 So, then we consider now a next example of a belt drive with and without no flexible belt ok. So, this is again a system wherein like you know that ok this is theta m here and this is theta l here . So, this part of this derivation without flexible belt we have seen in the in the previous class or we have seen previously. One can see from there the kinetic energy already be had expressed there . So, so, you again here also I would like you to kind of derive this in in now this Lagrange formulation kind of a setting. So, potential energy in this case is 0 ok without flexible belt ok . So, without belt your potential energy is 0 and under that case if you apply your equation with the equivalent inertia. See now the when you use a generalized coordinate as theta m here ok the motion of the motor side theta m that is my generalized coordinate if I define in that way then I express my kinetic energy in terms of generalized coordinate and then I get like you know this j equivalent naturally as a part of kinetic energy . So, half j equivalent which is equal to j m plus j l by n square that times like you know theta m dot square that that will give me the kinetic energy expression and this is now j equivalent . So, this j equivalent will naturally come and then when you apply Lagrange formulation for this case you will you will find the equation of motion coming up in terms of this j equivalent and thing like that. And b equivalent for example, here we is we will consider as external force or we have to consider from the similar kind of a energy principle that we saw in the last class ok last previously what we have seen for the damping term we introduced that as a as a external force . So, now the question is like you know how do you deal with the system when there is a belt which is flexible which is to be considered flexible. So, what we do is like now there is a we instead of like this kind of a rigid connection we say that ok there is a flexibility here ok this part we have discussed little bit in the last previous part also. So, then when there is a flexibility or there is a spring here ok then whatever force that is getting applied to the to the belt by the motor will result in the tensioning of this or extension of this spring ok. And that is how they you know the theta for the load will no more be connected with theta for motor directly ok it is not like no theta m upon l n will give you theta l that is not possible now ok. So, there these become like an independent degrees of freedom. Now if they are independent degrees of freedom say say theta m is given theta l is given as an independent degree of freedom then ok the question is can we find what is a extension in the in the spring ok. So, that will be we will be able to find the extension in the spring and once we have that extension in the spring we can express that as as you know say delta as a as a extension in the spring and half ok delta square will be our potential energy ok. So, this is the additional part that we have like no we will have US kinetic energies and this additional potential energy of a belt that is to be considered now and then we can use Lagrange formulation and derive the equations. So, this this part you can do as a part of your work here we can move on from here now. So, you you carry out this derivation and see what is the equation of motion that you get and one can interpret and once you get the equation of motion one can interpret them in terms of Newtonian kind of formulation way. Are these you know looking at the appropriate terms here that are to become coming in the in the final equation ok that that exercise you can do. So, this becomes like a under actuated system ok it has only one actuation for the motor, but two degrees of freedom motor and the load. So, we will we will see how this kind of systems can be dealt for the for the control. Now, we can further extend it for the other case which is again another kind of a interesting problem. So, you have a gear attached to the motor shaft and to that output of the gear you have a link attached ok this is typically this will be a like no configuration for robotic systems. You have a motor and gear and then further you have some links coming up there. You may have multiple links for the robot of course, but we are considering now this is the most simplest kind of a form for robotic manipulator ok. So, the kinetic energy for this system can be written in terms of inertia of the motor side ok. So, this is Jn theta and dot square and then like now you have a inertia of load coming up here which is now hinged here. So, the inertia of this link that will be coming up as transferred to the to the hub that will come up as a kinetic energy. One can write this expression of kinetic energy in terms of this load inertia or one can say ok I will see that say this as a rigid body and for this rigid body the center of mass is here for the center of mass mass into velocity square half of that will be the transition kinetic energy and then I consider rotational kinetic energy and then whatever that total I get in terms of theta L now alone you have to express the translational kinetic energy of this rigid link also in terms of theta L and then we can get this equivalent ok this or this J about this point ok. So, one can directly consider that or one can see it in terms of you know your translation kinetic energy and rotation kinetic energy separately. Then once you get this kinetic energy expression you use the connection between theta m and theta L. So, gearbox ratio will be in the connection to express it in only in one generalized coordinate theta m here ok. So, this is typically an important step in the gradient formation to see that the kinetic energy is expressed only in generalized coordinates ok. So, in this case it is has only one degree of freedom system you need to have only one generalized coordinate to be there in the kinetic energy whatever