 So, now let us move on to the new topic of our day today. So, we will move to work energy theorem, principle of energy conservation, momentum conservation and impulse momentum theorem ok. So, let us move in that direction today. Any questions ok, please send them to the chat ok, if you have any comments any questions please send them to the chat. If you have any additional question also post them on Moodle. Now let us look at what are the energy and momentum methods. So, about energy so yesterday Professor Shobig Banerjee he had briefly discussed about principle of minimum potential energy. So, what he had discussed is that if you have a spring ok, you push in the spring you can store the energy in the spring as a potential energy, but we did not talk about kinetic energy why? Because one step at a time. So, because we are going to discuss it today, what is the kinetic energy? What is potential energy and how does the inter conversion between these two happen? And that is the topic of today's lecture ok. Now this Pogo stick is one simple example ok, that if you press on this potential energy gets stored in this and when you release you jump ok. So, the mechanism is essentially as simple as that ok and you can add some potential energy release it, add release, add release and in this case the potential energy keep getting converted to kinetic energy and vice versa and that is how the mechanism keeps hopping. And this for example, these are the ideas we are going to use to describe what are the momentum methods. So, momentum methods typically ok they really work in the cases where for example there are impacts. What are different examples of impact? For example a cricket ball a bat hitting a cricket ball is where for example there is a impact between both of them for a very short period, but the forces are so large that due to this impact the momentum of the ball for example it comes rushing to the batsman and it goes rushing in the opposite direction ok. Large forces act on the contacting bodies for extremely short durations and that particular thing is called as impact or impulse and there are this impact momentum theorem ok which is nothing but a modification of Newton's second law of motion that we are going to discuss ok in the later part of the course ok. So, we will discuss work energy theorem, impulse momentum theorem, from this impulse momentum theorem we can discuss what is conservation of linear momentum, specifically we discussed central impact of two objects and the coefficient of restitution. So, these are the topics that we will cover by today 3.30 and tomorrow morning we will have a small tutorial ok and then we will start our kinetics of planar bodies of rigid bodies and kinematics and kinetics of rigid bodies is what we will start in tomorrow late morning session. So, just to remind ourselves ok we were discussing all the problems previously of the type f is equal to m a in this current lectures ok in this current lecture what we are going to do we are going to add two new components ok methods of work energy and methods of impulse and momentum ok. So, there are various problems ok in which for example we will see that in Newton's second law we say f equal to m a we will see in the work energy theorem that at any instant if the kinetic energy of the body is T 1 and from 0.1 you move on to 0.2 that point may be in space or time we move from T 1 to T 2 then U 1 2 is the work that is done by all the external forces on that particle then T 2 is the final kinetic energy. So, this is work energy theorem ok in the nutshell and ultimately we will discuss the impulse momentum theorem ok we will discuss the impulse momentum theorem where the momentum ok can be modified by acting on a impulse. So, this is the initial momentum when we act and impulse and impulse of this form on it what that impulse is we will discuss briefly ok then the momentum of the body changes. So, we have discussed this concept of work ok in great details when we discuss principle of virtual work also we discussed this thing in great details when we also discussed when professor Banerjee also discussed principle of minimum potential energy now let us discuss what is the concept of work of a force ok. So, what is work of a force ok so just briefly this brief discussion so we have a particle we move it from point a to a prime then the work of a force is nothing but f dot dr where is the f acting here this is dr. So, work is a scalar quantity that it has a magnitude but not direction dimensions of work quickly coming is joule ok sometimes feed point we will use but by and large ok we will not use this notation 1 joule is what 1 Newton force ok creating a displacement of 1 meter ok that is that is the unit of work. Now work done by a force during a finite displacement ok so this is infinitesimal displacement that a force acting on a body infinitesimal displacement of dr is done then what is the work done during a finite displacement. So, the work done during a finite displacement is ok if you take this particle from point 1 to point 2 ok this is point 1 this is point 2 this is the path along which the particle goes and when you move along the path this force may change direction may change magnitude we do not know but a total work done by the force ok when you go from 1 to 2 is nothing but from a 1 to a 2 f dot dr. So, f dot dr is the work done for an infinitesimal displacement and this integral is a total work done for taking this particle from 1 to 2 ok and this we can also represent in terms of scalar product at f x dot dx f y dy f z dz in other words work is also represented as the area of the curve f t plotted under ds ok. So, let us not worry about this what is this f t, f t is the force which is tangential to the curve because f dot dr is what dr is tangential to the curve and f dot dr is nothing but the tangential force to the curve multiplied by the arc length ds ok. So, this is the total work done of the force so this integral is what we need to know now second is work done for a force so in a rectilinear motion ok if the particle moves from a 1 to a 2 in a straight line then work done to take this particle from 1 to 2 is given by f times dx is not right because f has this direction so the work done is f cos alpha into dx ok because this force is constant the direction is constant the work done in moving the particle from 1 to 2 is straight away f cos alpha into dx. Now let us quickly look at a couple of sources of couple of sources typically observed in dynamics problems which can do work simplest is the work done by the force of gravity work done by the force of gravity is what you take this mass ok make it go from 0.1 to 0.2 the force of gravity acts in the downward direction so the work done by the gravitational force is how much from here to here is you are moving a distance of y2-y1 but the force acts in the downward direction so the work done is negative or u of 1 to 2 ok that the work done for taking this particle from 1 to 2 under the influence of gravity is just minus w into delta y this minus comes because from 1 to 2 you are going upwards but the force has always been acting downwards that is why the minus ok if on the other hand we take this mass and bring it from a2 to a1 then the work done to bring it from a2 to a1 is essentially positive y because from a2 to a1 delta y is negative you are coming downwards along the direction of the force ok so w delta y is essentially the work done by this weight. Second is work done by a spring ok which was briefly discussed by professor Banerjee ok so if you apply a force to the spring what will happen is that that infinitesimally ok when the spring is at position x when you stretch the spring the spring force acts in the outward inward direction whereas your displacement delta x in the outward direction so the work done is minus kx into dx and if you stretch the spring from a position x1 to x2 then the work done to stretch the spring from x1 to x2 is nothing but half kx1 square minus half kx2 square so if you extend the spring then the work done by the spring force is negative if you work done by on stretching the spring is negative that is u of 1 to 2 ok. Now what is the you can also think about this taking that if you plot that what is the force in the spring as a function of the stretch you will see that at x1 the force is kx1 at x2 the force is kx2 and clearly the overall work done by the spring force when you take the stretch the spring from x1 to x2 is nothing but the area under this and what is the area under this is essentially this integral but because this is a trapezium you can also think of this as minus half f1 plus f2 divided by 2 this is the work done to take the spring from 1 to 2. So