 we will start with the next part of the mathematical modeling topic. So, what we are going to look at is now if the motor is added with load. So, so far we have seen the modeling of motor and friction. So, different models of friction we have seen although those models were defined for a block on the on a flat surface those models can be easily like no seem to be transport like translated into rotary kind of a joint friction at the rotary joint as in the bearing case in the motor. So, that part is this friction here and now what we are going to do is like if see typically motors are not alone like they will have some load or some application to be driven. And when that application is getting driven by the motor this torque corresponding to that on the motor side will come. So, this is like a load torque, but which is seen on the motor side. So, that is why this torque tau ml. This m is stands for like note the torque on the motor side and l is like no it is actually because of the load. So, now since we have this load as we have seen the CD round drive like some kind of a belt drive operated here. So, there is a further that belt drive is driving gear and then gear is driving like some kind of a tray, but we are not kind of considering that for now. We are just considering simple load here, simple gear kind of a or the belt driven rotary disk here. For such a case how do you think about now getting the getting this belt connection modeled. If you plug in your kinematics and dynamics of machines fundamentals this is pretty straightforward. Remember this is like a rigid joint rigid body motion, but this is rotary motion for the rigid body. This is one rigid body and this is another rigid body and there is a coupling of bed in between. Now, this bed is considered to be some kind of a rigid coupling and it also has a is a massless kind of a coupling. Normally, beds if you see they are they are stretchable, one can stretch it and they will have some kind of a flexibility. We are not modeling that flexibility here. So, it depends upon how deep you want to get into the modeling. It depends upon what is your interest. Say if you are interested in like very accurately driving this output load and the belt is very flexible that it can get stretched when it is getting driven like the stretch is quite a bit then you may need to consider flexibility in the bed this as in like some kind of a spring here. But when it is now like let us consider to begin with a simple case where this is just driven in some kind of a rigid connection of the belt and belt is massless. So, belt mass and base belt inertia we are not we are ignoring because it will be much lesser as compared to the motor and load inertia. So, now with this like you can see that we need to draw a free body diagram to get these two rigid bodies considered. How will you draw this free body diagram? Just pause here for a while and then like no draw it yourself and then like no we will see further. So, we have to draw the free body diagram and then like see the concept of equivalent inertia. We will see what is this concept. But you draw the free body diagram here and then we can see in the next slide. So, you can see the free body diagram is up here. So, you have these tensions in the belt which are different here. So, if this is a motor which is getting driven on this in this direction motor which is driving the belt in this direction. So, the motor is rotating in this direction tension in the in this side is going to be higher than tension in the other side. This T1 is more than T2. Can you see that why? Pretty straightforward to kind of see that T1 will be higher than T2. So, we are assuming that you know here the belt is not slipping. So, this differential difference in the tensions is going to produce some kind of a torque here. So, you remember these are our like no motor equations on motor side of mechanical equations and in that we have this load which is we are considering right now load torque. So, this load torque now can be written. So, you can can you write these equations. How do you write this equation for tau ML here? So, in terms of T1 and T2 one can see that these equations are going to be like this. So, T1 minus T2 times our radius like the radius of the motor shaft will get you this tau ML. Now, if we see this expression this expression will be on a motor side. What is the connection to the load side? Same tensions on the load side will be existing T1 and T2 in the belt. Belt is continuous and like you know here this action and reaction process will exist up there. Now, one can write this tau ML is equal to tau M like tau on the load side is equal in terms of these two tensions on the load side. So, let us call that torque as tau L. So, tau L is a torque which will be applied by the motor to the load. So, it is like a driven like no driving torque the load. So, load is getting driven by this torque tau L and when load is given driven by the torque tau L on the motor side we see the torque ML as a load torque tau ML. So, that is how like we can consider this. So, you if you feel this is little I mean you can see the same thing without even considering this like this tau ML or tau L as a separate kind of quantities. These are just variables to kind of physically understand