 Ok, good morning and welcome all to today's morning session. So, yesterday we had a long long discussion about kinematics of particles and towards the end of the class we started our discussion on what we mean by kinetics. So, we discussed what is Newton's law for particles also later on what did we do? Why after discussing the Newton's law on particles we also discussed how when we have a collection of particles which is essentially a representation for a rigid body for example, how can Newton law be modified in such a way to say that f bar that is a force acting on the material is equal to mass times the acceleration. Note one thing is that I strongly urge you that if you want to have a good understanding of this dynamics in this tenth edition of vector mechanics for engineer is absolutely fabulous all the concept questions the kinetic diagrams the free body diagrams that for example, that are being shown to you are essentially a part of that book. So, if you want to get a nice understanding then this is a pretty good resource to have. So, it is a significant upgrade over previous editions and I cannot recommend it enough then we lastly came to this concept of free body diagrams and kinetic diagrams. So, we saw that free body diagram has the same meaning as what we had done previously for statics whereas, the kinetic diagrams now are drastically different here what do we do here is that the free body diagrams are drawn appropriately we can use Cartesian coordinate polar coordinate or the path coordinate which involves the tangential motion and the normal motion and then we isolate the body for example, in this case we isolate this body found out what are the possible forces acting on the body and once we realize that these are the possible forces on the body what we saw is that that these forces would lead to acceleration in this direction y direction and in the x direction. So, these two systems this is a kinetic system why is it a kinetic diagram because we are showing that what are the accelerations mass terms accelerations and these are the forces and the forces lead to acceleration is essentially what is predominant topic of kinetics that f is equal to m a this is the main equation that we are going to use for the entire lecture. Now, this is another free body diagram and kinetic diagram that we discussed yesterday. So, what we had is we had two masses so mass a or block a on which block b is resting and what we had seen is that that this entire assembly has gone through a complete system of pulleys and what we want to find out is for this system what is the corresponding free body diagram and what is the corresponding kinetic diagram and what we saw yesterday is that that we expose we make a cut here cut here cut cut remove this body and these essentially are all the forces that this body experiences and the corresponding kinetic diagram what we realize is that we draw an x axis which is parallel to this plane and y axis which is perpendicular x axis which is parallel to this plane and y axis which is perpendicular to the plane and what we realize is that that because this mass cannot lose contact or cannot penetrate inside this slope what happens is that that acceleration in the y direction is 0. So, this portion we realize from kinematics that because of this constraint that is being applied to this body because of the constraint that has been applied to this body we see that this is m that this m a y is equal to 0 whereas, m a x need not be 0 because there is a slope there is a there is a freedom for this particle or this body to move in this direction and the Newton's law says that some of all the forces should balance in such a way that sigma m times a is equal to some of all the forces. So, we know that some of all the forces in the y direction is 0 and some of all the forces in the x direction is nothing but m times a x where is the acceleration. Now, let us go for a simple problem again to draw what is the free body diagram and kinetic diagram for this problem. Now, this was the problem if you recall is what we had discussed all these problems are from BJ 10 I cannot recommend that book highly especially for dynamics excellent excellent problems excellent concepts and concept problems. Now, what do we have here that we have a rigid body O a we had discussed that earlier and one angle is theta. So, there is this rod is rotating in the clock anticlockwise direction this angle is theta. So, there can be theta dot there can be theta double dot we had been given an expression previously. Now, additionally we have a sleeve which can freely move in this direction what is problem say that assume there is a friction acting between the rod and the collar and motion is in the vertical plane. So, motion is in the vertical plane what means what does that mean theta is increasing motion is in vertical plane and. So, this gravity would act on particle B. Now, what are the free body diagrams and kinetic diagrams for the sleeve. Now, for the sleeve this sleeve green colored sleeve what we do we isolate this then we draw the axis. Now, because if you remember what we had done in kinematics that this problem has a natural geometry which is extremely amenable to the polar coordinates or we have a vector in the radial direction E r and vector in the theta direction E theta. So, we draw this axis first then we find out what are the applied forces. So, what are the applied forces the external force. Now, let me again reiterate the