 So, we now start a new topic which is the kinetics of kinematics of rigid bodies. So, as usual okay that most of the material is covered from Bear and Johnston 10th edition okay fabulous resources are available you can have a look at them. So, let us introduce now what is kinematics of rigid bodies now previously we were all discussing about points, points and points. Now, what is a big deal about points we had briefly discussed when we started our session on kinematics that there is nothing in this world which is actually a point okay, but then what is the point of taking something as a point okay the point being okay so too many points here okay, but the idea behind this thing is that that if a particular body completely translates without having any rotational component then what we say is that all those rotational components we do not bother about and essentially we can assume as if entire mass of that body is concentrated at the centroid center of gravity okay or center of mass there are various names that go with it most common name is the center of mass and we say that we can directly apply Newton's law for that at some of all the forces external forces acting on that rigid body is nothing, but the mass times the acceleration of the center of mass done. So, this was Newton's laws of motion okay for translation okay in the for rigid bodies without any mention to rotation that we do not really need to have a point okay what we can do is that we can just say that all the that if there is no rotation then the only thing we need to worry about are the two translations or three translation depending we are in 2D or 3D for that for the center of mass of that body and overall external force acting on it is nothing, but mass times the acceleration of the center of mass and done okay we are done with this and just to refresh ourselves what did we do we said that for a mass sum of all the forces vectorial forces is equal to m times acceleration okay where m is the total mass okay it is a very important equation we are going to use it again and again now even while discussing kinetics of rigid bodies and a bar is the total is the acceleration of Cm okay and how do we define Cm for a collection of particles Cm k x bar is defined as sum mi xi divided by okay so this is x bar divided by sigma mi and as a result x bar dot is equal to mi vi bar dot v bar divided by mi and acceleration of the center of mass is nothing but mi vi bar dot okay divided by some of the masses are all of these things okay so this is simply okay Newton's laws okay for a rigid body we can also which can also be thought of as a bunch of particles put together now let us go into actual details of what if we have a rigid body what kind of motions are possible okay and how do we take those motions understand the kinematics okay what is kinematics is what possible translations rotations what are the accelerations present in that rigid body and we use that information to figure out that if there are certain amount of forces and couples torques okay whatever you want to use the word okay those are acting on the body and they result in linear acceleration as well as angular acceleration now for a rigid body okay what are the possible motions okay so simple rigid body motions can be classified as just simple translation that a rigid body is present and the rigid body just move okay like this in a 3D environment or in a 2D plane okay there is no rotation whatsoever and that was kind of the definition of particle we had used that there is only translation and rotation is not playing any role into the quantities of our interest okay so we can have a simple translation just a rectilinear translation or we can have a curvilinear translation now what is a curvilinear translation the curvilinear translation is as follows so let us take it like this here okay no rotation no rotation again no rotation now what do I mean by no rotation the definition of no rotation is that that let us take 2 points A and B which are joined by a straight line on this rigid body now no rotation means that this AB always remains vertical this angle of AB it never changes okay it always remains like this this is what we mean when we say that there is no rotation but now note that in this problem that the trajectory which is traveled by the center of mass is no longer straight it is curved but still there is no rotation that any line AB which I draw on a rectangular mass will always remain straight okay that angle whatever it is at t is equal to 0 at any other arbitrary time that angle always remains the same and that is why we say that there is no rotation it is still a translation but it is curvilinear translation why because the center of mass now is no longer moving in a straight line it has a direction okay it has a change of direction and as a result the center of mass if you just track cm you will see that a cm also has tangential acceleration and it can also have a normal acceleration like we did for a particle that if you treat this center of mass as a particle then whatever we discussed about the curvilinear motion previously okay they all apply here also but the only difference here is that now there is an additional degree of freedom that we have to worry about which is that this line did not always remains straight okay it can undergo rotation and that additional component okay we had completely neglected when we had done kinetics and kinematics of particles. Now what is the other type of motion that we can have rotation about a fixed axis so what is rotation about a fixed axis is for example we take our rectangle okay our favorite rectangle which we had been discussing so far let us say that this rectangle is hinged about its center okay there is a pin here now let us say we draw this line here AB what can happen is that this rectangle can start having some rotation about this hinge O so this line AB is no longer straight now it can have further rotation this is B this is A so this is one motion okay and this point O need not be just at the centroid it can be at any other point but just for argument sake we took it to be at the centroid this move this is called as pure rotation okay about a point okay this is pure rotation why because the center of mass or one point is completely fixed and you can say that this is just a pure rotation about this point that we do not have to worry about any translations for the time being okay so this is pure rotation about point and in this case that point is O then what we can have is we can have a general plane motion now what is the general plane motion we definitely know what that plane motion is for example if you recall what we had done when we had solved variety of problem using principle of virtual works those ideas okay the kind of kinematic displacements we had given to various structures okay rotations translation combination of them they are going to come really handy when we discuss kinematics okay up to velocity of this rigid bodies take example of our favorite ladder leaning against a wall okay A B now this ladder can undergo a motion like this that A comes to A prime and B comes to B prime but if you remember what we had done we had said that this particular motion can be broken down into two parts we can say that we can pin this ladder about point A and let the ladder undergo rotation okay so this is the initial configuration of the ladder let the ladder undergo a small rotation about point A but what happens when this is done when this is done this point B double prime loses contact with the ground and we do not want that to happen because that is inconsistent with the overall kinematics of the with the overall geometry of our problem but then what we do is that further we apply a translation in the y direction and then what happens is that that overall translation when we apply in the y direction by how much amount by this much amount then we say that the final configuration of the ladder is like this so we can say now that this particular motion is a combination of translation and rotation that in