 So, welcome back. So, we discussed briefly about how to define for a curvilinear motion how to define velocity. This is position change in the position delta r it is a vector and divided by delta t is the instantaneous velocity vector and a corresponding instantaneous speed is given by dS by dt which is the change in magnitude is the change in the length divided by time. Similarly, for acceleration the acceleration comes not only because of change in the magnitude. It may come even if there is no change in the magnitude, but there is a change in the direction and we saw the simple example of uniform circular motion where the speed remains constant all throughout the trajectory of the particle moving along a circular trajectory, but the velocity direction keeps changing and as a result there is a normal acceleration. So, just one thing is that the velocity if this is the path of the particle velocity is always tangential to the path and the acceleration depends on the rate at which the velocity changes. So, there were a couple of questions about hodograph and so on. So, just please bear please keep in mind that these are not very important only thing you need to know really is how does the velocity direction change. That is how does the velocity change including both the direction and the magnitude and that will give us what is the corresponding acceleration or if we make sure that we understand how does the velocity vector change as a function of time velocity vector then that will automatically give us what is the corresponding acceleration. Now, the simplest way of defining the positions and velocity is in terms of rectangular components. So, what you can say is that you can fix the coordinate axis at some origin O what that origin is you can decide conveniently fix that thing. Now, at any point you can say that a position is given by x i hat where i hat is a unit vector in the x direction y j hat plus z k hat all these j k are the unit vectors along y and z direction. Now, the velocity vector straight away is nothing but d x by d t times i d y by d t j d z by d t k. So, this will get we will get a velocity vector this is x dot y dot z dot. So, v x v y v z are the corresponding components of velocity in x y and z direction. Similarly, to get the acceleration vector we only need to differentiate once more and what do we get we get this as x double dot i plus y double dot j plus z double dot k. So, that is the final acceleration. So, what we are doing here is that that position, velocity and acceleration we are expressing all in terms of the coordinates the coordinates of position coordinates of velocity and the coordinates of acceleration along x y and z direction. Now, the simplest example and the most common example that we see for such kind of motion in two dimensions at least is a two dimensional what do I mean by two dimensional that it is a planar motion the particle travels only in the in one plane and so that plane we without losing any generality call as the x y plane. Now, in this case suppose we throw a particle at some angle with respect to the horizontal fix our origin on the rigid ground and suppose and when the particle is thrown there is acceleration that acts on the particle in the z direction which is the acceleration due to gravity. So, what do we have if you break down the acceleration into three components and the acceleration in the x direction is 0 for this problem acceleration in the y direction this is x and this is y direction is nothing but minus g y minus because our y coordinate is fixed in the upward direction and g is acting downwards that is my minus and a z because the motion is planar we do not worry about what is the acceleration in the direction out of plane because it is clearly 0 at least for this simple problem. Now, suppose we have the initial conditions that x 0 is equal to y 0 is equal to z 0 for the timing let us forget about z because it is a planar motion. So, all the particle begins at the origin correspondingly we say that v x 0 v y 0 and v z 0 is equal to 0. So, these are the conditions that we are providing. So, integrating twice what do we get we get that v y is equal to v y 0 minus g t v z 0 equal to 0. So, these are the z motion we do not have to worry about it and x will be simply v x 0 into t y is equal to simply v y 0 y minus half g t square essentially what we are doing is it is as if the particle has velocity in the x direction velocity in the y direction. So, it has 2 independent rectilinear motions. So, one independent rectilinear motion in the x direction which has no acceleration another independent rectilinear motion in the y direction, but in that case there is a uniform acceleration of minus g. So, because there is no acceleration in the horizontal direction clearly what we had seen is that that a motion in the horizontal direction is uniform whereas the motion in the vertical direction is uniformly accelerated and these are 2 independent motion which now can be superposed and we can say that the complete motion can be broken down into 2 independent rectilinear motion one along the x direction and one along the y direction. So, this is how the trajectory will look like that v x will always remain constant v y will change in this manner v y is the motion in the vertical direction x position will linearly change with time depending on what is the component initially of the velocity in the horizontal direction correspondingly what is the component of y in the vertical direction and if we say that v z at 0 that is the velocity coming in the out of plane that is 0. So, what it says is that with this condition the motion becomes planar and this is the x coordinate as a function of time this is the y coordinate as a function of time. Now, this is a projectile motion what we have is that initially if we throw this particle okay from the origin at with a speed of v naught okay which is inclined okay. So, the vector v naught now is inclined at angle theta with respect to the horizontal we have x component given by v 0 cos theta y component v 0 sin theta and now we can think of this motion as this motion in the horizontal direction with no acceleration and this initial velocity whereas with this initial velocity and minus g as acceleration motion in the vertical direction and we see clearly that if you think about this particle only with a motion in the vertical direction the particle will go up and come down under the influence of this gravitational acceleration whereas it will uniformly keep moving in the x direction and the resultant curve that comes out of it okay can be seen to be a parabolic curve and this particular motion is called as a projectile motion and this is the motion we see for example when we throw a stone or when we hit a ball and see it is trajectory or for example when something is thrown from the top of a building okay so anything that is for example acting under the influence of a gravity we have some initial velocity in x and y direction in the resulting motion okay is parabolic and it is called as projectile motion. We will do a very simple problem okay to just see how this kind of trajectory is or how the kinematics of particles undergoing this projectile motion is understood okay. So a simple problem is that a projectile is fired from the edge of a 150 meter cliff so the origin now okay is now fixed here the projectile is fired from the origin of this cliff which is this much deep at an angle of 30 degrees with respect to the horizontal this is the angle and the velocity okay this is the direction