 So, now there are a few questions let me answer them before going on to the final topic of our day which is the angular momentum of a particle. So, first question is from Sir1313 and the question is how to calculate Coriolis acceleration component? There is one thing which I want to emphasize is that at strictly speaking Coriolis acceleration component the way this term is used is used when we are analyzing motion in a rotating frame of reference which we are not doing here but still if we are if we think about if you want to think little bit about Coriolis force this all these analysis that we are currently doing are from an inertial frame point of view and as far as this particular course is concerned we are good with that. But if you still want to have some interpretation of the Coriolis force how does it look like in an inertial frame of reference then this component to r dot theta dot you can think of this as the Coriolis force component when interpreted when interpreted from inertial frame of reference. So, this 2 r dot theta dot is a component which is very typical for radial and transverse components and we can think of this as something which is similar to the Coriolis force. So, the next is with friction on the pulley explain one problem is asked by center 1 0 6 0. So, if we have friction on the pulley then those problems they become reasonably involved especially even if the mass is present if the pulley has a mass and it also has friction but again now friction is of 2 types one is the internal axial friction which we had discussed. So, let me go on to the white board. So, the pulley friction can be thought in this way that this is the pulley this is the pulley and this is the tiny slot in the pulley by which it is attached to a bearing. So, the pulley is attached to the bearing bearing or the screw whatever way you want to think about it through which this pulley is attached and now the pulley is free to rotate in this direction and we have strings T 1 tension T 1 and T 2. Now note one thing if this pulley mass is neglected we say that the connection between the pulley and the bearing. So, this is the bearing is well lubricated or frictionless almost frictionless then in that case even though there may be a friction possibility of friction between this rope and the pulley. So, new as present that does not mean that this new this friction has to act just because there is a coefficient of friction present between this rope and pulley does not mean it will always act if this internal lubrication is very high that means this joint is almost frictionless and the mass of the pulley is neglected then in that case T 1 becomes equal to T 2 if the friction is present here then the problem becomes significantly more difficult why because we have to look at what is called as journal friction and I do not think we are going to discuss this problem here ok journal friction and the dynamics of journal friction we are not going to discuss that problem here. But if the everything is frictionless but still the pulley has mass then we will see that T 2 minus T 1 times R can be written as moment of inertia of this pulley ok about this axis of rotation times the angular acceleration it can be written like this and some problems can be solved accordingly. So, let me see if I can incorporate a problem of this type ok in tomorrow's tutorial ok let me see if I can do that but note that if friction is present here it is an incredibly difficult problem ok let us not worry about that for the time being ok. But if you have mass then this will how the equation will be modified but in the problem which we had solved earlier we had neglected the mass of the pulley and we had also said that all the joints are frictionless so automatically this I alpha ok because moment of inertia it depends on mass for a cylindrical disc it will look like M R square by 2 since mass is neglected this approximately becomes 0 and then T 1 become equal to T 2 is what we had discussed before fine. Third question is how we can use constraint relation for this problem the question is asked by 1 2 1 5 it is not clear to me that what is this constraint relation that what problem we are talking about can we visit that center 1 2 1 5. Yes sir sir the vag and the block problem sir can we use a constraint relation the acceleration in x direction of vag and x direction acceleration of block and y direction of acceleration of block with a constraint relation as we did in the pulley problem. We take the reference point and distance yes sir yes ok let me let me answer that question ok. So let me go to the corresponding slide let me go to the block problem look here in the block problem one thing which is very clear ok that this look at each of the blocks individually ok do not look at them like as a part of this complete system think of this block A independently now we know for sure that block A the constraint on the block A is that that it cannot have any acceleration in the vertical direction why because it cannot penetrate this it cannot lose contact. So this acceleration is 0 so but it can have a free acceleration in the x direction so that one acceleration that is block can have now look at block B now this block B ok if you look at it independently ok then what can it do it can have two particular accelerators it can have two directions of motion ok one ok it can have a motion in the vertical in a normal direction it can have a motion ok independently if this block were not present or it can have a motion in the tangential direction. But note one thing that because of the presence of this block what we have realized now that the presence of the block is constraining the motion for this one in the in the normal direction. So that constraint we are already using that we are using the constraint that this block A cannot penetrate so it has only acceleration in this direction similarly for block B there are two possible direction one is along the block along the incline one is perpendicular to the incline. But what we realize is that that it cannot penetrate in neither can it lose contact so this acceleration is still gone ok. So the only free acceleration it has is in this direction but also note that because these two are connected with each other if the block move A moves in the horizontal direction that additional ok because the contact should not be lost between A and B. So that additional acceleration also comes on to it but only thing that happens is that that there are two