is possible so for completing one cycle going from here to here and then coming back that's my spring motion right the particle will actually complete 2 pi rotation ok so it will actually revolve revolution will be just one circle ok so it will complete one circle therefore what we can say now from the time period of vibration e right that is completing one cycle going from you know some let's say positive to again coming back to a positive displacement that will be defined by 2 pi by omega n so now that is properly connected time period of vibration is properly connected with the natural circular frequency of the motion ok so that's the interpretation of why this natural circular frequencies coming into play remember there are other quantity this is my maximum amplitude so which is xm and there is a phase lag now what is the phase lag which part of it anyone which distance is phase lag how do I decide this phase lag this phi yes so you can clearly see that curve starts from right here so there is indeed a phase lag that is in time right which is defined by this much of time ok and a complete cycle is really from here to here one peak to the other peak one positive to the other positive now in a similar way therefore I know that the solution is really x equals to xm sin omega nt plus phi that is the displacement of the spring what is the velocity velocity will be therefore equals to x dot that is dx over dt and therefore my maximum amplitude of the velocity will be xm omega n right and the maximum acceleration is going to be xm omega n square so all I have to remember are these three parameters 2 square root of v0 by omega square plus x0 square ok now why did that come into picture because I had also given it a initial velocity see if I make it 0 right if let's say I just take the spring here and release it then my v0 is equals to 0 at a time t equals to 0 right so therefore v0 will go away from the solution so we can actually say xm equals to x0 in fact you can clearly see that in that situation that phi will also go away that means actually your motion will start from here and going like this right so phase lag will disappear from the system so it will start from x0 right at t equals to 0 and it will follow that motion now we are going to go through simple problems so let us assume that we have two types of you know spring mass systems I am using one is connected in parallel one is connected in series ok and we want to really study that if I really pull this by 40 millimeter down ok and release the system then what is the time period of vibration what is the maximum velocity and what is the maximum acceleration ok now before I go to even solving this problem I have to set up the equations of motion right so first take one problem at a time so let us take this problem first what I have assume I have already said that if I solve the problem from the static equilibrium configuration for this type of problem right you will not enter into the equations of motion right so I will assume let us assume that oscillation starts from static equilibrium configuration so I am solving the problem from the static equilibrium configuration as reference ok so how do I draw the free body diagram immediately let us say I displace the system by this way so this is my x in this case however you can change the direction just be careful with the directions of the forces and the inertia so I have taken it up and if I take it up then what happens these springs are going to be compression and we know if the displacement is x therefore we have k1x and k2x right and since the body is moving upwards so inertia force will be downward so equation of motion in turn becomes very simple mx double dot plus k1x plus k2x equals to 0 that is my particle equilibrium ok so now you can see that k1 and k1 and k2 can be combined to make an equivalent system so that will be k1 plus k2 that means I can simply forget about these two springs and I can just say k equivalent time x that is my spring force ok so once I do this all I have to get this k equivalent then I can definitely get the natural frequency of circular natural frequency of vibration k equivalent divide by m so that is coming out to be this then we can get the time period 2 pi by omega we can get the maximum velocity that velocity is simply the displacement that was given remember displacement given was 40 mm ok that is to you know maximum displacement can be simply found by xm multiplied by omega n and acceleration will be simply xm omega n square any question on this clear so let us move on to the next one how about this first of all understand one thing is it a 2 degree of freedom system or 1 degree of freedom system it is only a 1 degree of freedom system why because there is no mass attached here ok and what we can show therefore that let us say that this is being you know displaced by x1 this is going displaced by x so ultimately what we have to come to we have to find k equivalent time x for the system as that of before so ultimately remember in this problem what remains constant the force in the spring right that should be same ok so force is not changing if you do the simple equilibrium right the transmitted force remain same in both the spring and that is the key that is the key to solve this problem ok and what I have said here therefore just look here carefully that I am really taking here that what is the p that p is equals to for the spring 2 that means this one the k2 x minus x1 that is the net compression x minus x1 right that must be equals to the force in the spring one that is k1 multiplied by x1 ok and you can clearly see now x2 x and x1 are now properly connected so we get to this so x1 is solved once x1 is solved you again substituted back here right so once we substituted back here we get k1 k2 k1 plus k2 multiplied by x so I ultimately got my k equivalent for the system so I can now replace this to spring by a one spring