 So, let us start with the lecture on kinetics, so what we take now we saw as a derivation that for a rigid body some of all the forces is mass times the acceleration and this result I have not derived, I have not derived this result, but it is present in the slides it is present in the textbook, so let us take this as a gospel that if G is the center of mass then the rate of change of angular momentum of a rigid body about the center is for example that if we have a set of particles which are representing a rigid body there are internal forces and there are external forces this is external, this is external, this is external, this is external. Now for a rigid body what we have is that that we have G which is the mass center and with respect to G any point I has a coordinate given by rho bar whereas from a fixed frame of reference G itself has a coordinate given by R bar. So, what we can define is that we can define that angular momentum of this system of particles about point G will be given as rho I bar cross M i rho i dot and you can do a simple derivation and show that D H G with respect to D T is equal to sum of all the moments due to external forces on this system of particles. And the same logic will be true because a rigid body can be thought of to be composed of a lot of particles. So, D H by D T can be shown to be equal to sigma M with respect to G similarly if we want to find out what is the angular momentum with respect to O or this origin then we can write this as R i bar where R i is the position vector of point I M i R i dot and we can also show that D H 0 by D T will be equal to sum of all the moments about point O and then you can also show that D H G by D T is equal D H O by D T is equal to position vector of center of mass into M into acceleration of the center of mass M is the total mass plus D H 0 by D T. So, these are the results that we will use in our problem solving. So, keep that in mind. Now, coming to this now what do we have is that F bar is equal to M A and with respect to the centroidal frame that centroid is really a special point what we can do is that we can have a look at the slides later on offline and we will realize that central is a very special point and because of that sum of moment about centroid is equal to H bar G. If this G can be replaced by a fixed point of fixed point which is not moving then also this relation holds true. But if this point G is replaced by some other point P which has some kind of acceleration and motion in space then this relation does not hold true. So, always make sure that whenever this points are chosen for finding out moments or finding out what is the angular momentum and so on. Choose it to be a fixed point or center of mass both of them are equally fine but do not choose a moving point and it will become clear as we proceed further with problem solving. Now, these are the equations of motion of a rigid body that sum of all the forces equal to M A where M is the total mass A is the acceleration of the center of mass. This center of mass sum of all the moments about center of mass is nothing but H G which is the rate of change of angular momentum okay. What is H G is the angular momentum of all these particles about point O now in a planar motion coming to simple planar motion what we can see is that suppose we think about a motion our total motion can be thought of as translation of center of mass plus rotation about a center of mass. What we had seen earlier that any general motion of a rigid body especially in 3D also but especially in 2D can be decomposed into translation of motion of one point and rotation about that point. So, let us take that point to be the center of mass and so the center of mass will keep moving but the body also has some rotation okay around the center of mass. So, now what we define is that the angular momentum around the center of mass will be given by at any point will be R bar okay R I bar what is R I bar is the position vector of E with respect to R V I bar is the velocity okay what is this velocity is the velocity of this point relative to the center of mass okay it is relative to the center of mass but what is that that is nothing but omega cross R I so we substitute that here and do an integral over all mass particles and what we will see is that the angular momentum is nothing but this omega comes out it is common for all of them omega k and the only thing that remains is R square omega comes out into delta m for all these particles but what is this sum over all the particles R prime square delta m is nothing but I put an origin at this centroid and at this centroid when I put the origin this is nothing but the mass moment of inertia of the centroid this when this summation can be replaced by an integral and this will become R square dm and R square dm is nothing but the mass moment of inertia okay of this rigid body about about which axis about an axis in the z direction but passing through the centroid okay so this is our H g which is nothing but now I times omega so that is a mystery why this I comes when we want to find out the angular momentum of a rigid body for rotation about it centroid okay it is just I omega now after we differentiate okay this I remains constant so H g dot will be nothing but I alpha but we had seen earlier that what is H g dot H g dot is nothing but there are all these external forces acting on the body some of moments of all the external forces acting on the body about point g is H g dot so what we know that some of the moments of all the forces is nothing but I times alpha where what is alpha alpha is the angular acceleration of this planar body for rotation about an axis along the z direction but passing through the centroid note that these results are also valid for not only for this thin slabs but as I discussed earlier that is for this slab I take a sphere okay and the sphere has a nice symmetry about a central plane and the entire motion of the sphere okay happens along that plane then also this is fine okay so it is either fine for planar bodies or symmetric objects like sphere or cylinder where you can cut through the center and all the motions subsequently are happening along the plane okay which is cutting through the center now coming to what is called as the Dallamber's principle what we had seen is this that these are the forces acting on the rigid body what is the effect if g is the centroid if g is the centroid then what do these forces do these forces will lead to an acceleration okay some of all these forces vectorially is mass times the acceleration okay in some direction what direction will be the direction of the resultant forces what else in addition all these forces can also exert moment on this body about point O in that case the sum of the moments of all these forces about O is nothing but I alpha what is this I alpha if the moment is taken in the anticlockwise direction then correspondingly the I alpha also is equivalent in the anticlockwise direction so sigma fx is equal to max sigma fy is equal to ma y okay what is this ax acceleration of the center of mass m is the total mass same for a y and mg is what is the moment taken in either clockwise sense in this case it is taken in the sorry in the anticlockwise sense and alpha is the angular acceleration in the clockwise sense okay these are equivalent so what we think is this okay we can think about it in two different ways that forces acceleration moments of these forces about point O rotation okay I alpha ma now we can do another thing so what does the Lambert principle tells us that each of the small elements here how do they respond to the forces that are applied they respond by accelerating okay they respond by having some kind of angular acceleration about point G and this effect of all these forces is essentially okay is equivalent to the effective forces of the various particles forming the body that each body point on the body has a contribution and all those contributions will add to this and how are those contribution created they are created from this same goes for this okay that the external forces can create a torque about this but because each and every point now okay has some angular acceleration and that angular acceleration okay in terms of torque can add up about this point G which is the centroid and this is crudely speaking the Lambert principle but mathematically speaking what it means is that all the forces are acting this is equivalent to this that all these forces is equivalent to having mass times acceleration in a given direction and some of the moments of all these forces about point G is equivalent okay to having okay I alpha okay in the same sense but we can now do what do is that we can bring it in this direction what