independent degrees of freedom is this motion and this motion as opposed to in the pulley problem where we are we can reduce the two degrees of freedom to one degree of freedom problem here also we have used the constraint that is block B cannot penetrate this cannot penetrate but after using up all the constraints the four motions that these can have one two three four ok two for this two for this those four now they became two after using the constraint. So we have used constraints but only thing is that this system is more complicated than the other system and the four degrees of freedom that it can possibly have independently are because of the connections between them reduce to two degrees of freedom is that point clear. Yes sir. Sir there is another question sir E theta and E t such that radial direction in the radial direction and the tangential direction are similar to tangential and radial which we are taken in the radial components. So let me clarify this point ok this point is a very important point ok only when the motion is purely circular ok let us look at this a perfect circle the trajectory of a particle is perfectly happening along a circle only in that case note that E r direction is this because I choose the origin as my because note that for tangential and normal acceleration you do not need to choose an origin ok it just depends on the tangent to the path and normal to the path whereas for radial and transverse coordinate E r and E theta we need to choose a origin. So let us choose our origin as the center of the circle now this will be my E r and my transverse direction will be E theta but now if I want to go into tangential and normal coordinates what you realize that the tangent to this is also equal to E t ok and also note that if this is E t then this becomes E n the normal direction so E n is opposite to E r but this is true only when you choose the origin at the center for a motion along a circle. If you have any arbitrary motion like this ok and I choose the origin here this will no longer be true. So these are completely independent things but in this special case where the motion is along a circle this is definitely true ok that E r is equal to minus E n and E theta is equal to E t ok. Sir in means for E r we have to join that point to the origin along that line. Yes that is very important yes and perpendicular to this yes so in radial coordinates for this E theta. Yes yes yes so you need to have a origin ok you need to have a origin when we want to define say for this curve this will be my E r ok and this perpendicular to the direction will be E theta ok because this is my origin. Any curve other than circle what 4 components no no so those are all different they are not identical see I am not saying that all the accelerations are independent but you can no longer say that the radial acceleration is the same as minus of normal acceleration and the transverse acceleration or the transverse acceleration is same as the tangential acceleration you cannot say that anymore the acceleration is only acceleration it is only A bar you can write this as A n E n plus A t E t you can also write this as A r E r plus A theta E theta but now A r and A n they are not equal if they are not moving along a circle ok. So this equality is gone now because why these directions are now completely different the tangential direction will look like this the normal direction will look like this ok so this will be E t and this will be E n. So the acceleration is the same but only when you break it into components those components will not have any easy relation with respect to each other now ok but because acceleration is acceleration you cannot say it as 4 components yeah fine ok. So next question is by 1060 how to differentiate kinetics problems kinematics problem ok the definition is very clear ok kinematics only motion position velocity and acceleration is kinematics what is kinetics that given that certain forces acting on the particle or given that a particle has a particular kinematics what is kinematics it has some acceleration and so on what are the forces or given those forces what are the acceleration. So the moment we refer to forces then immediately we are doing kinetics when there is no reference to any forces we are doing kinematics. So kinematics has no reference to the forces where kinetics always has reference to the forces. So let me put it this way that we can trace the trajectory of a particle and do the kinematics of the particle ok but we cannot know what are the kinetics of the particle until and unless we invoke the Newton's law and say that f is equal to m times a and a is the quantity which is the kinematic quantity and m times a ok is the kinetic law ok which relates the kinematics with the forces in the system ok I hope that is clear. So what are dependent and independent motions the dependent and the independent motions so I believe that this is the kind of thing we are talking about for example in this case the motion of a and b are dependent on each other why because the constraint connecting them in this problem motion of b ok independently if a and b were completely free then in a plane the degrees of freedom that b has is motion in the y direction motion in the x direction ok if you choose x is this y x is this and y as this it has two possible motions a has two possible motions y vertical horizontal but because they are linked with each other ok those motions for example or because this a is constrained to move on this block we know that there is no motion of this a in the vertical direction similarly because b is constrained to move only along this block and along with the block we cannot have an independent motion relative to a which is going in this direction again relative to a which is perpendicular to the which is perpendicular to the direction of the plane. So these are simple examples of dependent and independent motion if this a and b were completely separated from each other and if there were no and there were no constraint provided then you would say that motion of b a are completely independent but because b cannot lose contact with a there is some component of b ok that we cannot have that b cannot have any relative acceleration with respect to a in this normal direction ok let me put it this way the only relative acceleration it can have with respect to a is along this way but it cannot have a direct a relative acceleration in the perpendicular direction. So that is I can say is