stuff more nicely actually. So, if you see that tau L will be T1 minus T2 into RL. RL is a radius of this radius of this load disk and RM was a radius on the motor disk here. So, this ratio between the two will be N. So, when one can see that this tau L is equal to N times tau ML. So, see since the load side gear is bigger in size the torque that is applied by the motor if it is on the motor side is tau ML then on the load side it gets like larger. So, N is a ratio of radius RL divided by RM. So, RL is larger here and RM is lesser here. So, this is like a number which is greater than 1. So, your torque enhancement happens when you have a bigger size this which you all very well know the same thing will happen with the gears also same way. So, you also can see that you know theta L will be equal to theta M upon N. So, the gear or this belt arrangement enhances the torque, but reduces the theta value or the displacement value that is the displacement value of theta L will get reduced by factor of N which is straightforward to see I do not see this is just a simple kind of a kinematics of this belt drive. Now, what are these complete equations yet? Yes, these are not complete equations yet. We need to further develop equations of motion based on the motion of this side. So, we have written the equation for motion on this side, we have developed some kinematic relationships between these two sides in terms of like you know both the torque and the theta. But still we do not have the equation of motion coming up on the load side. So, how do we do that? Again we go back to our kinematics and dynamics of machines fundamentals and we can see here that you know summation of moments is equal to j times alpha will give you equation of motion on the load side. So, apply this and let me see what all equations you can get and simplify them and see. So, you pause here for a while and then do this and then we will further see. If your equations are done now we can see here look you have jl theta i double dot plus bl theta i dot. Now, this bl theta i dot as we had on the motor side you have some kind of a bearing friction coming up on the load side also. That is what we have written here and that is equal to the torque which is which is driving the load tau l. Now, we know this tau l is nothing but n times tau ml and then now one gets one can get right equation expression for tau ml in terms of this quantities here. So, we can substitute that in our original equation. So, this we substitute in our motor side equation and we get now. So, here we are considering still friction torque here but we are not kind of really considering a model for friction yet. One can assume this to be zero also to simplify the analysis and see stuff without friction what is that we can see. So, now one can see that this is load theta double dot and theta l dot. Now, they can be transformed into the same quantities theta m here. How do you transform that? Theta l you know is equal to theta m upon n. When you use that you get this equation. So, you get this n square term here and now you collect the like terms which are multiplying theta m double dot and theta m dot and then you can see your equation gets to this form. This is very interesting you can see now. So, we get something like a equivalent inertia up here. So, this reads like j m plus j l upon n square. So, like you know see if n is really really very large. So, say n is 3 for example. Then your inertia l j l is reduced when it is seen by the motor by 9 factor of 9. So, if the gear reduction is really really high then as compared to the motor side like you know the load inertia value will be very very as compared to the motor inertia j m the load inertia seen on the motor side. This term is going to have a insignificant kind of value. Can you see that? So, for this value to be very significant you need to have a load inertia much much larger than motor inertia. So, this is very very interesting you know physical understanding up here. So, one can kind of say look I will just kind of account for my load side inertia by collecting like you know by by saying ok it may be like you know at most 4.2 or 0.5 times motor inertia and I can like directly say this j m equivalent is equal to some factor into j m that can be one simplification one can do depends upon like you know how do you have the inertia on the load side and how much is a gear reduction. So, you can work out some numbers for some specific cases and you will find that ok oh look this values of load inertia although like you know it looks very big that this is this motor is driving big inertia. When there it is transformed on the motor side it has not much of a you know value there ok. So, this is a very important understanding here. Now, we can reach the same understanding by energy way also ok how can you think about that. So, instead of like you know lighting all these you know free bird air I am safe especially if there are multiple connections. So, far we have considered only one bed connection then there are gears then there are some rack and pinion or some kind of a like you know gearing system is there. Whole of that gearing system would have a degree of freedom still one ok. So, motor position if it is specified all the positions in the system get specified ok