this case we can think of that to be a rotation about point O and a translation in the y direction equivalently you can also think of this as rotation about point B and translation in the horizontal direction both are perfectly equivalent so a general planar motion can be broken into two parts which is rotation about any point and a corresponding translation. Many a times it becomes very simple to take that point of rotation as a centroid but in many other problems as we will see okay you can choose an choose a point appropriately to decide about which you want to rotate and then decide how do you want to translate so that is completely dependent on us but the only thing is that at any general planar motion can always be decomposed into rotation about a point for that body and followed by a corresponding translation okay that is the most general motion in 2D and in 3D we can also have motion about a fixed point like for example a spinning of a top okay so these are the motion little bit complicated not really complicated but it requires more visualization and it is kind of beyond the scope of what we are going to do now so I will not go into details of that so this is motion about a fixed point like for example spinning of a top where it can spin it can do this what is called as a mutation and it can go around which is called as precession so it can do these three kinds of motion which is motion about a fixed point and a general motion is that that there are three degrees of freedom for translation three degrees of freedom for rotation and that is the most general planar motion so what we are going to focus on okay we had already focused in our previous kinematics of particles on translation both rectilinear and curvilinear okay in this today's lecture okay we are going to focus on rotation about a fixed axis and general plane motion now let us like look at these motions in detail now why I am going into all these things in detail the same reason why we had gone into lot of details of kinematics of particles why because unless and until we understand what overall kinematics is so what are the velocity components what are the acceleration components okay what is angular acceleration what is curvilinear acceleration and so on unless and until we have a decent handle over that it will be impossible for us to do kinetics where we for example relate the applied forces and applied torque with mass times the acceleration and moment of inertia times alpha okay what that quantity is okay we will come to that okay very soon now what is translation so rigid body translation okay is that for any two particles let us take I have particle A I have particle B then we say okay we can easily say okay this is a cliche or a standard definition that the position vector of B with respect to a fixed origin is given by position vector of R A plus position vector of B with respect to A now for a pure translational motion what we had seen is that that this position vector if there is no rotation then this position vector just keep translating as a function of time what does that mean is that that this position vector the relative position vector between any two points in translation does not change in time or in other words if I take derivative of this with respect to time then what we will see is that that this becomes zero why because relative position of B with respect to A okay for any arbitrary points A and B will always remain constant that vector will just remain this keep moving okay in 2D or 3D whatever dimension we want to work in planar motion non-planar motion does not matter this remains constant and as a result velocity of B will always be equal to velocity of A for any point and when that happens okay we say that the body is undergoing pure translation and since the velocity is also constant for every point we will see that acceleration of all the points okay is the same and now it makes perfect sense that when a rigid body is undergoing pure translation then we do not have to worry about anything about rotation and so on because all particles have the same acceleration then I can as well think that this is like just like one particular one small particle with all the mass concentrated at the center of mass okay and my total force is equal to F times M A this is pure translation now let us come to this concept this is again very important concept is rotation about a fixed axis because why we had just seen a few moments ago that any general motion okay for everybody for even for 3D but definitely for planar problems which we are going to discuss that can be thought of as a translate as rotation about a fixed point or a fixed axis and followed by translation so in 3D we have to talk about axis in 2D we can talk about a point why because the axis essentially is now the z axis which is out of plane so essentially there is no rotation about a point per say it is properly defined only in terms of rotation about an axis but when we do 2D we say rotation about a point why because essentially that point is nothing but z axis okay but because we are doing in 2D we say that it is rotation about a point but a rotation about a fixed axis okay what happens about the kinematics of that so you just go through this a small derivation here okay you have to do some visualization using calculus okay stare at this diagram for some time but what you will realize is that that for pure rotation about an axis okay that this is the axis over which the rigid body is rotating about okay we see so many times for example a spinning top okay if you think in 3D what is the axis it is rotating about for example okay we can always figure out if it is just vertically spinning then the axis going through the midpoint of the top is a spinning axis but then this axis becomes more complicated if the top is undergoing more complicated motions okay but still we can always figure out okay that there will be certain motions for example in which the motion is happening about a fixed axis and in that case the velocity of any point okay now this is the origin that we are fixing there is no translation why because we are seeing that it is a pure rotation about a fixed axis now let us put some origin okay one point along this axis and we say that we find out the position vectors with respect to that origin point what do we have v bar of any point okay can be written as dr by dt and this can be showed to be equal to omega cross r where omega is what omega is the angular velocity of that particle about the axis of rotation okay this is omega equal omega cross r and now when we are doing 2D planar objects then the axis of rotation is nothing but a z axis what is the direction okay and what is the direction of this omega vector it is nothing but if this is the direction of rotation clockwise or anticlockwise we just curl our right hand around it okay and the thumb the direction it points out okay if this is x this is y if this is the the is anticlockwise is the direction of rotation then the rotation is happening about the z axis okay and you can see that this will be nothing but equal to the theta dot quantity that we had seen earlier that theta dot is nothing but omega of rotation for that object now what is the acceleration for rotation about a fixed axis we just saw that acceleration is dv by dt but what was v v was nothing but omega cross r we do the derivatives when we do all the derivatives what we will realize is that that the total acceleration is given by d omega bar by dt cross r okay that is the acceleration plus omega bar cross v this is the most general expression now what we are interested in the time being okay is let us think about what happens okay when we are okay in the plane so what we can think about is that at d omega by dt is this quantity which is called as a angular acceleration so for example in planar motion okay what we see is that that the theta double dot quantity which we had seen earlier okay is essentially or it is equivalent to what is this quantity called as alpha that there is a pure rotation about a point okay and