a an example of constraint motion and if these were completely free then that will be an example of completely unconstrained motion ok. So I hope this point is clear the question is 1131 can the tension in the string connecting bob have a maximum value the maximum value ok will all depend on what is the velocity at that particular time ok what is the particular velocity because if you look at this pendulum problem ok let us look at this pendulum problem here ok let us look at this problem suppose at this point suppose we are not given the tension ok we know that there is some velocity for the bob here and there is a weight of the bob. So if you look at this diagram ok this figure what we realize is that that only forces that act on the bob is the weight which is the main force and the second is the tangent force which is a reactive force because it is a reaction to the motion and what we immediately see here is that that this acceleration ok the tangential acceleration in this direction is governed purely by mg sin 30 ok is governed purely by this component that is a tangential acceleration whereas this acceleration ok which is equal to v square by r is nothing but t minus mg cos 30. So clearly the tension depends on the weight of the block it depends on the weight of the bob and it also depends on what is the velocity of the bob it also depends on the length of the bob ok why because the acceleration correspondingly ok is given by v square by rho ok that this 2 point this tension minus mg cos 30 is equal to m times v square by rho. So the final tension in the bob it depends on the on the weight it depends on the velocity and it depends on the radius. So all these three components will decide what is the tension in this bob and depending on what those values are will decide what is the maximum tension that the string can have when it is connected to this bob which is moving ok with that particular velocity ok I hope this point is clear that it depends on mass it depends on velocity it depends on the length which is the radius of this path. There is a question by center 1060 is centripetal force and tension in rho the same or the different in pendulum problem ok. So this is a very valid question ok it is a very valid question and let me put it this way the centripetal force what is centripetal force m times an this quantity is the effective centripetal force that acts on the bob if you want to use the terminology and that centripetal force comes from a combination of tension and what minus mg cos 30. So the total centripetal force acting on this will be this tension minus mg cos 30 and that better have a direction acting inwards ok otherwise the block cannot sustain that acceleration means for example if the string breaks think about it if the string suddenly breaks ok I cut it then what happens is that the only force acting on this bob along this direction ok is mg cos 30 acting outwards. But what is the acceleration the corresponding acceleration is v square by rho but it is acting inwards and because there is no tension to take care of this ok this mass cannot sustain the bob cannot sustain the motion along the direction of the circle it will fly off it will fly off in the direction of the velocity that it currently has think about it if you are rotating like this ok in a vertical plane cut it suddenly then at that instant whatever velocity it has ok it will just fly in that direction it cannot go along that same curved path why because there is no tension ok to provide it the required centripetal force. So the centripetal force is only m times an and this m times an is the required centripetal force to keep this bob moving along this path with a given velocity v at any instant and t minus whatever the weight so weight is trying to pull it away tension is trying to give it the centripetal acceleration and the difference of them is the resultant centripetal force. So the centripetal force has component coming both from t and from the radial component or mg cos 30 component of this weight or the normal component of this weight. So centripetal force is just m times an and it comes from this various factors tension and the weight and so on and so long as the effective force coming from these the total centripetal force ok which is the effective force by all these forces along this direction can equal m an ok you are good ok the bob will keep maintaining that trajectory ok. So we cannot say that tension is a centripetal force or tension is different than centripetal force tension is a part of the centripetal force ok I hope this is clear last one how to identify the direction of Coriolis acceleration. So let me note this thing again that Coriolis acceleration ok is actually a 3D phenomena if you want to go to the complete ok non inertial rotating frames in 3 dimensions to get a full understanding of that but I mentioned that in passing is that if you want to have some interpretation of the Coriolis force because Coriolis force is a non inertial force in the strictest technical terms it will come into picture only where when we are working in non inertial rotating frames of references but all the problems that we are mentioning here all the kinematics is done in the inertial frames of references but if you want to have some intuition ok that in a non inertial frame of reference what will the Coriolis force look like then this quantity 2 r dot theta dot ok if I look here that this component 2 r dot theta dot and the direction is what the direction is this E theta direction is when you say you can think of that as a Coriolis force but strictly speaking this is not a Coriolis force ok to really get an understanding of Coriolis force we have to go to rotating frames of references and from that ok you will find out that what are the accelerations the non inertial acceleration and Coriolis force will be one component of that but if you are moving in a circle for example if the motion is motion along a circle ok where there is for example typically like you have like the merry-go-round problem that we have a merry-go-round ok and some particle is trying to move on that merry-go-round or we are trying to walk on that merry-go-round and we are ourselves stationed on the merry-go-round at the center ok and our frame now is the non inertial frame of the rotational frame of the merry-go-round and in that non inertial frame ok you can think of this 2 r dot theta dot as a Coriolis force term but it is not a good idea to interpret that as a Coriolis