which makes 30 degree with respect to the horizontal and the magnitude of the velocity or correspondingly the speed is 180 meter per seconds and what we are asked to find out is we neglect their resistance okay this is very important because for example if you would see that there are some cricket bowlers okay they bowl a ball but the ball does not follow a nice projectile trajectory whereas if it is a good spinner you see that the ball has sudden dip for example and it has a drift whereas a good swing bowler when he puts the ball when he delivers the ball the ball does not move in a perfect projectile but it undergoes swing it undergoes seam but these are special cases in which because of the air drag because of the forces coming from the surrounding air the ball the corresponding mass or the corresponding body does not have a uniform projectile motion but has a very very simple case okay if we kind of neglect that and it is a very decent approximation to the first order that we neglect that mass sorry we neglect the drag coming from the air then the motion is a projectile motion and now what we are asked to find out the horizontal distance from the gun here okay to the point where the projectile strikes the ground and the greatest elevation about the ground okay how high does the particle go okay what is the maximum height it reaches okay during the course of its flight the simple thing so horizontal and vertical motions are independent we as we had just seen now we apply equations of motion in the y direction what we have an initial velocity in the y direction given by 180 into sin 30 there is an initial velocity in the x direction given by 180 times cos 30 so those are my initial velocities in the y direction and in the x direction and then y direction may the motion has initial velocity and gravitational acceleration whereas in the x direction it is a uniform motion and then what we can do is we can find out the time for the projectile to hit the ground okay why because we know that what is the corresponding x y coordinate and use that time okay to find determine this time and then also find out what is the maximum elevation which will happen when as we had seen earlier okay the velocity component in the y direction becomes 0 to this what we do is that these are the initial speeds okay these are the initial coordinates x 0 is 0 but y 0 is equal to 150 this is the acceleration in the y direction minus 9.81 acceleration in the x direction is 0 okay so vertical motion is uniformly accelerated what are the equations of motion v y is equal to v y 0 plus 80 simple equation of motion putting all the values this is how v y looks like y or what is the y coordinate as a function of time can be written as y is equal to v y 0 okay v y 0 t plus half 80 square simply put in all the values you will get this is the y expression then the next expression we want to know is how does the y coordinate okay changes as a function of velocity so the third equation that v y square is equal to v y 0 square plus 2 a y so these are the 3 equations we get for v y as a function of time y as a function of time and v y as a function of y okay this is a mistake here okay it is a uniform motion it is not uniformly accelerated kindly make a correction it is a uniform motion there is no acceleration in the horizontal direction so v x 0 is equal to simply 180 cos 30 is the motion is the velocity in the horizontal direction okay that is the velocity and the corresponding displacement in the x direction is v x 0 into t and these are the corresponding expressions that we have now what do we want we want to find out the horizontal distance okay when the projectiles tries the ground now for that horizontal distance what do we have that y is equal to minus 150 meters now what do we do we use this equation and find out that for y is equal to minus 150 what is the corresponding time so we solve this and get that after time t okay is equal to 19.91 seconds okay by solving this quadratic equation okay and taking only the positive root you will see that at t is equal to 19.91 seconds we get this particle hitting the ground and then correspondingly take that time substitute here that this is the x distance covered by the particle in time t which is nothing but v x 0 into time t which is 155.9 into t this will give us that the x distance okay when the particle hits the ground is approximately this value. Now the second question we want to know is the maximum elevation okay when does it occur okay and at what value does it occur so to find out when does it happen okay at what elevation does it happen we simply use this equation where we have velocity as a function of the y position so we just substitute v y to be equal to 0 because at the maximum elevation v y has to be 0 and substitute in here and solve this simple equation and what do we get when we solve this we can solve this for y and get y is equal to 403 meters from here or from the ground it will be 403 plus 150 so that is the elevation from this level when the elevation will be maximum okay. Now one simple concept question you can prove to yourself okay you can prove to yourself that if we if you want to find out that if the particle is thrown okay from a bottom point like this without any elevation and you are asked to find out what is the maximum distance in the x direction that is covered by the particle that distance is called as a range of the projectile and when the particle starts from 0 end side 0 then you can convincingly show to yourself that and the logic is like this that when theta is extremely large okay let us say theta is equal to 90 degrees then what happens there is no x component and because there is no x component the particle will only go up and come down as a result the range is 0. Now let us take the other extreme case in which theta is equal to 0 okay which means the particle is thrown perfectly horizontally what happens in that case in that case the particle just stays on the ground it never takes off as a result the range of the particle is again 0. So in these two extreme limits we see that the range is 0 and by some simple common sense what you will realize is that that because there is a range there has to be some optimum value where the range will be maximum and for a problem like this you can see that the range will be maximum when theta will be equal to 45 degrees you can even see it mathematically okay and mathematically also just substitute in the values okay look in any standard book if you have not already done this okay the range will be maximum it starts from 0 end at 0 when this angle theta of launch is equal to 45 degrees okay whereas now here it is asked if you fire a projectile like this what launch angle will be the greatest horizontal distance so first is 45 degrees now 45 degrees clearly if it starts from here and end up here okay along this line then 45 degree but the question is what will happen now that logic is very straightforward okay the logic is something like this if theta is equal to 0 again nothing the particle will go up and come down okay if theta is now equal to 0 degrees unlike the previous case there will still be some range so what we are seeing here is that at 0 degrees also the range is not 0 so overall what we are seeing is that that the launch angle okay it is a simple okay you cannot rigorously show it with the simple argument but argument is like this that you are getting a range okay which is not 0 even at 0 okay whereas previously to get some range you needed to have okay at least okay