for any point on the rigid body okay if I draw a simple diagram here this is the planar motion this is the point about which the rotation is happening then if you connect any point here r bar then for complete rigid rotation of this body what do we know okay take this particle what we will realize is that that the acceleration of this point a will have two components that a bar will be equal to alpha r what is the direction of that the direction is perpendicular to this and another component will be which will be along the direction e theta okay and the second one will be a r okay what will that be equal to that will be equal to omega square r into minus e r now this can also be expressed in terms of tangential and normal components and how we do that okay we will come to that in a few moments normal directions so this a bar now has two components one is alpha cross r and other is omega cross omega cross r and what we realize this that for a planar motion that rotation about a fixed lap okay about point o so about o means essentially it is happening about the z axis what is that that this v velocity if you take this point o about which the pure rotation is happening we want to find out that instantaneously what is the velocity at this point p what is the velocity of this point p what do we do it is nothing but omega cross r what is omega omega is in this direction which is omega k k is what the unit vector along the z direction so it is omega k cross r and you will see that this direction will only be perpendicular to this line and the magnitude will be given by r times omega and the direction will be this similarly to talk about acceleration of any point what do we do it is alpha cross r okay plus omega cross omega cross r now what is this alpha vector this alpha vector is again okay its direction is out of plane okay so it is alpha the magnitude times k hat which is the unit vector in the z direction so we can rewrite this as alpha k cross r minus omega square r bar what is r bar r bar is essentially this okay so when we just resolve this okay we can say that this because this entire object is rotating purely about point o the trajectory that this point p will follow will be along the arc of a circle why because this distance is now remaining constant so the trajectory that is followed by p will be along the arc of a circle and then what we can say is that that now the acceleration is coming because of two parts why because the speed velocity magnitude or the speed can change while this particle moves along the arc of a circle because of this rotation okay because of change in the magnitude of the speed and that particular component is this component r times alpha where alpha is the angular acceleration which is nothing but omega dot okay that is the angular acceleration and the second component of acceleration comes why because the velocity is always perpendicular to this position vector because of it is moving in a circle and as a result okay the direction of this velocity keeps on changing and that will lead to the acceleration in the normal direction an and what is the magnitude okay it will be equal to minus omega square r bar okay or in other words its magnitude okay is r omega square in the normal direction so if I go back to this slide okay we can say that this point p okay this is point p because of this rotation this point p is now going to follow the arc of a circle and at any point p at any point along the trajectory this is a position vector its velocity is directed like this this is also the tangent direction this is the normal direction and what we are saying is that that the acceleration in this tangent direction the magnitude of that is nothing but alpha times r where r is this distance and the acceleration in the normal direction is nothing but minus omega square r okay where omega is also rate at which the angle changes and alpha is theta double dot or equal to omega dot so this is as simple as that this is the simple kinematics of rotation about a fixed axis now think about this okay let us look at it there are layers to look at various cases if this uniform rotation alpha okay if the if the rotation of this rigid body it is constantly moving at a at the same angular speed then omega dot in that case is 0 you have a planar body okay which is rotating at a constant angular speed then this quantity alpha is 0 so what we can say alpha is nothing but theta dot the rate at which this angle theta changes okay and so like we did for the rectilinear motion we can say that alpha is equal to d omega by dt okay which can also be written as omega into d omega by d theta recall we had said that rectilinear acceleration will be written as d v by dt which can also be written as v dv by dx so exact same analogy now for this rotation about a fixed point we just know that this is alpha and now we want to find out that if we know that the angular acceleration which is theta double dot is this alpha how can we find out that what is the angular velocity as a function of time what is the angular velocity as a function of rotation and what is the rotation as a function of time so these are all possible questions that we can ask and we can use this you we can integrate these equations using the same ideas that we had discussed for rectilinear motion and for uniform rotation when there is no angular acceleration it is uniform circular motion okay then theta will be equal to theta naught plus omega t straight away for uniformly accelerated rotation of a rigid body that this theta dot is now constant okay alpha so we can do the integration again and we will see that omega is equal to initial omega naught plus alpha t theta is equal to theta naught plus omega t plus alpha t square and same formula like in rectilinear motion for this omega so we can do all these quantities okay all the games we had played for rectilinear motion now we can also play for planar rotation okay of a rigid body about a fixed point now let us do this very simple problem okay with this with this much discussion what we have is that this is one mechanism this is a drive belt what is this drive belt used to do this drive belt is used to drive a pulley so professor Shobik Banerjee had discussed various ideas about belt friction and so on we are not going to go into intricate details of that but simple kinematics of a mechanism where a driving belt is driving this pulley now what we are told is that that this driving belt or this driving cable has a constant acceleration of 225 meter per second in this direction and an initial velocity of 300 millimeters per second both in this direction and with this information and also we are we are given that this rope maintains a constant contact with this pulley okay then determine the number of revolutions of the pulley that how many times this is pulley revolve in 2 seconds about this point A and the pulley is under we can undergo pure rotation about point O second thing is the velocity and in change in position of this load B after 2 seconds then what happens is that when this you are keeping pulling if this pulley keeps on rotating in the clockwise direction why because this point C moves sideways as a result this pulley will be forced to rotate in this direction okay in the clockwise direction and this point B will moving we start moving up how because this pulley rotate the external one is connected to the pulley that also rotates and this goes up so that is the work done by this driving cable and what we are asked to find out is the number of revolution that is pulley undergoes in 2 seconds the velocity and the change in position of B after 2 seconds that what is the velocity and how much it moves up by and the acceleration of the point D on the rim of the inner pulley that what is the acceleration for this point at time t is equal to 0. Now note one thing that point D okay will have 2 components of acceleration why because this inner pulley is rotating at with an acceleration sorry with an angular speed of omega and also with an angular acceleration alpha what that angular