force term I just mentioned it briefly so that you can have some correlation with this word that you had heard but strictly speaking this you cannot directly interpret that as a Coriolis force it is a non inertial force ok it comes when you are working in non inertial rotating frames of reference center 1 1 0 3 will there be any centripetal acceleration in the direction of radius ok so I think the centripetal acceleration the discussion that we had a few moments ago I think that will suffice ok so with this much discussion we come now to the end topic so we will for the next few minutes let us discuss about what we mean by angular momentum and how do we define angular momentum for one particle and how do we define angular momentum for a system of particles ok so this is an extremely interesting topic ok and it will become more clear when we take care of rigid body kinetics and rigid body kinematics tomorrow now how do we define angular momentum of a particle now this is a definition ok so you may ask that why do we define it like this so there is a reason ok one simple reason is that that we define angular momentum like this and whatever definitions we had defined about moments everything ok in for static scores similarly lot of empirical observations that we do real experiments and do measurements and whatever measurements we do and whatever we see whatever this equation tells us they do a very good match and as a result ok we define this quantity which is angular momentum of a particle ok this angular momentum concept is a concept when we define it ok we will see that we can understand a lot of things which we observe empirically and things fall consistently into place ok so let us define now what is angular momentum of a particle to do that ok let us define it in full 3D we have a coordinate frame x y z let us fix our origin at O now a particle P has a position vector of R and let us say it has a momentum MV in this direction the moment of a momentum ok or the angular momentum ok there are two different terminologies that is used it is moment of a momentum why moment so moment of a force is what R cross V so moment of a momentum what is momentum MV so moment of a momentum is nothing but R cross MV or it is a angular momentum of the particle about O it is just defined as R cross MV or R cross L so this is momentum which is represented as L so the angular momentum of this particle about O is R bar cross L now what is the direction of A0 because this is a cross product R bar and MV or the momentum ok are two vectors and two vectors can always define a plane so unless and until they are parallel to each other two vectors will always define a plane and what is this quantity if you recall our cross products then this quantity is perpendicular to the plane containing R and MV and we can use the cross product definition and can easily find out what this quantity is now what we can also do is this ok we can also say ok that this angle which MV makes with this R let us say that this angle is phi ok so then this quantity V sin phi ok what is this quantity V sin phi V sin phi is nothing but a quantity which is perpendicular velocity to this ok which is the component of momentum which is perpendicular to R and that can be written as M R square theta dot because this is V theta ok let me let me discuss this concept for a second on the white board ok if you recall let us look at the planar motion this is R bar ok this is MV bar let us look at only a planar motion for the time being this is X this is Y this is theta now what do we have this is E R and this is E theta ok so what is the angular momentum for this particle here P about point O so H O will be nothing but R bar cross MV bar but note one thing that R bar cross MV bar what will that component will be ok that component will be just some magnitude multiplied by k hat which is a unit vector in the z direction coming out of plane ok out of plane now what can we also do for this if you want to find out what is the magnitude of this if this angle is phi what is this this can be written as R M V is the speed multiplied by sin phi but what is V sin phi if you look at this E R and E theta so if you look at this E R and E theta this is V bar MV bar this is phi so MV bar sin phi is nothing but a component of velocity in the theta direction and if you recall our radial coordinates velocity V theta can be written as in radial coordinates as R theta dot and as a result we can say that the angular momentum is given by R into M R theta dot which can be written as M R square theta dot ok so this can be easily written in 2D we can write this as M R square theta dot now coming to this let us find out what is the derivative ok so this is just one additional derivation just remember that now let us look at this full derivation full expression that H bar ok that H bar O is equal to R cross MV bar now what if we take the derivative of this angular momentum with respect to time let us take the derivative what do we get we will get R dot cross MV bar that is the first derivative plus R bar cross M times V bar dot but what is R dot R dot is nothing but the velocity and what is V cross V when two are parallel to each other sin of angle between the angle between them is 0 so the cross product also becomes 0 so this terms become 0 or H bar dot can be written as R cross M times A or it can be written as R cross F why because from Newton's law we know that for a given particle M times A is nothing but the forces so H bar dot ok or the rate of change of angular momentum of a particle is nothing but the sum of all the moments ok because R cross F is what there are moments acting on the particle so R cross F is sum of all the moments acting on the particle and that is nothing but H bar 0 ok so this is as simple as that what we know is that that when we have one particle then the rate of change of angular momentum of that particle about a fixed point this point A should be fixed is nothing but sum of all the moments about this point ok now let us take a special case ok let us take a special case when only force acting on the particle is affected or from point ok let us say that this point P is here this point O is here now what happens is that the only force that acts particle is directed in this direction away from this direction then the body is in the case said to move under what is called as a central force for example two planets ok when they move under the influence of their gravity the force acting between them