 Good afternoon. Remember we had done area moment of inertia. So, what I am going to cover now is the mass moment of inertia. Remember so far in you know dynamics we have solved problem involving particles. So, now for tomorrow actually what will happen you are going to solve problems involving rigid bodies. So, basically rotational motion of rigid bodies will come into play. So, therefore, we have to understand the mass moment of inertia that comes into play for many engineering problems that involve rotational motion of rigid bodies. So, as you have seen that when we have discussed the area moment of inertia we have said that area moment of inertia is really proportional to y square dA ok. So, it is dA multiplied by distance square from the axis and what we have learnt also that for static problem in case of statics we need that area moment of inertia for problems involving buckling, bending, deflection and so on. However, we have not really looked at I when it came to the static problem solving the forces, but in strength of materials type of application we mostly try to look at I that is the area moment of inertia. However, in dynamic problem mass moment of inertia will naturally come into play when we think of that a body is trying to rotate about some axis. So, what happens in this case we can think of in a from a very simple perspective. So, look at this let us say I have a particle of mass delta m very small mass delta m let us say this particle is trying to rotate about the axis A A prime. So, essentially let us say this particle is attached by some arm of again very small mass compared to this mass and what I will try to do I will simply try to give it a couple. So, I am giving it a moment therefore, what comes to an our mind is that this particle will try to rotate with a angular speed and therefore, it will have an angular acceleration. Now, main thing that comes into our mind is that what will be that angular acceleration that will really depend on the distance from this axis as well as the mass of this particle. So, it will basically if you look at it that when we go back to the Newton's second law then we see that moment is equals to I times alpha. So, that is rate of change of angular momentum and therefore, this r square m that comes into play. If we think of in other way that I have an angular speed that means angular rotation is there therefore, I have a velocity that is r theta dot. So, therefore, if we look at it I will have r theta double dot that will be my acceleration. Therefore, a little force can act here that is m r theta double dot if you take a moment of that force that will be m r square theta double dot that will act on this axis. Now, therefore, m r square that is going to be the mass moment of inertia. So, what it is doing basically as we see here that it is really trying to put some resistance to the motion. So, therefore, if we think of a rigid body a complete rigid body instead of a particle. Now, rigid body will be composed of small small masses. Now, each of these particle will therefore depending on their distance from the axis they will try to put the resistance and that resistance will be r square multiplied by delta m. So, if you think of small small particle each one will produce and you know resistance effect that resistance effect is going to be I. So, I will be finally equals to r square dm which we call the mass moment of inertia. So, it is basically the tendency to resist the angular motion about an axis about which the body is trying to rotate. So, now you think of if I calculate the r square dm for the entire body about an axis. Now, we can also get the radius of gyration by saying that entire mass is located at some distance k. Therefore, now we have i equals to k square m. So, again what it is doing it is telling me that there is a resistance coming into play in the form of k square m. Now, depending on wherever or which rather about which axis it has more k square over m it will it is going to produce more resistance about that axis. So, logic is going to be the same as we have done for the area moment of inertia. So, again I have the instead of area moment of inertia I have now mass moment of inertia the form also looks similar. So, it is r square dm again I have a radius of gyration which is square root of i over m where i is an arbitrary axis about which the body is trying to rotate. So, i is mass moment of inertia about that arbitrary axis a a prime. So, with this now what we can do we can simply look at that if we really look at a coordinate axis that coordinate axis is defined by x y and z right. Now, I want to calculate what is the mass moment of inertia about each of these axis right. Then we can do integral. So, for example, let us say about y axis what is the mass moment of inertia that will be r square dm where dm is this small particle right that we are looking at. Now, look at what is r. So, r would be remember the r is this distance that is the perpendicular distance between these two axis right which is nothing but x square plus z square. So, r square is now x square plus z square multiplied by dm and we have to integrate over the volume let us say in this case. So, to get the mass moment of inertia about the y axis. Similarly, if we do mass moment of inertia about x axis then remember it will be again r square dm, but the coordinate of this particle if we look at then it has to be y square plus z square right ok. So, about the x axis we are going to get the mass moment of inertia which will be y square plus z square dm similarly for