you choose it can be theta L also. There is no that theta L also can be generalized coordinate is also a possibility here. So, we have chosen this theta m to be expressed here ok. Now if you see the potential energy the potential energy of a system is mg L 1 minus cosine theta L. So, this can again be expressed this is ok there is this L is likely different from this L because if you take a center of mass here the potential energy for that this should be for the center of mass here ok. So, you make that correction. So, this potential energy is in terms of theta L, but we need to convert it into in terms of theta m again. So, as to have a consistency in the generalized coordinate ok. So, you define generalized coordinate as theta m and then everything should be inconsistency with that that all your kinetic energy and potential energy should be expressed in terms of this generalized coordinate ok. So, that is a main kind of a thing that maybe you may miss ok. So, you need to make sure that all the time the generalized coordinate the energy expression should be in terms of generalized coordinate ok. So, you do this gradient formulation and then like you come back I think you first do this formulation yourself and see that what is the expression. So, you define the gradient to be kinetic minus potential energy and then apply this equation what we saw here. This equation if you apply you should be able to get the equation of motion for this case as well ok. So, through this exercise this is important that once you do you will understand how things work ok or some places you might kind of miss out some things that will get corrected ok and next time when you do you will not make those kind of mistakes ok. So, with that you will get this the gradient formulation like you know theta l replaced by theta m upon n get this kind of a equation of motion ok and for this tau is like your torque applied in the direction of generalized coordinate theta m. So, this is like a directly a motor torque here. So, this is motor torque or there are other torque that might be coming in. So, say for example, if you want to integrate this now with the with the motor equation how will you do that think about that ok. So, there will be friction torques and then there will be other torques that are coming. So, see we have already incorporated or counted for the motor inertia here ok J m is already considered here. So, so considering all these like you know one can see this torque can be consisting of missing torques that are not considered. So, this inertia torque is considered for example, damping is not considered here. So, those are kind of things that one can think about to incorporate into these external forces in the direction of motor generalized coordinate in the direction of theta m ok. So, those kind of forces will come incorporate that and then you may be able to get complete equation and this can be integrated with for example, with our motor equations ok. So, motor equations we have derived in the previous classes. So, that can be integrated with this equation to to get appropriately the entire dynamics. So, I will leave this exercise for you to do. You can see that in the absence of this torque ok there is no torque that is applied by the motor. So, this this this term will be 0 here when this term is 0 you can see clearly that this basically becomes and also there is no friction this torque is this torque consist of all these friction and other forces those all the forces are 0 ok. Then this gives you like you know the equation of a pendulum only right. So, say if you do not have even a gear box then like know it will be simply j m theta m double dot plus m g l upon n sin theta m upon n here which is now this n is 1 there is no gear box. So, this is these are 1 quantities and then this is j m plus j l it will become sorry it will not be this will not be 0. So, this will be 1 here gear box is not there means like you have a direct transmission. So, there is n is equal to 1. Then this be this this this falls into you know reduces to equation of a pendulum without gear box. With gear box you can see this n coming up here nothing like that. Like that one can have this physical interpretation of final equation formed and say that this makes sense. So, now this can be further developed by incorporating these other terms also into into this equation this damping term especially and you can additionally have you know the Coulomb and other kind of a friction models that can be incorporated ok. So, that that I will leave to you to look at. Then we can see this other application which is dynamics of these two arc manipulator ok. So, that is so I would suggest here again you can take this kind of a configuration pause here and think of your fundamentals from kinematics and dynamics and determine what are the energies here ok kinetic energy and potential energy. So, kinetic energy for that you need a center of mass of these linkages and at that point you need a translation and and rotational kinetic energy ok. So, it translate kinetic energy from the center of mass velocity and rotational kinetic energy from the real body omega angular velocity of that particular energy body. Notice here that this theta 2 is measured in the coordinate frame which is moving here. So, while writing energies you need to account for that and write proper expressions because this is not this theta 2 is is not defined with respect to the stationary x axis ok. So, the stationary with respect to the stationary x axis that the the angle will be theta 1 plus theta 2 ok for this link ok. This is done because like you know theta 2 is is deformation or like is is not deformation is the is the angle of rotation of the motor that is existing at at this point ok. So, if we have a robot which is like you know having these multiple motors attached to its different different joints then we typically need final dynamics in terms of