acceleration is we will figure that out okay so point D okay will have the same speed in the tangential direction as this cable C so velocity of this cable is VC in this direction is the same as velocity of point D why because these 2 are linked with each other okay point D is common both to the cable and to the pulley and if there is no slip between this pulley and the cable then both of them should have the same speed the point D on the pulley and point D on the cable are connected to each other okay they are common to both of them and in the absence of any slippage they will have the same velocity what is that 300 meters per second now what is the acceleration of D in the tangential direction it has to be again the same as the acceleration of this point C because that is a tangential component okay so both of them will have the same acceleration. Now what do we know from our previous discussion because it is a pure rotation about point D this tangential velocity should be equal to r times omega 0 where what is that omega 0 is the initial angular speed of this pulley now what do we want to do acceleration of point D in the tangential direction what will that be equal to again okay because they are all connected with each other this has an angular acceleration of alpha okay with respect to rotation about point A so the tangential angular acceleration will be nothing but r times alpha. So what we do we just put these equations together we realize that omega 0 is equal to initial velocity which is 300 divided by r and from that we can find out that omega 0 is 4 radians per second in the horizontal 4 radians per second what direction it is it is going in the plane okay use the right hand thumb rule okay what is the direction of the velocity is this so in the plane of the paper will be the corresponding vector. Now always note one thing that omega has to be represented in terms of radians per second okay if you have a question about that you ask me but omega should always be represented in terms of radians per second for safety purposes there are some applications in which you can immediately use that omega can be in degrees per second or number of revolutions but if you want to find out what is the distance that is travelled okay by the pulley okay or then in that case that omega should be in radians per second now what is alpha again just use this thing use these 2 equations we will find out that alpha will be this tangential acceleration divided by r which will again come out to be 3 radians per second square. Now we want to find out that after 2 seconds what happens what we do we use the simple equations for constant acceleration we are given that the acceleration is constant and what are the equations okay these are the equations for omega, theta and omega square as a function of time and as a function of theta or the rotation. So just substitute it in all these values here we can find out what is omega as a function of time at t is equal to 2 seconds substitute you will see that omega for this pulley will be equal to 10 radians per second again substitute okay this is theta what will theta be equal to omega t plus half a t square substitute everything you will realize that the total rotation that this internal pulley has undergone is 14 radians and then what is the number of revolutions number of revolutions is just simply this. So 2 pi times the number of revolutions will be equal to the number of radians okay and then just put them all together we will see that the number of revolutions after 2 seconds is 2.23 what is the velocity of point B after 2 seconds now note that as far as this point is concerned okay where this B is connected by a string to the outer pulley now the this point okay where the this mass B is connected to the pulley is common both to the string and both to this outer pulley and if there is no slippage between these 2 okay what we realize is that okay since they are there in the form of this spool okay this velocity the velocity of the point which is to the pulley which is connect which is a part of the pulley and the velocity of the point which is a part of this string should be equal and as a result the velocity of this point is nothing but r times the omega of this entire assembly that omega we already found out and then we can find out what is the velocity of this in the upward direction and significantly and subsequently what is the overall displacement in the y direction that how much theta it has undergone into r will be the total displacement of this thing in the y direction that is nothing but r times this q is essentially theta okay this is just typo r times theta will be the distance that this point B move in the vertical direction okay so all this quantities okay by connecting what is alpha what is theta what is theta dot all with the r with the corresponding dimensions of the pulley okay we can find out what are the corresponding accelerations motion displacements and velocities of various points in this structure if there are any questions okay please let me know so give two examples of general plane motion okay one simplest example of a general plane motion is for example if you have a cylinder rolling on a slope if you take a coin okay and make it roll okay exactly in one plane okay along a straight line or for example when a car moves or when a bike moves in a straight line okay then that particular motion okay of the wheel is is a simple example of a general plane motion okay or for example if I take a pen like this and exactly throw it like this okay with some acceleration without by taking care that there is no rotational component in the out of plane direction that I do not do this but just throw it in one direction and that is one simple example of a general plane motion what a general plane motion means general plane motion means that the center of mass of the particle okay it stays in that plane and not only the center of mass the entire object okay okay or there is one plane of the object which entirely or the plane of symmetry of the objects completely stays in that plane so either you have a very flat object like a paper and if this paper keeps on moving like this in one plane you can call that as a planar motion or if for example I have a ball but that ball also I take the central axis of the ball okay just cut it through the center and keep track of that central plane and if that central plane okay keeps moving exactly in one xy plane rather than having this kind of components then that is example of general plane motion that it does this rotation and it also does this translation but why is it planar because there is no component okay that no part of this object has any velocity okay or any displacement or any acceleration in the out of plane direction okay that is a simple those are the simple examples of general plane motion okay this is the question asked by center 1177 center 1224 they have asked that why do we study ladder as a 2D problem and not a 3D problem okay pure idealization okay clearly if you have a rod leaning against a wall in principle okay if I give it a motion like this then the rod can also undergo a motion in three dimension and this need not be a planar motion but if you take a ladder like this and give it a perturbation only in this direction so the ladder keeps on moving like this then the overall motion is happening in only one plane and we say that it is a planar motion so planar motion is one possible motion for the ladder which is in leaning against two walls okay but that does not mean that that is the only motion it can have it can also undergo motion like this okay and for the purposes of this course okay where we are not going to go into details of 3D or general translation and rotation we are solving simple problems in planar motion we are taking this specific case where the ladder is perturbed in such a way that it has all the motions in one plane okay there is no component of motion which goes out of plane that is the only reason but definitely a ladder can have a motion which is a general 3D motion no