we say is that rather than doing this we reverse the direction of this ma let us reverse it if it is in this direction let us say it is reverse like this okay this I alpha is acting in what direction I alpha is acting in the anti-clockwise direction we just put this I alpha in the clockwise direction and say rather than saying that these two that this is equivalent to this let us put the negative of that and then say that in terms of this torque okay in the pseudo torque what is the pseudo torque I alpha acting in the opposite direction and in terms of this pseudo force M alpha which is acting in the opposite direction we can think of this body as if it is in some kind of equilibrium and then what is one of the advantages of this okay it is little bit artificial but one of the advantages is that it will become very clear when Professor Shobig Banerjee discusses about simple harmonic motion that when we use the Lambert's principle all the virtual work principle and everything the simple ideas which we had developed for statics now can be applied to this equivalent dynamical system so rather than saying that this is equivalent to this we say this plus minus of this is equal to 0 okay that is simply the Lambert's principle okay so we reverse this sign put here reverse the sign of this put here and this should be in some dynamical equilibrium and that is nothing but the Lambert's principle. Now let us come back to what are the three body diagrams and the kinetic diagrams okay let us take this simple example okay what do we have we have this body around which a rope is tied and we have applied some tension to this rope okay previously if you remember we did not bother about angular acceleration angular rotation at all what we did is that we only bothered about acceleration in one direction acceleration in the other perpendicular direction done what is the vector for acceleration m times a is the force but now for an extended rigid body we saw that there is this additional equation which comes into picture okay that mg is equal to i alpha okay so either we can use the equivalence that this is equivalent to this or we can say that negative of this superpose over this sum of this plus negative of this is equal to 0 so this is the Lambert's principle this is like our regular usage okay standard use of dynamical equations. Now what are the various forces that act on this thing one is the external force w one is a tension t now what do we do is that we isolate the body of interest okay clearly it is isolated draw our axis system we decided it is x y Cartesian polar okay the same technique what we had done for using Newton's laws but now what do we need to do is that that we also have the applied forces which are the or the tensions which are the support forces we put everything together but now in addition to what we had done previously there is this component of rotation that also comes into picture which never came into the light when we discussed simple Newton's laws of motion for particles okay and now what do we do we draw what is called as a kinetic diagram that all these forces okay and effect of these forces what some of all the forces some of all the moments of these forces about point g which is a centroid later on we will do one simple example where we can also do this when g when this point g is not a centroid but it is some fixed point for the time being let us say that if g is a centroid then what happens okay but now what are the effects of these forces look here what is the coordinate frame we had taken x in the right y to the left and this rotation in the anticlockwise direction so what do we do these moments okay in that sense positive is this positive is this positive moment is in the anticlockwise direction will lead to ma x for centre of mass ma y for centre of mass and i alpha in the anticlockwise direction for this so that is the only component that gets added here that this is this i alpha which comes here some of all the forces is m alpha some of all the moments in the anticlockwise direction in this sense is equal to about point g is equal to i times alpha in the anticlockwise direction now let us take this another example we have ac gb we apply a force p we need to find out what is the free body diagram and what is the kinetic force diagram now note one thing that this is a mechanism if you apply some force p there is nothing okay there is no reaction which can balance for this free body diagram some of moments our point c so clearly this will start to go into some kind of acceleration what are the various reaction support reactions is cx and cy it is a pin joint this is the applied force this is the weight what is our coordinate axis x and y this is taken to be in the anticlockwise direction so this will produce an effect which is equal to what i alpha in the anticlockwise direction ax in this direction ay in the opposite direction if you want to use the Lambert's principle okay I am not going to use it in this class okay you will see that being used in process of energy class tomorrow what we can do we will say that this is not equal to this but put opposite of this minus ay minus ax minus i alpha and then say that the system is in dynamical equilibrium that sum of all the forces acting on it is equal to 0 okay you can do that also but what I am going to do in the rest of the class is this will produce these kind of forces will produce a kinetic effect which is acceleration in the x direction m times acceleration y direction i times alpha in the anticlockwise direction now let us come to this another problem what we are given is a drum of 100 millimeter radius okay radius is given attached to a disc of 200 millimeter radius all these are given to us disc and a drum has a combined mass of 5 kg because of the symmetry the mass will pass through the effective mass will pass through the center of gravity it is at g the chord is attached as shown the force of magnitude p is equal to 25 Newton is applied we are given the coefficient of static and kinetic friction between the wheel and the ground okay mu s 0.25 asked to find out the free body diagram and the kinetic diagram for the wheel. So what do we do we find out that what are the what are the given forces weight is a given force p is an applied force which is a given force what are the reactive forces friction can be reactive force normal reaction can be reactive force how to take the direction will become clear when we solve a few problems okay p w friction is this is the normal reaction now what is the effect okay all the dimensions are given to us x and y is like this. So the effect of these forces will be to create m times a in this direction for this mass m times y in the vertical direction okay and i times alpha in the anticlockwise direction okay so we can keep going on and on for the ladder problem also what do we have this was the problem we have briefly discussed but in terms of statics but suppose we are told that the ladder is now sliding down the wall okay this point a is coming down point a is sliding to the left. So what is the free body diagram and what is the kinetic diagram because this is sliding down friction force will be up sliding to the left friction force will be to the right reactions up reactions here the applied weight is the main force w here all these forces what will be the effect will be to create ma x in the x direction ma y in the y direction i alpha okay in the clockwise direction so why am I emphasizing this so many times is essentially everything that we want to do we have to just draw free body diagrams like this appropriately and we are essentially done okay now let us solve a simple problem what we are given okay is this is the assembly ea is inclined at an angle 30 degrees fd is a link which is inclined at angle 30 degrees to this we have a mass a b cd which is at which is pinned here at a pinned here at d so mass of this can be neglected very little mass very little mass this entire assembly is also attached to a chord hb what in the vertical direction at point b now note that if there is a chord also present and these two links are present then whatever force is present okay this assembly can be kept in equilibrium by the reactions here why because these two reactions are parallel to each other okay but the third reaction is not parallel to the three two reaction neither do all these three reactions intersect at one point and as a result any force that is applied moment balance force balance can definitely be done on this on the other hand what is told in this problem is that this link is this is broken okay that is wire is cut and we are asked to find out that immediately after the wire has been cut the acceleration of the plate and the force in