 in an arbitrarily moving reference frame. So in the last class we were discussing about the derivative of a vector in an arbitrarily moving reference frame and the result that we saw that if you have 2 reference frames one is capital XYZ another reference frame is small XYZ where small XYZ may be having an arbitrary motion in terms of an angular velocity in terms of a linear velocity which may even vary with time and so on. So if you have the vector A which is there in the small XYZ reference frame the derivative of that with respect to the capital XYZ which is like a stationary reference frame is this plus omega cross A. Clearly if this small XYZ reference frame is having no omega then the derivatives are identical. Now what we will do is we will try to utilize this to figure out that what happens with respect to the velocities and accelerations of fluid elements located in a control volume because of the movement of the control volume. So now we are going to consider that the control volume is not stationary it is moving and when we say that the control volume is moving let us say that we consider the control volume to be such that small XYZ is attached to the control volume. So the way in which the control volume moves small XYZ motion represents that. So small XYZ motion has 2 important aspects one is the translated motion another is the rotational motion. So the translated motion is like if you have the origin of the small XYZ as this one then if you have the position vector say let us call it r0. The rate of change of this r0 with respect to time gives the translatory velocity of the small XYZ reference frame. On the top of that small XYZ reference frame is having a rotational velocity in general. There may be special cases when rotational velocity is not there. Now if you consider say a point in the control volume let us say a point p this point in general represents a point where the fluid has a velocity acceleration and so on. So it is a point in the flow field. So this point in terms of its position vector is described by this position vector which we say call as capital R. But if we are trying to analyze everything with respect to the small XYZ reference frame for us the important quantity or the important vector is the position vector of the point p relative to the origin of the small XYZ reference frame and let us say that is small r. So capital R is the position vector of the same point p relative to capital XYZ origin and small r with respect to small XYZ origin. So directly from vector addition we can write capital R is equal to r0 plus small r. Now what we will do in the next step is we will try to find out the velocity. So the velocity of the point p in an absolute sense is dr dt that is d capital R dt with respect to capital XYZ. This is the velocity with respect to capital XYZ. So this is just in a shorter notation we may not always write d dt but write a dot on the top to indicate that it is a time derivative. So this is r0 dot plus it is d dt of small r relative to capital XYZ. d dt of small r with relative to capital XYZ is d dt of small r relative to small XYZ plus omega cross small r from this theorem. So that means this is as good as r dot small XYZ plus omega cross r. So if you try to figure out that what are the implications of these different terms let us just try to understand. This is nothing but the velocity of the control volume. The control volume is having a translational velocity. This is the translational velocity of the control volume because this is the time rate of change of position vector of the origin of the control volume of the origin of the reference frame describing the control volume. And this is what? This is the velocity as visualized from the reference frame small XYZ and this is of course omega cross r. So these 2 terms together may be sort of thought of as a relative velocity. So this is the absolute velocity. This is the velocity of the control volume. This is the velocity relative to the control volume like that. So the relative velocity has one component because of the translation another component because of the rotation. And if we want to find out the acceleration is what is the so important quantity for us because by that we can relate with the new terms second law of motion. So acceleration with respect to small capital XYZ that means you have to differentiate this again with respect to time. So this is so let us write different let us identify the different terms and write the first term it will become r not double dot. Next r dot small XYZ just substitute that with a in this expression. So the DDT of this one is what we are looking for. What is DDT of this one that is DDT of r dot small XYZ plus omega cross r dot small XYZ that is for the second term. For the third term again the same formula we can apply capital A replaced by omega cross r. So that will be DDT of omega cross r relative to small XYZ plus omega cross omega cross r. So next stage is just the simplification and collection of the relevant terms. So the first term r not double dot plus small XYZ double dot plus you have omega cross r dot but let us simplify another term before writing that let us simplify this one. This you can differentiate just as if it is like a product rule but maintaining the order because for the cross product the order is important. So it is omega dot cross small XYZ plus omega cross r dot small XYZ just like the product rule. So if you collect now omega cross r dot small XYZ and this omega dot omega cross r dot small XYZ so it will become 2 omega cross r dot small XYZ then you have omega dot cross r small XYZ plus omega cross omega cross r. So if we just give the interpretation as we did for the velocity for the different terms what is r not double dot it is the acceleration of the control volume linear acceleration. So this is the acceleration of the control volume then what is r double dot relative to small XYZ? This is the acceleration of the particle