whose equivalent stiffness is given by k1 k2 divided by k1 plus k2 and the procedure will remain same so rest we will just follow how to get the maximum velocity and the maximum acceleration ok so now we have discussed this now we are trying to come to really rigid body because we are going to you know study the motion of a rigid body planar motion of a rigid body therefore what I have here I have just explained here in a different manner first of all let us assume that this bar here it is being rotated about this pin so what is the mass moment of inertia about the CG we can simply calculate it based on the definition of mass moment of inertia that is integral r squared dm ok and that we can say that if it has a distributed mass the distributed mass is m bar so this is mass per unit length ok so m bar is the mass per unit length ok so when I do the r squared dm so what is my dm dm is simply m bar multiplied by dx and then I basically integrate it from negative l over 2 to positive l over 2 so I get the j so j should be given by capital M l squared over 12 that is the total mass ok so therefore what I can immediately write that now when I am going to apply the d'Alembert's principle how I am going to show this inertia force right so now I have shown the inertia force so what is my j CG about the center of gravity the inertia rotational inertia should be shown like this because it is trying to oppose the motion of the body so ultimately we have to remember this that mass moment of inertia that is the rotational inertia of the body about its own centroid is equals to m l square over 12 for a bar of length l with a total mass m now let us say what happens if I really move this one to here that means now I would assume instead of rotating about its own CG it is actually rotating about some point let us say at lift anyone has the answer m l square over 3 will be answer you can again follow the exact same you know definition here so only thing what you are doing is changing the integration limit from 0 to l and look at the solution because now your rotation point is actually at the extreme end right so I have to now change the integration limit from 0 to l not from negative l over 2 to l over 2 see that is how we get the answer the answer ultimately should be equals to about a m l square over 3 now when I show it in terms of d l number principle remember I have already stated that the translator inertia force should be concentrated at the CG its own CG okay and there also is going to be rotational inertia that we can put anywhere no issue but ultimately now to study this problem in terms of d l number principle we will resort to this what we are going to say this body is now trying to rotate about this point a now that motion will be decomposed into two parts as if the body is being purely translated and then it is rotated about its own CG so think of now we replace the entire body by just a lumped mass here and that lumped mass is equals to m bar l equals to m and that m what is happening to that this is actually undergoing a pure translation as well as that lumped mass is being rotated about its own CG that means there are two inertia forces coming into play when I want to define the free body diagram of this particular system so what I have done here I have shown the free body diagram where this is my translator inertia force right and look at why it is like that because the displacement right here is equals to if the rotation is theta then the displacement l by 2 multiplied by theta right okay so velocity will be l by 2 theta dot and the acceleration will be l by 2 theta double dot that is my translational acceleration okay l by 2 theta double dot so we multiply that by m so I get that translational inertia force plus it is also rotating about its own CG and that we already know the solution that we have already shown that it is going to be always m l square by 12 multiplied by theta double dot now look at how easy it is what I really want to do I want to now going back to very first class that is my equivalent force couple system yes no this one is for the total body this is rotating about its own CG no sir earlier we took the total length as l and that theta dot work out to be l square m l square by 12 into sorry j that is m l square by 12 theta double dot if the total bar is length l l now l by 2 only it is no it is still l look at this thing I have written here so what the body is doing you see the body that it is actually purely translation plus it is rotating remember you have to you have to be consistent with the constraint at the left end body can be detached from the constraint at the left end so it is actually rotating about that point right here we are simply decomposing so I take the bar up then again I rotate it and put it back that's all very simple so as if now I said ok it has a pure translation which is simply you know what is the translation amount of the center that is l by 2 times theta that's the translation and then it is actually rotating about its own mass center so that is m l square 12 theta double dot now what is the definition of equivalent system now I want to transfer so now this system right here and that system they are going to be equal so equivalent so that means what I will do I want to find out what is my rotational inertia about this a right so I simply can say that I take the moment of this about a right that will give me the equivalent moment right so that equivalent moment is what I am looking at so what is my equivalent moment it is written here this is actually a tuple moment so that can be transferred anywhere in the body and this one will go with a multiplied by l over 2 is that clear so I ultimately get the same answer as I do using the formula in fact I am here even not talking about parallel axis theorem all I said that I want to create an equivalent system and I can transfer the equivalent moment to at some point a about