for example as the block moves from a0 to a1 the work done by the spring is negative as the block moves from a2 to a0 ok what is happening here is that this is a2 the force on the spring ok is always inverse ok because we have stretched it the force is acting inverse but we are also moving inverse in this direction so if you stretch a spring from rest length ok the work done by the force in the spring is negative. Now if you start slowly releasing it the direction of the displacement and the direction of the force both are in the same direction and so the work done in this case is positive. This is work done by the gravitational central force let us not worry about it this simple question does the normal force do work as a block slides from b to a so when the block sides from b to a the displacement is along the slope whereas a normal reaction is perpendicular to the slope so the normal force does not do the work when the block slides from b to a. Second question does the weight do the work as the block slides from b to a clearly when the block goes from b to a it has a downward movement and the work done by the gravity ok will clearly come out to be positive why because the displacement in the downward direction is positive whereas the gravity is also downwards which is positive so the overall work done is also positive ok. Now just quickly we remind ourselves that what are the forces the forces which do not work are when displacement is zero or cos alpha is equal to zero what does that mean that the displacement and the force are perpendicular to each other. So the forces which do not do work are the reaction at frictionless spin supporting a rotating body why because there are no internal forces no internal torques ok so in that case there is no resisting force so no work will be done. Reaction at frictionless surface then the body is in contact moves along the surface at a frictionless surface the reaction is only in the normal direction so the normal direction will not do work when you have a displacement in the horizontal direction perpendicular to it reaction of a roller moving along its tracks same reason weight of a body when its center of gravity moves horizontally for obvious reason that the gravity is downwards center of gravity moves sideways so no work done. Now this is the crux of the topic today is the kinetic particle kinetic energy and principle of work and energy so what we realize is this let us say we have a particle which is taken from 0.1 to 0.2 ok so what is the force acting on it the force acting on is MAT ok and so on ok so this is a detailed derivation ok so what we figure out is that this is the force acting on it we then write down ok that this force can be written as MV DV by DS and so the work done FT DS ok will be nothing but MV DV ok and this MV DV by DS is nothing but the overall acceleration in the tangential direction ok S is the arc length DS by DT is the tangential speed and MV DV by DS is the derivative of the velocity with respect to this arc length ok this is the derivation ok do not afraid about it too much in the case it is not clear go and go back and have a look at the text so FT is the tangential force which is essentially going to do the work when the particle moves along this trajectory why because this normal force will not do any work because the displacement is perpendicular to this force so no work is done by the normal force but a tangential force is actually pulling the force along its direction so it does the work so if you integrate from A1 to A2 what we will see is that from S1 to S2 FT DS ok is nothing this integral is the work done when we move this particle from 1 to 2 against the application of this force what is that equal to that will be nothing but half MV2 square minus half MV1 square so this is T2 this is the kinetic energy at position 2 T1 is the kinetic energy at position 1 and what we are telling is this work energy theorem that if we move a particle ok along whatever trajectory ok from 1 to 2 then we find out that under the application of all possible forces when the motion happens along that trajectory what is the work done when you take it from 1 to 2 that is denoted by U of 1 to 2 ok work done when you move this particle along this trajectory under all possible forces acting on it should be equal to nothing but the kinetic energy at point 2 minus kinetic energy of at point T1 if you bring this T1 on the other side what it means is that that if you have some kinetic energy to begin with then that kinetic energy will be modified at point 2 by how much amount by the work done by all forces acting on that particle when it is moving along a given trajectory from 1 to 2 so this simple statement is the principle of work energy or the work energy theorem. Now how do we apply this theorem ok we will start from some very very basic examples and then we will move on to little bit more complex examples so if we have a pendulum which we for example lift it up and just release it we want to know ok for example what is the velocity of the cylinder when it comes down very very simple problem at textbook high school problem but very illustrative why because if we take this ok if you look at the cylinder position here the kinetic energy is 0 why because we are leaving the cylinder from rest when the cylinder goes from A1 to A2 what is the only force acting on it in the what are the forces acting on it one force is the force in the tangential direction ok which is sorry one force is coming in the normal direction which is the tension in the string other force is the force due to gravity. Now what is the work done by the force in the string look that the path of this can be only arc of a circle because this length of the string cannot change the only way this particle can move can be along the arc of a circle now what happens when this particle moves along the arc of a circle the velocity or the infinitesimal displacement at any instant is tangential to the applied tension and so the work done by the tension is 0 but only other force that does the work is the work done by the gravitational force and how do we find out gravitational force at downwards which is constant it is not changing with position so what we can do is that we can just say that we have a constant force acting in the downward direction so the work done is nothing but how much displacement happens in the vertical direction when this pendulum moves from 1 to 2 how much is it is equal to L in the downward direction work done from 1 to 2 by gravity is how much is nothing but W into L is the work done and what does our impulse movement or what does our energy movement work energy theorem tells you that initial kinetic energy plus work done to take it from 1 to 2 which is u 1 to 2 which we just saw is W times L should be equal to the kinetic energy at 0.2 and what is that equal to half mass into V2 square just solve this immediately find out what is the velocity at the bottom point. Now note one thing that from mass energy from this energy work energy theorem we can only find out what are the velocities that initial velocity is this much work done is this much what is the final velocity but we cannot directly find out what are the accelerations that is not obtainable but once we find out what are the velocities okay once we know what are the various forces acting on it then we can find out what are the corresponding accelerations by using the kinematics ideas that we developed yesterday. Then next one is about power and efficiency how do we define the power power is defined as du by dt the rate at which the work is done but what is di by dt du by dt is nothing but f okay into dr by dt because work done is f dot dr so f dot dr by dt okay is the power consumed or but dr by dt is velocity so f dot v is the power consumed and this is the efficiency for example it is extremely important in mechanical engineering that you that there is some input work for example we run a motor okay it is consuming some electricity electricity is a form of energy so we know how much electricity is consumed at the same time we can figure out that that electricity can be used to run the motor and we can figure out that the motor is used to run some work okay lift some water okay or for example it is a turbine okay then the turbine is rotating by converting gravitational potential energy into electricity and so on okay so we have one form of the energy converted to the other so we know how much is the input energy or input work we also know how much is the output work we get and whatever is the ratio is defined as efficiency that output work divided by input work is efficiency. Now let us look at this very simple problem with all the introduction