is only the gravitational force between them and that is one example of what is called as a central force ok now since the line of action of the central force passes through O what we can say is that that sum of all the moments of those forces is equal to H bar dot and it has to be 0 why because those moments pass through the origin and they cannot and they cannot have any moment on this force about point O so in that case we will realize that when all the forces acting on the particle pass through the origin then the rate of change of angular momentum is 0 now what does that mean that dH by dt of H bar 0 is equal to 0 it means that the angular momentum of this particle remains constant and we will solve a few problems tomorrow about what this thing means now if you are doing a planar motion for example as we had discussed just a few moments ago if you are looking at a planar motion then in that case what happens the magnitude of the angular momentum is simply R this is the R this is mv is the velocity into sin phi and if all the forces acting on this particle are moving only in the direction ok are acting only in the direction which pass through this point about which we have taken the angular momentum then we can say that the angular momentum now is conserved ok that dH by dt is equal to 0 so H bar remains constant and as a result R mv sin phi is my angular momentum of the particle along this trajectory about point O and this quantity will hence remain constant so for example if we know what is the angle what is the radius then we can find out how does the velocity move so if we can keep finding out that what this theta and phi how does it change as a function of time then if the only force acting on this p acts in a direction such that it cuts to O ok that it passes through O only force acts is only in this direction then knowing the position of the particle and knowing the trajectory ok then we can immediately find out how does the velocity of the particle vary as a function of as a function along the trajectory so this in simple words is conservation of angular momentum so there are a few questions so let me take those questions and then I will discuss briefly about angular momentum of a system of particles ok so first question is by center 1313 is asked how to calculate a gyroscopic couple for a car ok so first of all let me briefly mention that this gyroscopic motion we are not going to cover in the syllabus it is given in great details in for example in Bayer and Johnston 10 but what I will briefly mention is this ok I will not go into the details of that ok because gyroscope ok it is a it is not a straight forward topic you have to think a lot about it it is a 3 dimensional rotational motion I am not even gone into the rotational motion portion at all but let me just briefly discuss what is the gyroscopic motion the gyroscopic motion is as follows ok but again I am emphasizing this that I am not going to go into the details of that ok you need to do appropriate amount of mathematics ok lot of visualization that is not a part and parcel of this course ok in most engineering syllabus is ok this is not discussed ok so I am not going to discuss this but briefly let me mention what is the gyroscopic moment so let us say ok you have a wheel ok which is rotating like this ok let us say you have a fly wheel ok forget about rolling on forget about even rolling on the ground let us say that there is a axis like this and this wheel has a constant rotation omega above this axis ok now this axis is x this is y and z comes out of the plane now if I look at this wheel from the side how does it look like this is z this is y ok z y this is the angular acceleration and this is the vectorial representation ok this is the vectorial representation of omega again this is a highly specialist topic I am not going to go into details of that but just let me briefly mention since you have asked ok let me just briefly mention that so that is how it will look like now typically the gyroscopic moment is what the gyroscopic moment is as follows now you take this wheel ok and you try to rotate it ok your wheel is rotating like this ok now you tend to rotate it about the y axis you tend to apply a rotation about the y axis here and additional rotation for example in a gyroscope what happens is that you have also and another axis ok not only this so there are these 3 different axis over which the rotation can happen so if you also try to apply a rotation along the y axis ok this is the spinning wheel you try to apply a rotation like this then what happens what happens is that that this was the direction of the angular momentum ok because omega i omega you can think of that as angular momentum we discuss that briefly tomorrow that direction of angular momentum tends to change because when you are trying to rotate it like this you tend to change that direction and when you tend to change that direction then what happens the rate of change of angular momentum is the applied torque and you will see that the rate of change of angular momentum happens in this way and there is a torque that happens in the third direction that you try to rotate it like this you will see that the torque acts on it in the third direction and that torque is typically called as the gyroscopic torque it is nothing but if you look from the top ok now if you look from the y direction then what happens is that if I look now this is y this is z if I look from the top direction how does it look like z x what was the initial direction of angular momentum like this but if you rotate if you rotate this wheel then the direction of angular momentum becomes like this even though omega remains constant and this is a change in the direction okay what is this direction change in the direction what does it do okay this change in the direction will create a torque like this why because vectorially this is l 1 bar this is l 1 bar prime and this is delta l bar what is the orientation of this delta l bar the delta l bar has orientation around the x axis and what is this torque this torque acts like this okay so this is a much detailed question so wheel rotating like this you tend to rotate like this it is very counter into to you will see that you will try to topple in this direction and that effect is called as a gyroscopic effect okay so let us like I will only discuss this today let us not discuss about this question okay tomorrow because this is completely outside the purview of this course but because