which the body is rotating so I do not have to even explain what is my parallel axis theorem you can do that also using parallel axis theorem is that clear because that is the basis of whatever we are going to discuss from now so let us look at this problem again I have a spring mass system but look at the arrangement one spring is upward another spring is downward and what I want to first make sure that the bar let us say assume that there is a mass here let us try to assume this bar is mass less to begin with so first we assume that there is no mass attached to this bar but there is a lumped mass at the right of this bar which is mass less so in other words the mass of this lumped mass is larger compared to the mass of the bar so how do I develop the let us say equation of motion first of all and then I can definitely find the natural frequency of vibration and also the time period of vibration so again can we think of like this way once I give you this problem first of all what is going to happen it is going to be under let us say gravity right and it is going to actually get a static equilibrium configuration right so let us say what is my static equilibrium configuration so first problem is simply a static problem so what happens due to the gravity it is going to come down let us say it came down by delta s t okay and remember so this is my displacement diagram so what is the force diagram force diagram will be simply this spring is going to be under tension and this spring is going to be under compression now look at the direction of forces if it is under tension then it will be upward right and if it is under compression it is also going to be upward because in both cases one is creating the pull on the body and another one is the push in the body so in the free body diagram they are actually going to be in the same direction okay now you have the gravity so this is mg so once I do the moment about point a I get this equation so that is my static equation and we can get basically the static displacement here now what I said that I am always going to do release that now I am going to oscillate that system from the static equilibrium configuration so if I do that what happens so that is my static equilibrium configuration and now I am saying that okay the motion is now being measured from the static equilibrium configuration so it is basically going to oscillate like this about its equilibrium configuration what I want to really prove in this problem is that if the body oscillates about static equilibrium configuration in this case again the gravity will not enter into the equation of motion can we show that okay so now take a look at the you know this is my displacement now now from the displacement I could again get the forces right and I can create this force diagram so what is my force diagram look at the net displacement of the spring right so it has came down and then I have taken it up therefore the spring top spring is under compression right okay so now top spring under compression so I have put the sign downward in this case similarly this one will be however in tension right so in this case also it will be downward on the body right so it is creating the same effect as before so ultimately I have two downward forces coming from the spring force and then I have mg and then I have the inertia force which is being acted downward that is how I draw the free body diagram okay again take the moment about a equals to 0 what is my final outcome final outcome will be equals to this yes sir in this second FUD it is written over there or on the upper side it is 1 by 3 of xt so xt is measured from the static equilibrium configuration to upward okay it is the that particular distance and 1 by 3 because you are taking moment about some point no no 1 over 3 is the if you look at the span all are LLL okay so that was as you know so ultimately you have 3L as a span right okay so therefore it is at the one third distance from the so that distance is L over 3 similarly for the second spring it is at 2L over 3 it means that for the lower ground we are sure that it is delta xt but for the upper side it is a function of x in terms of t x in terms of t exactly because I am now so first is the delta xt is my static displacement now the system is oscillating from the static displacement because distances are same that is why it is 1 by 3 2 by 3 index exactly right okay is that clear I mean this is what the virtual work you know yesterday we talked about it okay because these are all small oscillation so we can do that way so now if I do moment about a then I am going to get this equation remember this is already a static equation that is what I have derived in the previous slide that what is my static is this right so therefore again this is going to go away so ultimately I get this equation which does not have any gravity effect now in this problem why there is no gravity effect actually entering because remember the action of the gravity is always these springs but if you think of inverted pendulum or a pendulum you have always seen that gravity is going to act gravity will enter into the equation of motion why because gravity will produce the disturbing torque and spring will always produce in this case what is that is balanced the disturbing torque is balanced by the spring but in case of a pendulum even if I attach the spring what happens as soon as I oscillate the system there is simply going to be a angular momentum that is created by the gravity which is not going to be balanced by the spring so there will be always a disturbing torque that will try to you know rotate that system trying to make it unstable so any problem that concerns with a inverted pendulum or a pendulum you will always see that gravity will enter into the equation of motion you won't be able to nullify it there is no way so we will just go to that problem later on now what is shown here is that