about the work energy energy momentum theorem okay that a automobile of mass 1000 kg is driven down a 5 degree inclined at a speed of 72 kilometer per hour when the brakes are applied causing a total braking force of 5000 Newton that the braking force is what that when you apply brakes the wheels are locked and then the car starts to slide okay as a result the kinetic friction between the wheels and the road gets activated okay and that friction is given to be 5000 Newton and what we are asked to find out is determine the distance by the automobile as it comes to a stop this is a very simple problem in Newton's law we can just apply the constant force acting on it so we know our the equations of motion we can use them and immediately find out what is that final displacement how much time for it take to stop and so on but we use a simple trick here we know that the initial kinetic energy is nothing but mass into the velocities half mass into velocity square we know that the final kinetic energy is equal to 0. So work energy theorem tells us that T1 plus the work done by all the external forces to move from 1 to 2 should be equal to the final kinetic energy. So essentially T1 plus U1, 2 should be equal to 0 in this case now what do we do to find out what is the work done we need to find out that what are the forces acting on this that are doing work there is a normal reaction will not do work why because motion perpendicular to it there is tangential friction force which is 5000 which can do work okay in what direction velocity is downwards force is upwards so the work done will be negative now additionally there is a component of gravity okay which is doing work okay because when it what is that component of gravity one component of gravity acts along the slope which does work one component perpendicular to it does not do work. So the component parallel to this is what mass G into sin theta sin phi degrees so that does work apply force in the downward direction so the work done is positive so positive work done by gravity negative work done by friction so the total work done the total force acting on it will be friction minus gravity which will come out to be 4145 initial kinetic energy we can immediately find out and difference is equal to 0 this is the force that comes from all other external sources they act parallel to the plane x is the displacement along the plane some should be equal to 0 because at this point the vehicle has stopped so no kinetic energy and so finally x is equal to 48.83 is that distance where this T1 plus U1, 1 to 2 becomes 0 so this work done essentially destroys what are initial kinetic energy the vehicle has. This is another very simple problem okay I will not go into details of that because there is lot more syllabus that I have to cover but I will quickly mention this problem one mass of 300 kg other mass of 200 kg what is given is that assume that the coefficient of friction between block A and a plane okay is 0.25 and a pulley is weightless and friction less find out okay determine the velocity of the block after it has moved a distance of 2 meters note one thing in this position okay let us take this as our first reference position since this is 300 okay let us first figure out if this system is in equilibrium or not 300, 200 what is the friction force max that can act on it nuke is given here okay so 300, 200 the difference of forces how much is this 300 will be transferred to this one okay so clearly the kinetic friction will not be capable of balancing this okay now let us see how do we use the work energy theorem now when this particle is moving what are the various forces acting on let us look at the combined system on this total system on this total system what are the various forces acting on it okay look at the top block the force acting on it is the internal tension but the internal tension is common to this and this so all the internal work done will be positive here negative here and they will get compensated so all this internal strings okay when they are inextensible they do not do any work so work is only done by the frictional force here and by the gravitational force here so what is the frictional force here 200 kg okay into 9.81 Newton into 0.25 is the kinetic friction force the slip direction is downwards resisting force in this direction this is the direction of motion to the right friction force is to the left so the work done from 1 to 2 for a distance of 2 meters by the friction force is minus 0.25 into 200 into 9.81 this is the force into 2 meters that is the work done by the friction force okay when this block moves from here to here by distance of 2 meter now let us look at this block this block moved under the action of gravity what is the work done when you move from 1 to 2 the work displacement is downward force is downwards so 300 into 9.81 is the force acting on it into 2 meters is the work done by the gravity so the total work done by the system the internal forces cancel out external work done by this is this quantity external work done by the gravity on this one is this quantity sum them so the total work done is 4900 joules for this to come down so what is the kinetic energy of the system if it starts from rest initial T1 is equal to 0 so what do we know that T1 plus U of 1 to 2 should be equal to T2 so U of 1 to 2 is 4900 joules that should be equal to our final kinetic energy so final kinetic energy is what both velocities are the same so half m v square plus half m b v square so you put in this values kinetic energy is known from here we will find out what is the velocity of this after 2 seconds okay so we can get the velocity but note that from this work energy theorem we cannot get any acceleration okay. Now let us move on to this third problem okay it is a little bit of an interesting problem let us pay some attention to it in this problem what do we have but a spring is used to stop a 16 kg package okay this is a 16 kg package which is sliding on a horizontal surface this package is sliding towards the spring okay at a speed of 2.5 meter per second the spring constant is given to us 20 kilo Newton per meter and it is initially compressed okay the spring is actually compressed and hold in position how much is the initial compression 120 millimeters the package has okay a velocity of 2.5 meter per second in this position shown and when it hits the spring when the package hits the spring the spring gets deflected okay by an amount 40 millimeters so from this much information okay so this is the observation that we are throwing this bracket at a speed of 2.5 meter per second onto the spring which is already compressed by an amount 120 millimeters what we observe is that when this block goes and hits the spring the overall compression in this is of 40 millimeters and from this information we are asked to deduce what is the coefficient of kinetic friction between this package and its surface and second the velocity of the package as it passes through the position shown so it goes from here and it will rebound back again okay so we are asked to find out the velocity as it goes back hits here gets compressed and go back again okay so those are 2 parts of the problem that we have to do. So let us quickly solve part A what do we have here look at this what are the forces acting on it weight normal reaction force coming from the friction now what are we given that this hits the spring and compresses by 640 millimeters now what do we know that when it hits this at 640 millimeters okay the total distance traveled by this package before it comes to rest because that is the maximum distance it goes into and then it gets rebound back again so we know that before this package stops the distance traveled is 640 millimeters okay so correspondingly what is the initial kinetic energy we find out initial kinetic energy is half mv1 square we find that out is equal to 187.5 joules final kinetic energy is 0 now what is the overall work done by all the forces so the overall work done by all the forces in this entire operation is the work done by the friction force and work done by the spring now the spring is initially in compression okay how much that it has some compression here of distance 120 millimeters the spring constant is what 20 kilo Newton per meter okay this can be written as so from this spring constant K into this displacement we can find out what is the initial force in this spring before the block hits it and when the block hits it and move inverse by 40 millimeters what is the corresponding force in the spring what is this corresponding force is nothing but 120 plus 40 into the spring constant that will be the second force. So now before the mass hits the spring okay