you needed to have a large enough angle okay if you put theta equal to 0 you never got any range so with that logic okay you can just think that even at 0 I am getting some finite range so I would have some angle of projectile angle of launch less than 45 degrees which will give me a maximum range okay so this is a very hand waving answer okay so do not expect any rigor from it but a simple logic is like this that even at 0 you see some range whereas previously at 0 you saw no range so just extrapolate and say that so because at 0 you see some range the maximum will be reached somewhere before 45 degrees okay very hand waving answer but you can also prove it rigorously now another extension about motion relative to a frame of translation previously what we had seen is that that if you have some position vector along a straight line then we had seen that R B was equal to R A plus R of B comma A now here it is just straight forward if the motion is happening not in a plane okay but it is happening along a curve and not in a straight line then the X B and Y B just are to be replaced by the corresponding vectors so what we will see is that that if I have another frame okay so this is one fixed frame of reference in which for example you can measure what is the position of A what is the position of B what is the velocity and what is the acceleration of A and B in that frame but suppose I am moving with this new frame okay which is with a moving observer A then what we can say about the velocity of that from this frame what will I see as a position of this point B velocity of point B and acceleration of point B and what we will clearly see is that that from the this frame which has origin at O the position vector of B will be given as R B but that can also be written as R A which is this position vector and using triangle law this is R of B with respect to A and what is this R of B with respect to A means that in this moving frame okay in the frame of this particle A R of B comma A is nothing but what is the position it will see of B from this moving frame okay only whatever we had done about relative motion we are simply generalizing it to a general 3D motion okay if you differentiate this once and twice you will again see that velocity can again be written as VA plus VA of B with respect to A same for the acceleration the acceleration from for B will be nothing but acceleration of A but acceleration plus the acceleration of B will relative to A okay and this is called as a velocity of B relative to A and this quantity B slash A is the acceleration of B relative to A you may ask that what is the big deal about these things okay why do we need to have this relative velocities okay so even though we are not going to go into extreme details of these questions okay but what you can see is that there are many many problems okay which are solved very well if you do not use some absolute frame but you go to a moving frame and these problems for example are especially important when we for example are in some are in a rotating frame for example okay if you want to for example analyze how weather pattern move or for example what is the because earth again is a moving frame so one gentleman had asked me a question about the moving frame of reference of earth so our earth actually is a moving frame but we take that as a frame which is for us and move and measure everything with respect to that frame so as a result because of a variety of problems in a variety of problems choosing not some absolute frame but actually choosing a frame which is moving can actually become very easy for solving some problems okay hopefully we will get time to solve some examples related to this okay so a simple example that we have here is that automobile A okay is travelling in this direction okay which is given at a speed of 36 kilometer per hour okay automobile B okay is coming in this direction what is it doing okay this vehicle is coming in this direction at an acceleration of 1.2 meter per second square okay what we are asked to find out determine the position velocity and acceleration of B relative to A 5 second after A crosses the intersection okay so currently this is what the position is okay these are the this is the acceleration for B it starts it starts from rest this is the position this is the initial velocity of A and we are asked to find out that what is the position velocity and acceleration of B relative to A this is a very straight forward problem what you can do is that that from the frame of reference of this road which we now take to be the stationary frame we can find out what is the position of A you can take the origin here you can say with respect to this origin what is the position of A what is the velocity of A what is the acceleration of A similarly with respect to this origin we can find out what is the position of B what is the velocity of B what is the acceleration of B after 5 seconds okay just because using all the equations of motion that we have derived earlier and once we get all these positions from this stationary frame to find out what are the relative velocities acceleration and position we just do RA-RB VA-VB and acceleration of A-acceleration of B okay and the simple thing to do is that VA okay look at this one after initial velocity is 36 kilometers per hour acceleration is equal to 0 initial position is 0 this is the origin okay from the point of view of the road so what do we know okay so VB0 is 0 AB is equal to for the for the second motor B velocity is 0 at this point acceleration at B is minus 1.2 Y because we have taken a frame in this rigid frame as X and Y like this is X and this is Y because acceleration in this direction we take it to be negative minus 1.2 meter per second square Y A0 is positive 35 degrees what do we do we find out okay what is the motion of automobile A what is the velocity straightforward the velocity is given in kilometers per hour we convert that into meter per second there is a uniform motion for A so acceleration is 0 with respect to this fixed frame velocity is just this whereas the position okay after how much time time is equal to 5 seconds is just given by X plus VT will be just okay this okay so what we know now now vectorial pay attention here now that acceleration is 0 velocity is 10 meter per seconds in what direction in the horizontal direction and R A the position is 50 meters in this direction okay so this is a vectorial representation now we do the same thing for B B what do we have we have the acceleration given to be 1.2 meter per second square okay we have velocity of B given as 0 minus 1.2 T T 1.2 T Y minus 1.2 we just saw because acceleration is downwards with respect to this frame similarly this is the position we put in the time we get that acceleration of B okay after time T is equal to 5 seconds is 1.2 meter per second square downwards velocity of B is 6 meter per seconds downwards whereas the position of B okay put in the value T here will be 20 meter still upwards okay so all these values are given to us and this is the acceleration velocity position these are all vectors these are also vectors for B and there are 2 things we can do we can solve this problem geometrically okay so what we can do is that to find out what is R of B slash A means position of B with respect to A what do we do this is the position vector for A this is the position vector for B just note here 50, 20 and what is R of B comma A finish the triangle law R of A plus R of B slash A this is nothing but the position of B with respect to A for the