question about that okay so center number 1059 the question asked is please explain the concept of degree of freedom for a rigid body by considering n number of particles okay see this n number of particle is a simple idealization okay so let me discuss that okay what simplicity for the time being let us take the case of a planar motion okay we have a notebook like this okay and suppose this book falls from our hand by mistake now what can happen is that that this can undergo a motion like this where if you track this point O this point A O has moved so there is clearly a translation also if you take A and B as another two points then we will see that this A and B the distance does not change for a rigid body but they can undergo rotation like this now what we can think of is that that this entire rigid body we can think of as a bunch of points and infinite number of points and the same rule that we have applied for this centroid O and for these two points A and B we can play the same game with these two other points let me call them P and Q and then what will happen is that after the rotation you will see that P and Q will look like this after more time this P and Q okay may also look like this but for a rigid body what is true is that that the actual distance okay this length P Q does not change but the orientation of the P Q and the overall translation of the P Q can change so any rigid body can be thought of as a infinite number of points and what is the concept is that you choose any two points a general planar motion of those two points okay because when you join two lines that line can undergo a general motion and the general motion can be a translation what is the translation in any 3D if I take a bottle like this what are the motions that I can give it to it what are the degrees of freedom one is translation in the X direction other is translation in the Y direction translation in the Z direction now XYZ are purely chosen as some nomenclature okay you can have this A B C P Q R X prime Y prime Z prime whatever but it can have this three independent degrees of freedom okay but not only that there can be a rotation about Z axis there can be rotation about Y X axis and there can be rotation about the Y axis so these are the six degrees of freedom that are possible for any general 3D motion but when we take the case of a 2D planar motion what do we say that a 2D plane okay that the motion that are possible are like we did in 2D equilibrium that it one translation is possible vertical translation is possible and this third rotation which is rotation about axis Z is also possible so for planar motion okay what is a planar motion either the object is very thin like a play paper and there is no motion that is happening out of plane the only motions happening are is one two and three we say that this is a planar motion or the object may be okay something like for example a sphere but the sphere okay if you cut the sphere through the center the center of symmetry of the sphere if the sphere is moving in such a way that it has only motions are the rotation about the Z axis translation in the X and translation in the Y axis even though the sphere is not strictly a 2D object okay or not a thin object what we can say is that there is one plane that I can find on the sphere where all the particles on that plane undergo only a planar motion okay that they do not undergo this kind of motion like for example swing of a ball those motions are non-planar motion they only undergo three motions in plane and that is another example or a cylinder rolling along an inclined cylinder has a object okay it has a thickness in the third direction but if we cut the cylinder through the center what we see is that that particular plane the projection is like a circle and we can as we will say that it is like a disc which is rolling down the ground or which is rolling on a flat surface so that is a concept of degrees of freedom and what do we mean by saying that a motion is a planar motion or a 2D motion okay I hope that this explanation is clear so now let us move on now this is one example of the general plane motion okay as we had just discussed okay that what is a planar motion that this object is very thin okay and so we feel that the motion is essentially planar and that it does not have any translation or rotation out of plane so we say that this is a planar motion but we can also have that this need not be a disc this can as well be a sphere and sphere is a 3 dimensional object but what we are doing essentially is that that we are cutting through the center of the sphere and that projection is essentially a circle and if the sphere does not undergo lateral motion only undergo motion in this plane okay that the center plane it does not undergo some motion like this just undergo motion in one plane then we say that this is a plane motion so one of the most general plane motion is rolling of a disc or rolling of a sphere on the plane ground a general motion as we had seen is neither a pure translation not a pure rotation and even from the example that we had discussed about the ladder okay slipping in one plane it can also slip outside the plane but one simple example is when the ladder keeps slipping in one plane that is one example of a planar motion and as we had seen that a general motion need not be a pure translation on pure rotation it will be a compression it will be a combination of translation about one one point or one axis and rotation like we did in the ladder case okay now let us define this a very important concept which is the absolute and relative velocity in planar motion. Now what we do is this that in a rigid body what we had seen earlier that only when a rigid body undergoes pure translation then the velocity of each and every point on that rigid body is the same but as we just saw okay that any general planar motion can be a combination of a rotation and a translation and because of that what happens because of that any point A and B even though the motion will have a course in such a way that in the course of its motion the distance AB will not change why because this is a rigid body and by definition this distance cannot change but point A and B can keep having different velocities which have different magnitudes and different direction. Now there is a relation between them okay but they need not be the same now how do we relate take two generic point A and B and how do we take the relation between what is the velocity at point A and what is the velocity at point B. Now let us see what do we do is that this motion we break it into two parts we had seen that any planar motion can be broken into rotation about a point plus a translation. Now how do we break it we break it into two parts one is a complete translation okay a complete translation with A okay at this entire rigid body is translated plus a rotation about A what is this rotation this is the rotation. Now what is happening here that in this particular thing the overall velocity of point B will be nothing but look at this component this component velocity of point B is what is the translation okay or the translational velocity that we have undergone here which is nothing but the velocity of A so it is V A plus V of B with respect to A. So this is a standard statement that velocity of B is nothing but velocity of A plus velocity of B with respect to A. Now what do we do okay so here what we have done is that that we have taken this entire object and given it a translation okay or given it a velocity in the direction of A okay and plus rotation about A. Now how do we then how do we replace that now the velocity of B with respect to A will be what that what is the velocity that point B see when with respect to A but we just saw okay that one component is this total V A so this part okay so velocity of B has two parts one is this component plus the second component that comes because of the rotation of this rigid body about point A okay. So first part is that we have broken down this motion into two parts one is this translation with