until it comes to rest the total distance traveled against the friction force is 640 so the work done to take it from 1 to 2 will be this kinetic of friction force into displacement x y negative because the displacement is positive whereas the friction force is negative you put in you will see that this much energy is dissipated okay when you go from here to here now how much is the work done by this force okay note that the force here acts in what direction force acts in the outward direction because the spring to begin with was in compression and it is exerting more and more compression and you are going against the compressive force to the work done by the spring force is also negative and what is that clearly it is P min plus P max by 2 which is the area under this trapezium minus y because the force is negative and the displacement is positive so this is the overall force in this spring sorry the overall work done by the spring force against this displacement of 40 millimeters overall work done by this so total work done by all the resisting forces is minus of this and ultimately what happens is that that T1 plus this work done plus this work done should be 0. Now what is unknown coefficient of friction is unknown here plug in in this for this to become 0 when this comes to rest you can immediately figure out what is the coefficient of friction between this and this second part of the question is that that when it rebounds back okay it starts from here velocity is 0 you wants to know that when it again travels that distance of 640 millimeter what is the speed and that again we use principle of we again use the work energy theorem what do we do work energy theorem now is what you go from here to here work done by friction force again will be how much it can only be negative why because when it is going from this side to this side okay the work done by this okay is mu k times n but the direction is opposite to the velocity again so this work done okay will again be negative but note one thing when this spring is relaxed and when it is moves moving forward what will happen is that the spring will again follow this pattern this is the force to begin with and this is the force to end with and so the work done again will be the area under this curve but note one thing because now the displacement is happening in the right direction okay the force under displacement at the same direction so that work is 112 joules only okay but now it is in the positive direction so just this is the work done okay by the friction force this is the work done positive okay u122 is the work done from the spring which is actually putting some kinetic energy into the sum of all this is coming out to be actually positive which means that actually this mass can go and reach all the way there why because the amount of energy provided by this is more than the amount of energy dissipated by the friction and you see that this is the final work done to go from here till here and what is the final kinetic energy this is 0 this is the work done to come from here to here which is 36.5 and that should be equal to the final velocity which as we can see is dissipated and clearly we can observe that if this friction coefficient were 0 then that velocity would be exactly opposite okay this will be same as 2.5 meter per second only in the opposite direction so friction is a dissipative force and it is eating up your kinetic energy till for example if the entire velocity will become 0 because the entire kinetic energy is dissipated by that friction force now there is another concept that we can use rather than using the work energy theorem okay suppose okay if we have conservative systems what do you mean by conservative systems that simple example of a conservative system is this is for example if you take a slope like this have a mass here make it go upwards from 0.1 to 0.2 the work is done against the gravitational force okay this is W for the mass so if you bring this mass take it from 1 to 2 what is the work done to take it from 1 to 2 simply equal to displacement upwards this is delta y work is downwards so that will be minus W delta y now you take this mass and bring it down again what is the corresponding work done to bring it from 2 to 1 it is W is the force acting downwards but the work done is also positive y because displacement also is in downward direction so this is W delta y so overall to take this particle from 1 to 2 and to bring it from 2 to 1 the total work done is equal to u 1 to 2 plus u 2 to 1 and it will be equal to 0 and such systems are called as conservative system why conservative because it is opposite of dissipative there is no dissipation but on the other hand if you are doing this work against friction then what you will see is that the particle moves like this friction acts in the opposite direction so the work done by the friction is positive when you go from 1 to 2 you come from 2 to 1 also the work done by the friction is positive so you only dissipate so 1 to 2 and 2 to 1 the friction force will not act to 0 so such systems are called as dissipative systems and where you take it from 1 to 2 and bring it back to that position and the overall energy is still 0 you will see that the system is called as a conservative system note one thing that rather than bringing it along this path if I take a second path okay path 2 and bring it from 2 to 1 you will note that even in this case the work done to bring it from here to here is still equal to W delta y and the total work done to bring it from 1 to 2 and 2 to 1 is equal to 0 such systems are called as conservative systems the spring force for example which we had discussed earlier is a conservative system why we saw in the last problem that the mass compresses the spring okay does some work okay the spring does negative work against the mass but then when it comes back again the spring does positive work on the on the mass and when the mass is released of the spring the overall energy that you put in the spring is the same energy you get back so again spring is another example of a conservative system now for conservative systems okay as professor Banerjee had explained okay what we can do is that we can define that work done from 1 to 2 we can define what is called as a potential energy that we define that this is our datum and if you take this particle from 1 to 2 then the work done okay is nothing but the negative of the potential energy so you say that at this point the potential energy of this particle okay with respect to this datum okay is what potential energy at party at point 1 which in this case will be simply if you leave this particle and take it from 1 to 2 what is the work done work done will be equal to minus w into y 1 the potential energy is negative of the work done okay so you can think about it in this way work done from to take it from 1 to 2 is nothing but the potential energy of the particle at point 1 which in this case with respect to this datum is m g y 1 minus work done okay minus the potential energy of this particle at point 2 how much is that with respect to this reference datum which professor Banerjee had discussed this is equal to m g y 2 and work done from to go from here to here is nothing but the difference in the potential energy at 1 minus 2 so if we define an appropriate datum then for conservative forces okay it is also appropriate to write down that the work done from 1 to 2 is nothing but the difference in the potential energy at 1 minus the difference in the potential energy at 2 okay we can also do it the choice of datum is arbitrary okay you can decide here here anywhere you want to but you have to be consistent with the datum same for the potential energy if you take a spring okay start from a deform start from a initial position x 1 go to position x 2 then the then the work done to go from 1 to 2 okay so the potential energy stored inside is half k x square okay if you take it from what is x x is a displacement okay when you take it from the fully undeformed configuration okay the spring is undeformed here you stretch it by distance x 1 then with respect to this undeformed configuration which we take to have 0 energy the potential energy of the spring is half k x square now if the spring is stretched at a distance x 1 then the potential energy at this position is half k x 1 square you go here the potential energy at a distance x 2 is half k x 2 square and then what is the work done to take this mass and stretch it from x 1 to x 2 the work done is nothing but okay clearly the area under this curve okay this is the force acting on the spring at this point which is k x 1 force acting on the spring at this point which is k x 2 okay force acts in the invert