the z axis. So, now again as we see we may have to do a double integration or triple integration depending on the problem. So, going back to when we are going to solve this we are going to see that how we are going to solve certain class of problems, but before that again important concept that comes into play is parallel axis theorem ok. So, what it is basically we know the let us say we know the mass moment of inertia about the centroidal axis that means the center of gravity. So, they are actually the center of gravity is g. So, we know the mass moment of inertia let us say about x prime y prime and z prime ok. Now we want to find out what is the mass moment of inertia about x y and z axis. So, they are parallel to the coordinate axis or rather the centroidal axis right. So, let us say I know the centroidal axis I want to shift here. So, how I am going to do let us say I do one of this. So, let us say I want to find out the mass moment of inertia about x axis. So, therefore, what we have to do we have basically y square plus z square d m right. Now, what is y? y is nothing but y bar. So, what is the y bar? y bar is distance of these two parallel plane what are those planes x bar z bar plane and x z plane ok. So, if we look at this. So, this is my y bar plus y prime square. So, y prime is the distance of that mass from the x bar axis ok. So, I have y prime plus y bar square plus z prime plus z bar square. So, that is the mass moment of inertia about the x axis. Now, remember when I expand this and write it what we are going to get I have y bar you know y prime square plus z prime square d m plus now this term what happens here this term right here these two terms will disappear what they are they are again the first moment of the mass about the centroidal axis. Remember we talked about in case of area moment of inertia the first moment of area about the centroidal axis they will become 0 right. Similarly, what we see here the first moment about the first moment of the mass about the centroidal axis. So, these two term will be again equals to 0. So, what we will eventually get we are eventually going to get now this one is basically about the centroidal axis. So, this is the mass moment of inertia about the centroidal axis that is I x prime bar plus you have m multiplied by y bar square plus z bar square ok. So, what is y bar again I repeat. So, this is basically this you can think of as if distance from the global centroid to the local centroid. So, this is the global centroid let us say about the you know axis O we are trying to find out and this is the local centroid of the body ok. So, we have to find out y bar square plus z bar square. So, these are basically distances of two parallel axis similarly we can find out what is the mass moment of inertia about y axis and mass moment of inertia about the z axis ok. So, the main idea is look at this is in you know kind of 3D configuration, but in 2D configuration it will be much simpler. So, if we go back to the area moment of inertia for planar bodies basically one of these components are going to be dropped ok. So, for example, if I just think of planar bodies in x and y ok then what we will not have basically this z bar term on those calculations ok. So, ultimately what we can say in general so in general what will happen let us say I know the mass moment of inertia about the axis B B prime. Can I transfer to the A A prime the answer is yes. So, I A A prime should be equals to I B B prime plus mass times D square where D is the perpendicular distance or these are two parallel axis. So, we have the distance between these two axis ok. So, this is in general what the parallel axis theorem is all about. So, now what is most interesting if I want to get the mass moment of inertia of thin plates ok. We have already obtained the area moment of inertia of thin components that means let us say we have a rectangular plate or let us say we have thin disks ok. So, we already know that what is the area moment of inertia right. Now how do I get the mass moment of inertia. Remember when the body is getting thin with a uniform thickness then ultimately if I want to obtain the mass moment of inertia about A A prime then again remember the formula is still valid so I have r square dm ok. Now remember since I have uniform thickness going on here therefore what we can suddenly do dm should be equals to rho, rho is the density multiplied by T multiplied by dA. So, dm is simply equals to dA multiplied by rho. So, therefore what we essentially have rho T is coming out of the integration and we have r square dA. So, now what is r square dA remember integral y square dA. So, this is nothing but area moment of inertia about A A prime axis. So, if I can calculate the area moment of inertia about A A prime axis I will simply multiply that by density multiplied by thickness rho T to get the area moment of inertia. Now the problem is very very simple because I already know how to get the area moment of inertia. So, for thin components it is going to be very simple as if we know all the answers, but when it actually forms a volume then we have to really work on r square dm integral and that we are going to see. So, similarly let us say I want to get the mass moment of inertia about b b prime axis. So, what will happen I will again have rho T i b b prime of the area. So, area moment of inertia of this plate has to be calculated with respect to b b prime axis and we know how to do it already. Finally, what would be my moment of inertia mass moment of inertia about the c c prime axis