that is what happens in this case also right. You see that you know when the motor position is specified like you know load position is specified time upon it. So, that kinematics is very simple for such a cases here. So, in that case like we can do like know some energy analysis like know little bit of energy analysis and reach the same conclusion how we can do that we can see it here ok. So, total kinetic energy if I want to write here one can say that see from the motor side the kinetic energy is j m theta m dot square half and load side it is j l theta l dot square ok. There is no translational kinetic energy and there is no nothing translating here. So, we are again ignoring the bed effects ok. The bed inertia is ignored here ok. So, this kinetic energy again one can see this transformation of load side theta position versus motor side position and after that transformation is applied you can get the same kind of a equivalent inertia term here ok. This is like equivalent motor equivalent inertia as seen in the motor side ok. So, by using the same kind of a principle for damping energy ok which is half times damping factor times c time dot square ok or theta dot square in general. Then one can write like the damping also in a similar way and you get equivalent damping then once one can see that this whole thing translate into the same kind of a governing equation as we had seen for the case before ok by using a Newton's analysis by drawing paper diagrams ok. So, this is a very important kind of consideration that one can directly use total kinetic energy and get to the same point ok kinetic energy and of course, the damping energy and things like that ok. So, you can reach this equation by using the energy way much in the simpler form without drawing the field per diagram. So, this T1 T2 does not come into picture when we when we write this kinetic energy ok. So, now this can be applied to a little bit more kind of a complicated cases also. So, we have now here like one gear driving the other gear then other gear driving the third gear then they are driving the rack and pinion and some kind of a mass is attached to the rack and pinion which is our CD-ROM head mass. So, this is a kind of a system where we can apply this now and you know one can see some kind of a simplified form of expressions coming up there ok. So, you may draw some schematic and try this out yourself if you want to. I am just kind of giving you some kind of a small schematic here to just prompt how things will happen ok, but really you need to draw your own kind of a expressions ok, you derive your own expressions and then like you know come to this stage and you know see if they are matching and things like that. So, you can see here there are this is a motor inertia jm here, load inertia I am considering only one kind of a gear instead of two gears two reductions here and then you have the rack and pinion which is driving this mass is a mass of the CD-ROM head. Now, if you see the kinetic energy again you can add the same kinetic energy for jm and jl, but the kinetic energy for the mass will be now half x dot x dot square ok. Now, we need to know the only the the kinetic relationship kinematic relationship between the x and theta ok. So, theta l for example is theta m upon n and once you know the theta l then you know this gear radius say rl2 then rl2 times theta l will be x ok. So, r theta is equal to x that that kind of a relationship we can use. So, this is rl2 is like a pitch radius of a pinion which is attached to the load side and driving the rack. So, this is how one can get this kinematic relationship. Once the kinematic in the relationship is known one can write substitute that in the energy expression x dot will be equal to now rl2 times now theta l I can substitute here theta m dot upon n ok. And once you once you use that into writing the total expression of kinetic energy you are going to get this equivalent inertia on the motor side. So, this now the mass inertia here gets transformed to the to the motor side in proper dimension form which is required at the motor side is m times rl2 square ok. So, this is like a inertia term this is having same kind of a dimensional quantity as jl or jm and divided by n square still comes here. Ok. So, this is very interesting way one can get to the simple systems very easily to resolve like you know how many different kind of moving elements are there in the rigid system kinematically most of many different many of mechatron systems would fall in this kind of a category and one can write nicely these form of equations. One can extend these to the screw drive also and things things like that. So, you can see now that. Now, other important part is like you know friction in this case. So, friction as you see you can see here exists at these points you remember now right like we had this two points b and c underneath where the contact happens with the rod on this side and there is some kind of a contact happening on the other side as well ok. So, so friction would exist mainly at these points. Now, if this mass is moving here and mass is having say inertial force you know x double dot times m in this direction passing through the cg of this mass. This force is going to have some kind of a torque as we see that if we