each link is what we want to find out now first let us think about this problem okay a little bit okay what may be the kinematics of this problem how will the system deform this is a parallelogram distance ad cannot change distance ef cannot change fd is constant ea is constant so kinematically what can happen that this angle okay can decrease or increase but this parallelogram will always remain a parallelogram so ad line okay it was vertical to begin with it will always remain vertical and as a result ab will also remain horizontal bc will remain vertical this everything will just translate but what will happen is that if you look at the center you will see that this point a just concentrate on point a because it is a translational motion each and every point will have the same motion okay including the center of mass all of them are moving in the same way so look at point a how will point a move okay when this link ea moves this point a will move along the arc of a circle so even center okay we will move along the arc of a circle okay and what direction in this direction will be the radius of curvature okay the same as this direction and what do we have okay the radius of curvature will be 50 millimeters so if you look at the center what is happening okay it is as if if you just concentrate on the center then center will try to move along the arc of a circle but instantaneously okay at t is equal to 0 when this system is is left from this position okay when you cut this the system was initially at rest so if you recall our pendulum problem okay that there was a problem which was present in the tutorial what did we have we had a bob okay which is tied here but there is a horizontal string which is attached to the bob and this is a weight what we saw is that if you cut this string that at that instant the velocity of this thing is equal to 0 and as a result the centripetal acceleration okay the acceleration in the normal direction a n also is equal to v square by rho equal to 0 the only acceleration it has is in the tangential direction okay and what is that that will be equal to w cos theta same logic is working here why because the center of mass because the center of mass is only going to move in the arc of a circle and what is the direction of the normal direction what is the normal direction a n is going to be along this okay just think about this the normal direction will be along this and the tangential direction will be perpendicular to this so what we see from our logic is that that instantaneously because v is 0 v square by rho is equal to 0 so for the center of mass what is the acceleration the acceleration will only have a component in a direction which is perpendicular to the direction of the link so this is 30 degrees so the direction of acceleration will only be in a direction perpendicular to the direction of the link the second component of acceleration will be gone now that is great so what we can do we draw the free body diagram of this this is the center this is the force acting in the a from link a it is a 2 force member we are neglecting the mass of that f df is the force acting in this link df now what do we have this is the weight acting this line is parallel to f a this line is perpendicular from the logic that we had just discussed because the system starts from rest velocity of g is 0 so v square by rho is 0 this a n is 0 there is only 80 but what is 80 now these 2 forces are not contributing to that at all so the only contribution coming okay of force coming in this direction is w cos 30 and so what we see is that some of all forces in this t tangential direction is nothing but w cos 30 is equal to m times a okay is the acceleration in this direction from that we can find out that the acceleration of the center of mass is how much is 9.81 cos 30 okay in this direction inclined direction so we know what is the effective acceleration of point g the moment this is released now after some motion happens some velocity is accrued then the acceleration will also have a normal component but because we are starting from rest center of mass will not have the normal acceleration at all now what do we want to find out we also want to find out what are the forces in these links so let us do so we have used one equation of equilibrium we are still left with 2 equations of equilibrium one is for moment moment and i alpha and other is for another force now let us say we do find out that along this n direction what is sum of all the forces it is nothing but f a plus f df and what is the another force acting in this direction it is minus w sin 30 and sum of all those forces okay will be the acceleration in the normal direction but we just saw that the acceleration in the normal direction is 0 so that is one equation that we get now let us take moment about point g of all these forces and what do we know that sum of moments of all the forces about point g in this case convention taken to be in the anticlockwise direction should be i times the alpha in the clockwise direction okay but note one thing that what is happening is that that this rod has no alpha because we just saw that this is a curvilinear translation that because of this linkage mechanism okay all points will just move translation like this so the center of mass there will be no rotation about it and as a result alpha is 0 so we take moment balance about point g only moment component will come from this and this but the weight will not add to the component and so we can find out two equations two unknowns they can be solved to give us what is the force in this link a and what is the force in this link df okay so we know this force this force and instantaneously when this link is cut cord is cut from the top what is also the corresponding acceleration. Let us look at this problem what do we have that a cord is wrapped around a homogeneous disc of mass 15 kg so mass of this disc is 15 kg this cord is pulled upwards okay it start from rest okay and just start pulling it upwards with a tension of 180 newtons what we are asked to find out we are asked to find out what is the acceleration okay of the center of the disc okay the center of the disc is here g what is the acceleration what is the angular acceleration of the disc and the acceleration of the cord okay so let us do that problem what are the various forces various forces are tension w now what are the possible accelerations that it can have x direction ma x y direction ma y and i alpha okay or angular acceleration in the anticlockwise direction now what are the forces acting in the x direction nothing there are no forces acting in the horizontal direction so ma x will be equal to 0 no acceleration in the horizontal direction but now what are some of the forces in the y direction t-w that should be equal to mass times the acceleration in the vertical direction so t-w is equal to ma y t is given to us w is also known to us so we know that instantaneously ay will be equal to 2.19 meter per second square now the last part is you want to find out what is the angular acceleration what is the angular acceleration straight away take torque about this center take moment balance about moments of all the forces about this system the sum of all the moments of this force and this force should be equal to i times alpha if this is anticlockwise this should also be anticlockwise so the only force that can cause a moment of can cause moment about point g is this tension t so t times r okay what is the direction direction is anticlockwise of this moment so that should be equal to i alpha for a cylinder okay Professor Shobig Banerjee would have discussed that the moment of inertia about the center of mass or the centroid is half m r square okay half m r square is for a disc or for a cylinder half m r square is the center of mass about this axis so alpha will come out to be minus okay or in other words that alpha is not acting in the clockwise direction but it is acting in the anticlockwise direction now the second question for us is what is the acceleration of the rope now the acceleration of the rope again by the same logic which we had used previously that this is an inextensible rope so for every point here okay for the every point here the tangential component of the acceleration okay will be the corresponding acceleration for the rope now point a here is common both to the both to this and to the rope but from this disc what is the angular acceleration okay is equal to alpha and alpha times 0.5 will be one component of the angular acceleration alpha r plus the angular acceleration plus the total translational acceleration of G the sum of these two essentially will be the total acceleration of this point a which also because of the inextensibility