direction from the spring okay so the work done by the spring is negative but you have 1 to 2 is nothing but potential energy at 1 minus potential energy of the spring at 2 what is this half k x 1 square minus half k x 2 square note that x 1 is less than x 2 so again this will come out to be negative and note that half k x 1 is nothing but the force here the k x 1 is the force here k x 2 is the force here and this expression will again be nothing but equal to half f 1 plus f 2 by 2 which is the area of this trapezium fine so this is the definition of potential energy and conservation of energy then what we can do we can define that for a conservative system when there are no dissipative forces then u of 1 to 2 okay is nothing but v 1 minus v 2 as we had discussed then what do we have here is that previously our work energy theorem told us that t 1 plus u 1 to 2 is given by t 2 so we bring t 1 on the other side so u 1 to 2 is given by t 2 minus t 1 or in other words what do we say we use this 2 together and we will see that t 1 plus v 1 is equal to t 2 plus v 2 so what is this t 1 is the kinetic energy at position 1 v 1 is the potential energy at position 1 and t 1 plus v 1 is the sum of p e plus k e or the potential energy of the kinetic energy which is nothing but a total energy of the system which for a conservative system if there are no frictional forces and other thing this remains constant okay and this is what is called as a principle of conservation of energy so more fundamental is this work energy theorem where we say that a change in kinetic energy is nothing but a work done to remove this particle from 1 to 2 against all forces but when the forces are conservative according to definition that we have just discussed the total energy is given by t plus v and for a conservative system the energy remains constant or at point 1 t 1 plus v 1 is equal to t 2 plus v 2 and we can directly use for example for this pendulum if you take this pendulum throw it from here what is the maximum height it will reach it will reach this point a to y because here and here the potential energy remain constant okay now here note one thing that we make the assumption the assumption that we make in these things is that that when you go from here to here that tension in this in the string is enough okay is appropriate that the spring never goes into the string never goes into slack that the particle can always follow this trajectory which did not always be true okay we can always find some crazy examples okay where the gravity where the tension can actually become negative but for this simple problem for the time being okay let us realize that if we assume that this is the path okay that we need to know what is the path that the system takes and if this is the path of the system we know what is the initial condition then we can find out what is the velocity of the particle at any point along this path and once we find out what is the velocity once we know what is the path then we can immediately find out what is the tangential acceleration what is the normal acceleration and once we know that we can immediately find out what is the tension acting in the string and so on by knowing the weight of this particle but principle of conservation of energy makes this one assumption that this trajectory which we believe that the particle follows is perfectly followable that this tension always remains positive and this never goes into a slack on the other hand for example we can see that if we apply some positive velocity here then what may happen is that this may go up and then go in a slack okay it may lose tension okay that example if you have the time we will discuss later okay you look come look up example solve problem example problems in BJ 10 you will see that there are one or two problems in which they see that if you apply some velocity here okay then not necessarily the particle will always follow a trajectory like this somewhere it can go into a slack okay but for all practical purposes if you can always figure out if you can always make sure that the tension remains positive then this trajectory is perfect and using principle of conservation of energy we can find out what is the energy what we can find out the energy and using that energy we can find out what is the velocity of the particle and what is the direction of the particle at any point along this trajectory okay so let us discuss this really simple problem we discuss is very simple problem but very illuminating problem so whatever I have discussed now let us this example will illustrate that there are also many interesting examples in BJ 10 okay I suggest that you can have a look at that so what we are doing is this at 250 gram pellet okay so this mass is of 250 grams is pushed against a spring now this is a conservative system you push against this spring okay and release from rest okay you keep pushing it pushing it pushing it release now the spring will get coiled up when you release the spring the potential energy of the spring that you have input imparted to the spring will get released and exert some kinetic and give kinetic energy to this block okay till it releases it loses contact with the spring and then this go along this trajectory and come down but idea is that okay you are asked to find out determine the smallest deflection of the spring for which the pellet will travel along the loop okay is what we want to find out we want to find out that the pellet will follow all the way along the loop now a very naive thinking okay if you don't if you just don't think about this problem properly what we say oh for this pellet to start from here go all the way up and come down it needs to reach this maximum point so at the maximum point we can find out what is the gravitational potential energy this is the bottom point this height is 1000 millimeter we can find out the potential energy here and the potential energy that is stored in the spring here should be equal to the potential energy here and bravo victory but this does not work why because what can happen is that that this loop can provide a reaction like the string only in the inward direction it cannot provide a reaction to this string in the outward direction but because of the velocity of this particle along this path what may happen is that it has to maintain okay it needs to have an acceleration which will be it will have an acceleration which is equal to v square by by row and that v square by row acceleration okay will be provided by two components one will be by gravity and other will be by the normal reaction but if the gravitational force okay which is providing this is significantly larger okay the component of the gravitational force is significantly larger as compared to whatever centripetal acceleration or the radial normal acceleration that it requires then it will lose contact so what we need to know is that this top point is a crucial point because if everything because at top point if the reaction just becomes zero or in other words that the centri the normal acceleration is balanced by the corresponding gravitational force okay then we are good okay then the particle will not lose contact and fall down but if that does not happen then you will lose contact with this you will lose contact with this soup somewhere between and the ball will fall down so how do we do this what we need to know is this that at the top point if we want to find out what is the minimum velocity it should have at the top the minimum velocity be velocity should be such that there is no normal reaction and then what is that minimum velocity okay that minimum velocity should be equal to mg that this force okay weight balances this acceleration because the normal reaction here is zero so mg is equal to m vd square by r and from that we can find out that the minimum velocity which you need at the top portion should be equal to 4.905 meter square root of 4.905 okay so that is a minimum velocity that you require now we are done that we know that principle of conservation of energy has to now work true because everywhere there is no dissipation we know what should be the kinetic energy here so now what is the only question we need to know is how much should you compress it by such that the total potential energy here okay only because of the spring will be equal to the kinetic energy at point D which is governed by this velocity plus the potential energy energy of this mass at point D so what is that v1 initially is equal to v1 plus vg if you take this as the datum potential energy to gravity becomes 0 what is otherwise the potential energy due to the spring is how much compression we give in the spring half stiffness constant into this so 300x