under consideration relative to the control volume and then if you consider these 2 terms these are angular acceleration sort of that is if you have an angular acceleration effect so this is not directly angular acceleration effect but because of the rate time rate of change of the even a fixed vector because of the angular motion within the element. So if you have even a fixed vector because of the rotation that vector is getting changed with respect to time and this is directly because of the angular acceleration of the reference frame. So these are directly as a consequence of the angular motion of the control volume so the angular velocity and angular acceleration these are direct consequences of the linear this is the direct consequence of the linear acceleration of the control volume this is because of the linear acceleration of the particle relative to the control volume and this is the combination of the linear and angular motion effect. So as you all know this is called the Coriolis component of acceleration. So why where from this Coriolis component of acceleration comes? So if you have a reference frame that rotates and relative to that reference frame something translates. So the combination of that gives rise to an acceleration. So the 2 things which are necessary for this one is the reference frame should be a rotating reference frame. Another thing is that there should be a translatory motion relative to the reference frame and then it will experience a sort of acceleration and that acceleration will try to deflect the particle from its original locus and it is very common in particle mechanics also in fluid mechanics just think about ocean currents. So earth is like a rotating object and on the top of the and over the earth the ocean currents are moving. So you have the water moving in the sea with a particular velocity on the top of a reference frame which is rotating and that is why and the rotational sense is different at different places. So you will see that there will be a certain deflection of the ocean current in the northern hemisphere and in the southern hemisphere. These things you have studied in junior classes of geography but these are these are very important examples of the implications of Coriolis effects in fluid mechanics. Now what we can do is we can just write it in a bit of a more compact way. We can write that so this term we will write as a small x, y, z and all the remaining terms like this we know this is like a Coriolis and this is the angular velocity effect angular velocity and angular acceleration effect. So we will just write a capital x, y, z as if is equal to a small x, y, z plus let us give the all other terms and name a relative this is just a name that we are giving. So in the a relative we have a control volume plus 2 omega cross v xy small x, y, z plus omega dot cross r small x, y, z plus omega cross omega cross r, r is again small x, y, z. So you can clearly see that this type of this is basically a transformation from reference frame which is stationary to arbitrarily moving reference frame and when you do that the transformation terms appear in the form of the quantities which you are visualizing relative to the moving reference frame because when you are writing your equations of motion related to the moving reference frame all quantities you are measuring relative to the moving reference frame. So that position vector, the velocity vector all those things you are measuring so you can clearly see that when the transformation is made in the right hand side everything is written relative to small x, y, z. So small x, y, z has become your reference. Now how will this be important in the context of use of the Reynolds transport theorem? Let us try to go back to that. Let us say that we are interested to write the Reynolds transport theorem for linear momentum conservation. So Reynolds transport theorem applied for linear momentum conservation. Let us say that we are interested to write the theorem relative to this control volume. This green colored contour is the surface, control surface of the control volume and that is our basis for which we are writing the theorem. Let us first write the general theorem dn dt system. We recall that capital N is the quantity that we are interested to conserve. It may be a scalar, it may be a vector, small n is capital N per unit mass and this eta is the unit vector normal to the area that elemental area that is considered for writing the area flux term. Now one important thing is that we have no restriction on the choice of this capital N and small n. That is we are not restricted to write this only as quantities relative to an inertial reference frame. We may even write this relative to quantities in a non-inertial reference frame because when we derived this theorem we never had any consideration for a specific inertial or non-inertial choice of the reference frame. So we can as well write say capital N as say m into v small x, y, z as an example. Nothing restricts us from doing that because it is just some quantity that we were looking for. It may be linear momentum relative to any reference frame. So when we write that left hand side becomes d dt of m. Let us complete first write the right hand side and then we will see that whether the way in which we have written the left hand side is the proper way or not. We have to keep in mind that this v relative is nothing but v small x, y, z that is the velocity of the fluid relative to the control volume. So this is v small x, y, z dot. Now let us devote our concentration on the left hand side. When you look into the left hand side, see when we write this what is the assumption? The assumption is that the entire system is having the velocity v small x, y, z relative to the small x, y, z but that is how we have written that is the mass into this one because this is the total linear momentum of the system. So total linear momentum of the system if we write that