difference between the applied force and the reactive forces whatever forces this rod gets this sleeve gets from the rod those are the reactive forces why because on its own the only external force that act on this is this Mg. Now, this is Mg and all the other forces are support forces. So, we have to replace supports with forces. Now, what are the forces because we had seen that there is a possibility of friction there can be a friction force as we look at this problem further we realize is that this sleeve has a tendency to slip outwards and because it has a tendency to slip outwards even the expression that we were given yesterday for R that also had an increasing R dot that R keeps on increasing. So, because it has a tendency to slip outwards the frictional force will act inwards because it will try to resist that motion. So, this is the force of friction this is the normal reaction. So, this simply is the free body diagram for this sleeve straight away nothing else. Now, dimensions we put everything theta all are put here and now what do we want we want the kinetic diagrams and what is the kinetic diagram that sum of all forces in the radial direction should be equal to mass times the mass of the sleeve times the acceleration of the sleeve in the radial direction. Whereas, sum of all the forces in the theta direction is the acceleration of the sleeve in the theta direction. Now, note one thing that from an inertial frame point of view what do I mean by inertial frame if for example, at this origin O you fix a frame X, Y. X, Y is just a name you can as well name it M, Q and whatever. But, X, Y because it is a convention you put a frame like this here and for that frame in this frame at rest or in the lab frame what you will see is that this theta keeps on it changing this R position of the sleeve keeps on changing. So, as a result this acceleration AR and A theta are in this inertial frame of reference. So, that is one point that we should always remember that Newton's law work only in inertial frame. So, we have to make sure that all the accelerations are with respect to an inertial frame which for example, which means a frame which is at rest or which moves with a uniform velocity. So, that is an inertial frame. Now, a problem like this for example, I think we will be attempting sometime soon. Now, this is one simple sample problem from BJ 10 what do we have here? We have a mass of 80 kg which rest on a horizontal plane find the magnitude and force p required to give the block an acceleration of 2.5 meter square to the second. The coefficient of kinetic friction between the block and the plane is mu k equal to 0.25. Now, note in this thing that the problems in Newton's laws of motion and if you look at the text, if you read the material in the text and if you look at some other standard text also. Then what you will see is that that the problems are typically of two types. One is that that we are given that this is the acceleration of the particle and then ask to find out what are the forces acting on the particle that is one case. And second case we are given the forces and we are asked to find out that what is the corresponding acceleration. So, the type 1 problems where for example, all the accelerations are given and we are required to find the forces are typically much easier. Why? Because once we know the acceleration then done then we know what are the forces acting on it. Whereas, what happens is that we will see that if a system for example, let me draw on the white board that if you are told that a particle has to follow a trajectory like this in the sense that this is a constraint that acts on this particle and we are told that a force acts at a particle at an angle theta p. And then we are asked to find out what is a final acceleration of the particle as a function of time, velocity as a function of time and position as a function of time. If you are asked to do all these things then there are unknown reaction forces that we have to solve for and make sure that the particle has this trajectory and those unknown forces balance these forces such that sum of all the forces which is nothing but if you draw the free body diagram of this particle it will have some reaction r bar then sum of all these forces is equal to p bar plus r bar and that will be equal to m times a and then we have to use various constraints and then find out what is this reaction and what is this acceleration. So, typically the problems of the type where the force is given and we are asked to find out what is the acceleration are typically the difficult problem types whereas these kind of problems where the acceleration is already given to us and we just need to find out what are the resulting forces that are little bit easier as compared to those problems but ultimately when we represent the problems with appropriate free body diagrams and kinetic diagrams and realize that what are the appropriate constraints on the system then any problem actually will become reasonably straight forward. The idea is to recognize what is the proper concept what are the forces acting on the body and then just realize that sum of all the forces acting on the body will be equal to mass times the vector acceleration of the body. So, the entire rigid body mechanics the crux is essentially this and later on as we will see is that the sum of all the moments is equal to the rate of change of angular momentum of a body. So, these two equations essentially constitute