position that the B will see from his frame okay this is R B comma B A we can use Lame's theorem for example and solve this we can use triangle law whatever and can solve this geometrically second if you want to find out what is the relative velocity what is the relative velocity of A 10 meter per seconds okay in the horizontal direction relative velocity of B is 6 meter per second in the downward direction so R A sorry VA V of B with respect to A this triangle is completed this will be the corresponding relative velocity of B with respect to A and ultimately we want to find out what is the relative acceleration the acceleration is 1.2 okay only for B which is in a downward direction so we immediately know that the acceleration of B with respect to A is 1.2 meter per seconds downward square so essentially what it means is that I am if I am the passenger sitting in vehicle A then the motion of B from my point of view will look like this that this will be the position after 5 second this is what will be the velocity I will measure from my frame okay like this and this will be the corresponding acceleration which I if I am moving in car A is what I will measure for the car B or we can also use these vectors so we can also write in terms of vector forms okay that horizontal can be written as 50 I vertical can be written as 20 J write everything vectorially solve it and you will see that the answer can also be obtained in terms of I J K okay because it is a planar motion there is no K so either you can do this geometrically or you can do it vectorially depending on what is convenient for you okay because some things are more convenient for some people but they are both equivalent okay and a simple example now that if we are sitting in a train okay B okay looking out of the window in which direction does the train B does a train A is moving okay so essentially what we want to know is that from my frame of reference okay what is the velocity of train A and this can be very easily done using the vectorial approach so what we see is this is the orientation of vector okay this is my velocity I am sitting in train A VB bar okay this is the velocity of train A from a from a fixed point of view for the point of view of the ground and what we know is that relatively this is V of A with respect to B a vector okay so this angle okay is given to you some 25 degrees so what we know is that this triangle is completed and VA bar is equal to velocity of B plus velocity of A in the frame of reference of B and note that this angle has to be less than 25 degrees clearly from this figure and this cannot be possible this cannot be possible this horizontal cannot be possible and just we saw from the triangle okay that the only way it can have is that it can the direction has to be inverse like this and it has to be little bit less than 25 degrees why because this magnitude is not 0 it cannot lie along this it has to be lower than this so the simple answer for this concept will be this. Now we come to the very interesting part of kinematics okay which is tangential and normal components that this was rectilinear motion previously what we did is that just a few moments ago we just said that the motion of a particle can be expressed let us say take planar motion in the XY plane it can be expressed in terms of the X coordinate plus the Y coordinate and the corresponding unit vectors I and J in the X and Y direction but there are many alternative ways of solving these problems and as and as we solve the problems it will become clear that for example like for framing and XY X and Y coordinate frame what we need is we need some origin okay we need some origin and with respect to that origin we need to have a displacement vector X a position vector XY and then differentiate with respect to that and find out what is the velocity and the corresponding or acceleration but suppose okay that I am a rider for example and I want to figure out that what is the corresponding acceleration that I have just by looking at a local trajectory just by looking at this local movement as opposed to fixing some origin and finding out all the motions with respect to that okay those I have clearly have some uses but in many cases I do not care about what is origin I just need to know that when the rider for example is driving like this what is the corresponding velocity and what is the corresponding acceleration relative okay not to some coordinate system here but relative to what trajectory that I am seeing here for example if a train is moving if I want to find out not put some coordinate axis here and figure out that X and Y coordinate of this train but at this instant okay what is the corresponding velocity and what is the corresponding acceleration with respect to the trajectory that the train is following without worrying about what some origin is somewhere here or not same goes for this roller coaster I just need to know that if this is the trajectory then is here instantaneously what is the velocity and what is the acceleration without any reference to any origin system here okay just by looking at this trajectory and in order to do that one most effective way is to define what are called as a tangential and normal components of velocity and acceleration. So what is the tangential direction so the tangential direction okay so if this is the path of a particle of a train of a mass okay or of a cart anything you can think of or of a ball if this is the trajectory okay this is Y this is X and a particle is moving in this plane now the tangential direction is defined simply as a tangent to this curve okay we had seen earlier what is meant by tangent so you draw a tangent to this curve and that tangential direction okay draw a unit vector in that direction unit vector means ET come magnitude is 1 and that unit vector is defined by this tangential direction which is tangent to the path of the particle and now it is very clear we had seen earlier that a particle keeps on moving along this path and as a result by definition the velocity has to be in this direction why because how is velocity defined the particle is here it moves a little bit distance of here the displaced the corresponding displacement divided by delta T when delta T goes to 0 is the velocity but that is nothing that delta R bar in the limit when delta T goes to 0 is nothing but the tangential direction and the magnitude is the speed so what we realize is that that when the particle is moving along a trajectory the velocity vector is always directed along the tangent or it is along this unit vector directed call ET and what is the corresponding velocity velocity is the magnitude V of T what is T is the tangential velocity because why because velocity is always tangential so V can be written as V which is the magnitude of the velocity times E of T what is that is the direction so once we say that the velocity vector is speed okay multiplied by E of T we know what is the corresponding velocity vector now we need two coordinates for example in our Cartesian coordinate or XY coordinate frames what do we have we have I and we have J correspondingly what do we have here we have ET we take another unit vector which is perpendicular to to this now how do we choose do we choose an EN in this direction or do we choose the EN in the opposite direction so the sign convention we follow is this that I cross J is K how does that K comes out okay the K comes out out of the plane using the right hand thumb rule similarly we choose our EN in such a way that ET cross EN is also