of the string the total acceleration of this cord which is nothing but the tangential component here which is alpha times r plus okay a because this tangential component is nothing but the acceleration relative to the center of mass but from an in a from a general frame point of view there is also an additional acceleration for the center of mass which is a so to add up here okay and that is the overall acceleration for the cord now this is one really nice problem okay so this is one very practical problem okay now what is given is this very nice problem that a uniform sphere of mass m and radius r is projected on a rough horizontal surface with linear velocity v that this cylinder has some velocity initially it is not in contact with this surface which has some kinetic friction okay coefficient of kinetic friction is k what we are done doing is that that we give it some velocity and let it fall on to the ground now what happens is that because this has velocity v initially all the points have the same velocity because we are not giving it rotation to begin with but so when this point comes in contact with the ground there is a finite slippage this point on the uniform sphere now note this uniform sphere actually is a 3D object but why are we taking this as a 2D object because if we look at the central plane of this sphere cut it this will look like a circular disc and all the motions are happening in the plane of that center okay that is why we are taking this as a planar motion now what we are told is that this is initial velocity because of which there will be a finite slippage between this point and a disc and there is a kinetic coefficient of kinetic friction mu k between this and this as a result what will happen there will be a returning force how much will that be equal to that will be equal to normal reaction times mu k in the opposite direction now physically when we think about it what happens think about it this mu k force can cause a moment in the clockwise direction about point o because of which this will start having angular velocity and angular acceleration in the clockwise direction but what is happening this force f what does it do this resisting force f it also tries to apply a force in the opposite direction so if the velocity of the center of mass was in this direction this is creating a force in the opposite direction so it tries to stop this sphere tries to decelerate this sphere the acceleration of the center of mass or if you just look at the center of mass the speed of that will try to decrease but what is happening is that that because of this force a torque is created or a moment is created about point g as a result of which this sphere will tend to rotate about the center so what do we do we draw the free body diagram what are the applied forces m the w is the weight of the cylinder n is the normal reaction f is the force coming from the friction between this and the subsurface and what we see is that that this using the laws okay using the balance laws will be equal to ma only why because this sphere cannot lose contact with the ground neither can it penetrate the ground so there is no acceleration for the center of mass in the y direction this distance should always remain equal to r and as a result there is no acceleration in the y direction so n will become equal to w straight away now what about sigma in fx we know that there is a finite slippage here so f in this direction will be equal to minus ma because the direction of h taken positive here f is negative so minus f is equal to ma but f using friction law because there is a slippage between this and this is also equal to minus mu kmg that will give us what is the acceleration so we see that the acceleration is minus mu kg in the opposite direction or we have a deceleration in of the magnitude mu k times g now let us write a third equation moment of this force is about point g now the only force that cause a moment on this body about point g is f times r what is the direction direction is in the clockwise direction and so if you take the clockwise sign convention this i alpha is also taken the clockwise direction so f times r is the clockwise moment about the center ok we are taking about the centroid will be nothing but the forest sphere what is the moment of inertia it is 2 by 3 m r square ok we can directly take from the tables 2 by 3 m r square is the moment of inertia of the sphere about an axis which is coming out of the plane and passing through the center of the sphere so mu k g r is equal to 2 m r square alpha so what do we get we get that this alpha ok what is the alpha the angular acceleration will be equal to 5 by 2 mu kg by r now note one thing that there is no particular relation right now between this a and alpha because if there are pure rolling then a and alpha will be related to each other ok so these two are independent quantities now a and alpha are two independent quantities but now what we are asked to find out we are asked to find out determine the time t at which the sphere will start rolling without sliding and a linear angular velocity of the sphere at time when that happens what do we do we write down the equation of equilibrium for this one there is a deceleration of how much minus mu kg so we write down that there is a constant deceleration so the velocity of the sphere as a function of time is v 0 minus this deceleration times time t same equation can be written for omega but for omega this omega will keep increasing how much will it increase by omega plus alpha t what was alpha we just found from the previous equation ok which is equal to 5 by 2 mu kg by r put that in here and at the instant t 1 ok now how do we figure out that the sphere will start stop sliding it will stop sliding when at some instant the velocity of the center of mass will be equal to r times omega 1 why because if v is equal to r omega then relative to this to the center of mass the velocity of this point c is equal to minus omega r in the opposite direction but if v itself is omega r then minus omega r plus omega r cancel out and there is no relative slip between c and the ground so what will be here is that just put v 1 is equal to r omega 1 substitute everything what will be find will find that that after this time t v will become equal to r omega and substitute that in omega we will correspondingly find out what is that corresponding omega ok and from that omega we can find out v is equal to r omega and what we find from here is that after what time and with what omega and with what velocity will this sphere stop sliding and come to rolling ok such that v omega is equal to r omega and you can convince we can convince ourselves that when that happens in this idealized behavior ok that when the contact is purely point contact when the sphere gets this uniform angular velocity then all the frictional forces will cease to exist and the sphere will now keep moving constantly at an omega ok and uniform velocity v of the center of mass and that is the paradox that for example if a wheel is rolling with a constant speed then in principle this coulomb's law will tell you that a wheel will never stop but we see in reality that a wheel stops and the reason being what is called as a rolling fiction that the point of contact is not perfectly a point there is an extended area like for example tile deforming and as a result this final reaction ok doesn't the contact is not a point contact and the resultant reaction does not directly pass from this point of contact to a but it has some moment and that moment arm ok decides what is the rolling coefficient of friction we are not going to discuss that. So, what I want to do is that this problem I want to briefly discuss in terms of conservation of angular momentum ok the way I will do is something like this. Now let us say so I will show you that how different techniques can be used to solve a same problem this one has velocity v bar 0 to begin with the angular momentum of this entire system about some point o fixed point ok about a fixed point o is H o bar which is simply m r bar cross v bar plus i omega bar ok. Now if this sphere ok look at this sphere what do we have we have its initial angular momentum ok and what we had also seen is that d H 0 by d t is equal to sum of all the moments ok about that point o ok and if this is equal to 0 what does that mean that means that H 0 bar is a constant vector. Now what is the initial angular momentum for this system ok let us say we will take this point o now this point o need not be at the point of contact it may be somewhere here along the ground. Let us say that this is point o here note what are the various forces acting w n friction what do we know that about this point o because of