square is the initial potential energy what is the final total energy is the elastic energy plus the energy due to gravity but there is no contact with the spring here so the only part of the potential energy is due to gravity okay and then what is the other part okay initially kinetic energy is 0 what is the kinetic energy that we should need at the top point to maintain contact everywhere is this velocity so second kinetic energy is equal to half m vd square which you already know so that kinetic energy will be after you put all the numbers will be 0.6 the potential energy here will be equal to 2.45 joules so the total energy at point D will be some of these two total energy here will be 300x square which is at 0.1 because this entire system is conservative no dissipation due to friction we will just equate the kinetic and potential energies and then from that we can find out that this is the minimum compression you should given in the spring so that this ball can travel and come back again okay so that is the simple that is the logic that we have used now what is impulsive motion okay for example okay very simple examples is forget about some simple like if for example in cricket a ball comes to a batsman okay he hits the ball so ball comes with one momentum you hit the ball with the bat okay and the momentum direction completely changes and we know that for example everything happens so fast even for example if you look at all the slow motion cameras which for example try to figure out when does the ball hit the bat when does it come up you will see that it is like a fraction of seconds within which the ball touches the bat and leaves the bat so the overall contact time that the ball has with the bat okay which we are which is evident to us okay there are the slow motion cameras from which we see for example that the ball hits the bat and very fast okay within a fraction of seconds is leaves the bat so the overall contact time is very small but on the other hand we see that the total force okay that is exerted by the bat on the ball has to be huge why because there is a huge change in the velocity because the ball is for example especially in the straight drive for example the ball comes with one speed okay it goes out in the other direction the momentum is in one direction when it comes the momentum is in the other direction when it goes so there is a huge change in the momentum why because a huge amount of force is exerted on the ball okay in a very short time so such kind of forces are called as impacts okay and the motion which happens because of the small contact or small time of contact but extremely large forces okay is called as impulsive motion okay and many of these impact problems okay can be reasonably understood okay at least to the first order when I mean first order means at least to the most basic level by the simple ideas that we will learn in the next few minutes okay second thing is for example when you hit when somebody hits again when the car hits against a wall train hits against a wall two cars have collision even in those cases the overall collision time is pretty small but we can see that even though the collision time is so small the overall damage that is caused to the car object is extremely large the reason being that even though the collision time is small the amount of forces that are involved are extremely large and as a result the amount of damage that happens is very much is very huge now what is the principle of impulse and momentum okay so this is very beautifully done in BJ 10 okay so let me briefly explain what that principle is so look at okay let us look at it from the point let us look at just one particular particle which has a momentum of MV 1 okay the particle as a momentum initially given by MV 1 then what happens this particle comes suppose you hit it with a baseball bat or hit with a with a bat or stick whatever this this particle this ball this mass we hit it there is a quick impact what is that impact okay that quick impact is denoted by this integral okay T1 to T2 FDT now why are we doing that why is this integral to look at the origin of this integral let us look at Newton's second law what does Newton's second law tells us in the way we discussed yesterday evening and today morning at a force acting on a particle is given by rate of change of its momentum so F bar is equal to D by DT of L now note one thing take this T on the other side do an integral from T1 to T2 what is T1 to T2 suppose okay I have a slow motion camera okay and now they are quite common okay and we see that the first contact happens at T1 and the contact is lost between the two impacting bodies at T2 so essentially the force that is acting on the ball or on the mass because of the impacting for because of the impact starts from T1 ends at T2 where this interval delta T which is T1 to T2 is very very small and if I do that integral then what do I get T1 to T2 F bar DT is nothing but MV2-MV1 which is the final momentum minus the initial momentum so what is this telling us that this overall force integrated over time okay so force exist and a force in most of these impact cases okay is a function of time for example okay if there are some sensors that are put then the sensors will sense some force like this that at the beginning of the impact okay the force is small the force shoots up and then it comes down and this is the delta T or impact and this force can become very very large okay whereas for example typical weights due to gravity maybe somewhere here whereas the maximum impact force okay can be pretty high 2-3 times more now okay start from here the impact goes like this this is delta T very high force maximum peak force will be reached and the area under this curve is nothing but this is T1 this is T2 so integral from T1 to T2 of F bar DT is the total impact and from this simple theorem okay which is nothing by integrating Newton's law you will see that the change in momentum and again note that is a vector change in momentum is nothing but T1 to T2 F bar DT or what we can say is that in other words take this MV1 on the other side that initial momentum plus the impulse what is the impulse from time 1 to time 2 is equal to the final momentum and this is a very very powerful theorem this is very very simple theorem why because most problems of impact where the impact force is of very short duration this T2-T1 is a very short duration and the impact forces would be very large but the impact duration is very small but we can clearly see that a cricket ball for example where I think weighs what 4 ounces is pretty heavy but we see and the velocity and the speeds are extremely large okay so both the speeds will be of the order of say 100 120 kilometers per hour okay which are extremely large speeds and the direction can change for example in a square curd okay the direction can change by 90 degrees in a straight drive the direction can change by 180 degrees so there is a huge change in the momentum so the necessarily the impact that the batsman has to put on the ball has to be very high okay and that is why they say that you need very strong wrist to do that okay because the maximum peak force that you exert is what we had seen here has to be very large now note one thing so impulse motion impact now what happens is that so this is a picture of a baseball ball you can as well because the slide is taken from BJ 10 okay it's an American book they will of course talk about baseball but it is not a big deal even if you talk about for example okay a standard for example any other any other ball cricket ball tennis ball whatever it does not matter so what we realize here is that that suppose this is the this is the direction is for example which the ball is traveling towards the batter now the batter hits it what force does he exert it he exerts a force of average force okay what is this average force this average force is an interesting concept okay it's not a representative concept but integral f bar dt from t1 to t2 is nothing but equal to mean force f bar average into delta t so that is another way of representing impact okay we say that this equal to f average t so that's what we are denoting here that this is the overall impact that acts on this ball okay from the bat now you will ask me well when the impact acts on this ball okay there is also weight acting on it okay the weight is also acting on it so why do I just say that mv1-mv2 it can also start moving down but at that instant okay this delta t is very small and within that time delta t the overall impact of this force okay the overall impact that is weight produces will be significantly lower okay than the non-impulsive force or the impulsive impact okay