equal to K so that is called as defining a right hand coordinate system so we define ET which is tangent to the curve and EN which is normal to the curve and by definition okay by definition of velocity since this is the particle trajectory the velocity or the particle at any point is nothing but the speed V multiplied by E of T so this is what my V is but now note one thing when we dealt with the Cartesian coordinates you see that I J K or I and J if you have a planar motion that I and J always remain I and J I always remained horizontal okay J always remained vertical okay so one remained horizontal one remained vertical but note here this ET is always a tangent so what will happen is that if you keep proceeding along the trajectory ET will keep changing and because EN is perpendicular to ET such that ET cross EN is K okay you will see that ET EN always keep on moving and so these are called as moving coordinate systems okay now what do we do we want to find out what is dV by dt because we want to find out the acceleration because many colleges they had the question that what is the difference between kinetics and kinematics kinetics is just telling you that this is a trajectory I want to follow but I may want to drive a car in a particular direction but if for example it is a rainy day okay first train where there is for example there is some petrol spillage on the on the ground the ground is extremely slippery so no matter whatever I want to do the ground friction will not allow me to do that and will it allow me to take that turn what are the forces involved that can only be taken that can be only understood from dynamics and in order to dynamics to understand dynamics we need to understand how the Newton's laws are applied which is m times acceleration so whatever we have position velocity we need to ultimately find out the acceleration and acceleration how does it come we thus use chain rules so acceleration will be dV by dt E of t but we had just seen that this coordinate frame can keep moving okay and because this coordinate frame can keep moving what you will see is that that E of t will also change with respect to time the direction of this unit vector okay it will keep changing and so we have to use do it in the differentiation by parts so acceleration will be dV by dt E t plus V square by rho times d E t times E n okay now this derivation not difficult okay but not you have to just do some visualization will come to that in the next slides but just take my word for it for the time being you can go into the derivations into detail in details later but what happens is that that this can be shown that d E t by dt okay times V will look as V square times rho times E n what is E n this is this E n so what do we know from here okay just take this on the phase value that the acceleration has two components as we had expected that one component comes from the change in the speed what is the direction of that component of acceleration that component of acceleration has a direction which is tangential to the curve okay the curve that the particle is taking whereas the acceleration because because the particle is moving okay not only there is a magnitude of speed change but the direction also change and the change in the direction is reflected in this term okay which is called as V square by rho rho will go to come to that in next second is nothing but the radius of curvature here or what is the radius of this trajectory okay times E n which is in this direction okay so you can look at all the derivations okay let us not go into the details of the derivation because that will eat up a lot of time now what is the radius of curvature okay let me briefly discuss that that at this curve for example the particle moves from P to P prime in distance delta A in time interval delta T the distance it moves is delta S now as I will quickly show a brief demo what you can see is that you can fit a circle okay you can fit a circle here okay like this what you will see here is that this is some arbitrary trajectory okay and this is the particle which is moving along this trajectory now what is the radius of curvature the radius of curvature is that at this point you can fit a circle okay which approximates the local part of the curve and this particular thing is called as the and the radius of this curve is called as the radius of curvature which is the quantity rho that we have used here so just look at this trajectory okay look at this trajectory here and our intuition will fit like this okay just look here at this point of the trajectory it is pretty flat the trajectory is very flat okay so when the trajectory is highly flat the second derivative of the curve with respect to the arc length coordinate will be very small okay and this is the radius of curvature okay so when the curve is very flat the radius of curvature is very large but when the curve now note here okay so when the curve becomes very sharp note this is a very sharp point and when the curve is sharp the radius of curvature decreases again now the curve is becoming more and more flat so the radius of curvature increases similarly when the curve becomes sharp the radius of curvature decreases and that is the reason for example when we are think about it so when we are traveling in a very sharp curve and since the radius of curvature is very low look here this rho goes in the denominator and the acceleration is very high so we see that when we want to take a very sharp turn we have to decrease our speed because if we keep our speed also very high then the acceleration becomes extremely large and the friction of the ground okay will not be sufficient to do that okay and also we will get dizzy if the acceleration becomes very high so this is the logic that this rho is the radius of curvature if the curve is very sharp okay the turn is very sharp this rho becomes small and if the turn is if the road is completely flat then this rho becomes infinite essentially and so if you are going on a flat road this becomes 0 if you are going on a very very curved road this becomes very small and acceleration drastically increases and another name for this thing is called as a centripetal acceleration which is something which we know a lot about so just to keep in mind okay so there is a there are a lot of derivations but a simple thing to keep in mind that the velocity okay is simply given by v e t e t is that unit vector along the tangent and the acceleration is simply given by this okay the tangential component is the because of the change in the speed and the normal component is because of the absolute value of the speed and the radius of curvature or the sharpness of the corresponding turn that we are taking okay so this is one thing you have to keep in mind you have to change in speed you have to change in the direction of motion okay so this is a sample problem what we are asked here is that that a motorist is traveling on a curved section of a highway of radius 750 meter and at a speed of 90 kilometers per hour okay so kilometer every is all everything is given the radius of curvature of the speed is also given to us now we are given additional information that knowing that after 8 seconds the speed has been reduced to 72 kilometers per hour so it starts with 90 and it reduces to 72 kilometers per hour determine the acceleration of the automobile immediately after the brakes have been applied means right here so this person sees this curve coming okay if you are if we if we are sensible drivers then what we do is that we are coming