respect to A V A V A and so what is the total velocity of A is just V A plus because in the second case we are doing pure rotation about A this point okay okay has no velocity no acceleration nothing so the total velocity of point A is equal to V A plus this but the total velocity of point B okay V B will be what this V A which is broken into one part V A plus the velocity component that comes because of pure rotation about A. So we can say that my total velocity is nothing but this V A plus what is this that with respect to point A what is the velocity with respect to point A which is nothing but this component now what is this component these components come purely from rotation about point A and that can be written as omega k cross this relative position of B with respect to A and this can be written as R omega the magnitude is clearly this is the R omega is the angular velocity about this point A and so we can write down that the overall velocity of point B is nothing but V A okay which is given to us plus omega k bar cross what is k bar k bar is the unit vector along the z direction R of B with respect to A or the position vector of B with respect to A. Now if you recall okay if you recall what if you just discuss this with the help of a simple example the point will become very clear and this is precisely the point that we had discussed when we use the principle of virtual work where we had under where we had provided a combination of infinitesimal translation and rotation okay to rigid bodies. Now let us take our favorite example okay which is equivalent to a ladder that this object okay AB can have velocity at point A in this direction V A can have velocity of point B in the downward direction. Now what we can do we can say that the overall motion that this object can undergo it cannot undergo it is a 1 degree of freedom object why because this V A and V B cannot be unconnected because if V A and V B are unconnected what will that mean that point A and B the distance between them will either try to decrease or try to increase which is not allowed for a rigid body. So we break this motion into two parts what are the two parts one is pure translation okay what is that translation we give it the same translation as A same velocity as A but in addition what we say now that the overall motion of this object is a combination of this pure translation plus rotation about point A. Now what is that rotation about point A when we are doing this motion A is fixed so there is no velocity for point A but what is the velocity for point B velocity for point B even our simple geometric intuition will tell us that we want to rotate in this direction okay if this is the direction of rotation then what will happen is that that the velocity will be perpendicular to this rod which is velocity of B with respect to A. Now what do we do we form this triangle that this is V A this is V B and what do we know that V A plus velocity of B with respect to A will be equal to V B now this is little bit complicated okay we are doing vector algebra in the next okay in the next few moments I will also briefly discuss how can we easily do this without resorting to any vector notation okay but just bear with me here why because this is systematic way of doing these problems so this is V A this is V B with respect to A. Now how does our geometry play a role what we know what we clearly know is that this relative velocity should be perpendicular to the position vector of B with respect to A since we know this angle we immediately know that this is theta this is 90 minus theta so this angle also has to be theta so we know what is this angle theta if now we know what is V A suppose V A is given to us what we know that V A divided by V B or V B divided by V A should be equal to tan theta so if we know what is V A we immediately know that the relation between V A and V B is equal to tan theta doesn't this look familiar if you recall when we did principle of virtual work that for that ladder problem we gave it a virtual displacement in the horizontal direction and vertical direction such that there is no loss of contact between A and B for that ladder also there is no change in the length of ladder A and B and what we had seen there we had seen that delta x by delta y okay or delta y by delta x is equal to this tan theta similarly that delta y is now replaced by V B why because delta y divided by delta t will be nothing but V B similarly at this point okay what is V A in that case there was delta x and delta x divided by delta t so we divide and subtract we divide both the numerator and denominator in our virtual problem by delta t and what will we get that V B by V A will be equal to tan theta so we know from this vector algebra we also if you recall our virtual work discussions properly we will see that both of them are consistent with respect to each other so V B by V A is also equal to tan theta but now note one thing what else do we know we also know okay from this expression that V A divided by V of B with respect to A the magnitudes should be equal to cos theta but what is the magnitude of this the magnitude of this is nothing but the length times omega and isn't it the same as saying that if we give this thing in our virtual work discussion a small rotation of delta theta about point A then the virtual displacement here will be how much will be equal to L times delta theta now just divide that by delta t what will we see that this is nothing but L delta theta by delta t which is omega so what we had discussed in principle of virtual work that same kinematics we are also seeing here okay albeit we are seeing it from a more formal discussion but what we see here is that that V A divided by V B, B says B with respect to A is nothing but L omega and this is equal to cos theta now we are done what is the omega what is the acceleration what is the angular rotation of this we immediately can find out and say that this will be equal to V A divided by L cos theta and once we know V A we also can find out what is V B so straightforward so the idea is that we just break our overall motion into combination of two motion one a translation and other rotation of what point now you may ask me that what point should I choose why should I choose point A why cannot choose point B it completely depends on what you are comfortable with and many a times it depends on the geometry of the problem you can also repeat this entire problem by a translation about B okay translation in the vertical direction and a rotation about point B and you will see that you will get exactly same answer by exactly similar consideration only difference is that in this case V of A will be equal to V of B plus velocity of A with respect to B and velocity of A with respect to B is nothing but minus of velocity of B with respect to A that will be the only difference otherwise you will see that this problem okay we can break down the overall motion why because there is a constraint that the length of A B cannot change so we break it down into two parts one is pure translation and rotation about A or we break it into a pure translation in the vertical direction okay following what is the velocity of B plus a rotation about point B and both are equivalent as far as the overall kinematics of this rigid motion is concerned. Now this is this famous piston problem which we had just done earlier okay if you recall the problem we had done earlier so what we will do is that this piston problem okay this is the crank okay a crank and a piston problem that this is something which so this is an assembly okay if you remember even when we did principle of virtual work we had a similar problem which we had solved and now what we are asked to find out okay that is crank A B has a constant clockwise angular velocity of 200 rpm okay so the angular velocity is clockwise in this direction what is the what is it it is 2000 rotations per minute and now what we are asked to find out that for that angular speed for find out that in this position when all the dimensions and all the angles are