the question was asked I briefly explained that what is the gyroscopic movement one rotation is already happening you try to change the direction of that you will get a torque in the third direction that is one simplest example of what is a gyroscopic torque and like when a car is rotating for example a wheel is rotating like this when you tend to turn okay a similar effect is happening that the wheel of the car is rotating like this okay when you try to steer the car what you are changing is that you are changing the angular momentum of the direction of the car wheel and so the wheel will have a gyroscopic couple or a gyroscopic moment that acts on it okay so I hope that this is good enough but I will not go into any more details of this problem so the next question is can you please repeat the concept quiz okay sure okay just give me one minute let me go through the other questions also this is simple because that quiz okay I have to go explain that in some more details there is a third very simple question is asked by center 1165 for a block kept on a rough horizontal plane how do we apply a constant velocity the way you apply a constant velocity okay is such because on a rough surface when the block is moving the force acting on the block is what okay the force acting on the block is this mu k times the normal reaction so this is the weight if I draw the free body diagram this is the normal reaction mu k times normal reaction this is the direction of velocity and to keep this block okay moving at a constant speed now what does constant speed means that no acceleration if that is the case they should know up then you should apply a force P which is equal to mu k n then the block will keep moving at the constant speed okay that's all it means okay for a rough surface how do we achieve constant velocity what is the difference between centripetal and centrifugal force I had discussed that yesterday the question is asked by center 1071 the question is very straightforward centrifugal force is almost a misnomer centrifugal force is used okay when you go to a non inertial frame of reference where you see that apparently there is no acceleration because if I am rotating okay if I mean a rotating frame of reference I will see that everything around me is stationary but still I can measure the force for example if I am holding on to a block okay and I am moving okay then I will see that there is a tug that is coming on to the string but I apparently see no acceleration then how do we explain that we say oh in that non inertial frame of reference there is a non inertial frame which is called as centrifugal force that is acting outwards okay so centrifugal force let us forget about it it comes only when you discuss non inertial frames of references from an inertial frame point of view okay centripetal force is the appropriate thing to use okay so let us stick to inertial frames and we use centripetal force for a point moving in a circular direction or moving along a curved direction center 01047 okay they are asking what is holonomic and non-holonomic constraints okay first of all okay so this is a highly highly sophisticated advanced question typically these kind of questions come in what is called as classical mechanics in physics okay but since I know the answer to this question let me show off by telling you what is holonomic and non-holonomic constraint but again the gyroscopic couple holonomic non-holonomic don't form the portion or the syllabus for this course what is holonomic constraint okay this is typical this is clearly show off okay that there is nothing else because this is not relevant for this course whatsoever holonomic constraint is what that suppose we have a block which is constrained to move in this direction okay then what do we say we say that the position of this block okay this is x I can say that it has a position x a it has a position x b this is one example of holonomic constraints look here here what we saw is that the y b or the y coordinate of b is equal to half x a by from this inextensibility of the roof so this particular concept which is explicitly mentioned okay in terms of position coordinates okay is one simple example of what are called as holonomic constraints now what is non-holonomic will become clear if I tell you that suppose we have two spheres like this this is a this is b this is omega a is angular velocity of this this is omega b these two gears are rotating about their respective centers now if we are given that there is no slippage at this contact point then what do we know that the relative velocity here should be 0 and what will you say that a theta dot a will be equal to b theta dot b or if I think in terms of infinitesimal then for any tiny rotation a delta theta a is equal to b delta theta b so the constraints are given in terms of infinitesimal motions and not direct coordinates okay so those are typical examples of what are called as non-holonomic constraints okay but these kind of concepts okay I kindly request that we will stick only to the syllabus okay these are highly they are not highly advanced but for example when you go to variational calculus or when you go to classical dynamics then these are the concepts that we discussed there and these becomes very important for example when we discuss what are called as Lagrangian or the Hamiltonian approach we write what is called as a Lagrangian approach using generalized degrees of freedom and there is what is called as Lagrangian's equation of motion which is very similar to Newton's law but is completely generalized and in such equations of motion we have to input these constraints okay so as to solve the problems appropriately. So one question 13 say again question by a center 13 or 2 in the problem 12.6 it is mentioned that the weight of the sleeve acts in the z direction can you justify that with the help of a diagram okay it's not a justification it is a statement of fact okay but I can only illustrate it more carefully okay in this problem what we are told is that this is a horizontal plane okay so if I draw a simple diagram for this how will it look okay so this is the r theta plane or x y plane x y plane this is the sleeve and z axis how it acts like this okay outwards now if I look it from the point of view from the x z plane okay let us look from the x z plane how does it look like this is x and this is z and this is how the sleeve projection will look like okay sleeve projection will look like this and gravity acts in