that mean body is now rotating about c c prime axis. So, what is the mass moment of inertia remember we talked about the polar moment of inertia that came under the area moment of inertia that was nothing but i x plus i y. So, basically we have i a a prime plus i b b prime that is going to be polar moment of inertia right which is j c of the area. So, we did that in the very first day. So, ultimately what we essentially have we essentially have rho T multiplied by i a a prime plus i b b prime of the areas. So, that will give me or in other words that will be equals to i a a prime that is the mass moment of inertia about a a prime and mass moment of inertia about b b prime. So, that is going to give me i c c prime. So, therefore for thin components I need not to do any extra calculation I can simply take help of the area moment of inertia that we have done in the past and we can get the mass moment of inertia about all the three perpendicular axis. So, let us take a quick example and this is known to all of us. So, suppose I want to find out the mass moment of inertia about three perpendicular axis of this thin rectangular plate. So, the rectangular plate let us say having a length of a and height b and it has thickness equals to T. So, ultimately I want to find out what is the mass moment of inertia about a a prime which is passing through its centroid. So, I know for sure that what is the area moment of inertia about a a prime that is going to be b a q divided by 12. So, I take that multiplied by rho T. So, I get m a square over 12 why because remember the mass of the entire plate is now going to be rho T a b that is the mass. So, volume multiplied by the density. So, similarly if we look at i b b prime what it is it is simply going to be now when you are looking at the area moment of inertia about b b prime what is the area moment of inertia about b b prime that is a b q by 12. So, if you take a b rho T that is equals to mass. So, ultimately m b square over 12 now it is following a nice pattern that pattern is mass moment of inertia about a a prime that is actually m a square over 12. That means the length which is perpendicular to the a a prime that square divide by 12 similarly what is i b b prime I am getting b square over 12. So, that is the length again perpendicular to the b b prime. So, what is i c c prime is simply going to be sum of these two which is nothing but m by 12 a square bar plus b square and we are going to see this very often tomorrow during the planar motion of rigid bodies. So, let us say for a thin disc. So, it is a circular plate very thin circular plate how I am going to get the i a a prime or i b b prime because I know that the area moment of inertia about any of this axis a prime or b b prime should be equals to pi r to the power 4 by 4. So, therefore, we have 1 by 4 m r square remember the area is pi r square. So, the volume is pi r square multiplied by T. Therefore, the mass is volume multiplied by the density. So, we have to factor that. So, rho T pi r square if I take that that will be mass that is the m total mass of this circular plate. Now, why these informations are very important for us is that when we are really going to study the volume mass moment of inertia of body having some volume then we are again going to take help of this small small you know formulas that we are developing. That means, we have to really integrate when we are trying to integrate we will see that the way we will choose the element is going to be either a thin disc or a rectangular plate such that I can create a volume out of this and I am going to get the mass moment of inertia of those volumes. Is that clear? So, if we move to the next one let us say very common problem that we see in many applications would be let us say what is the mass moment of inertia of a slender rod of length L and mass m with respect to an axis which is perpendicular to the rod and passes through one end of the rod. So, it is a slender rod you can say that mass is uniformly distributed. So, you can say that you know you can take a mass per unit length actually. So, if you say have the density rho and the area is known. So, rho multiplied by area that will be mass per unit length. So remember therefore what we can do here I can I can just integrate it along x axis. So, I am going to take a small dm you see that element. So, that element is small dm. So, I am going to apply the formula now r square dm. So, in this case therefore it is x square dm. So, what is my dm? dm is nothing but if I know the mass per unit length that multiplied by dx. So, therefore what will happen? I can calculate either I z or I y it does not matter about which axis it is rotating. So, the question was any axis that is perpendicular to its own axis. So, therefore either you take I y or I z it does not really matter for this problem they are going to be same. So, if I choose I z let us say. So, you have r square dm. So, that is going to be x square m bar dx. So, this is the dm. So, m bar is the mass per unit length let us say total mass is m since the bar is uniform. So, we can simply say total mass is m divided by l that is the mass per unit length. Now this is going to be very helpful throughout from now onwards. So, m l square over 3 that is the mass moment of inertia about either y axis or z axis. Now remember if I want to get the mass moment of inertia about its own center of gravity about these two axis of course. So, this what I have I have to apply the parallel axis theorem. So, the point is I would have written I z should be equals to