the joint motions rather than like you know these otherwise we will have to find out the joint motion from the other coordinate. If I define theta 2 differently I will have to find out what is the joint motion ok. So, at this point let us you just pause and like you know derive the equations for velocity first ok. So, we just need velocity we do not need acceleration analysis here we just need a velocity analysis here and get the expressions for total velocity and further further further kinetic energy in terms of those velocities ok. So, velocities of the center of masses of these two linkages is what we need to get from our kinematics. So, how do you go go about that you can like you know write the coordinates of x x and y coordinates of this CG point here ok given all the all the geometry parameters and same thing you can do for the CG point here for second link what are the coordinates x and y for the CG point for the second link. And based on those coordinates once you get the coordinates you get say x CG 2 and y CG 2 then you differentiate that to get x dot CG 2 and y dot CG 2 and your total velocity of CG will be square root of x square plus y square x x dot CG 2 square plus y dot CG 2 square ok. So, that is how we have one can get total velocity and once you get that velocities or velocity velocity square you can use that in the expression of kinetic energy or writing kinetic energy of this. So, please do this derivation yourself and then you can proceed to look at the further thing further developments that we are discussing ok. So, I will pause here and discuss this further in the next part of the lecture ok. So, now we will continue with our topic of discussion of Lagrange formulation with this 2 R manipulator and then we will continue further about some of the properties very interesting properties that these mechanical systems yield with the Lagrange formulation ok. So, let us see the screen let me get the pointer now right. So, for this formulation I had asked you to do in the as a part of your own activity I hope you would have done that please see that you have done yourself before getting into like you know seeing the further thing because that will kind of give you the real feel of what is happening and also you any mistakes you have kind of made along the path they will get kind of rectified and that will kind of open up some concepts please do that activity before we move on. So, now for such a manipulator you can use your kinematics fundamentals and get to these equations for the velocity ok. For example, the velocity of the CG C 1 V C 1 is given here in terms of the coordinate Q 1 here ok. So, this is a Jacobian this is a new concept called Jacobian which expresses velocities of interest is velocity is C 1 in terms of the velocities of generalized coordinates ok. So, this generalized coordinate Q here. So, Q 1 and Q 2 are generalized coordinates. So, Q 1 and Q 2 dots will be formed by this vector and this matrix when it is multiplied by that you will see this velocity getting represented in this kind of form. If you just work out it based on using our kinematics and dynamics fundamentals, then also you can see the same thing let me get to that. So, if you see here these are the equations that you will start writing and you can see this Q 1 here and Q 2 here based on that you can write the coordinates of the CG's of the links X 1 Y 1 and X 2 Y 2 for link 1 and 2 and you are seeing the here there are the coordinates written here and X 2 Y 2 are written over here X 2 and Y 2 and then further differentiation of that you will give you X 2 dot Y 2 dot and X 1 dot Y 1 dot and you just take out this Q 1 dot out of these Q 1 dot out of these and Q 2 dot is not there in the X 1 case. So, that those terms will be 0 and you can form this Jacobian matrix as we have seen in the last part. So, this part. So, that is how like now you get this the velocities in terms of a Jacobian matrix and this Jacobian has some value we will see what value it adds especially for the robotic systems it has a significance when the Jacobian becomes singular matrix then we will have a problem in operating that robot ok. So, maybe that much is now enough for the discussion of Jacobian, but this is interesting way of like no concept and like write the equations of velocities of linkages especially there are CG's here for the sake of writing kinetic energies. So, that is how like now you then develop the kinetic energy and buy in the Jacobian form and further you get a rotational form added to it and then this rotational form becomes based on the angular velocities of linkages link 1 and link 2 you add that and here they are kind of added in some kind of a you know form which can be like the matrix multiplication representation here. So, this rotational kinetic energy I mean you can see that this is same as what this expression gives you ok. So, it is just written in the by considering this q 1 and q 2 is in this kind of a form. So, this is this may be useful for some cases when you kind of start developing things for a more complicated systems. Then potential energy of the system is straightforward you can write based on the relative CG locations and mass multiplied by the length and the location and by the gravity ok and you get this potential energy of the system. And then you crank in like you know the Lagrange equations you do it by yourself like now I will show you some part of the derivation here you can pause and look at it if at all if you have to cross check your own derived stuff you can do it in this fashion here. So, look here we have written expression for kinetic energy now not in the Jacobian form, but like complete expressions as you get and then they are simplified here b c 1 square b c 2 square there is some simplification that happens there. And then further you go and develop this complete form of kinetic energy here and start now you know collecting some terms to