the crux of entire rigid body mechanics and will solve a few of these problems in the coming in the coming hours. So, what we do how we solve this problem we resolve the equation of motion for the block into two rectangular components unknowns what are the unknowns that there is a friction force that can act here there is a normal reaction that can act here. So, there are two unknowns but what we are told we are told that the acceleration is 2.5 meter square meter per second square to the right. So, what we do is we draw the free body diagram first weight normal reaction friction force why because the body tends to move to the right that is the condition of the problem and. So, now the body keeps moving at a finite velocity because of which there is a finite slip that happens at this point and if you recall our lectures on friction when there is a finite slip between two surfaces then the corresponding friction is a kinetic friction which is nothing but mu times f where mu is the coefficient of kinetic friction. Now what is the kinetic diagram that this will be equal to m times a in the horizontal direction these all these forces will lead to m times a what is the vertical component there is no vertical component of acceleration to m times y is equal to 0. So, writing sigma f y equal to 0 we immediately find out that this equation in terms of n p sin theta will be the component of the force in the vertical direction and in the horizontal direction we will have p cos theta is the horizontal force and that will become m times acceleration and acceleration is 2.5 meter per second square. Now these are two equations and there are two unknowns n and p we solve them very easily and we can find out that the corresponding force that is required in this case turns out to be 535 newton's very simple problem. Now let us look at this another problem from BJ very simple problem but it is a nice concept of principle problem. So, what we have is this we have two blocks who start from rest the horizontal plane and the pulley are friction less. So, this is friction less the pulley is friction less and the pulley is this small pulley at c is assumed to have a negligible mass this pulley is also supposed to have a negligible mass and what we are asked is that that determine the acceleration of each block and the tension in the chord. Now note at first we figure out how many degrees of freedom does this problem have. So, this is the kinetic diagram that we are talking about now look at pulley a it has one degree of freedom it can move in the horizontal direction this can move in the y direction the x direction there is in the horizontal direction there are no forces. So, we do not worry about the acceleration in the horizontal direction here in the y direction or the vertical direction there are no this block can neither penetrate this surface nor can it lose contact. So, there is no acceleration in the y direction. But now note that the horizontal position or the horizontal velocity and horizontal acceleration of a is not independent as we had seen in kinematics of the vertical acceleration of b. So, what we do is we write down all the equations of motion we immediately realize the what is the length of the spring the spring that if you take a origin here then this is x a this is 2 times y b. So, total length of the spring if we recall is that that 2 times x a sorry x a plus 2 times y b is a constant we differentiate twice once when we differentiate with respect to time what do we get that acceleration at b will be equal to half times acceleration of a it is just straight forward that we recognize that because of this constancy of the length of the string the acceleration of point b will be half the acceleration of point a. Now we draw the free body diagram for this now on this mass what are the various forces acting on it this is the tension T 1 normal reaction n we had assumed that this is frictionless this is weight. So, this is the free body diagram for mass a and what is the corresponding kinetic diagram that that this sum of all these forces will be nothing but mass of a times the acceleration in the horizontal direction. So, immediately we find out that T 1 this tension will be equal to mass times the acceleration of a. Now note one thing because this pull is frictionless this entire tension remains constant this pull is frictionless and massless. So, these two tensions also remain constant. So, these two tensions again remain constant now what we write this equation for the equation of acceleration for mass b what are the forces acting on it we have a tension acting from here T 2 the link connecting the c and the mass and what else do we have we have the force acting on it from weight in the downward direction and this entire system of forces what does that do that brings about the acceleration of this mass in the horizontal direction. So, note that this is 300 kg. So, the acceleration in this direction is taken to be a b. So, what is the equation of equilibrium that m b times a b is sum of all the forces we write down the equation of equilibrium what do we see we see that T 2 or this tension between this portion this portion c and the mass is simply this equation 2 4 9 0 minus 300 kg is the mass times acceleration. But now what do we have we have 1 unknown we have 2 unknowns we have 3 unknowns we have 4 unknowns. So, we have 4 unknowns in this problem, but note that this acceleration a and b are related by this equation. So, essentially