that will be provided by that f dt and so in many cases where this is an impulse for example you are hitting with something two balls are colliding with each other then the corresponding weight component for example we can throw that away because this delta t is very small and so w delta t can be very negligible quantity whereas this f delta t quantity because delta t even though it is small this f can be very large and this is and this okay overall quantity then integral f dt or f average into delta t can be very appreciable so typically okay so non-impulsive forces for which f bar delta t is small okay may therefore for example be neglected okay and again this depends on the problem because small compared to what if you have impact forces then the impact forces are very large compared to that and maybe we can neglect but we will see that there are certain problems in which we have to take these forces also okay and that will become very clear when we solve some problems okay let us solve two problems okay okay so look at this automobile okay our favorite problem so an automobile of mass 1800 kg now we are trying to solve it in a different way now note one thing that every approach okay a same problem we can approach it in various different ways but the answer that we get are also different for example if you recall the same problem when we wanted to know that this car is has subjected to a break when you break the car the wheels stop rotating they start sliding with the ground friction force acts on it slows the car down and the car comes to rest when we did this problem using the work energy theorem what did we did we found out the initial kinetic energy we found out the final kinetic energy which is 0 because that is the statement of the problem and what do we want is that that overall work which is done by the resisting force that work should essentially nullify the kinetic energy and from that we found out after what distance did the car stop so we got a distance well here we ask a different question that for the same problem okay if the braking force is 7000 Newton okay the mass of the car is 1800 kg incline is 5 degrees determine the time required for the automobile to come to the stop there it was distance here it was time so same problem but we can solve it a different approaches but when we solve them with different approaches we get different quantities directly and you can use those quantities indirectly use Newton's laws and again get your final answer but immediate quantities that we get are different so here we are asked to find out time required for the automobile to stop what do we do we just note that mv bar is equal to impulse okay from 1 to 2 sum of all the impulses acting from 1 to 2 and that should be equal to mv2 now note that this problem is not an impact problem in this case what is the impulse okay so if you come here this is the typical definition of an typical way and impulse loop but it is nothing like that I can also have a force acting on this particle okay from t1 to t2 where delta t is pretty large I can think of that also as a impulse but this is difference why because this comes from impact and this comes from steady applied forces but nevertheless the principle still remains the same okay the principle still remain the same and how do we deal with this we deal with this by saying that initial momentum plus the impulse okay coming from all the forces now what are the forces acting on it there are normal forces acting on it okay so velocity mv bar okay has this direction okay what there is no component of momentum in the vertical direction okay what are the forces along this direction okay this is momentum you have normal reaction you have weight okay weight has 2 components one is the component normal to this one is the component which is tangential to this so what we realize is that and there is a friction force so the normal direction momentum is 0 so you 0 equal to 0 we do not worry about anything that weight the normal component mg cos 5 degrees will be balanced by n but now let us look at the impulse momentum theorem in the in this direction what do we have total force acting on this is how much friction force minus mg cos component okay so that we subtract it off okay so this is the friction force okay you subtract that from this component what is the direction along the direction of the momentum okay what is the impact the impact is along this direction w okay sin 5 degrees minus because this is upwards minus 7000 okay is the force acting from the friction in the upward direction so the total impulse that acts in the duration t till the car stops will be this force minus this force okay and effective direction is what in this direction because our momentum convention we have chosen this so this is the momentum convention impact convention also should be in this direction convention should also be in this direction so mv1 is known to us this force multiplied by time t is the total impact this should be equal to mv2 we just substitute everything here v2 is equal to 0 mv1 is known to us we immediately see that the t will be equal to 9.1 second and by just a simple application of impact momentum theorem we found out that what is the total time of duration for the car to stop okay we have to do another problem but we are running out of time and there are some questions okay so let me answer some questions difference center number 1068 ask difference between impulse and impact okay very nice question okay we will come to this slide directly okay which I discussed one minute ago that typically impulse okay impulse means you have a force acting for a particular duration and integral f dt over that duration is impulse okay straightforward the force can be acting over a very long duration time varying force constant force does not matter the integral from t1 to t2 of that force within that duration that total integral is the impulse for that duration whereas impact in impact what happens is that impact is a very special case of impulse okay to the best of my knowledge where the overall duration of the contact is very small okay that the force acts it shoots up it again comes down and overall duration is very small and the impact also causes an impulse okay but that impulse happens over a very short duration as opposed to any generic impulse which can happen over a finite duration okay fine next second is this okay question is asked by 1 2 1 5 the question is why not take potential energy in work energy theorem as we know in conservative forces work done is also equal to change in p okay either you do u1 to 2 and find out what is the work done or you do it in terms of potential energy if you take both of them then you are doing double counting okay because the basic theorem is what t1 plus work done which is u from 1 to 2 is equal to t2 that is the most basic theorem and only for conservative systems for which we can define potential energy u of 1 to 2 can be defined as v1-v2 so we either do u1 of 2 or we define there is v1 and v2 because the system is conservative and then use a different form of work energy theorem for conservative systems which is t1 plus v1 is equal to t2 plus v2 but you cannot do them both okay because from one the other thing comes most fundamental form is still this t1 u1 to 2 is equal to t2 but for conservative system it reduces to this so 1067 asks what are the practical or general application of work energy principle so clearly we have what our applications we have discussed are the practical applications okay that ultimately we should know that whatever you do the total kinetic energy plus the work done okay should be the kinetic energy at last point so for example there are many problems okay in which the particle for example can take a very complicated trajectory okay for example a car takes a very complicated trajectory while the course of its movement so if for example you want to find out what is the initial velocity what is the final velocity okay so work energy theorems for example in many of these practical problems make our life very easy because we do not have to do the complete integration but we just realize that what is the work done which in many cases as we had seen is very easy to figure out and especially for conservative forces if you want to solve the complete problem okay some complete complicated geometry problems say for example this problem if you want to solve using full Newton's second law application that what is the tangential acceleration what is the normal acceleration okay and the question that we need to know is that and this is a very practical question why because for example if these are some these are the rides for example if you had been