the downward direction. So the only thing that gravity does is for example that is mass think about it this mass can slide only horizontally it can have only planar motion because of the constraints and as a result the acceleration of this mass in the z direction is equal to 0 straight away and the only thing that the gravity does if you draw the free body diagram is that that this is the weight which the gravity exerts on the mass and the sleeve which was and and a rod to which the sleeve was attached that will provide a normal reaction to balance this weight as simple as that so for the dynamics is concerned this is not doing anything interesting and so we neglected it but on the other hand if this were not like this and gravity acted like this then clearly that will be a part of the free body diagram and it will contribute to the radial motion and also to the theta motion of this entire object. There is a question 12-03 how to convert dynamic equilibrium to static also explain the D'Alambert's principle. So D'Alambert's principle the professor Shovik Banerjee will discuss on on fifth it is nothing okay see D'Alambert's principle what is the original D'Alambert's principle original D'Alambert's principle is that like in the past people were more accustomed to doing force equilibrium moment equilibrium okay when all the dynamic laws okay were not appropriately fleshed out okay and as a result it was a it was some thinking that okay we are very comfortable thinking in terms of force balance movement balance but we are not very comfortable thinking in terms of acceleration that forces result in acceleration. So what do we do we just say that the acceleration can be thought of to be m times a is minus m times a is something like a force acting on the body and let us do that but that is not the best way to do problems sometimes it may simplify it but typically if you can write down equations of the form that force is equal to ma and the torque okay is equal to rate of change of angular rate of change of angular momentum then the problems become much much more simpler okay. So D'Alambert's principle essentially just that it is just that in the historically there has been some thinking that it is very easy for us to think of equilibrium so let us convert the inertial forces also as some actual forces and that essentially is the D'Alambert's principle not a big deal about it but if you want to be very consistent with our approach then D'Alambert's principle need not be the best way of doing it but of course it can be helpful for quick thinking explain what is path coordinate okay path coordinate okay let me briefly discuss about that okay path coordinate why do we need okay or why is it useful let us say the particle has a trajectory like this that particle is moving along this and by definition if the particle is moving along the trajectory then this arrow that we I am drawing is essentially the instantaneous direction of velocity which is the tangent to this curve now why do we need path coordinate and what do we mean by path coordinate so if we had seen previously that if you use radial coordinates or radial coordinates then I need to fix an origin then I have to define position vector then I have to define that for that trajectory how does the position vector change and so on but for many problems for example when we want to find out what is the acceleration instantaneously of a car at a given point of its trajectory or what is the acceleration of a plane okay in its given trajectory we do not really need to always want to refer to some origin and then discuss what is the acceleration of velocity and in that case this path coordinates become very handy because we immediately realize that velocity vector is nothing but the speed into E tangent what is E tangent E tangent is nothing but the tangent unit tangent vector to this curve okay that is ET and then what do we do immediately we know that my speed and the direction is tangent and then dv by dt is then immediately can be written as d speed okay ET plus you can do some manipulations look up BJ 10 or BJ or any standard textbook okay in also it is given very nicely in BJ 10 what you will see is that this will become V derivative of ET with respect to T now why does ET change because here the tangent is like this if I move a little bit further the tangent will change direction so ET will not remain constant okay tangent keeps changing so ET unlike your I and J where ET keeps changing with respect to time and this quantity can be shown to be equal to V into V by rho where rho is the corresponding curvature that we had discussed here that if you try to fit a circle here then the corresponding radius of that circle is the curvature or the radius of curvature and then this becomes V square by rho into En what is En? En is the normal in that direction okay so that is how we can mathematically we can show but what is the good thing about this that in the path coordinate we immediately understand that acceleration as two components one is the rate at which the speed changes but second is the rate okay it is also comes because the direction of our path changes and that is essentially the centripetal force that acts on the particle and at that given instant so it is very transparent okay and it has no reference to any coordinate frame or origin so that is why it is attractive and depending on what problem we are doing okay this can be very helpful okay. What is the difference okay is the question asked by center 1073 that what is the difference between moment and momentum okay now this is one of the common misnomers the student have moment okay let me again okay discuss this okay this is a very simple thing but students many a times are confused by this in fact even our students they use this word moment and momentum interchangeably and nothing can be further from the truth what is moment moment is nothing but a short form for this is force this is r so r cross f is called as moment of a force and as a short form we say just a moment okay no pun intended okay so moment of a force and as a shortcut we just use this word moment whereas momentum as we had just seen is a particle okay it is a kinematic it is a dynamic quantity but a particle as velocity v bar it has mass m then this quantity l bar which is m times v bar this is called as momentum okay so they are only similar sounding words but conceptually they are quite different from each other so moment is nothing but short form for moment of