if I just assume that if I ignore the gravity effect let's say if I don't take the gravity effect from the very beginning so I am going to get the same equation as before so I need not to complex my life what I am trying to say I need not to you know kind of worry about where is the static equilibrium configuration for these type of problems so I can simply assume I do have a static equilibrium configuration in fact I can assume static equilibrium configuration is my original configuration and I can simply try to calculate the oscillation from that static equilibrium and my answer will not change so you can clearly see what is happening in this case I have precisely dropped the mg effect that is not there so mg effect is dropped and I am going to again solve this problem and I get the same equation so now let's move on to the next one so far we have assumed that there is no distributed mass here yes like we can write the energy for the springs and differentiate it this will give the force that will be equal to that is also approach in fact what I will come to that just wait I will come to that in fact why I am taking moment here we have already studied virtual work right we have already studied virtual work right do I have to really take moment I can simply say the virtual work done by these forces equals to 0 and I will get the same equation back the reason being these are actually my active force system what are my active forces here spring forces, inertia forces right see if I now I am already said this is my equilibrium configuration now what virtual work principle states that if I give it a infinite a small displacement which is consistent with the constraints then the virtual work done by the active forces undergoing the virtual displacement will be equals to 0 right that means the work done by the active forces should be equals to 0 and that is what you know you spend the whole morning yesterday we can simply derive this equation of motion based on that what I will do I do not take the moment I will simply distort this system by a small amount let say delta theta or something and I can simply calculate what is the work done of all the forces acting and I will set that equals to 0 I will definitely get this equation back okay yes so can't we use d number principle straight away for this problem what d number principle for what for solving the solution straight away by taking that angular deflection that has been used this is what d number principle we are doing because we are using the inertia force which is opposing the motion what is that angular because there is angular displacement can't we take i alpha i alpha is already there in this case why i alpha is not coming into play see remember I have not taking the mass here right this is this bar is mass less now look at the next problem therefore that will actually remove your worry so let us assume now this bar has a distributed mass so if that is so then what d number principle states first I put the translational inertia force that is at its mass center so what is the mass center for that bar it is exactly at the middle so that is 3l over 2 right okay so remember therefore what is being put here you can see clearly the total length is 3l right so if you do that way so what is the displacement of this point l over 2 multiplied by theta capital l over 2 multiplied by theta therefore what is the acceleration l over 2 theta double dot so that will create the translational inertia force and then I have the rotational inertia about its own mass center so that rotational inertia force is ml square over 12 that is already proven so ultimately whenever I see system like this and it is actually disturbed at any point except for its mass center that means I am really talking about rotation about any other point than mass center then what we can do we can simply have the lumped mass at the mass center and that lumped mass will undergo both translational motion as well as rotational motion and that will give me translational force as well as rotational inertia force okay so that is how I draw the free body diagram now if you do that for our exercise sir one small doubt why do we use the symbol tau in place of t for time what is tau time period no no no this is what what is happening I am assuming this not here in general time period is tau so far you have used the symbol for time as a tau small that is for time period of vibration that is 2 pi by omega n is there any specific reason there is no specific reason you can also use capital T okay so capital T is also normally used but that is just a notation here but this problem that we are discussing now is really I am creating the static equilibrium at an instant of time t okay and that is why I am saying that at instant of time t what is the displacement that is x t x as a function of t this is x as a function of t okay I think that is the confusion produced why did I put this bracket was that the question okay no that is okay so you can keep it capital T okay so now in this case we get this equation what I want to show you quickly suppose I do the virtual work can we do the virtual work and you know get the same answer how do I do it so okay so what we have done basically now this is my equilibrium configuration right and what we have done we have given it a virtual displacement delta theta now remember that delta theta is artificial in nature and has no relationship with the real displacement or real theta that is taking place so we can clearly show that we can calculate the virtual displacement where the concerned location where the forces are acting and we can get the work done you know for each and every forces so we can clearly see how much work is done by this force let's say the work done is equals to k113x l delta theta okay similarly we can look at the other spring force right here so that is done so likewise we can get the for the inertia forces what are