from a straight road okay we see a curve we start slowing down by applying brakes so the curve starts from here brakes are applied here and for after 8 second the speed is 72 kilometers per hour so now what do we know is that the speed is now the tangential speed so what we know is that that what is the tangential acceleration is what we know but also another thing in addition to the tangential acceleration because this is a curved motion even we are moving at a constant speed we will have a centripetal force or a normal acceleration which is given by v square by rho so what do we do we define our coordinate system calculated tangential velocity and tangential acceleration normal acceleration and overall magnitude of the acceleration so how do we find out the tangential acceleration that this is 90 okay this is the initial speed which can be converted into 25 meter per second and 72 kilometer per hour is 20 meter per second and the deceleration is how much okay the deceleration from here to here but nothing but change in the speed divided by time okay which is minus 6.625 okay which means that the speed is actually decreasing if this is t the deceleration is in the direction which is opposite to the direction of the tangent now immediately after the brakes are applied okay immediately the speed is still 25 meter per second which is 90 kilometer per hour so the acceleration is v square by r r is given to us okay is how much r is 750 meters so the tangential the normal acceleration is nothing but v square by r which is 0.833 meter per second square the tangential acceleration from this figure you can see because deceleration the speed is decreasing this is the tangential deceleration in this direction at this point correspondingly the normal acceleration is 0.8333 in this direction okay and the total acceleration can be found out the magnitude by just square root of a n square plus a t square and direction by tan inverse of a n by e t with respect to this point okay so this is one simple way where for example we do not have to refer to some coordinate axis which is x y and then do everything there and then by knowing what is the trajectory we can find out what is the acceleration and what is the speed at what is the velocity at that particular instant okay there is this one kind of a interesting question this is from Bear and Johnston 10 slides from McGraw Hill they mentioned this interesting fact that is one of these race was schedule because the normal acceleration because the the curves were so sharp for though velocities that the racing was happening there was huge amount of normal acceleration and what happens is that so this is called as a g effect say for example when fighter planes for example okay this I looked up from Wikipedia you can look up to okay that in fighter planes where for example like suddenly at very high speeds when sharp turns are taken what happens is that the brain doesn't receive enough amount of blood and so the fire and so the pilots for example get dizzy and in order to prevent that from happening for example one thing they can do is reduce the allowed speed or increase the turn radius which cannot be done because the track is set okay but or have research where G suit and what G suit is is you can refer to in Wikipedia is very interesting that it applies pressure on the body and make sure that the blood doesn't drain to the lower extremities and brain still has enough supply of blood so that the passenger does not pass out okay so now we move on to the final topic of today's kinematics let us look at radial and transverse component now sometimes what happens is this okay that the XY coordinate system okay we need to have an origin this for example if for example when flights are tracked okay this is another example from Bear and Johnston the flights are tracked they are tracked with respect to the control center okay with a radar so what we know is that okay so if you have seen enough movies okay even I have seen only movies not a real demonstration that we see that with relative to this the control center you can see what is the position of the plane okay it means the distance and the angle okay so there is a origin which is the control center for example and the corresponding distance is the radial distance and the corresponding angle is theta so now in this case instead of referring to XY coordinate frame okay or for example if you want to understand what is the acceleration of this member which can rotate about this point so this is a fire ladder we need to set some coordinates system here and this is an alternative way okay so instead of expressing that expressing the position vector and the velocities in terms of XY what we do is we express in terms of what is called as a unit vector in the direction okay so this is the origin so rather than saying that this point P has a coordinate X and Y we say that it is at a distance of R and at an angle theta which is measured clockwise why clockwise because infinite similar angle clockwise the vector direction is Z which is coming out of plane so we are maintaining the right handedness of the coordinate system so we say that if this theta is measured clockwise with respect to our axis okay then this is R bar and this correspondingly okay the unit vector along this is called as ER and the corresponding vector which is perpendicular to it such that ER cross product with E theta is again okay coming out of the plane so the right handedness is mentioned okay now you may ask that why not doing X and Y what we will see is that sometimes the geometry of the problem is such that if you write expressions in terms of X and Y coordinate then the derivatives become extremely cumbersome and it is very difficult to understand what is happening whereas in some problems where we need to have an appropriate origin but the geometry of the problem itself is nice and circular okay then this coordinate system becomes very easy this particular aspect will become more clear when we solve problems in kinetics okay there are a bunch of problems we solve them and you realize that this coordinate frame is much nicer okay for those particular problems with circular geometry now what do we have here straightforward by definition by definition the position vector R bar of this particle with respect to a fixed origin now we need the origin so with respect to this fixed origin the position is given by R which is the magnitude or the distance multiplied by ER bar which is a unit vector along this so R bar is equal to nothing but R ER bar now what do we do note here again that if point P moves from here to there again ER does not remain constant so this is again a moving coordinate frame because ER now will become in this direction and so correspondingly ER also changes the direction changes E theta car direction also change so what happens is that if now we want to find out what is the velocity of this particle we do DR by DT which is R dot plus we also have to do DER by DT and there is a derivation given here we do not need to go into those details but what we will see is that that this DR by DT can be written as DER by DT R theta dot E theta now what we have here that the velocity as we had expected it has a radial component why because the particle tries to move outwards and that is the R dot and the component is in the this direction but without changing this R okay keeping this R constant like we when we move in a circle we can also have velocity why because the particle changes direction okay the position is this the next