by using these 3 equations we have 4 unknowns now. So, we need 1 extra equation and what is that extra equation we draw the equation of equilibrium for the pulley itself. Now, what we had said here is that this pulley has very little weight and also the pulley is frictionless. So, because the pulley is frictionless we realize that this T 1 should remain constant all throughout on top of that because the pulley has very little mass 2 times T 1. So, the sum of all the forces on this pulley which is what 2 T 1 minus T 2 will be equal to mass of the pulley times acceleration since this is very small you throw this away and say that T 2 is equal to 2 T 1. So, now we have 4 equations and 4 unknowns and they can be easily solved to get to give us values of acceleration at a acceleration at b, tension 1 and tension 2. So, this is another simple problem that we are solving. So, let us look at this problem quickly what do we have. So, now because we have discussed 2 problems in great details let us only gloss over just only gloss over what is the methodology and you can look the book or later on some of some parts of the slides to see how the problem is being solved. So, what do we have here is that a 6 kg block b starts from rest and slides on the 15 kg wedge a. The mass is given for this, the mass is given for this and this is a horizontal surface and what we are given is that neglecting friction everywhere. There is a possibility of friction at this interface, there is a possibility of the friction at this interface. So, if you neglect all the friction determine the acceleration of the wedge and acceleration of the block relative to the wedge. So, now note one thing that this is definitely a 2 degree of freedom problem that this acceleration that this b can have a displacement in a direction or it can have acceleration in a direction parallel to this slope whereas this mass individually can only have acceleration in the horizontal direction. So, there is one possibility of motion for b is this and overall translation whereas for mass a there can be only overall translation in the x direction. Now what do we do that because the block is constrained to slide down the wedge therefore their motions are completely dependent is what we had seen. So, what we have to do is that that when this mass moves when there is a when this block a has an acceleration it will also take block b along with it. But additionally this block b can also slide down the slope. So, these are the various kinetics that we are now seeing in this problem. So, we write down the equations of motion for wedge and the block and solve for acceleration. First if you look at the free body diagram for this wedge the sorry the kinetic diagram for this wedge it can only have acceleration in one direction. Let us choose the direction for this x direction in this way. So, this is the acceleration of the block whereas for the top block what can happen because the block b is connected with block a it is not losing contact with it neither is it interpenetrating the block. So, what will happen is that it will have one component of acceleration same as the acceleration of block a. Additionally it can also slide along this block with an acceleration what a of b relative to a. So, it can also have an acceleration b relative to a. So, note now that all these concepts about relative velocity relative position relative acceleration we are really using them now in kinetics. Now what do we have totally? What are the forces acting on it? On the bottom block the forces acting are the normal reaction weight and the normal reaction from the bottom surface and this should be equal to mass times the acceleration whereas for the top block the forces acting are the main forces acting are the weight and the corresponding reactive forces is what? This normal reaction. Now what we can do is for this block we can have a coordinate frame in this direction x and y what do we know that the total acceleration of the block in this direction and this direction is known we can resolve the acceleration in along x and along y and equate that with some of all the forces in that particular direction and we have one equation for this one equation for this one equation for this. So, the unknowns are acceleration at a acceleration of a is an unknown acceleration of b relative to a is an unknown and this normal reaction 2 are unknowns but we have enough equations and we have 4 equations y because equilibrium for this block in the y direction will tell us that sum of all these forces in the y direction will be equal to n2. So, we have the number of equations now become equal to number of unknowns equilibrium in the y direction will give you one equation equilibrium in the Newton's laws of motion in the x direction will give us second equation Newton's law of motion in this direction for mass b will give us third equation and equation of motion in the y direction for mass b will give us the fourth equation and the number of unknowns are 1, 2, 3 this acceleration and 4. So, 4 equations 4 unknowns we can solve this problem and ultimately obtain what are the accelerations and what are the reactive forces. Now, let us go on to kinetics using normal and tangential coordinates. Now, we had seen previously that when you have a mass or a particle which does not travel in a straight line or for example for which the geometry is such that the use of normal and tangential coordinates is preferable then we can also solve the