I c g z plus d square m. So, what is d that d is equals to l by 2 square and the mass is the total m. So, ultimately in this case now I am going from z axis to the centroidal axis. So, ultimately the mass moment of inertia about the centroidal axis is equals to I z minus m l by 2 square. So, m l square over 2 l. So, both are going to play a very important role when we look at the rotational motion of rigid bodies. So, for example, you know sometimes you may see some kind of pin connection here and body will try to rotate about the pin or you will may be find that there is some pin connection at the end body will try to rotate about that point. So, those problems we are going to study. So, ultimately remember that is how I have chosen the small element. Now you can think of more complex problems. How do I get the moments of inertia of a 3D body? So, that is our final objective because some of the same shapes we always encounter. For example, it could be a rectangular prism that means a rectangular block that is one thing. It could be a cylinder that is other thing. It could be a cone and it could be a sphere. So, for all of these we really want to find out what is the mass moment of inertia about the 3 perpendicular axis passing through the center of gravity. So, let us look at how we are going to solve that class of problems. Usually the bodies that we are trying to examine they will have symmetry about 2 planes. For example, this shape has symmetry about what are the planes it is exhibiting the symmetry xy plane or xz plane. So, if the body exhibits symmetry about these 2 planes then what we can do suddenly we can actually instead of doing a triple integration or even double integration we can actually resort to just a single integration. Now, how that is possible? That is possible by choosing a thin slab perpendicular to the planes of symmetry. You see this thin slab will be chosen that is a disc thin disc of length or thickness dx. So, dx is going to vary along x and remember that this you know its surface is always going to be predicted by some xy function that we always know. So, we can basically analyze the mass moment of inertia of these type of shapes with respect to x, y and z axis. So, what is most important that when I understand that what kind of you know disc or let us say you know what kind of element I am going to choose likewise we have done the similar way in case of area moment of inertia we have really chosen thin strips and so on and so forth. So, in this case since it is a volume integral and I want to convert it to a single integral along x. So, I am choosing an area that area multiplied by dx is giving me the dm. So, now let us look at it that what I know I know already the mass moment of inertia of this thin disc about its own axis, own central axis. So, as we have done in the previous slide we have di x. So, di x is the mass moment of inertia about x axis of this thin disc what the what is that that is m r square over 2 that means dm r square over 2. Now similarly I also know what is the di y prime that means when the thin disc is trying to rotate about the y prime axis. So, what is di y prime we have done that we have already done in the previous slide if I go back right. So, that is going to be m r square over 4. So, m r square over 4 so in this case therefore what we have is dm r square over 4. So, that is the mass moment of inertia about its own axis that is the y prime axis. Now I want to find out what is the mass moment of inertia about other axis which is y axis here. So, therefore, I apply the parallel axis theorem. So, the parallel axis theorem will be di y prime plus x square dm. So, ultimately we have dm r square by 4 plus x square dm. So, as if I am looking at i y prime plus m d square. So, that d square here is x square similarly what is my d i z d i z is simply d i z prime plus again x square dm. So, m d square. So, once I know this and I know what kind of shape it is I need to now know the mathematical function involved to define r let us say I have to now define r with respect to x or some other parameters that is given. So, once I know that what is the r in terms of x I can simply integrate it from 0 to some length. So, that is how the procedure goes in case of a three dimensional body. So, let us look at to start with a very simple example find the mass moment of inertia of a rectangular prism or rectangular block. So, the rectangular block has a length a breadth c and height b. So, the total volume is a b c and the mass is rho times volume. So, that is known. So, therefore, the answer that I am expecting let us say I am trying to calculate what is i z the final answer is going to look like this. Now, how do I proceed to solve this problem again how I am going to choose those thin slabs we said thin slabs right. So, what would be my thin slab here again that thin slab is going to be a rectangular element right with a thickness dx. You see again it is a rectangular area multiplied by dx. So, that is my thin slab now. The question is for this thin slab I already know what is the mass moment of inertia about its own centroidal axis. So, for example, I know what is i z prime of this slab. So, therefore, what we have here if we just go step by step what is d i z prime? d i z prime is going to be b square 12 multiplied by m right m b square over 12 remember I we already studied that for a rectangular area if I want to find out about the z prime axis it is always going to be b square by 12 b is the perpendicular you know length with respect to the z prime axis. So, m