position is in the other direction okay you have to move along the tangent so that change in position will also induce velocity and that velocity is given by R theta dot and the magnitude is in the direction E theta that is a physical meaning now go one step further as we all know that in when we are applying Newton's laws when you are doing kinetics we cannot just stop at position and velocity we have to go for acceleration and acceleration is what DV by DT just do all the math together okay the mathematics is given here what we will see so let us only understand what this means okay so acceleration will have two components so one component in the radial direction other component in the tangential direction okay so the tangent or radial direction so ER and E theta sorry I should not use our tangential direction it is in the theta direction so sorry sorry please pardon me this is not in the tangential direction ER is in the radial direction E theta is in the perpendicular direction to this radial direction so what are the two components so R double dot clearly that a particle if it just starts moving outwards okay with some acceleration R double dot we will have acceleration in this direction but not only that you also have minus R theta dot square now what is this minus R theta dot square remember we had V square okay divided by rho so this R theta dot square acting in the inward direction is very similar to the acceleration that comes just because even if the velocity remains constant R dot remains constant like the particle is moving okay that you are taking a a stone tying it to a row to a string and turning in a uniform circular motion even in that case R remains constant so R double dot is 0 but because you have theta dot or angular velocity that theta dot is not constant it is omega which is the angular velocity you will get this centripetal acceleration now coming to these terms this R theta double dot now what is R theta double dot R theta double dot just means that for example if I am not rotating this particle at a constant speed but the angular velocity itself is changing that theta dot is not constant there is also theta double dot now what happens when the E theta direction the particle speed keeps changing and as a result okay for a uniform circular motion E theta and the tangent direction are the same but for any general motion that is not the same okay but when the motion is circular motion then E theta and the tangent both are in the same direction okay and normal direction and E R are also the same but think in that case that the velocity is constantly changing in the tangential direction and as a result this is the component okay that adds to E theta whereas the last term this 2 R dot theta dot it is a little bit complicated to understand okay so what happens is that if you try to move outwards okay then what is happening is that think about it you try to move outwards means your means your velocity is changing in the outward direction okay so that is your going out but when you move out because of this theta dot component you will also see that your velocity in the tangential in the normal in this theta direction also changes okay and so the combination of those two changes give rise to what is called as 2 R dot theta dot it is not very straightforward to understand okay but just think about it okay there is a figure that is given in the next few slides think about it but this is you are trying to move in the radial direction but when you try to move in the radial direction not only are you having velocity there but because your velocity in the perpendicular direction also changes because the velocity in the perpendicular direction is what R theta dot okay and since you move outwards the R position changes so the velocity components get added and so this particular term is another in non-inertial frame this term is called as Coriolis force and this can be easily be understood that for example if you have a if you have a rotating wheel for example like there are there in parks okay that you have those wheels which are rotating about the circle and you try walking from the center outwards you will see that you will need to exert some force to create balance otherwise you move sideways okay you will fall down sideways so it is that force that you are trying to move in the radial direction but this force is trying to push you in the direction which is perpendicular to the radial direction okay so all these derivations are given here and you look stare at this figure okay stare at this figure and you will see that what are the various acceleration components that come here okay but the main thing you have to just realize that these are the various components that can come okay when you have radial and transverse components and now we have our last sample problem what do we have here at the rotation of this arm now again this problem makes it clear why we do not want to deal with XY system in this problem but it is much better dealt with in R theta system or in ERE theta system so what we have is that that the rotation of the arm about O okay the arm rotates okay so we are looking it from the top forget about like well do not worry about any gravity nothing nothing this is pure kinematics and this is an observation that that I measure the rotation of the arm as a function of time and what I see is that that the angle theta okay goes as 0.15 t square okay as a function of time okay this is given to us now why is this like this will come when we do kinetics now for this collar B keeps sliding outside okay so this is again why this will happen can will come from kinetics okay so the collar B keeps sliding along this arm in such a way that R goes as 0.9 minus 1.2 t square where R is in meters now what we are asked is that after the arm has rotated to 30 degrees determine the total velocity of the collar okay after the arm has rotated to 30 degrees okay so what you want to find out total velocity of the collar this collar total acceleration of the collar and the relative acceleration of the collar with respect to this arm okay so these are the three questions now what do we do we first find out that this theta as a function of time is given to us so what do we do first we find out that when theta is equal to 30 degrees what is the time we put theta equal to 30 degrees we get after that time okay after how much time will that happen second evaluate radial and angular positions and first and second derivatives at t now we are given theta just look at this expression what do we want we want R double dot we want theta dot we want R dot and we want theta double dot so essentially we want R dot R double dot theta dot theta double dot these four components is what we want are we given those components we are already given we are given theta as a function of t so theta dot will be 0.3 into t theta double dot will be 0.3 R is also given to us so we can find out R dot and we can find out R double dot okay so all the quantities can be found out by so R R dot R double dot theta theta dot theta double dot now everything is there either as a constant or a function of time now we see that theta is equal to 30 degrees at time t is equal to 1.8 seconds so we substitute all those values here to find out what is R dot R double dot theta dot theta double dot and from that when theta is equal to 30 degrees one thing we should note that when theta equal to 30 degrees we also know what is ER just along this what is E theta is also we know so we also know then that AR is acceleration in this