problems for such type of problems. But now how do Newton's law look for such problems this equation does not change f bar is equal to m a bar it does not change note one thing. So, this is a point which typically is a point of contention for many students that this acceleration is a vector I can break this acceleration I can write it into various in various ways. For example, I can use I can write the acceleration as a bar can be written in Cartesian coordinates as a x i plus a y j in normal and tangential coordinates I can write this as a t which is a tangential acceleration e t which is a unit vector along the tangent plus a n which is the magnitude of the normal acceleration times e n which is the unit vector along the n direction and in polar coordinates or in radial and the normal in the radial coordinates what we can say is that that a bar is equal to a r e r plus a theta e theta. So, we can have the same acceleration written in all these ways and note that all these things are equal to each other. The only difference is that the way we are representing the components depending upon the problem and the corresponding geometry of the problem the representation change. But the problem the acceleration remains the same we either represented like this or like this or like this and Newton's laws just tells us that sum of all the forces will be equal to m times a bar and this f also subsequently can be written in whatever direction we can write it as f x i plus f y j and same can be written as f t e t plus f n e n and this can also be written as f r e r plus a theta e theta. So, whatever we do ultimately the way we represent this forces the way we represent this forces when we represent the acceleration it just depends on the problem that we are solving. Now, coming to the problem which we want to represent using tangential and normal force what do we write we write f bar is equal to m a bar and so correspondingly sum of all the tangential forces is mass times acceleration in the tangential direction. So, these kind of problems become very useful when we want to find out what are the forces acting on a car acting on a plane and so on. Sum of all the forces in the normal direction is nothing but mass times a n. So, f t is equal to m dv by dt f n is equal to m v square by rho where v is the speed of the particle at this point straight forward. Now, we go to this coordinates for a simple pendulum problem. So, what we are given is that that a bob of this 2 meter pendulum the length of the pendulum is given to us which is essentially the radius of curvature for the motion of this pendulum because this pendulum can only move along the arc of a circle and the radius of the corresponding circle is 2 meters. And what we are asked is that that if tension in the chord we are given additional information that if tension in the chord is 2.5 times the weight of the bob for the position find the velocity and the acceleration of the bob in that position. Now, again this is an easier problem that we are given that we are given that this is the tension and then we are asked to find out correspondingly that what are the velocity and acceleration for the bob in that position. So, what do we find out we first find out that what are the corresponding equations of motion we resolve the equation for the resolve the equation of motion for the bob in the tangential and normal components and then we solve the component equations for the normal and tangential directions as we had discussed here. So, what are the forces acting on the bob the forces acting on the bob are the weight and the tension. Now, we are given that the tension is equal to 2.5 mg because if you really want to solve this problem from start to finish for a given position we should be provided that what is the corresponding initial velocity then we solve the differential equation or use energy conservation principle to find out what is the velocity here and then finally, find out the tension, but here the tension is already given to us which is 2.5 mg we are also told that this is the weight acting whether the mass of the bob is also given. So, this force is the main force acting on the bob is given to us now we resolve we define our coordinate system along the tangential direction this and the normal direction this. So, what do we do we say that this is mass times acceleration in the tangential direction all the forces some of all the forces in along this direction will be equal to mass times a of t some of all these forces 1 2 along this direction is mass times the acceleration in the normal direction. So, what is this some of all forces in the tangential direction will be nothing, but mg sin 30. So, mg sin 30 is the only force acting in the tangential direction and that will be equal to m at. So, we can find out that the acceleration of this cylinder of this bob in the tangential direction is nothing, but g sin 30 which will come out to half of g by 2 or 4.9 meter per second square. Then we want to find out the normal acceleration normal force how does that normal force go that the normal force is equal to mass times the acceleration in the normal direction. So, what do we get this is tension t 2.5 mg minus weight cos 30 mg cos 30 this should be equal to the effective force in the normal direction which is nothing, but mass times a n. So, we can immediately find out what is the acceleration for this cylinder in the normal direction. So, really straightforward we knew these forces and