as mass of the system times this velocity that means we are implicitly assuming that the entire mass of the system is having this velocity which is not correct because fluid is a deformable medium and in general you can have different velocities at different points. So it is not appropriate to write it in this way but write it as an integral form of like integral of dm v small x, y, z over the system. The reason is quite clear that at different points in the system it might have a different velocity. So when you want to represent the total linear momentum, physically we are representing the total linear momentum. The linear momentum consideration should keep in mind that there could be different velocities at different points in the system and then we may write this as dm as how do we replace dm in terms of the elemental volume. So this is the low into an elemental volume dv okay. Next step, so if we put the d dt outside then you have the d dt also. Next step, can we put the d dt inside? Yes or no? We have discussed about these things earlier. See what is the variable with respect to which integration is done? Volume. So if that volume is not a function of time then we may put it inside the integral without requiring to put any correction terms. So we are assuming that it is a non deformable control volume. So with a non deformable control volume we may write it as integral of say rho v small x y z d dt of this small dv. See this assumption is not a very restricted assumption because for most of the problems that we are considering this it is not so common to have a control volume which is arbitrarily accelerating plus deforming but I can give you a nice example if you are fascinated with mechanics try with this example. Say you have a balloon in the balloon you fill it up with water just throw it with a spin and make the water come out of that and try to figure out how the velocity of the water coming out is changing with time. So you require everything deformable control volume arbitrarily accelerating because the balloon when the water is coming out the balloon might be having arbitrary rotation and arbitrary linear motion and in an elementary level I mean this problem experimentally one can do but theoretically I would say it is one of the very tough problems in mechanics which will require the combination of the understanding of the fluid mechanics and solid mechanics and deformable control volume arbitrarily accelerating control volume it will test your understanding of mechanics to the best of abilities. Let us not go into that complication here so we assume that this is what we are writing next step is that we will assume that rho is not changing with time. So if rho is not changing with time we will just write this as integral of rho d dt of v small x, y, z. So this the other thing is that see magically we were as if we transformed it from the system to the control volume that we will not do immediately but we should keep in mind that when we are writing it in the limiting sense that is we are dividing it by delta t and in the limit as delta t tends to 0 we can write it in the limiting sense tending to a control volume because in the limit as delta t tends to 0 system tends to control volume that is what we have experienced earlier but that we will do in a bit of a later stage till now we will preserve it as a system sense why the next step will be giving you a clue of why we are trying to do that. So in the next step what we will do we will write the left hand side simplification this is again system you have to keep in mind that what are the assumptions that we made so you have the rho and what is d dt of v small x, y, z this is the acceleration small x, y, z what is appearing here. So this is rho acceleration small x, y, z dv system what is our trouble the trouble is that when we are having acceleration relative to small x, y, z we cannot directly apply the Newton's second law of motion for that because Newton's second law of motion is valid for an inertial reference frame small x, y, z in a special case when it is moving with a uniform linear velocity is still an inertial reference frame but in a general case not. So to address the generality we have to keep in mind that we need to first transform it to capital x, y, z so that for that we can use the Newton's second law of motion. So we will write this as integral of rho in place of a small x, y, z we will write a capital x, y, z minus a relative. So when you take this integral of rho a capital x, y, z dv for the system and minus then what are the implications of these 2 terms the implications are very great forward. The first term represents the mass into acceleration of the system as an effect the integrated the total effect mass into acceleration of the system as viewed from an inertial reference frame. So by Newton's second law of motion this you can write as resultant force acting on the system and since we consider the limit as delta t tends to 0 this is as good as resultant force on the control volume and again in that same limit you have in that limit this also the system tends to control volume keeping in mind the limit that we are considering for this time derivatives that automatically implies that the delta t is tending to 0. So what is the final form of the equation that we are getting let us write it in the final form. So the final form is the left hand side becomes what the left hand side becomes resultant force acting on the control volume minus let us say integral of rho dv you can say as dm elemental mass. So that is just like here also so it is like integral of a relative dm is equal to the right hand side whatever is there. So if you just look it look into it from a user view point or a formula user view point what is the correction that has occurred because of the transformation the correction has occurred mainly with one important thing that you have this correction term and this correction term is because of the