start doing this free body diagram for the for the case of this you know rack and mass system ok rack is connected here and then we assume that ok this b and c points are somewhere here and here. So, can you see now how the friction will come up ok. So, so especially like you know the if there is any connection between friction and the inertia is what we want to see here. So, so for that you you really need to draw the free body diagram I will just kind of like now give you some small insight into how do we proceed and then like now maybe you can proceed from there ok. So, let us do that ok. So, if you see here I will draw this this rack ok. So, I will just draw some schematic for the rack ok and sorry this here and then you have this big mass coming up here ok. Now, this mass at this cg of mass you have some x to my dot ok or this this this mass here and this is x here. Now, these are the points where like you know some some kind of a a contact is happening with the rod. So, there is a reaction that is going to get developed at these points ok. So, some some reaction will be there at we call these points as b and c and there is this reaction here and there is this reaction here. Now, so say let us assume this reaction to be say in this direction and this reaction is in this direction ok. So, if I if I want to write the equations of motion how will I how will I write here is what you need to think ok. So, so summation of forces in x direction is equal to mass into x double dot ok is what I would write. Now, at these points where this this normal reactions are produced there will be some friction that will be produced which is in the opposite direction the motion ok. So, there is some kind of friction which is in the opposite direction with the motion that is going to happen there ok. Then summation of forces say y direction the perpendicular direction ok are going to be 0 because there is no motion in y direction ok. So, let us say this is y direction here ok I just put this because like you know write and thumb like this y y direction that you get ok. So, can you see now what is happening and I will also have equation for the moment right. So, this is one rigid body on which some forces are getting applied and for that this summation of moment also is equal to 0 because there is no alpha there is no rotation of this body happening here this body is getting only translation in the x direction ok. So, use these equations of motion and one can see that this this reaction forces say whatever this let us call this as n b here and this is a n c here ok. So, you will get from this equation you will get n b is equal to n c ok and then from this equation one can write. So, your friction force here say this is f b or f f b and at this point we have say force f f c ok let us say capital F ok. So, this friction is always opposing this direction of motion. So, and of course, you have this driving force coming from the rack somewhere ok. So, let us say driving force point we can assume somewhere here ok. So, this is a driving force which is coming from the pinion the rack ok. So, this is f let us say call this as f pinion ok. So, f p is your driving force which is driving the mass in the same direction as x here. And so, you say that if you sum the forces up in the direction of f p ok f p minus like you know your f f p minus is f f c is equal to your mass into x. Can you see that now f f b and f f c are our friction forces. So, f f b is equal to mu times n b and f f c is equal to mu times n c. But we do not know what is n b and n c yet but we just know that ok they are both equal. We need to plug in this moment equation here ok. So, now, summation of moments you take about what point say if you take the summation of moments about the center of mass ok. So, if you take about center of mass then you get the equation. So, m c g here ok. So, then so the moment created by n b is in this clockwise direction. So, you have say some distance between n b and n c. So, let us first name the distances here. So, you say this is x 1 or x b and x c. So, x b is the distance up to the point b from the center of mass ok and likewise you will have x c distance coming up here. So, so you will need to kind of consider n b times x b ok in the clockwise direction plus n c times x c again clockwise direction will be opposed ok. You will have this friction say friction moment up here also. So, that is in anticlockwise direction. So, minus now your f f b plus f f c ok. Now, I can consider this f p also acting as a similar kind of a point. So, minus f p ok f p is in opposite direction times say y some distance y p let us say ok. So, this moment is equal to 0 ok this is what I will get from here ok. Now, one can see that this f f b plus f f c minus f p ok is minus f m x double dot ok. Like that you can substitute that and then like now you get this n b and n c are equal. So, this will happen as n b times say x b plus x c ok which is this total distance between these points b and c. So, this distance b c will be x b plus x c and then one can write directly like you know n b times this distance b c here ok and this is going to get. So, you have this y y distance y p is still there and then like now this is this is equal to minus x double dot ok as f p plus f c minus f p is equal