the virtual work done and you can carry out this straight forward procedure answer will be same but remember one point virtual work does not give any advantage if we are just considering a single rigid body because you have seen that virtual work is mostly used for interconnected bodies and you can calculate certain forces very easily using the virtual work so it does not provide any advantage whatsoever in case of a single rigid body okay so I just wanted to show that that is also approach in fact D'Alembert principle was you know kind of takes into this virtual work into account okay now so in this problem we have a rectangular rock resting on a frictionless surface and we want to derive the equation of motion for small oscillation in its own horizontal plane so remember the problem is like this so the reason I have taken like this is because we want to eliminate the gravity effect okay so let us not think about the gravity effect so it is actually in motion in a horizontal plane can we solve the equation of motion for small oscillations problem is clear so again this and this pin and we have a spring there so see if we can very quickly draw the you know free body diagram for this system so let us say how do I go about it first I get the displacement diagram okay so displacement diagram will look like this there is again a miss plane there is no static equilibrium again so just remove this one and just think of it that it is rotating about its own horizontal axis okay and therefore what we have if we calculate the displacement diagram what happens to the mass center the mass center will now undergo horizontal translation as well as vertical translation right okay so that means what are the inertia force coming into play what are the inertia forces coming into play remember the displacement here is b by 2 times theta and displacement here is equals to a by 2 multiplied by theta therefore the translator inertia force if we look at it so I am going to get a translational inertia force upward here translational inertia force leftward and as well as it is rotating about its own mass center so there are three inertia forces that it coming into play if I assume that body is lumped at its mass center okay so with that then what we can do remember I have used this gamma so gamma is the actually mass per unit area I have assumed so let us think of a uniform thickness okay otherwise what will happen if I say gamma equals to specific weight that thickness is coming into play so it will be you know abt will come everywhere so t will come everywhere in your equation if I just use gamma as specific weight so what I have done here gamma is being taken as mass per unit area okay so just look at this free body diagram and see how easy it is now for us to solve this problem all I have to do is take the moment about 0 sir in the first FPD will there be the opposite a by 2 and b by 2 theta opposite means for vertical displacement vertical displacement is given by you can go back to virtual small displacement so therefore what is happening if you keep looking from this so b over 2 theta will push that mass center here right plus it is going to come down by a over 2 theta right so there are two displacement taking place one is horizontal one is vertical and therefore I have inertia force on both ways translational inertia force is coming both ways and then I just add the rotational inertia force of the mass center which is equals to capital M a square plus b square by 12 that's my inertia force about the mass center okay so we solve this problem again so it is actually going to be equals to you see clearly that capital M a square plus b square over 3 that is what we are going to get if we really transfer all these three forces right here so what is my mass moment of inertia let's say about 0.0 see for a bar for a straight bar it was m l square over 3 right so for this plate it has two dimensions a and b so it is basically m a square by 3 plus m b square over 3 that should be the answer if I am talking about the inertia force mass moment of inertia about o okay so we can verify that by logic also but what is most easy here is that we can quickly create this free body diagram and try to show all the effect of inertia forces on that body so final answer will be this is your natural circular frequency and this is time period of vibration okay now this is again a very nice problem here now we are going to think about a horizontal bar let's say and we have also a mass on the top now remember in this case what is again assumed that the bc is mass less that is the whole idea here so assume bc is mass less okay and if bc does not have mass then how do I get the equation of motion for this problem so it is very similar to the one that you know spring system that we have studied we have spring system in series where there was no lumped mass at the middle of two see in this case you remember we are able to convert that problem a single degree of freedom this system so we have started with a two degree of freedom system but we are able to show that one degree of freedom is related to the other degree of freedom statically okay just because there was no mass in one of those spring attached to that spring so in this case it is a similar problem so how do I start it so let us assume I have a xa is the displacement here and xd is the displacement here okay so these are my two degrees of freedom okay so if I do that then what is going to happen now let us look at the force diagram so free body diagram of both so ultimately this spring is being so this is the net effect so k2 minus xd right and then we have here from here we have k1 xa right on this body and here we have k3 xd okay so now if you look at the equilibrium of this body then we can clearly say that what will happen is that this will be statically linked okay so I will get a solution between xa and xd so what is being done I take the equation