given the angular velocity of the connecting rod BD okay so what will be the angular velocity of the rod BD and the velocity of piston P at what is the velocity of this point D. Now we can do this problem in two ways one is the formal way in which we had done the problem earlier but before I do that we had just done a few moments ago let me discuss this problem very easily using the virtual work kinematics that we had done previously the problem which we had discussed previously what we had is this this is essentially our assembly A B why I am using this various approaches is that what we are seeing is that where that we use vectorial approach we use the approach which is more geometric which we had done previously okay whatever way you do you get the same answer and that is to emphasize that and also to relate that the whatever concepts we had learnt in virtual work okay they are not waste okay that even if you do not use the principle of virtual work those concepts will give us a very good understanding of kinematics okay or at least how does the rotation of a rigid body happen when you are multiply connected rigid bodies this is B and this is point C what is this point C this point C is free to slide so we replace that with a roller this is length L1 this is length L2 this is theta 1 this is theta 2 now what we want to do is that if this assembly okay rotates in the clockwise direction by an angle delta theta 1 rod A B what will happen if I draw this rod A B separately what do we get this is the initial position of the rod after rotation you will see that this is how the rod position will move from B to B prime okay this is A what is this distance B B prime this angle is delta theta 1 so this distance B B prime is nothing but L times L1 times delta theta 1 how much is the vertical distance okay delta B in the downward direction how much is that equal to okay if you do the geometry if you recall what we had done in principle of virtual work this will be simply but L1 cos theta delta theta 1 how much will be delta B in the horizontal direction will be L sin theta delta theta 1 now let us say that I take this rod BC and I do not move this point C to begin with let us keep the point C here but let us give it a rotation now in the anticlockwise direction about point C okay this is the anticlockwise rotation this is the original position of BC so this point B okay sorry this is the original position of BC we give it a rotation like this in the anticlockwise direction so what happens is that the split will happen like this okay at this point will go inwards okay this point will also go in the inward direction and so finally after giving these two rotations what do we see that this was the initial position A B C after the rotation what do we see like this and like this we had seen earlier that delta B in the downward direction is equal to L1 cos theta 1 delta theta 1 and delta B in the sideways direction is equal to L1 sin theta 1 delta theta 1 similarly delta B coming from rod BC in the direct in the downward direction will be L2 cos theta 2 delta theta 2 and delta B coming in the inward direction here from rod BC will be L2 sin theta 2 delta theta 2 now what we want is that we do not want this gap or this split to happen okay this vertical and horizontal split it cannot happen so first we do one simple thing we make sure that the vertical displacements from both sides are the same so what do we get L1 cos theta 1 delta theta 1 is equal to L2 cos theta 2 delta theta 2 so what do we learn from this we learn that L1 by L2 cos theta 1 by cos theta 2 into delta theta 1 is equal to delta theta 2 now this rotation was in the clockwise direction okay and this rotation is now in the anticlockwise direction now if we just say that we divided by delta T on both sides then this is omega and we immediately find that if this is omega okay if this is omega the rod AB is rotating with a clockwise in it with an angular velocity omega then correspondingly in a given configuration theta 1 and theta 2 we will see that the angular velocity of rod BC has to be equal to L1 by L2 cos theta 1 by cos theta 2 into omega so whatever kinematics we had discussed in the virtual work method they again work now the last question is that we want to find out what is the overall displacement of this thing in the horizontal direction now how we decide that that there is a split now that has come in the horizontal direction we want to clear that split now how do we clear that split that split we just realize is that that to clear that split this entire assembly has to be translated to the left by how much distance by this delta B plus this delta B and what is that equal to that will be L1 sin theta 1 delta theta 1 plus L2 sin theta 2 delta theta 2 now what we know we know delta theta 2 in terms of delta theta 1 so we just substitute that value here okay let us bring this delta theta 1 outside this goes here this becomes delta theta 2 by delta theta 1 what is that ratio that ratio comes from here and divide everything by time t we will see that this has to be the velocity of this point c in the horizontal direction so the idea that we had used in the principle of virtual work okay you can use the same ideas here only thing you realize is that that there we had discussed in terms of delta x delta y delta theta now we have to just divide everything by delta t and delta theta will be replaced by omega delta x will be replaced by v delta y will be replaced by v in the vertical direction okay so this I realize is a tangent but why I really wanted to bring it up is to make connection that whatever we have learnt in the principle of virtual work that same kinematics is actually applicable here also now we can take a more formal approach we can say that the velocity okay at point d okay will be nothing but velocity of point b plus velocity of d with respect to b what is omega omega is given to us so what is this velocity this velocity of b will be perpendicular to the to the rod and the magnitude will be nothing but ab into omega so straight away we know what is the magnitude of this velocity which is nothing but r into omega or 75 into omega now what do we learn from this what we learn is that that for this motion okay for the direction of the absolute velocity vd is horizontal and the direction of relative velocity of vdb is perpendicular to db okay so we compute the angle between the horizontal and the connecting rod from the law of science so what we see is that the way I write is that this is v of d this is v of b okay and this has to be velocity of d with respect to b and now what do we know from here straight away is this angle is given to us this angle comes from here how much is that this angle is 40 degrees okay so what do we use we use rule of science here and we say that sin 40 okay by 200 why do we say 200 here because this was rod bd the overall motion was taken to be the translation okay in the direction like this okay which is which is bd plus rotation with respect to this what is this distance this is 200 okay and as a result this relative velocity v of b, b is nothing but this 200 okay into omega for rod bd okay so we then use this rule of sin here because the omega will get cancelled out okay so sin 40 by sin beta okay what is the no no so this is just to find out what is the vertical what is the sin beta here we are finding out the vertical distance from here to here is the same as this so a b sin 40 is the same as 75 sin 40 but now to find out overall okay what we have done that the planar motion can be broken into two parts translation along this direction vb plus rotation about point b okay and what is the direction of vdb is what we had discussed here so vdb magnitude is nothing but omega bd into l and this magnitude here for vb is already given to us which was okay from this previous slide given as omega into a b so we form this triangle here this is known to us this vd has only to be horizontal so this angle is also known to us this is known this angle is also known to us on top of that what we also know we also