direction is nothing but R double dot minus R theta dot square just put it in acceleration theta direction is R theta double dot plus 2 R dot theta dot just put that in and you will see that this is acceleration in the theta direction minus what does that tell that this is E theta the acceleration in the theta direction is actually in this direction okay velocity is nothing but R theta dot okay if you look here what is velocity is R dot plus R theta dot R dot is in the direction ER R theta dot is in a direction E theta put that in we will know what is the velocity in the this direction and in the perpendicular direction and then we can find out what is the resultant velocity and so on everything we get now the last part of the question is the interesting part the relative acceleration of the collar with respect to the arm now note one thing that what is happening that this collar has the same motion as the arm and in addition it has this motion along the radial direction with respect to this so think about it that the acceleration for this will be nothing but at this point what is the acceleration for this collar okay which is already there okay plus what is that extra which comes in this direction so essentially what do we have that the motion of the so A of B with respect to this O O A which is this arm is nothing but R double dot because why because everything else okay everything else okay is already a part of the collar so it is only R double dot okay which will decide what is the relative acceleration because if I am sitting on this rod okay if I am rotating on this rod then the only acceleration I will see is that in my frame of reference which is rotating with the rod I will see that the collar is only moving outwards okay and the only component for that is R double dot and that is what we will see here okay so with this much of a preamble what we have done is that with like lot of kinematics essentially that are required okay so we took a lot of time with this but it was very important why because all the kinematics is absolutely mandatory for any kind of kinetics or using Newton's laws because once we know kinetics appropriately we are done Newton's laws become really straight forward once our kinetics once our kinematics is absolutely fine okay so center number 1 2 0 3 they say would you please explain the difference between centrifugal and centripetal forces okay we will discuss this in great details when we go to kinetics note that when we are in a inertial fixed system okay in an inertial system what is an inertial system the inertial system is a system which moves okay at a uniform velocity whereas an accelerated system or a non-inertial system is like for example when you have a train which keeps on braking every now and then so if a train keeps moving uniformly and I am present in the train then I am present in an inertial system but for example if the train keeps braking or it keeps like going around curves okay then the then the train is accelerating and then that frame of reference is called as a inertial reference frame of reference for a spinning particle for example if you take a stone and spin it in a circular motion then if you are in a coordinate frame which is rotating along with the stone that is called as a non-inertial frame of reference and in that frame of course I will not see any acceleration for the stone but then I am seeing that there is also tension in the string so how do I explain that so to explain that what do I say is that that there is a force called as centrifugal force which is acting and which is pushing this thing outside so centrifugal force we say to that when we are in the moving frame of reference with the rotating disc or with the rotating stone okay whereas if I am on a fixed frame of reference okay and I am saying seeing that the stone is being rotated around in a circle then in that case I will see that to keep that stone in the circle the tension in the string is applying an internal force and that force is the centripetal force it is going into the center so from a inertial frame point of view you have a centripetal force whereas if you are in a non-inertial frame and we are not going to deal with non-inertial frames in this course then what you see is a centripetal force it just depends on what frame of reference you are in so to 1, 2, 1, 5 say please explain hodograph I think you just have a look at that it is not a crucial concept okay so it just that like we have position vector we have a velocity vector and then correspondingly whatever velocity vector velocity is like what is the curve that is stress is called as a hodograph not important okay not important at all in a projectile motion why do we take okay the question is by 1172 it is by TCEBEM 1172 they have asked the question in projectile motion why do we take the negative value of acceleration due to gravity that negative value is purely because our coordinate frame is chosen that y is upwards x is sideways you can take your y downwards okay and then your g will become positive just because of the way we have chosen the coordinate frame and nothing else so center number 1108 they have asked okay why the horizontal component of velocity in projectile motion is constant okay so the it is constant because there is no acceleration the acceleration is only due to gravity in the x component for example there is unless and until there is some air drag which in this first course okay we just do not worry about it okay we just say okay it is small okay the air drag is small compared to the inertial compared to all the other forces air drag is small we forget about it in that case there is no acceleration that acts on the particle in the horizontal direction there is no reason for us to say that the particle will have any deceleration in that direction so velocity has to remain constant because v is equal to v minus v plus 80 where a is the acceleration since a is 0 v remains constant then there is another question which is asked by 112 what is the role of jerk in kinetics okay it is a very nice question so note one thing jerk is what the rate at which the acceleration changes now jerk is not a fundamental quantities in kinetics why because Newton's law force is equal to m times a at least for particles okay or particles or for only translational motion f is equal to m times a so jerk does not enter into that picture at all but why is jerk important jerk is important because if you want to design for example a track okay if you want to design a road if you want to design a path in which for example a vehicle should go then in that case if you want to have a motion which is jerk free then you have to make some arrangements such that those components are nullified okay so jerk in kinetics it does not directly come but it comes for us that we want to make sure that for our ride for vehicles for example that if you want to have as comfortable ride as possible then we should have as little jerk as possible and in that case for example we want to make sure that our trajectory such that acceleration does not change appreciably when we move along the path the role of jerk in kinetics is from the point of view of design from the point of view of Newton's laws okay it is not particularly relevant but how do we do appropriate designs for cars for roads okay or for example what are the appropriate forces we should be applying to change the acceleration and so on okay so in that sense jerk is important but it is not a fundamental quantity so with this one okay we will take a break