we are done and ultimately what do you want to find out we want to find out the velocity in terms of the normal acceleration we know that the normal acceleration if it is this then a n is equal to v square by rho recall the equations we had derived earlier that a t is equal to dv by dt where v is the speed whereas a n is equal to v square by rho we know rho what is that 2 meters that is the radius of curvature of this part and from that we can immediately find out what is the speed of this cylinder speed of this bob at this position. So, really straightforward problem. So, this is another problem what is this problem is this is the car which is going on a turn this car is going on a turn typically when the car goes on a turn we had seen that there is a acceleration of the car in the inward direction. Now, what can provide the acceleration in inward direction if the road is flat then it is a friction that acts between the car tires and the road that provides the acceleration in the inward direction, but friction is a very dicey thing because friction can change drastically depending on the condition of the road. So, if the for example the road is slippery because of rain for example then the car is liable to slip and many accidents happen in rainy season because of that because the car tries to move around a curve if you are speeding along a curve then v square is very large and if especially it is a short turn a sharp turn then rho is also very small. So, the acceleration in the inward direction is very large v square by rho. So, the force that the friction needs to provide is also very large, but if the road is slippery because it is wet then the friction that the surface can provide is only limited. It cannot provide enough friction and as a result you may think you may want to turn the car along that road, but the friction does not allow you to do that. So, the car does not turn, but it will just try to go it will go straight and hit the barricade and there will be an accident. So, that is typically the reason why in slippery in higher any season the speed limits are typically lower, but there is another way in which we can get this friction completely out of the equation by doing what is called as the banking of the road. So, the road is inclined a bit at some prescribed angle and the road is prescribed inclined at such an angle that for that given angle there is something called as a rated speed that rho is equal to 20 in this case is the radius of curvature it is the radius of curvature of this turn and this road is banked with an angle 18 degrees and what we are asked to find out that what is the rated speed of this banked highway curve. Now, what do we mean by the rated curve rated speed the rated speed is that is the speed at which the car should travel if no friction force is required to be exerted at its wheel. Now, again recall that just because friction is present does not mean friction will be always mu times n friction will only act as much as is required only thing this mu s times n tells you is that that the friction cannot exceed more than mu s times n in this simple Coulomb friction. There are more complicated friction laws as we have discussed, but in simple Coulomb friction law the friction cannot be more than mu s times n or mu k times n depending on if there is relative slippage or not. In this case the friction may also be 0 depending on what is a particular geometry and this geometry is given to us and we are asked to find out that for this entire geometry what should be the speed at which this vehicle should travel such that the cars do not exhibit. So, the cars do not have any friction force acting on them this is very simple. So, what do we do is this we resolve we take this we take this car this is the weight acting on it we say that we do not want the friction force to act. Suppose we do not want the friction force to act completely we do not have it say no friction force we do not have it this is the normal reaction that can act on the car. But now the normal reaction is acting on the car note one thing if the road is flat then the normal reaction cannot have a component in this n direction. But because the road is banked at an angle this r this reaction can also have a component in this direction and what we want is that we want a speed to be in such a way that only the horizontal component of this reaction can provide us the necessary force that the car requires to make this turn okay make this turn along the curve of a given radius. So, what is the corresponding acceleration that we require the acceleration that we require is equal to v square okay we write down some of forces in the y direction and from that what do we find out we immediately can find out what is this reaction this reaction will be W divided by cos 30 okay straightforward just write equilibrium equation of equilibrium of this in the y direction r is equal to W cos 30 cos theta theta is equal to 18 in this case what is the what is the force acting in this direction n it is nothing but r sin theta and which we substitute this value okay. So, r sin theta will be equal to this but what should r sin theta be equal to r sin theta should be equal to W by G which is mass times the acceleration in the inward direction. So, what do we want is that this is the reaction r so W cos by cos theta r into sin theta should be equal to W by G v square by rho why because a n is equal to v square by rho. So, cancel everything off and then you