a force and this is momentum so 1182 is there are two normal reactions for wedge in problem number 12.3 weight of block is also acting on a wedge okay please elaborate let us go to that problem 12.3 okay so weight of the block also on the wedge so this is one clarification which I want to make this is a concept in free body diagram that this block a is lying on block b so we say that the weight is acting on the block but note one thing that the weight is acting on the block but the effect is felt only by the normal reaction that this block is present on this can only be seen when this normal reaction can be exerted means think about it in simple way if I create a hole in this block such that this block a can no longer like step on that block but can collapse inside then in that case even though the weight of the block a weight of the block b is still trying to act on block a because it has no support there will be no that weight cannot be exerted on that block so what can be seen here is that there is this normal reaction and this normal reaction okay it acts from this block to this of course the weight is also present but the weight only makes itself felt on the bottom block through this normal reaction and that normal reaction is very clearly mentioned in this free body diagram so the weight does not directly come and act on this but the weight come and acts why are the normal reaction and then this block okay why that normal reaction is required because to prevent the acceleration of this block in the vertical direction why because the block cannot penetrate this something needs to balance that weight because otherwise that weight will keep accelerating the block along the direction of the weight because that force is acting there force is equal to ma so acceleration will keep acting in that direction and that acceleration is prevented how by the presence of this normal reaction so you see there is an intricate connection between the forces and the degrees of freedom that if there were no if there were no proper connection okay that there will be a there may be if there were a hole between this and the connection between this then there is nothing to prevent that acceleration and so no normal reaction is required but on the other hand to prevent that kinematic motion in the direction perpendicular to plane A you need to have a normal reaction which is exerted here okay and the weight of the block makes itself felt to block B okay only by this normal reaction okay it is fine so there is so there have been a bunch of questions about what is centrifugal force what is Coriolis force I explained that those are the forces that come in non inertial frames of references non inertial rotating frames of references but there is just one last thing okay I just thought that I should mention that how can we think about those forces now let us think about it that there is a rotating table okay and at the center of the table okay exactly at the center of the table assume that you are standing okay you are standing upwards like this at the center of the table and this table is rotating in this direction clockwise direction now suppose I keep a coin here somewhere here and let us for simply simplicity assume that a friction between the coin and this table is negligible now what is happening is that to begin with I move along with the coin okay if this for example if the coin were glued to the table perfectly glued to the table what would I see I would see that if you are perfectly glued then the coin moves with me so from my point of view this is me this is coin so from my point of view I will see that the coin is at rest because after sometime the coin will go here okay then next sometime the coin will go here but I these are my hands outstretched okay I will also keep rotating so as far as I am concerned okay if I do not look outside I will see perfect that a coin is just staying there but then what happens is that suppose something happens okay and this initially the coin was glued but because of heat okay the glue dried up for example and a coin came off and now there is no friction between the coin and a table then what will we see what will I see is that okay that is coin okay will have some velocity okay it has some velocity omega r the coin will start moving out and the coin will also start moving sideways and then I will see that suddenly that what is a force acting on it there is absolutely no force acting on it okay why is the coin suddenly moving out and it is moving sideways if you analyze this problem perfectly okay if you perfectly analyze this problem from an inertial frame of reference you will see that the moment the coin loses contact with the table whatever velocity it had okay it will just keep moving in along that direction but from my point of view what I will see is that that a coin just keeps moving okay outwards as well as sideways and then in a non-inertial frame of reference I am suddenly at loss that there is no force acting on it why is the coin moving and then I say that that particular force acting on it okay outwards force I call that as a centrifugal force okay and this force which acts sideways I call it as the Coriolis force but strictly speaking okay if you do everything from an inertial point of frame point of view then these kind of components okay they will automatically be taken care of okay in the in the acceleration terms for example there will be a term which goes as theta dot square r okay that I will interpret that as a centrifugal centrifugal force okay but the centrifugal force is purely coming because of the non-inertial rotating frame of reference and what is that frame of reference is the frame which I am in okay I am on that rotating table and I see what is happening to the to the coin so from my frame of reference I will suddenly see are a something is happening to the coin but no force then what is that force that force is nothing I call that as a centrifugal force for the the radially outward movement and for the sideways movement I call that as the Coriolis force so these are the forces that only arise when we look into non-route into rotating non-inertial frames of references these forces can be interpreted when we also write down equations of motion okay from a inertial frame of reference okay but strictly speaking these names okay only makes sense when we are referring to non-inertial rotating frame of reference so I hope that this clarifies this okay so we will now take a brief tea break