of motion of this so equation of motion of this is given by here then in the next slide I take a moment about c so moment about c when it is taken then I can see that xd is related to xa okay and once I do that therefore it is converted to a single degree of freedom system so whole purpose of doing this problem is to show that there are certain problems where we can convert it to a single degree of freedom system we start with a two degree of freedom system because just that we do not have a mass in one of the system therefore it is statically correlated okay so we get the equation of motion and remember this will be my equivalence stiffness as such for this particular body and therefore we have the natural frequency of vibration so we are just going through series of problem sir in most of the cases we are not taking the mass of the spring my point is I mean spring is taking all the restoring forces so why aren't we taking the mass of the spring that is usual assumptions we make because its stiffness is very high compared to the gravity effect that you can actually take into account so we assume that spring constant is relatively higher so that k times u that will actually cancel or that is actually very large compared to if you just consider the mass of the spring and that is always engineering practice this problem xa is taken up so your xa minus xd so that will be net tensile that is a net tension going in so xa minus xd so if you think that way that means that spring is stretched so that is given as a upward whereas in this case this is also spring is stretched this is also stretched but it will come down on the free body right clear now here this rod a b is attached to a hinge a and we have two springs connected so what we want to do now this is a very interesting problem determine the value of k for which the period of small oscillation is once again an infinite so there are two questions asks what would be the value of the k such that time period of oscillation is either infinite in the first case and in the other case it is going to infinite this is a very very good problem to look at and we will explain why see first of all let us just try to get the equation of motion so again it will be very straight forward so let's say I give it a small oscillation theta so if I give it a theta let's say then what is my forces coming into play so remember there are two springs attached here this spring is going to be stretched so tensile force so that will act outward from the body and this spring is going to be compressed right that is again going to act towards the body but net effect is always going to be left toward right so we have that here now we have k d theta k d theta and if we look at the inertia force right here of this lumped mass that was a mass attached here then we can clearly see the height was h so displacement was h theta therefore the translator inertia force is m h theta double dot okay and now you have mg also now there is a disturbing torque which will never be balanced by the springs so the static equilibrium configuration that we have in this case is indeed vertical itself okay and once we distort that then we are going to get a disturbing torque which is going to come into play so ultimately you know the answer would be easy now we just take the moment about a and we find the solution okay now what it said is that we have two questions now one is the time period of vibration was one second right and another case it was infinite so what happens if I say time period is one second then I can just say t equals to 2 pi by omega n and I get the value of k from this right k was unknown what is the spring constant in the second case I get k value to be equals to this so we can compare this to values in one case first case high right that means system is how do I define the system it's a rigid system right okay and it was a stable equilibrium now this problem you can actually solve through the potential energy approach what happens what what does it mean by t equals to infinite infinite means it is oscillating very slowly coming back let's say that means in other word it is not coming back actually what happens once it is gone like this it is just unstable so that means what we can show you can calculate backward let's say I assume k equals to this and can I show using potential energy approach that the system is indeed unstable for k to be less than this okay so we are going to solve that type of problem in the tutorial so remember omega n equals to 0 is a static condition right okay because time period of oscillation is very infinity so we have actually trying to resemble the static condition here so what we can show for this k value less than you know this whenever we have this situation arises therefore the system will be unstable and you can do simply potential energy forget about dynamic potential energy approach just proceed with that k less than equals to this your system has to be unstable for theta equals to 0 degree that's the answer clear okay now this will be the last problem to work on a bit complex but again now remember this is kind of frame system so I have a frame system in hand I have two bars attached to it so this is well right here so we have basically you know that's a L shaped bar this attached by a spring I am also considering the distributed load everywhere so we have a mass per unit length for these two bars and I have a lumped mass at the end can I solve it equation of motion now we have to think of you know very systematically in this case that how different masses are actually participating in inertia forces and that we can do simply by look at if I give it a small rotation theta as you have done also in case of virtual work what are the different displacement that is being taken place okay at the concerned location how system is displaced and once we identify the displacements at each and every individual point of our