equilibrium in the y direction sigma w minus n should be equal to 0 implies n is equal to w. Now what is the moment by w and n about point o both of them cancel each other out what is the moment of this force about point o 0 because it passes through this line. So what does that mean that as far as point o is concerned ok the angular momentum of this body ok there is no torque acting on this about point o. So the angular momentum of this object ok about point o should be conserved. Now how do we use that we note this is v 0 to begin with this is o. So initial angular momentum h 0 in what direction in the clockwise direction simply equal to m v 0 r the final angular momentum where v is equal to omega naught r is how much this is h 0 initial h 0 final will be equal to m v r plus i omega but we want the point where v is equal to r omega. So substituted what will we get m v 0 r is equal to m v r plus i v by r and solve for this the velocity you will get will be exactly what we had obtained by solving this problem ok in a longer way. So this is the way in which you can use principle of conservation of angular momentum for solving this problem. Now second is we talk about constrained motion ok constrained motion one of the examples is a non-centroidal rotation ok what is that example ok the non-centroidal rotation one simple example is when this sphere for example moves such that there is no slipping then we know immediately that x is equal to r theta a is equal to r alpha and for rolling and no sliding this friction force here should be less than or equal to mu s times n a is equal to r alpha rolling with sliding impending f is equal to mu s n a is equal to r alpha and rolling and sliding in which case now it no longer remains constrained that a or the acceleration of the center of mass and r alpha which is radius of the cylinder or radius of the sphere times the angular acceleration are completely independent ok. So let us solve one problem like this. So what we have we have a sphere of weight w it is released with no initial velocity and rolls without slipping on this incline the incline angle is 30 degrees. So what we are asked to find out is determine the minimum value of coefficient of friction at this contact point the velocity of g after the sphere has rolled 10 feet ok or 10 meter ok you can take this 10 feet is converted into 3 meters. So the velocity of the g of the center of mass after the sphere has rolled 3 meters and the velocity of g or the center of mass if the sphere were to move 3 meters down a friction less incline ok. I mean if this were completely friction less then what will be the velocity it will get ok. So these are the 3 parts of the problem. So let us look at the first problem where we are asked to find out that for no slippage to happen between the point of contact c between the sphere and the surface what is the minimum value of coefficient of friction. Now note one thing ok what can be the friction ok let us say the sphere is trying to move down ok when it tries to move down what happens ok it tends to roll about this point c which is the correct instantaneous center of rotation at that point ok because there is a torque by the weight caused about point c. So it will tend to rotate in this direction and now what do we do what we do is that it will have an angular acceleration about the centroid and to create angular acceleration in this direction ok what is that it is in the clockwise direction the only way the friction can act will be in the upward direction. So only when friction acts in the upward direction we can have a clockwise rotation and clockwise angular acceleration for this sphere. Now what do we know we know that the sphere does not have any acceleration at its center of mass in the y direction. We also know that i alpha is related with this m alpha ok this alpha and i are related with each other why because we know that there is no slippage. So a is equal to r times alpha and so what do we do first thing ok. So this was one of the questions that was asked is when there is a point of contact between the sphere can we ever take a moment of this force about the point of contact and then write down the equation the force and the moment balance equation and the answer is this that strictly speaking ok strictly speaking the equations that we have written that the rate of change of angular momentum with respect to a point is equal to sum of all the moments about that point acting on the body that equation is true strictly only if the point O is a fixed point it is not a moving point or it is the center of gravity or the center of mass of the body. But what we think here is this that we want to find out what is the instantaneous acceleration ok that in this configuration what is the angular acceleration and because there is no slippage correspondingly what is the linear acceleration of the center of mass of this sphere. So what we do is that this point C ok even though it keeps moving the point of the contact keeps moving what we do is this we say that this point C is say it is stationary on the ground. So at that instant this point C is one fixed point on the ground and so what do we do we find out what is the moment of all these forces about this point ok. Now what is the relation that we get take sum of all the moments about this point what should that be equal to if sum of all the moments were taken about the center of mass then what happens this will be only equal to I times alpha. But note because now we are all we are not taking the moments about the center point we are taking it about an arbitrary point. So what quantity will come a quantity that will come will be m r cross a ok and in this case r cross a will be what a r in what direction the same as the direction of moment it will be in the clockwise direction. So m a r ok in the clockwise direction is the additional torque is the additional component that we have to add here ok and so just note one thing that this sum of all the moments about a point is only equal to I alpha when this C is the center of mass but when it is not the center of mass there is an additional contribution and additional contribution comes from m a what is m a m a is mass times the acceleration of the center of mass. So in vectorial language ok in vectorial language that contribution will be H o bar dot is equal to mass r bar with respect to o for the center of mass times b bar dot which is a bar and plus I alpha ok this is for 2D plane motion and what we see is that and in this case all the components we are taking in the clockwise direction. So r cross v in the clockwise direction this looks as just m a r ok in the clockwise direction r cross a ok the direction will be in the clockwise sense. So moment w sin theta into r is the moment of this force weight acting from this free body diagram about point C which is equal to m a r so this term should not be forgotten because this is not the center of mass plus I alpha and you substitute in here what do we have this becomes m r m r square alpha plus 2 by 5 m r square alpha substitute everything together and you will see that w sin theta r will be equal to this into alpha or what you will see is that instantaneously at that point the acceleration in this configuration alpha is 5g sin theta divided by 7r and correspondingly the linear acceleration of the center of mass is r times alpha which just comes out to be this r goes away 5g sin 30 by 7 ok which comes out to be 0.3.5 for after substituting the corresponding angle 30 degree here. Now note one thing that we got alpha we got a we also we also realize is that that there is a constant acceleration ok constant angular acceleration and constant linear acceleration. So if you want to now find out that how will the position and velocity of this evolve then what do we do we do not need to take this moment balance again and again but we realize that at any point if this were at any point we will see that no matter what we do the contribution will exactly be this we can do we can do whatever we want to ok but the contribution that comes for this alpha will always be the same and alpha cannot change from one point to the other. So let us choose this alpha so we know that for this cylinder the alpha is given by this a is given by this. Now let us do further force balance let us put our coordinate axis x in this direction y in the perpendicular direction what do we have some of all the forces in the x direction should be equal to m times a in this direction what are all the forces in the x direction w sin theta minus f it should be equal to ma so we find out from this simple equation see we have not invoked it that mu is equal to f is equal to mu times n we can directly find out we can directly find out what