to one of these entertainment park then these are the rides where they go up like this now when they go up like this okay in those cases okay it is very clear that for example they have guided rails because of which the guided rails can also exert force in the one direction and the other direction okay but at least for those cases in we want to find out that what is the overall velocity or what is the overall speed as a function of its position along that trajectory if you use Newton's laws if the trajectory is very complicated then all the calculations become very complicated because your equations now you are solving along this complicated trajectory whereas in this case you just need to match the energies in the beginning and at the end and you can immediately find out what is the velocity and once you know the velocity you know what is the curve you can find out what are the corresponding acceleration so in many of these kind of problems where the geometry is very complicated but only thing we need to know is the corresponding speed there okay things become extremely straightforward 127 hours in sample problem 13.1 okay 13.1 sample problem from BJ10 what is the effect of slope clearly there is a effect of slope because if you note here that the 13.1 what we have used this effective force we have used is 4145 whereas the braking force is 5000 so 5000 braking force is acting upwards and minus we do which is the sign component of the slope okay so the weight into sign of 5 degrees is the force acting along this direction parallel to the slope we subtract that okay because the effective force is this upward force minus downward force and then it came out to 4145 so we have taken that into account okay centre number 11 century hours how do you distinguish work in thermodynamics and work in engineering mechanics okay this is a very nice question okay but it is a very deep question why for example in thermodynamics okay thermodynamics is a extremely extremely complicated strictly speaking because for example take case of ideal gas in ideal gas what do you have in ideal guy you have bunch of gas particles in the simplest form like kinetic theory take kinetic theory of gases even in the simple kinetic theory of gases what do we have we have all these particles we take them to be elastic particles okay they collide with each other such that their mean kinetic energy is equal to half kbt okay for every particle so half mv mean rms square for any gas particle in any direction is half Boltzmann constant times the temperature okay so that is for example a single even a simple thermodynamic system like an ideal gas is a highly complicated okay multi body system in thermodynamics what do we do in in in dynamics we are worried only about for example we have a bunch of finite particles and we are looking to understanding what are the energies of each and every individual particle what is the trajectory what is the acceleration and so on and we can track energy kinetic energy potential energy of every particle whereas in an ideal gas okay you cannot try the kinetic energy of every particle so in that case okay what we say is that that work done is some kind of a mean kinetic energy a mean energy of the gas and we say that for example for an ideal gas at temperature t energy is given by 3 by 2 nrt for example or 3 by 2 nkt so so that is once or where n is the number of moles or the number of particles depending on how you look at it so an ideal gas in thermodynamics so thermodynamics essentially a collection of many many many many dynamical systems okay and on an average sense we define what is the mean energy of the system and so on and do thermodynamics with it whereas in dynamics okay so now for example there is something called as analytical dynamics in which you have many body systems now those problems are extremely complicated to do so to bypass that problem okay we go to thermodynamics where instead of working into kinetic potential energy of every particle we just say that there is some average energy of the particles which I say to be equal to 3 by 2kt for an ideal gas and then according to that we do some external work against some pressure so we define a quantity called as pressure now that pressure comes because of for example many many of these particles colliding on the piston for example if you are compressing against it so pressure again is a dynamical quantity but it is some kind of an averaged out quantity so even though thermodynamics strictly has its origin in the dynamics of individual gas particles overall it is very difficult to do the dynamics of this combined system so you look at what is called as an average description and for example it is a very deep topic goes under what is the concept of statistical thermodynamics so we say that the system is in some statistical equilibrium as opposed to thermodynamic equilibrium or statistical non-equilibrium as opposed to thermodynamic non-equilibrium okay I really like this topic so this long answer but it is completely irrelevant for this course so can the variation of force during impact with respect to time can be determined it can be determined okay but for example if you develop a highly sophisticated finite element model so you for example have a bat have a ball you give all the material properties for the bat do all the material properties for the ball okay simulate the initial conditions and you have to run a very very sophisticated finite element simulation and in that case okay you can for example figure out that what happens in that FEA simulation how does the ball deform what is the contact point and so on but overall this is a contact problem and contact problems are extremely difficult but you can do it in finite element analysis you can do this there are standard as softwares like abecca's ansies which you can use and you can easily develop these kind of things and as far as like real observation is concerned there are extremely sophisticated high frames per second cameras by which you can track the deformations of the substrate okay how does the ball deform how does the bat deform okay you can find all those things okay and then see if the simulation that you have done matches with the experiment this is by center 1202 it is asked difference between the conservation of momentum and conservation of energy very different that if you have two systems for example if you have two balls that impact with each other like this then this ball exerts an impact on plus direction on this this ball exerts an impact in the opposite direction on this and the total impact component of these two cancel each other out so we say that even after the impact the total momentum m1v1 plus m2v2 okay will be equal to m1v1 prime plus m2v2 prime so momentum conservation theorem is a vector theorem now what can happen is that during impact okay it is a beautiful problem that when two balls for example impact with each other you will see that a noise is made tuck noise is make okay so some part of the energy gets dissipated okay as sound okay there are vibrations that are created so some part of the energy is dissipated in the form of vibration because of which the ball will heat up some part will get converted into permanent deformations which are called as plastic strains but because of all these dissipations you will see that in most of the impact cases the energy okay the kinetic energy of the balls before impact is not equal to the kinetic energy of the balls after impact there is a loss of kinetic energy and at loss of kinetic energy is to some extent okay taken care of by what is called as this coefficient of restitution okay but the momentum will always be conserved if there are no external forces your energy may not be conserved but momentum will always conserved and energy conservation theorem is a scalar theorem whereas momentum conservation theorem is a vector theorem okay how to find force applied by the bat on the ball we cannot directly find the force but we can find the impulse for sure okay we can find out the impulse but not the force okay we can only find the average force by finding out what is the average time of contact from very high frame per second cameras how does relative velocity relate to impulse motion in any impact okay so impact okay for example acts on one particle and the relative velocity for example okay for example what we can figure out is that at the relative velocity okay may lead to some high relative momentum and that may lead to high impact but essentially they all come under impact momentum theorem okay and once you figure that one out everything else become apparent from that okay so with this much answers okay I think now is the time to take a tea break