will find out that v is equal to around 20 meter per second or corresponding to approximately 70 kilometers per hour and what this problem tells you this problem tells you that if you have a road which is banked at an angle of 18 degrees then on that particular road if you take your vehicle okay on that particular road where the radius of curvature okay the radius of curvature is 20 meters okay so the turn radius is 20 meters then for that particular road if we go at a speed of exactly 70.4 kilometers per hour or at least we travel around this speed then we do not need any friction forces okay if you try to go ahead of this okay if you try to increase the speed limit beyond this okay then the friction may not be enough and the car can run into trouble. Now a simple question is that is for example okay just to understand that when the car moves along a road like this what are the direction of the car's acceleration at each and every point. So, suppose the car is moving in this direction okay we most of us drive cars so what we realize is that that we are moving from a straight road okay then we go into this curve the moment we go into this curve okay so what do we do we try to go at a constant speed we had just discussed here right that for example if you want to keep avoiding lateral friction at the avoid frictional forces then we need to go at a constant rated speed so the driver will go at a constant speed now if the driver is going at a constant speed okay then what is the direction of the car's acceleration in that case the acceleration is almost normal because there is no acceleration in the tangential direction. Now when we go to the straight road we see a flat road in front of us we start accelerating because this road is flat the acceleration is in the in the straight direction now we again see a curve so what do we do at sea okay we start stepping on the brakes or slowing down the car when we start slowing down the car what happens there is a curve so there is also an acceleration in the inward direction and there is also a deceleration in this direction so overall acceleration of the car will be somewhat inverse here okay which will look like this and ultimately we come to this part of the road okay take a turn like this and start increasing the speed so what will be the what will be the corresponding acceleration this is the inward acceleration okay there is a tangential acceleration also so the overall acceleration will be approximately in this direction so typically given the state of the curve and given what we are doing we are going at constant speed or we are stepping on brake or we are accelerating we can have a decent idea of what the total acceleration direction will be okay so ultimately okay so we are coming to essentially the last part of this topic is we had discussed about kinetics in Cartesian coordinates we had discussed about kinetics in tangential and normal coordinates and ultimately now we are going to discuss in radial ER and transverse E theta coordinates and these kind of problems are very important why because if we know that this particular this arm okay this arm of a crane if it is for example we know that there is a rotation rotational motion that this arm can have around this now in this particular problem it becomes very straight forward why because this is the origin that we can choose and a motion acts in the and a motion of this will be in a circular direction so the coordinates are perfectly fit okay for radius and so the geometry of this problem is perfectly suited for radial and transfer coordinates okay now what do we have in this problem the same thing we had discussed earlier that sum of all forces in the radial direction okay on the particle will be equal to m times the acceleration of the particle in the radial direction but if you remember what we had done yesterday what is the acceleration of the particle in the radial direction it is nothing but r double dot minus r theta dot square now come to the theta direction or the transverse direction what do we have m a theta what is a theta is the acceleration component in the transfer direction that will be nothing but sum of all forces in the transverse direction which is nothing but m times a theta which is r theta double dot plus 2 r dot theta dot so these are the concept that we had discussed earlier and even in this coordinate system idea is really straight forward that on this mass sum of all the forces in radial direction is m r double dot minus r theta dot square and m a theta is equal to m r theta dot double dot plus 2 r dot theta dot so it is as simple as that once we recognize this is the acceleration from kinematics this is the corresponding force that a particle has then m times a is equal to the corresponding sum of forces acting on the particle now let us look at this simple problem a block B of mass m can slide freely on a frictionless arm away the problem is simplified a bit the arm is frictionless which rotates in the horizontal plane that the mass rotates in the horizontal plane y so that the gravity component can for the time being be neglected so this is r this is theta and z axis is coming outwards so essentially in this problem the gravitational force acts in the z direction and we can totally forget about that force we can only concentrate on this r theta direction without having to worry about the gravity now in this case what is known is that knowing that B is released at a distance r 0 from O that B to