is the value of f and the value of f comes out to be 0.143w for whatever parameters that we have here similarly some of all the forces in the y direction there is no acceleration in the y direction as a result n minus w cos theta will be equal to 0 or n will be equal to w cos 30 degrees and we find out what is the normal reaction. So now what do we have we have that when this sphere is rolling without slipping then the corresponding frictional force is this and the corresponding normal reaction is this so what do we want that this ratio f divided by n should be less than the coefficient of static friction here because there is we do not want to have any slippage here and as a result the minimum coefficient of static friction will be nothing but this f divided by n and we can find out that in this problem the static friction coefficient comes out to be 0.165 and note why we have used static friction coefficient is that even though the center of mass is nicely moving with some velocity but at the point of contact there is no relative slip between c coming from the from the sphere and c coming from the surface. So at the point of contact the instantaneous velocity is 0 and so there is no slippage and that is the reason why we are not using mu k and we are using mu s for this problem. Now we want to find out velocity after 3 meter of uniformly accelerated motion initial velocity is 0 the acceleration a we found out is what from the previous slides the acceleration is found out to be 3.5 meter per second square we substitute in here v square is equal to v naught square minus 2 x minus x naught we already know. So we know that after moving a distance of 3 meters the velocity that it will get will be 4.59 meters per second. The next part is assuming no friction calculate a linear acceleration under corresponding velocity after 3 meters. Now if there is no friction then what happens then there is no rotation at all because it is only the friction which is causing this angular rotation. On the other hand if there is no friction then this entire ball moves as if it is a particle of mass m without any rotation and as a result some of the forces in the x direction the only force is W sin theta. So acceleration in this direction is W by G is equal to W by G into a W sin theta equal to W by G a or what we find out is that acceleration in this direction is simply acceleration due to gravity into sin 30 sin 30 is half. So this become G by 2 and then as a result we find out that the acceleration of this is nothing but just G sin 30 and from that we can easily find out what is the velocity after time after it moves by distance x and what we realize here is this that the velocity here is lower when the when the rolling is allowed without slipping the velocity is lower actually than correspondingly this. Now how do we understand that? We understand that by noticing that if we use principle of conservation of energy for extended bodies then this is the sphere it moved down okay by a distance s is equal to 3 meters this angle is given to us. In first case what happens is that initially there is no velocity here there is both velocity for the centre of mass and there is also acceleration there is both omega coming from angular rotation whereas in the other case when there is no friction it starts from here it reaches here the same distance but it has only velocity. Now note one thing that in both problems the change in potential energy is the same initial kinetic energy is 0. So delta V should be my kinetic energy at the bottom here and delta V should be equal to T1 prime at the bottom here but note that because there is no angular rotation here the kinetic energy at the bottom is simply half mass of the sphere into velocity of the centre of mass whereas in this case because there is also angular acceleration that there is also angular velocity the kinetic energy will be equal to half m VC square plus half i what is i this i is the moment of inertia about the centre omega square where omega is equal to R VC and because of this there is some part which goes into this as a result this one okay this sphere when the rolling is also allowed will in general have a lower acceleration because it has two resistances one resistance is coming from moment of inertia other resistance is coming from the mass whereas this one will have less resistance okay only resistance is coming from the mass and some part of the kinetic energy in this case can go into the centre of mass velocity some part can go into the rotation into i whereas here the entire potential energy get converted into half m V square and as a result okay with this simple argument we can realize that the kinetic energy or the velocity okay in the first case will be lower than the velocity in the second case when there is no rotation now we will solve this one last problem now note here is a very nice problem what is told to us is this that knowing that the coefficient of static friction between the tires and the road is 0.8 for the automobile shown determine the maximum possible acceleration on a level road assuming a rear wheel drive now this was one of the discussions we had posted on Moodle okay that if something is a real wheel drive then there is a motor which can which is connected via the engine is connected to the wheels okay through the axle via the gearbox okay as a result on this wheel there is an internal resistance because of which it cannot freely rotate and there can be a friction that can act on the back wheel whereas in the front wheel assuming that it is a it is a perfect circle okay and the line of contact is only at one point tangential to this we note that because there is no engine here if the inner wheel is properly lubricated then this is free to rotate and as a result it cannot take any horizontal force now note here what can happen what kind of forces are possible if this car is accelerating the forward direction okay the frictional force has to act in the forward direction this is the normal reaction on the back this is normal reaction on the front this G is the position of the center of mass of the vehicle it is at a distance 1.5 from the back wheel 1 meter from the front wheel and it has a height of 0.25 meter from the bottom now let us see okay what we are asked to find out that when mu is equal to 80 degrees what is the maximum acceleration that we can get for this now how decides the maximum acceleration the maximum possible value of FR decides the maximum acceleration why because the only force acting on the car in the forward direction is a force coming from the friction between the tires okay and that will lead to acceleration of this car so note what is the effective kinetic diagram that in principle this car can have I alpha rotation our center of mass MY AY and MAX but we do not want this car to have a rotation okay which means that the car is actually lifting up from the ground neither do we have neither do we want to have an acceleration in the y direction y this also means that the car is lifting from the ground so what we want is that that without lifting from the ground what is the maximum acceleration that can be achieved when mu is equal to 0.8 so what do we say that MAY is equal to 0 I alpha is equal to 0 so we write down equation for this car in the x direction that sigma FX is equal to MAX okay what is the only force in the x direction force r friction the frictional force coming on the rear wheel is equal to MAX some of all the forces in the y direction is equal to MAY so nr plus nf minus mg equal to 0 y because we do not want this car to lift up okay and it is not possible for it to penetrate down so this acceleration in the y direction has to be 0 ultimately we do not want this car to topple up okay because we do not want any alpha so sum of all the forces above the center of mass should be equal to 0 so we solve this okay we get three equations and now what we want to find out is a what is the maximum possible force that can be exerted so we say that fr is equal to mu times nr okay we want to look at the limiting case we now solve these four equations so there are four unknowns fr nr nf okay this is a normal reaction on the rear wheel normal reaction on the front wheel friction on the rear wheel okay and acceleration of the car so four equations four unknowns we can solve for them and what do we get that the acceleration of the car okay will be given by g divided by 5 by 2 mu minus 0.5 and when mu is equal to 0.8 the maximum acceleration that the car can generate is equal to 3.74 now there is something pretty unusual about this equation nively what we will think is this we say okay if the car okay so there was one one professor who had asked a question about the optimal value of