b square over 12 would be the mass moment of inertia of this thin slab. So, therefore, now it is very simple. So, ultimately we have to transfer this d i z prime to i z. So, I want to find out i z. So, what is d i z is equals to d i z prime plus x square d m. So, m d square now we are ready to do the integration along the x. So, all we have to do really we have to integrate this from 0 to a. So, once we integrate it from 0 to a. So, what would be the final answer? Final answer will look like this remember rho times a b c is always going to be the mass. So, the ultimate idea is that although it is a 3 d problem I am converting to a 1 d problem by choosing a thin slab. Now, that thin slab could be a rectangular block small rectangular block or may be a thin disc. Usually you know most often for the common shapes we have to handle the problem based on this kind of thin slab such that a one dimensional integration can be carried out. Now, if we go to a little bit of complicated problem let us say we take a bit complicated problem of that of a sphere. Now, can I do this? See remember the sphere has axis symmetry about any axis it has really axis symmetry about any axis you choose. So, I have again chosen a very thin disc that thin disc let us say is located at a distance z from the y x plane x y plane it is located at a distance z. So, remember if I can find out what is my let us say i z then i z should be equals to i x should be equals to i y. So, now we know that what is the i z of this thin disc i z of this thin disc i z of this thin disc is nothing but m r square over 2. So, that means d m y square over 2 again I am simply trying to operate it along the z axis. So, my integration limit will go from minus r to r and I have to integrate you know with respect to the d z. So, now let us look at what is my d m d m is nothing but rho d v. Now, what is my d v what is this volume? Remember that volume is determined by pi y square d z. So, pi y square is the area of this thin disc multiplied by the thickness is d z. So, therefore, we have rho pi y square d z. Now, remember I have to also express now y in terms of z that is very simple because you always get a circle in this case. So, ultimately what relationship exist between y and z? I have always r square should be equals to y square plus z square. So, r square equals to y square plus z square that is the relationship you know exist between y and z. So, therefore, my integration becomes very very simple. So, now look at the integration half rho pi comes out integral y to the power 4 d z. You go from minus 4 to 4 minus r to r. So, now y is replace as I said y square should be equals to r square minus z square. So, we do this integration the ultimate outcome will be this 8 by 15 rho pi r to the power 5 that is the i z or i y or i x because it has its axis symmetry problem. It is a 3 fold symmetry. So, in this case what we have found out now suppose I can replace now rho density can be now replaced. So, the density is nothing but m by v. So, the mass I want to always I always want my answer in terms of the total mass. So, total mass by volume volume is also known 4 by 3 pi r cube. See if I simply substitute for the density right then I am going to get the answer 2 by 5 m r square. So, it remains very simple problem now to work out. So, if we understand that I have to always take this thin disk or let us say rectangular you know elements of thin slabs then we can always perform this integration in one direction. So, what we have really studied I have looked at remember with the concept that is developed we have learnt how to find out the mass moment of inertia of these kind of shapes with respect to its own central axis. In fact, now we can look at any other axis because any other axis can also be obtained by using the parallel axis theorem. So, if I apply parallel axis theorem I can get the mass moment of inertia about any other 3 axis. So, knowing this remember once I know these solutions I can always now go to some of the you know shapes that encounter in machine components. So, knowing the you know solution for this I can now go to composite section and we have done in a very similar way remember for the area moment of inertia. We have also studied how to do these things for composite sections right composite plates. So, therefore in this case we can also you know take the similar concept and apply the parallel axis theorem to get the mass moment of inertia of any complex shapes. So, just look at this problem now this is body usually you will see you know in machine components it may rotate about the x axis. So, let us say rotation is happening about x axis and I am planning to find out what is the mass moment of inertia about x axis. Similarly, I can also find out what is the mass moment of inertia about y and z axis. So, how do I go about this? As I said that we have to really divide it into 2 cylinders because I know the solution for cylinders about their own axis own centroidal axis right. So, 2 cylinders I can choose and I can choose this rectangular prism. So, I can break this into 3 different volumes and each of these volumes centroidal axis mass moment of inertia about their own centroidal axis is known. So, I just have to transfer those to the x y and z axis that is some other axis about which the body is let us say rotating. So, I will apply the parallel axis theorem in this case let us go step by step. So, once this kind of problem is given first thing that we will do again break it into parts and for each part I will do it separately. So, for each