know that what is the angle of this relative velocity with respect to d why because this has to be perpendicular to this so by using simple geometry what we realize is that that angle of relative velocity of point d with respect to b is also known to us so then we can use rule of signs that vd divided by sin 53.95 this is this angle is equal to relative velocity of d with respect to b okay this divided by sin 50 but what is relative velocity of d with respect to b is nothing but this length into omega bd and from that we can immediately find out what is this vdb is this and vdb is nothing but omega bd times l and from this expression we divide vdb divided by l and find out what is the rotation of this entire rod okay so that is the final rotation that we will get and I urge you to solve this problem both using vectorial notation and the virtual work notation that we had done before I realize that the explanation I gave was a bit choppy okay so if you have doubts okay I will be glad to entertain them okay the one question is asked by center 1 2 3 8 how to choose the fixed point okay choosing the fixed point is completely up to you but in this case okay especially in this problem why is what a good idea to choose b as this fixed point as far as the rod bd is concerned it was a good idea to choose b as the fixed point why because we immediately know that this a b okay and bd have a common point b and because this rod is rotating this crank is rotating about point a we realize we immediately know that the velocity at this point b okay the absolute velocity of this point b is nothing but omega times this length okay so this velocity is known to us and since this velocity is known to us we try to take the second motion in such a way we say that it is a translation in this direction for b plus rotation about point b okay so that is why we split it into two parts we took b we took this point b as our anchor but we can as well do this problem by saying that the total how does the total motion happen that it is translation with respect to velocity of a and rotation about point a but because we do not know rotation velocity of a to begin with in this case it becomes very nice if you look at this point b and refer everything okay with respect to this point b that velocity of point d is referred in terms of b as opposed to b referred in terms of point d okay so that is a simple logic here because we know what is the velocity here second question is by center again 1 2 3 8 they have said please explain how omega cross omega cross r is equal to minus omega square r is valid okay let us explain this in planar motion this is r bar okay omega bar let us say this is x this is y this is theta and omega is nothing but the magnitude omega k hat what is k hat is the unit vector in the z direction now what is the velocity okay v bar is equal to omega is equal to r cross omega now what is r cross omega if you think about it what is r cross omega what will be the direction the direction of this will be perpendicular to this what will be the magnitude r and omega are both perpendicular to each other why because omega has a component only in the k direction whereas r has a component only in the plane omega has a component only out of plane so this can as well be written as e tangent okay into r omega that is equal to v bar where this is unit vector in the tangential direction now if we do omega cross r cross omega then what do we get we get omega into k cross r into omega e t this can be written as omega square r k bar cross e t and if you do k bar cross e t this is e t okay k bar is out of the plane you will realize that this will be nothing but equal to minus omega square r and the resultant direction will be perpendicular both to e t and as well to the unit vector in the k direction and this will be nothing but e n or the velocity or the unit vector in the normal direction that is why we say that omega cross r bar cross omega can be written as minus omega square r e n or its magnitude is minus omega square r and it acting inwards okay it is very similar to our motion in the curvilinear direction but now what do we have is that that there is a relation between the tangential direction okay and the normal direction and everything can be linked together with this quantity omega and alpha so that is why you get this quantity minus omega square cross e n okay just do the geometry this will become even more clear there is a question by 1314 saying sir can shall be considered jerk and impact are same not necessarily okay because jerk is rate of change of acceleration so clearly sorry the jerk is rate of change of acceleration whereas impact what is changing it is changing the momentum okay so jerk and acceleration need not always be equated there is a question by 1305 how does reversing take place while bowling in cricket okay so please look up okay so there is something which is called as a Magnus effect which happens because of fluid drag acting on the ball the ball has two roughs one surface which is rough one surface which is smooth so the velocity of the air is different on two and so that is why there is a differential drag and it moves the ball okay for details look up on various articles that will be available online last question what is a top spin about a vertical axis what is the mechanics behind this motion so top motion is a motion which is extremely complicated top doesn't spin just in a vertical way if you realize that only when the speed of the top is very large it just spins like this about a vertical axis but you realize that sometimes okay if you don't spin the top in the appropriate way the top gets inclined and when it gets inclined you will see or even if the speed is low the top gets inclined like this and while spinning about its axis it can also tend to rotate like this so that motion is called as precessive motion of the top whereas now you also realize when the top starts slowing down it does this motion okay it starts to wobble and that wobbling motion is called as nutation so there are three motions for the top one is spin one is this rotary motion rotational motion which is called as precision and other motion is a motion about the axis which is called as the spin okay and all these three motions happen in a top okay or it can happen in a top depend on the angular speed of the top depending on how is the initial inclination of the top okay this is way beyond what we are going to do in this course okay if you are interested in looking into that look at any standard textbook in dynamics and these problems for people who are more mathematically inclined these are topics under analytical dynamics so look up for example the standard books in physics like Goldstein okay where these problems are discussed in great details using what are called as Euler angles okay so this is way beyond the topic of this course okay but if you are interested you can look in standard text last question okay before the lunch is by center 1 2 8 3 right side roller support can the right beam okay move leftward okay so let me come back to this problem again can this move leftward okay it can also move leftward but just think about it if this moves leftward then what happens then to accommodate this b this rod has to move okay in the anticlockwise direction okay but because we are given that the driving rod is going in the clockwise direction okay the overall geometry that we had discussed will mandate that this d has to move horizontally only in the right direction so think about the geometry if you can think in terms of the vector notation or you can think in terms of the virtual displacement that we had given or that we had discussed during the principle of virtual world whatever way you think about it because this is given that this is clockwise this d has to move inverse in the right direction and overall if this is clockwise this will automatically come out to be anticlockwise okay so we will finish this portion okay if you have more question about the kinematics okay we can discuss that we will reconvene at 2 o'clock