the friction I think some that time I could not answer it properly but what this expression tells us is this expression tells us something quite funny our naive guess will be that if we keep on increasing the friction coefficient this is great okay our acceleration will keep increasing increasing increasing okay but what we realize is that that if we make mu equal to infinity let us take an extreme case where mu is equal to infinity well what happens this x becomes negative what does that mean that a car is trying to move backward this makes absolutely no sense then what is the mystery the mystery here is that we have to realize that what is this frictional force doing this frictional force it is trying to create a moment about the center of gravity in what direction in the anticlockwise direction and that moment what what tendency does that moment have the tendency of that moment is to lift this reaction up and if you solve this problem appropriately what you will see is that in a hypothetical scenario if the coefficient of friction is 3 then this normal reaction becomes equal to 0 so if we keep increasing the coefficient of friction okay make it like really really large beyond a critical value what will happen is that that this car front will lose contact with the ground and the car will tend to topple off okay so as a result okay we cannot have a friction which is arbitrarily large there has to be some optimum value of friction okay if the friction value becomes very large then the car will lose contact from the front wheel okay and will no longer be a stable drive okay it will try to lift up so that is the the point I wanted to put forth from this problem okay now with this okay just one last problem and then we can solve the tutorials we have this assembly we have an inner we have a pulley of mass 6 kg and having a radius of gyration 200 okay and connected to two blocks now what is the radius of gyration okay professor Banerjee would have discussed that m times radius of gyration supports its k so m k square is the moment of inertia of this entire object about center c now this problem becomes so nice and simple that the total sum of the moments acting about the center of mass okay is equal to I times alpha we use this concept this problem becomes amazingly good now what we do here is this point c here okay point c here is what point c here is a fixed point so we use the concept here is that that since this point c is the fixed point let us look at this all are connected to each other so what is the overall alpha for this system okay we want to find that alpha out now what we do is that let us take point g what is sum of all the moments about point g sum of all the moments about point g is if we draw the free body diagram of this entire assembly removed from here there is a weight acting here 5 kg there is a weight acting here okay 5 kg so what is sum of all the talks about this is 5 kg into r b in what direction in the anticlockwise direction okay minus 2.5 okay this 2.5 about center what is the force it will exert what is the torque it will exert it is r a multiplied by 2.5 in what direction in the clockwise direction so the total moment of all the external forces about point o which is a fixed point is simply equal to this difference of these forces and we know that what is the unbalanced moment coming from the external forces about g okay which is the fixed point about which the pulley is free to rotate okay which comes out to be 1.227 Newton meter now the rotation okay we take it to we take the counter clockwise the rotation is now counter clockwise okay and mg in the counter clockwise through the counter clockwise direction is taken to be positive so both rotation and mg in the counter clockwise direction we now take it to be positive what is the moment of inertia okay for this pulley is i k square now what do we know we know that the acceleration of the block okay of this block a is what r a times alpha because why they are connected to each other so the tangential acceleration at this point will be essentially the acceleration of this block similarly the tangential acceleration at this point will be the acceleration of block a so we know that if alpha is one quantity okay which is an unknown which is the angular rotation for this pulley about point c then a or the tangential acceleration at point a will be r a times alpha at b it will be r b times alpha so we know this is a acceleration of point a this is acceleration of point b now what do we do what we do is that we take moments about point g what are the some of moments about point g we had discussed that in the last slide that these are the only moment components that are acting okay we discussed here on this free body diagram about point g but what should that be equal to that should be equal to okay what we had seen is i alpha okay about this g what is i alpha plus what is happening now is that this is m a b into a b so if you remember the term which we had discussed that h bar dot about any point o is equal to m r a okay plus i alpha okay everything has to be in the appropriate clockwise or in the anticlockwise direction so here what is that contribution the acceleration is acting in the downward direction so m b into a b okay we are taking clockwise to be the positive one so m b into a b into r b is that extra moment that we have to put which is 1.5 minus m a into acceleration of a into this point 2 phi now what is that quantity if you take that this is the acceleration of a okay what direction will be the what will be the corresponding r cross a it will be in the opposite direction now sum all this thing together sum of all the moments about point g will be equal to i alpha plus the extra components coming from the masses of these two solve this and we can find out what is the angular acceleration of this pulley and once we know the angular acceleration we can find out what is the acceleration of a and b it is simply equal to r a times alpha for a and r b times alpha for b okay so we are stopping our session here if you have any questions you need any clarifications okay I will be happy to answer now so there is a question from center 1331 in the place of cylinder if we take a block can you please discuss the free body diagram and the kinetic diagram okay so let us discuss that if now having a cylinder for example we have a block okay so what are the forces acting the forces are w sin theta w cos theta the normal reaction and the friction force now what is the difference between this and this the difference is as follows in this case for example the cylinder has no control okay it has no control okay over what should be the point of action of the reaction this has to be the point of action of the reaction okay it cannot help that so as a result f has to act here only in the tangential direction normal reaction has to act at this point only in the vertical reaction and because of this unique capacity what happens is that if for example for a cylinder okay if there is if for example this w sin theta acts on the along the direction of the plane then what happens is that this w sin theta can cause a torque about this point and what is that torque that can cause a torque in the in this direction and there is no reaction to balance that torque whereas in a block what happens is this that suppose okay you have this w sin theta we have this w cos theta now what you will say is that sorry this is w cos theta what you will say that the w cos theta is passing here friction is passing here okay what if I take a moment balance about this point o well there is nothing to balance the torque which is created by this w sin theta about this point but that is not right why because this is an extended contact okay this is an extended contact and as a result the normal reaction that can come from the ground on this block it can either act here it can act here or here or here or here it has a complete freedom to change this direction change its line of application and in the case of a block what will happen is that unless and until for example the toppling problem that we had discussed that it is a huge block and to this huge block we are applying some weight here okay what will happen is that that the direction the point of application of the normal reaction coming from the ground can always change and adjust itself in such a way that that moment that is w sin theta causes okay about this point can always be balanced by adjusting the normal reaction line of action and that is a major difference between having a cylinder here and having a block here cylinder has no freedom to choose what will be the point where the reactions will act whereas in a block because the contact is extended the resultant normal reaction