concern then we can basically put all the inertia forces correctly so just look at it carefully what happens to this mass let's say this bar so as per the alarm bar principle I am going to consider a lumped mass right here at the center of it so that lumped mass how much it is translated translated by L by 2 multiplied by theta that's the only translation in small displacement okay therefore my inertia force is m multiplied by L by 2 theta double dot that's my translator inertia force clear I do have also rotational inertia force that is my j about the CG because it is actually rotating about this point right here right and therefore the lumped mass will make its own rotation right about the center okay so I have j is equals to how much j should be equals to m L square over 12 okay for a bar now when I go to this point right here if I assume that I have a lumped mass right here now what is happening in this case this point is displaced both horizontally as well as vertically so I have two displacements attached to that point okay right similarly this mass will also have two displacement attached to it if I just go by the small oscillation small displacement then I just make sure that I calculate all the displacements correctly so therefore the inertia forces if I look at it remember this body is able to rotate about its mass center as well as it is actually making adjustments for the displacement that is being taken place so therefore in this mass I am going to have three inertia forces one is coming from the horizontal translation another is from the vertical translation the other one is about its own axis it is rotating is that clear for this mass however I am going to have just a translator inertia force because this is kind of you know lumped mass right here so that is properly attached with this bar so I have a horizontal translator force and I have a vertical translator force so it is very easy practical it is very easy as long as we understand that small displacement and how each and every point or rather the lumped masses are being moved in a rigid body so we get all the inertia forces and we take either we do virtual work and we take out the zero now remember what I have not considered in this problem I have assumed already that ignore the gravity effect do you think in this problem gravity will come into the equation of motion it is actually an inverted pendulum problem it is an inverted pendulum problem because that what is happening once this mass is shifted they are all going to create a disturbing torque about the pin the gravity effect will create the disturbing torque about the pin therefore what happens the gravity actually comes into the equation of motion but I have not considered in my solution so that will be you know you can go back and kind of exercise this that really if I look at a static equilibrium configuration first so take into account the gravity effect then you displace the system from the static equilibrium configuration then what is the equation of motion it is going to change so I have one suggestion relating this kind of problem sir the approach that you have done is alright no problem but earlier slide you have proved that the center of rotation if you are considering individual bars in that case you have to considering both translation and rotation in fact this is making a bit difficult but that is the proper dialam but principle see if we cannot move the eye if you are doing this let's say I calculate the mass center about this yes I am talking about about A it is taking pure rotation because A is the center of rotation so if you take moment of inertia and everything about that point it will you can actually figure out the moment of inertia about that point will be equal to D exactly so if you take the whole system about that point moment of inertia this is less prone to from an understanding point of view that is ok but for solving that will be much more easier if you take about if you want to go that way I have no objection to that but that is not a proper dialam but principle just keep in mind but we can prove that it will be same about that point what happened dialam but said the individual masses needs to be lumped so we are going to first consider the where is the mass center where is the mass center of the individual system and we are looking at the displacement of those centers and we are simply lump you know taking the effect of the inertia forces onto that but what you said yes we can solve the problem by you know solving the mass moment of inertia about that point itself but this is more visible I would say you can get immediate feedback that ok what is happening to the system but difficult to solve considering both motion no not at all I don't buy that at the free end where the mass is there you have considered the two displacement x and y where at the free end mass yam displacement diagram yes at the free end you have considered the two displacement x and y at the corner where there is only one displacement along the x only where now I look carefully here junction point this m small m go to the diagram displacement diagram yeah so what happens at the junction when the bend is moving to the right that one small m or which one you are saying small m just see at the corner when you are moving the body to the bend to the right yes you have given the displacement we have x and y both no y is not look at this no at the corner I am talking about corner corner junction point junction why do I have to consider the junction in displacement diagram in displacement diagram yes that is only l theta that is small oscillation right we have done this in virtual world these are all small oscillations you cannot get a vertical component if I have a body like this and I rotate by theta very small then what is the displacement in first order that is l theta horizontally only there is no displacement in the first order vertically okay