kind of express this kinetic energy in the form something times q 1 dot square only then something times like q 1 dot q 2 dot terms here. So, these are we are kind of generating some kind of a quadratic form for the kinetic energy expression here ok. So, this has also a value one has to kind of see that we will see what is we will discuss what is this value that we had by using this kind of a form ok. So, one can write in kinetic energy directly not directly I mean by simplification one can write the kinetic energy in terms of this vector here and then like you know this matrix multiplying application and again this other vector here ok. The vector is of the velocities the joint velocities. So, these are the velocity this is a total kinetic energy of the system expressed in the form of a generalized coordinate velocities. And in general these internal terms are functions of q 1 q 2 and so and so forth. They are not functions of q 1 dot and q 2 dots that is to be point to be noted. Once you get this kind of a kinetic energy expression then you further start developing the Lagrange formulation you write the expression for potential energy here and then express the Lagrange and by kinetic energy minus potential energy and further you start differentiating and this differentiation gives you you know different different terms in the expression ok. So, while differentiating you if you are expressing in terms of like this d 1 1 coefficients here we need to make sure that this coefficients are further functions of there may be further functions of q ok. So, wherever this full differentiation may come like this term need to be considered accordingly. So, say for example, this d 1 1 term has this cosine q 2 part inside it. So, when you are like you know differentiating in with respect to time then like you know your additional part of the terms will be coming there ok. So, this is like you can see here d 1 1 dot terms is coming there ok and you can account for that accordingly. So, like that you keep on doing this formulation and you know simplification of algebraic simplification of the terms considering all different derivatives that exist in Lagrange formulation and then you can for q 1 part you get this particular equation for q 1 alone variable q 1. Then you get another equation for q 2 part by considering the differentiation or a generalized coordinate q 2 for the Lagrange equation and then you get this second part of equation q 2 ok. Then you assemble these equations in certain kind of a form ok and why we do that we will see it later. But the form is basically you consider this matrix d q to be having this kind of a form then c q q dot to be having some form here ok. And then these g q terms are basically gravity terms coming because of the gravity forces in the potential energy and you have a torque vector here. And with this expression like you know your complete Lagrangian equation will get developed ok. So, this will get developed in the matrix form ok. So, we will further see what are the properties of this matrices little later. But you are expressing this particularly in these forms for some reason here ok and we will see that reason in a in a while ok. So, this is a formulation of two R manipulator that we have seen here and the same thing is put up here. Now with this formulation we further move on to see some of the interesting properties of these Lagrange formulations ok. So, the equations are expressed in this form and as you have seen this d 1 1 d 1 2 terms are defined as here also and in the other long end derivation also. And these are the c terms mentioned here. And with this you have these complete equations of dynamics of a manipulator derived. Now if you notice here this d 1 1 d 1 2 d 2 1 and d 2 2 they form a some kind of a matrix multiplying q 1 double dot q 2 double dot in the matrix equation form. So, this is a d matrix here and then this c matrix we define out of these terms here which are basically not having any double derivatives terms ok. And not we are not considering damping in the system no loss of the energy in the system and then there are these gravity terms that are coming here. So, with this there is this there is a very interesting property called skew symmetric property of d dot minus 2 c matrix which will which will come which can you can verify that or validate that property. But for this d matrix itself like you know you have this property coming up here that this d matrix is a symmetric matrix and it is positive definite matrix so this is a invertible d matrix that will come here. So, we will look at this little more detailed as a generalized thing, but you need to notice that you know this d matrix is coming out of the energy expression ok. So, one can without even Lagrange formulation one can write this d matrix as a as once you write the kinetic energy of the system in the form that was expressed in the equations here. You can see this form of equation of kinetic energy was written in this form ok. So, d 1 1 like now this was a d 0 term here which can be split into two parts like you know one part can be written here other part can be written here which will be both same parts ok. So, this d 0 by 2 we can come actually. So, so this is this form of the expression this terms directly appear in the expression of the of the Lagrange formulated equations of motion. So, that is what you is important thing to notice here and one can understand very easily why they would come once you have this energy expressed in this particular kind of form ok. So, maybe we will stop here for now and we will continue the discussion of this generalized form of this rigid body equation in the next part of this lecture ok. So, you go ahead and make sure that you understand this application to 2 R manipulator. So, that this generalization can be understood in a relatively easy way. So, that is what I would suggest you to do and we will stop for now here ok.