for example can act here here here here such that the moment which is w sin theta creates about this point will be appropriately balanced by that particular force okay so that is a difference okay between a block and a cylinder and as a result for example this is the simple reason why we use for example why wheels roll freely whereas a block does not roll why because a block there is a resistance to rolling because of this fact now from center number 1085 a question is asked that friction force is generally opposite to the motion and in this problem friction force has the same direction as the motion and it was explained that a friction force is the one that accelerates the car so very nice question and the idea will become apparent if you look at closely okay that this is the back wheel of the car suppose you have a rear wheel drive okay this is car this is the front wheel of the car now note that that this back wheel okay what happens that this back wheel has an axle axle is connected to the gearbox okay in layman's in very simple language and the gearbox okay is connected to the engine okay via flywheel and clutch assembly okay so but those are the extra details but what we have here is that that if we see that when we want to start the car what do we do when we turn the key on okay engage the clutch okay and give accelerator the wheel starts to rotate okay like this now what happens that when the wheel starts to rotate like this it will tend to slip against the ground at this point it will tend to slip in what direction tend to slip in this direction so the relative motion that the wheel tends to have against the ground will be in this direction and it is that motion okay that the internal torque is being provided here so if I draw the free body diagram of the wheel what do I get this is the internal torque which is provided to the axle from the engine this internal torque now becomes active this is the weight this is the normal reaction and what happens is that the frictional force how can it act this is the internal torque M okay in this direction so about the center sum of all the moments in the clockwise direction about the center O should be equal to I times okay what is I is the moment of inertia of this wheel times alpha where is the it is the angular acceleration about the center now note one thing that what are some of the moments some of the moments is the internal moment that is applied in this direction okay minus okay there has to be some force coming from the friction let us for the timing assume that the friction force is like this okay let us say it is this then what will happen is that that the second force is F times R where R is the radius of the wheel and this is equal to I alpha but note one thing is that that this M is extremely large and as a result because this M is extremely large you will see that this I alpha is a negligible quantity and this quantity has to be approximately equal to 0 okay to the first order okay it is a 0th level approximation that this I is very small compared to the moments that are being exerted and as a result this F cannot have a positive direction because then this cannot be nullified and so to resist this moment you will have M minus FR which is approximately equal to 0 and what is this F so F should have a direction in this direction because otherwise the torque cannot be in the anticlockwise sense and there is nothing to take care of the internal torque that the engine okay is exerting on the axle and that has to be balanced only by the friction force and so the friction force comes in this direction and now that is a friction force and that is the reason why now you have a car that is getting pushed forward because the wheel is trying to do this that is one component friction force acts on the wheel because of the free body diagram here okay which is drawn here and as far as the complete car is concerned the total force now acts in the forward direction and that is the reason why the car accelerates the same principle will also act for example you are trying to move a bike okay so I hope this is clear okay in the last two questions sort will there be any gyro couple generated okay there is no gyro couple generated why because for a gyro couple to be generated we had discussed the other day okay this is beyond the syllabus that we are discussing now but for a gyro couple what should have what should happen what should happen is that if you have a wheel which is rotating in this direction unless and until you apply a third torque okay in a perpendicular direction like this the gyroscopic moment okay or a torque like this because if you tend to rotate it like this it you have to provide a torque like this so unless and until you apply this out of plane torques okay there cannot be any gyro couple but in this case all the moments okay or in other words all the torques they all act in the plane so there is no possibility of having a gyro couple in this problem then the next question is if slip is present then only translatory effect will be considered okay so you are right that for example if you say in this problem that these for example are not tied here okay this is like a spool that when this rotates this is compulsorily will go up and this will compulsorily go down and when this rotates in the clockwise direction this will compulsorily go up and this will compulsorily move down this is like for example a spool but suppose this is like for example a belt and when it is like a belt for example there may be no friction okay and in such a case like that okay that these two motions may be completely independent that this may start slipping okay and this may rotate independently and in that case we have to draw the free body diagrams for this and this separately and analyze them separately even this problem can be attempted in that way we can draw the free body diagram for B and A take tensions, tensions okay find out what is the appropriate tension that is required what is the appropriate tension required then this TA-TB okay will create a couple about 0.0 not couple a moment about 0.0 and that should be equal to I times alpha so you can do that also okay but this problem becomes very simple why that this is like a spool okay that this is tied at one end here okay so this rope goes here through the groove and it is tied here so once you start rotating this this has no no recourse but to go in the upward direction if you rotate it in the anticlockwise direction it has no recourse but to come down and same for point A so that is why this problem is a bit simpler and we can solve this as one complete problem modify the mass of inertia okay by taking into account these two components but you have to also add the components coming from the acceleration of A and acceleration of B and then this problem can be very easily solved by noting that acceleration at A is equal to rA times A and acceleration of B is equal to rB times alpha okay not A alpha there is a question by center 1314 can we consider the motion of a cylinder in a plane a constant motion definitely we can if there is no slippage okay the friction is enough such that there is no slippage between or forget about friction you know that the conditions are such that the velocity of the center of mass V is equal to radius of the cylinder times omega which is the rotational velocity of the cylinder if that is the case then what happens is that at the point of contact okay if I draw this here this is V this is omega so at the point of contact C what is the total velocity now relative to the center of mass the velocity will be minus r omega in this direction plus the velocity of center of mass will give us what is the absolute velocity of the cylinder at this point now if this is not equal to 0 then there is slippage and omega and V can be independent or A and alpha can be independent but when there is no slippage then what happens is that that minus r omega plus V there is no relative separation between these two this is equal to 0 and then V becomes equal to r omega then it becomes a constraint motion in which case it is a one it is only a one degree of freedom that is that is left that either omega or V and they are linked with each other okay so this is the reason why a cylinder okay moving on a flat surface okay without without having slippage at the contact okay then this is one example okay where V is equal to r omega and then this becomes a constraint problem that the alpha or the angular acceleration omega angular velocity are not independent of the linear velocity and linear acceleration they are related by V is equal to r omega or acceleration is equal to r times alpha so I hope these discussions are fine okay so thanks for the bearing with me