acceleration of the control volume because if you see all the terms are related to acceleration this is directly related to acceleration of the control volume this wherever there is angular velocity you expect that it is related to that so angular velocity angular acceleration and again angular velocity. So these are all related to the nature of an accelerating reference frame. So if the reference frame is not of accelerating nature then or if the reference frame is not having an angular velocity also because having an angular velocity automatically makes it of accelerating nature at least the last term is important. So when none of these terms are important then this will be 0 we will see when and in the right hand side you see that nothing great has happened we have replaced in all the terms the absolute velocity by the relative velocity. So when you are writing the Reynolds transport theorem for linear momentum conservation then basically all the absolute velocities are replaced with the relative velocities this was anyway always relative velocity but the other terms these are now replaced by absolute velocity relative velocity and in the left hand side there is a correction because of the non inertial effect of the reference frame just like minus mass into acceleration this is just like a pseudo force. So with that understanding let us try to solve a few problems to understand that what we do for a moving reference frame in reality. Let us start with the problem of a similar type which we attempted in the previous lecture. So you have a cart like this on a friction less ground and there is a water jet which comes on it and which leaves the cart the angle change is theta relative to the horizontal and the surface over which it is changing its direction the water that is considered to be a smooth surface and this is also considered to be smooth the friction is 0 friction less. Now what we have seen earlier we have seen that because of the impingement of the water jet on the system and because of its change in direction there will be a force and because of the force the cart will have a tendency of moving. So we saw that one way of like restraining it from moving is like maybe making a pulley mass arrangement so that it is restraining from its motion. But in general that will not be the case so in general it will be free to move and therefore it will start moving. So we will consider a first case when it is moving but moving with a uniform velocity just as a simple case we will then move on to more and more complicated cases when it may move with an acceleration. So first let us consider that it is moving with a uniform velocity so this cart is moving with a velocity vc towards the right. The water jet comes out from an area A with a velocity v and it gets deflected like this and we are interested to find out what is the resultant force exerted by the water on the cart. So only difference from the previous problem that we considered is that now it is moving but this is constant it is not moving with a variable velocity. So let us try to write it relative to the moving reference frame. So the left hand side resultant force on the control volume minus the relative acceleration see what are the terms which are there. First is acceleration of the control volume. Here the control volume we attach the control volume on the cart. So that is we attach the reference frame on the cart small x, y, z reference frame on the cart. So small x, y, z is moving with a uniform velocity vc. So it is not having any acceleration, linear acceleration. There is no angular motion of small x, y, z that means the A relative term is totally 0. So that is 0 and the right hand side if you consider it to be a steady situational acceleration where the velocity is not changing with time anyway the density we are considering to be constant. So this term will go away and you are left with the last term that is rho integral of v small x, y, z into v small x, y, z dot eta. So if you consider say the flow boundaries say this is the flow liquid film that we are considering. So in this liquid film we know what is the velocity here v because that is what is the velocity on which it is impinging from a nozzle or some place. Question is what is the velocity here? Again you may neglect the height here by considering that the kinetic energy effects are much more important. Question is and a very big question is that if all the assumptions of the Bernoulli's equation are valid still can we apply the Bernoulli's equation from here to here directly. Remember it is a moving reference frame. Yes or no? See more and more these types of questions you answer yourself your fundamental understanding of that is clear. Many times it is commonly understood that Bernoulli's equation is one of the easiest things in fluid mechanics. To me it is one of the toughest things in fluid mechanics. It is very easy formula at the end but its restrictions are not well understood in many times. So the question is can you directly use it keeping in view that all the assumption that we have learnt otherwise are valid except we are now in a dilemma why we are now in a dilemma because this is itself moving. See although this is moving we are saved with one important thing this is moving with a uniform velocity. So this is still an inertial frame. So we have seen that in the Bernoulli's equation all the quantities are relative like pressure is relative you may express it relative to something. Potential energy is relative you may express it relative to some reference or datum the height similarly kinetic energy is also relative provided you are still using an inertial reference frame. So here since this reference frame is inertial you may have the velocity like the kinetic energy the relative velocity that is what is only of importance. So what is the relative velocity here? Relative velocity here is v-vc and that relative velocity will be preserved because the kinetic energy