part I am going to first calculate what is the mass. So, to calculate the mass I need the volume I know the specific weight density if density is given multiplied by g. So, I get the specific weight you can also calculate in terms of kg per meter cube no issue. So, you can either do the density or specific weight if you want to go to Newton but you know it is always advisable to do it in terms of mass. So, mass is always taken. So, mass is there. So, therefore you have the total mass is obtained and then what is my I x. Now I x to get the I x of this cylinder remember I already know what is the mass moment of inertia about its own centroidal axis which is parallel to the given x axis. So, that is m a square over 2 from the formula that we have learnt for this step. So, m a square over 2. So, a is nothing but r square right which is 1 plus then you have to transfer this axis this is the centroidal axis to this axis. So, what is the perpendicular distance remember in this case I do not have any z prime because these 2 axes are parallel to each other. So, I have only y bar. So, y bar is simply this distance. So, that is 2. So, therefore I can get the I x of this shape about this axis. Similarly, let us say I want to calculate I y. So, first thing to understand what is the I y about its own centroidal axis that is passing through you know that is parallel to y axis. So, I am trying to calculate what is the I y of this shape which is already known. So, m 3 a square plus l square divided by 12. Now again we have to transfer this. So, apply the parallel parallel axis theorem remember we just have m d square. So, d will become just x bar square because again z bar is 0 because these 2 axes are absolutely parallel to each other. So, I again get the I y, but remember when I want to do the I z. So, if I want to calculate the I z remember the centroid of this body and the you know the coordinate that is given the origin that has a offset of z prime sorry that has an offset of x prime and y prime. So, what will happen in this case you now have m x bar square plus y bar square y x bar square plus y bar square. So, you have x bar here then you have y bar. So, we have x bar and y bar. So, that both components are coming into play. So, there is no z bar. So, z bar is 0 therefore, it did not appear in this for I x and I y, but for I z I have both x bar and y bar. So, which is nothing but m d square that d is basically distance between this individual centroid to the origin. So, finally, I get I z. So, now this is for just one cylinder similarly how about the prism that is very simple because x y z itself is passing through the centroid of this prism the way the problem is given. So, we can do this very quickly. So, we know what is I x I z and what is I y. So, those are all known. So, then what it remains it remains just to add all of this whatever we have done. So, we are going to really add the mass moment of inertia of the prism let us say I take the I x. So, I x of the prism plus 2 times of what I have obtained for this why 2 times because there are 2 components attached. So, 2 times of I x of this remember that I x means it is calculated already about this axis. So, therefore, I can do also for the I y and I can do for the I z. So, ultimately the problem is solved. So, just using the concept of parallel axis theorem. So, remember everything kind of remains same as that what we have learnt in case of area moment of inertia. All the features remain same, but here only thing to remember understand that when it gets to the volume in the parallel axis theorem always as m d square where d is the distance between the centroid to the origin of that coordinate axis that we are referring to. So, therefore, we will always have 2 components we may have 2 components either x bar y bar x bar z bar or y bar z bar. So, we have to be careful about that when we are doing the mass moment of inertia for volume. So, I think with this we can have discussion. So, let us see. So, we have completed the mass moment of inertia for simple shapes and as well as for the composite section. Now, we will go to the tutorial before that let us have few questions as you have. There is a question came up please explain the product of mass moment of inertia. Remember the product of inertia we have not done also for you know in case of the area moment of inertia. So, we are not going to really touch upon the product of inertia at this point. 1 2 6 0. What is practical significance of radius of gyration? Please explain. See radius of gyration we have studied in terms of the area moment of inertia also. So, think of this let us think of state static that will be much easier to understand. I am taking something that I have in my hand just excuse me for that. Let us say I take a body this is my body. I am trying to understand think of this is kind of long and let us say it is a column. Now, I am trying to apply a load and I am trying to understand which way it will buckle. So, buckle means it will tend to bend this way or that way. So, what radius of gyration helps me understand is which way it is trying to buckle fast. Can you tell me from this which way it will try to buckle? See it is always square root of i over a right and what we assume that mass sorry the area let us say area is concentrated at a you know certain distance right. We are saying that square root of i over a as if the total area is concentrated about some point some distance right from the x or y axis. So, if you calculate what it immediately tells you that it will have tendency to buckle