begin with at t is equal to 0 the position of this B was at a distance of r 0 somewhere here from the origin around which the rod is rotating what we are then asked to find out the component v r of the velocity along away okay so what is the radial component for velocity and correspondingly we are also asked to find out what is the magnitude of the horizontal force exerted by B on arm away if this is arm away okay there is no friction so this B cannot exert any force on this rod in the radial direction but it can exert a force in the transverse direction and we are asked to find out that what is the magnitude of that transverse force which the arm which the which this sleeve exerts on arm away so we write down the radial and transverse equation for the motion of this block we integrate the radial equations to find expression for radial velocity okay substitute known information to find out the expression of force on the block so what we need to first find out is we need to first find out there is no friction in this direction so this motion is purely kinematics okay this motion is purely kinematics and then once we figure out what is this kinematics we can then correspondingly find out what are the accelerations okay what are the forces acting in the transverse direction and we also know that the mass of the sleeve multiplied by the accelerations in the transverse direction is essentially the horizontal force exerted by this mass B on the sleeve OA so what do we do we write down equations because there is no friction okay we have seen that there is no friction between B and this arm the only force that can exist between this sleeve and and the rod OA is this normal force so let us say that the normal force acts in this direction we choose this direction now what are the accelerations that this can have so it can have two accelerations one acceleration is mA theta in this direction and the second acceleration is what AR so the corresponding force will be m times AR along the radial direction now just note this these equations what do we have f in the radial direction along this is equal to m times acceleration in this direction but there is no force acting in this direction so this will be free acceleration that 0 will be equal to m r double dot minus r theta dot square what is this quantity this quantity is nothing but the acceleration in the radial direction now look at it the second equation what is the second equation it is the equation in the theta direction what do we say that m times the acceleration in the theta direction is the force and what is that force f is equal to m r theta double dot plus 2 r dot theta dot now we do not know r double dot okay but do we know theta dot do we know theta double dot we know that why because we are told okay we are told that this arm moves okay this rotates at a constant rate of theta dot 0 so theta dot 0 is constant and whatever is theta dot for this arm has to be the theta dot for this mass b because it keeps moving okay along the theta direction it keeps moving along with the roll only motion that this one has different from the mass OA is the motion in the radial direction okay so we just write down that r double dot now is equal to vr dot okay what is vr dot is the velocity in the radial direction for this mass and that is equal to dvr by dt and we can just rewrite this as dvr by dr into dr by dt or vr dvr by dr okay just write this recognize that this is nothing but vr or the velocity in the radial direction so what we know is that that r double dot is vr dvr by dr but we also know okay what do we also know that this quantity is equal to 0 so what do we know is that that r double dot minus r theta dot square is equal to 0 we substitute that here okay that this r double dot which is vr dvr by dr okay is nothing but r theta dot square into dr just know this equation and this equation put together but we also know that this assembly OA okay the rod OA is rotating in the clockwise direction at a constant angular speed of theta 0 dot theta dot 0 so just substitute everything here what do we know we have a simple equation this is a constant we take this out integrated from 0 the mass starts at rest so what do we get that the velocity vr square will be nothing but theta 0 square into r square which is the current position of this mass minus r0 square so we find out what is the velocity of this mass as a function of the position and when r is equal to r0 the mass b initially starts from r0 we see that the velocity is 0 so we have what is the velocity r of b along OA as a function of r okay straightforward now what do we want to know we want to know that what is this force what is this force in the acting in the horizontal or the transverse direction it is nothing but r theta double dot plus 2 r dot theta dot but what is theta double dot this this is rotating at a constant angular speed and the mass has the same angular speed okay and as a result this theta double dot is 0 and we are left with f is only equal to 2 r dot into theta dot now r dot is this okay vr is nothing but r dot and theta dot is nothing but theta dot 0 so substitute everything here and we can find out that as a function of position r what is the corresponding horizontal force that this that acts on the sleeve and equal and opposite will be the force that sleeve exerts on the rod by Newton's third law okay so this is the problem okay in radial coordinates that we have radial and transverse coordinates that we have solved