is preserved even in a reference frame which is moving but non-accelerating. So if you apply the Bernoulli's equation here then that v should be replaced by v-vc because you are writing it relative to the reference frame. So the thing is that when you write this expression here what will come out? So there are 2 boundaries flow boundaries 1 and 2 first let us write for 1. So for the flow boundary 1-rho okay first let us write plus-sine we will see inside rho then what is v small x, y, z here let us write it in a vector form. So that is v-vc i cap what is that remaining term again what assumption we are making that it is a uniform velocity profile over the thickness that we are considering. So this is as good as some relative velocity into area with a proper sign. So that means relative velocity is v-vc the area is a the proper sign is negative. So we put a minus here then for the outflow boundary what you have v small x, y, z it is v-vc in magnitude because kinetic energy only bothers about the magnitude. So that is preserved means v-vc is preserved but it has changed its direction. So when it is coming out the relative velocity is v-vc with this normal direction say n1 and what is this n1 cos theta i plus sin theta j. So this is cos theta i plus sin theta j. So keeping that in mind it will become v-vc cos theta i plus sin theta j and what about the v relative into a that term positive a into v-vc okay. Remaining work is easy you can separate the x and y components by taking the coefficients of the i cap and j cap and write what are the components of the forces. These are forces exerted by the cart on the fluid by Newton's third law the force exerted by fluid on the cart will be equal in magnitude of this one but opposite in sense right okay. So we have seen an example where the control volume is moving it will not make us very happy because the expression that we have derived we are not able to exploit that fully the entire relative acceleration term vanishes. So let us look into some example where it does not get entirely vanished and then what happens. Let us say that there is a block which is moving on a surface or which may move on a surface from a nozzle a water jet comes strikes this one and gets deflected in the 2 sides okay. The area of the velocity jet area of the jet that comes out is given the velocity of the jet that comes out vj that is also given aj vj area of the jet and velocity of the jet. The coefficient of kinetic friction between this surface and the block that is given and our objective is to find out an expression of if let us say u is the velocity of this block or let us give it a different name may be ub for the u block then how this u block is changing with time the mass of the block is given let us say mb is the mass of the block. Let us try to look into this problem with the framework of an arbitrarily moving reference frame. This is also a simple form of arbitrarily moving reference frame because this cannot have angular motion I mean if this may have angular motion again that becomes a very interesting problem. If you consider the extent the dimensionality of this and consider the toppling possibilities because of the force which is acting on it but if you just idealize it like a point mass type of a thing then it is having just a translation but it may have a variable translational velocity. In general it should be like that because before the jet is striking on it it is stationary when the jet is striking on it it will try to make it move so its motion from 0 velocity will come to a non-zero velocity over a finite time so there will be a finite acceleration. So it is expected in reality to be a accelerating reference frame if you choose your reference frame attached with the block that is your small x, y, z attached with the block. Now let us consider a control volume maybe let us say that we have the control volume like this which contains the block and whatever water is adhering to the to this vertical plate. One important thing we should understand what is that important thing the thing is that now let us say we now at least we are confident in one thing that we should be bothered about the relative velocity when we are writing with respect to the reference frame small x, y, z say which is moving. So all the expressions that we write in the Reynolds transport theorem should correspond to the relative velocity that much we have seen. So when we are bothered about the relative velocity now if you say that velocity here is vj there should be some relative velocity here as it is the water jet deflected in 2 parts one will go to the top let us say it goes with the top as vt v top and another goes to the bottom with the velocity v bottom okay just 2 parts of the jet. Question is is it so that v top in a relative sense is vj-ub assuming that other assumptions other restrictions of the Bernoulli's equation are somehow satisfied answer is no because it is now an accelerating reference frame. So kinetic energy in an accelerating reference frame cannot be in a straight forward way brought out from the Bernoulli's equation as preserved. So we do not know what are these in the previous problem we could exploit the situation that the control volume was moving with a uniform velocity now it is not moving with a uniform velocity. So we do not know actually what these are even in a relative sense but the good thing is that since these are vertical these will not have any consequence on the horizontal momentum transfer. So our ignorance or lack of knowledge on these will not be magnified this is a good thing for solving the problem but it is important to know that this is a very important ignorance it is not so easy to tell what are these velocities because this is an accelerating reference frame. Now let us try to apply this equation for the resultant force on the control volume. So for the resultant force on the control volume let us write it only the x component because that will