about this axis. So, it will have always higher tendency to buckle about this axis compared to if you choose this axis it will have more resistance to buckling. Remember this is always going to be square root of i of there is a competition between i and a. Now area is constant in this case right area is same. So, which way you have the more i there you are going to get the less you know more resistance. More i means that is that direction is always going to give you the more resistance. So, I am always going to get a more resistance about this. Similarly very similar way if you think of the mass moment of inertia okay. So again if you think of 3 axis right which way it is going to give you more resistance. So, it is basically which way it is going to more resistance that has to come in natural sense okay. So, it is basically dictating which direction it has it is going to give me more resistance. So, in static problem it is bending static problem it is bending, but in dynamic problem it is rotation okay. Yes area is irregular shape then how we will take that area is squeezed it is at some distance from reference axis and what is it is a practical significance. See there are practical that is why I am trying to understand you if you just try to visualize it is very difficult you have to do some computation that computation is square root of i over a and I cannot help it you have to do it okay. After doing that computation you immediately know that what is you know which direction you are going to get the more resistance. Visually if you have a irregular shape it is very difficult, but if you have a regular shape like the way I have you know shown this problem it is very easy to understand which way it is going to give me more resistance either to buckling or either to rotation. Is that clear? Sir you define last time moment area moment of inertia as resistance to bending and then you also said it is resistance to bending and resistance to deformation, but if you look at that expression m by i is equal to sigma by y there stress is equal to m by i into y that i term is a constant even you take stress at the top or neutral axis wherever so if it is resistance to deformation so that value should vary according to the position. Better what see your question is is it really resistance to deformation or resistance to bending right see ultimately what you are doing here see remember m is proportional to y square dA right why it is moment of inertia first of all why it is moment is proportional to y square dA now remember if you think of even a bending problem right ultimately what is that bending moment I have a beam I apply the load the bending moment at a cross section is what I calculate the bending moment from the load right so that bending moment is balanced by the y square dA effect right so ultimately we say it is a internal moment beam so ultimately it is coming through bending so I have been both if you look at the deformation as well as bending i is always coming down in the denominator sigma equals to m y over i you think of bending it is let us say cantilever beam with a tip load p you have p l q over 3 e i that is the deflection i is always coming at the bottom no question about it and remember bending is leading to deformation it is the bending that is causing the deformation though the both are interrelated so do not mix this mix them up but ultimately what is happening in both cases right y square dA is the moment that is controlling the both because if you look at the deflection equation e i d2 v dx2 equals to mx that is how your problem start right your problem start really e i times curvature equals to m is in that so that m is always y square dA is that clear now okay so it is the same I mean meaning see unless otherwise you have bending you cannot have deformation also so it is always y square dA okay. One more question block is kept on a horizontal plane how to achieve constant velocity your question is how to achieve constant velocity I have no answer I mean somehow you have to apply the forces you do if you do not have friction force right then you know it is it is not going to be I think it is very challenging how to achieve the constant velocity see at the first part you know first part you may have different resistance but ultimately what you have to do the resultant force equals to 0 ultimate what is it the resultant force if I can show that is equals to 0 right it will achieve a constant velocity right think of this way let us say you go to the shopping mall right you have those carts in the shopping mall we have the carts right and on below you have wheels right and I do not know if that problem was studied in friction actually what happens that you have those wheels comes with some kind of locking mechanism okay so when you push the cart let us say one problem could be you push the cart right wheel is locked and wheel is on wheel is unlocked when wheel is unlocked and once you actually you know let us say your friction barrier is gone you can actually go under a constant velocity so it is not that it cannot be achieved it cannot be it can be achieved but that scenario is bit complex okay to physically explain that why how the resultant force is 0 because your caster if you look at those wheels they are not going to get any friction force actually and I think that was studied by you know professor enam that must have done this in the friction problem we had actually a problem like this okay see ultimately resultant force has to be 0 on the body okay to maintain a constant velocity.