to minus m x double dot this is minus x plus and times y p ok. So, this is the kind of a relationship you will get for n b and you substitute that n b up here ok I will substitute n b here or here whatever and you get this in this equation and you get some simplification then. So, you will find here now like this friction forces are dependent upon this mu mu and n value and this n value is itself is dependent upon the x double dot term ok. So, this is the reactions are dependent upon x double dot inertia here. So, you can see in this on this block here see this if this distance b and c distance between b and c is very very small ok then this n b and n c are going to be very large. See this b c value if it is very small ok this value is smaller then for the same m x double dot here ok m x double dot here your n b value is going to be very very large ok. See if n b is higher these reaction forces are higher then that is not a very good situation because correspondingly your friction forces are going to be higher ok. So, if the friction forces are dependent upon these reaction values ok. So, that is that is why we want to kind of keep this distance b c larger. So, as to reduce the friction force in the system you see that at some point you know you may get into a situation that the these reactions are so high that will have very large difficulty in moving the block. So, this is how one can see the friction coming up in the system and the need for these points b and c to be like you know far away from each other then your friction forces are going to be lower. So, if you if you think that ok you you will support only at one point ok. So, say say you have you you have this mass and it is guided only at one point and on this side here ok I did only one point single point here then only whatever distance over which it is getting guided the ends of that those will get like know will be points b and c ok and since they are so close together you may feel that you know there is a situation of jamming coming up there ok. So, this is how one can think and like know analyze the friction or such a case ok. So, clear this part. So, these are the ways to think about and like to generate the equations of motion that that you want to use for your system and further use them in the in the control development later or in the in the synthesis of a system say in this particular case we want to push these b and c points far away from each other to reduce the friction that is happening in the system and then you get this equation of motion here. Now, from here it is not very direct connection as in the energy base connection to the to the motor side how the friction happens to the motor side that is not very apparent here we need to write this equation down and then one can kind of transform that to the motor side ok. So, when the when you have friction forces to be accounted for energies may not help you really better, but you can still see that ok the forces are getting still translated in terms of some kind of a gear ratio ok gear ratio and you know rack and pinion kind of a drive kinematics ok. So, those kind of parameters may scale your your friction forces in some way ok. So, I will leave it to you to explore that further and you know write develop complete model for such a system and see ok what is a what are the equivalent forces coming or how the friction is coming to the motor side this friction what we are talking about here. So, that you can derive yourself and look at that in more detail later ok. So, next come back to our slides. So, now this friction force whatever you are considering in here can be now considered as one of the models that four models that we have developed for the friction and then like know your equations will be fully ready for for other simulation and analysis purpose or for control development purpose ok. Now, let us see little more complicated cases. So, when you have a slider crank or like know some kind of a mechanism that is driven by your motor then you need to see ok think about say this question ok what is the dynamics for such a case in terms of this question one can see that for a constant torque what would what would be time evolution of the position. So, one of the quiz problems that we had ok where like know you are given as a simple link attached to the motor and it is driven and what is the time evolution of the position of the of that kind of a system ok. So, it may. So, if you have a slider. So, this was a simple single link now if you have a slider crank kind of a mechanism then like know there will be little more complications although again those systems have single degree of freedom ok slider crank mechanism or pore bar kind of a system you know those are again simple systems in terms of degrees of freedom there only one degree of freedom, but there are complications in terms of a non-linear terms that are coming and one can get to the energy methods then to handle such things ok. So, we will then like the next topic that we will see is the Lagrangian formulation which is completely energy based there we do not need to consider the free body diagrams ok that is a plus point for the Lagrangian formulation. So, that is what is the next topic that we will start with ok all right. So, we will stop here for now.