give rise to the variation of velocity because we do not expect that this block will be lifted from the ground because of this. So now if you write the equation the force then what are the forces which are acting on this? So if you draw the free body diagram of the control volume or may be free body diagram of the block plus water whatever is contained there. So you have one frictional force. So in the vertical direction there is equilibrium of forces because the block does not get out of the surface. So you have n equal to mb into g now you have to keep in mind that you are not considering it just the block block plus some water but the water mass may be neglected it is very very small as compared to the block mass. So whatever mass of water is just like flowing in contact with the in contact with this plate that water mass it is so small as compared to this capital mb that for weight or mass calculation aspect we will not consider the water weight that is a very that is an assumption that we have to keep in mind. Otherwise whenever we write the total weight it should be the weight of the water in the control volume plus the weight of the block which is there okay but we are neglecting that weight of the water in the control volume which is a very valid assumption because it is a very small amount of water which is there that is very practical. So then the normal reaction will be capital mb into g because of the equilibrium along the y. So the friction force will be-mu k into mb into g there is no other force along x then what is this correction term this correction term here is important because you do not have the omega related things but you have acceleration of the control volume and here this integral will just become a relative into the mass because it is just moving like a rigid body. So all points have same acceleration. So this will be-mass of the block times acceleration is du block dt right. So it is clear why we could take this a relative out of the integral because all points on the block are moving with same acceleration and we could have been in trouble if there was lot of water in the control volume because then for water we cannot have such a rigid body type of assumption. Then at each and every point the acceleration would be different but we are neglecting the mass of water present in the control volume for writing these expressions. The right hand side again there is a very important thing the unsteady term is fundamentally not 0 but we can approximate it by 0 because again if you see the volume of the water that is present in the control volume is small. So it is not because that v small xyz is small but because of the volume of water which is there in the control volume is small then its time rate of change with respect to time is also very small okay. So these are certain important things see we drop this term for solving the problem in exam fine. When you drop the term and solve the problem in the exam if you solve it correctly we have no way to deny you from the credit but your conceptual understanding is not satisfied there. You must be convinced that why this term is dropped despite the fact it is an unsteady flow it is not because that this velocity 0 that is why you are doing it clearly it is coming out with some velocity but just because you neglect the effect of that volume of water which is there inside and the time dependence associated. And the final term will be what see here only the x direction is important because in the y direction whatever it goes the out flux so there are out flow boundaries but those are not along x. So this velocity if it has component along x then only that will remain for force calculation along x. So that will be rho A into what you can straight away write it what is v small x y z so vj-ub and then – vj or if you want to put the – sign properly then this times – of A into that is the flow rate vj-ub okay. So once you have that expression now it is quite convenient because it is a simple differential equation with variation of ub as a function of time what are the constants in this expression the constants in this expression are this vj so maybe let us write one more step. So what you have here is mb dub dt so if you cancel all the – signs in all both sides – rho A vj-ub square plus mu kmg that is equal to 0 keeping in mind that at time equal to 0 you have ub equal to 0. So it is a straight forward differential equation even if you do not solve it you can find out that if when it attains a steady state what should be ub when it attains a steady state see the initially the velocity is 0. Now because of the striking of the jet it will be pushed to a velocity and it will come to a stand still it will come to a uniform velocity that is that change in velocity will not be there anymore when the dub dt term will be 0. So it will come to a limiting uniform velocity which then will be considered by equating these 2 terms. So it is possible to find out that what should be that equilibrium velocity and that equilibrium velocity is quite obvious it is a situation that occurs when the driving force on the block is just sufficient to overcome the kinetic friction and then it is in equilibrium but it takes some time for the driving force to be in equilibrium with the kinetic friction till the time the velocity will be changing with time and that how it changes with time is governed by this equation. Now when we have considered this problem one important thing that when we have considered this theory as such one important assumption that we have made is that the fluid in the control volume is of constant mass but it might so happen that the mass itself is a variable within the control volume. So what we will do if the control volume is accelerating plus the mass within the control volume is a variable it is not a constant. So we will see that in the next class in the next class we will take up the cases of variable mass in the control volume thank you.