 Hello students, myself, Siddeshwari B. Tulsapure, Associate Professor, Department of Mechanical Engineering, Walsh and Institute of Technology, Sholakpur. So today we are going to deal with the topic design of flywheel, learning outcome. At the end of this session, students will be able to derive a formula which will be useful for deciding the dimensions of the flywheel. Now, in case of the earlier, we have seen that the flywheel it consists of the outer part it is the rim and then at the center we are having the hub which is having the hole and then the hub and the rim these are connected with each other with the help of the spokes or the arms. Now in case of the flywheel, so the mass moment of inertia it is required and now in case of the rim as it is placed on the outer side, the contribution of the rim to the energy storage it is more. So in case of this one, so here in case of this diagram, we are only shown the rim. So this part it is rim, so from here to here it is and now the rim it is having thickness equal to T. Now let us first define the different parameters and then we will go for the formula which we are going to make use for the determining the dimensions of the flywheel. Now for a flywheel, let R is equal to mean radius of flywheel then we will have A is equal to cross section area of rim, B is equal to width of rim, T is equal to thickness of rim. Now here we observe that cross section area of the rim it is A, so it is going to be perpendicular to this one, the perpendicular to this plane we are going to observe that one. So here in this plane the thickness will be seen, thickness of the rim and perpendicular to this one we are going to observe the depth of the rim and then we are going to have the depth it is nothing but it is the width of the rim and in case of the area cross section area of the rim. So we are going to get this one as it is nothing but it is B into T's. Now next to that one we are going to have say omega is equal to angular velocity then we are going to have say it is sigma is equal to tensile stress then we will have rho as equal to this density of material. Then we are going to have now the consideration of one element. Suppose this hatched portion it is now the element and now it is this element is lying at an angular position of theta from the xx line which is horizontal here. So the element it is making an angle equal to it is d theta say it is this angle. Now next to that one we are going to have the cj of this particular element and then the centrifugal force it is acting on this particular element. Now as the element it is smaller let df be equal to it is centrifugal force on element dm is equal to mass of element. Now from the formula of the centrifugal force we are knowing that df is equal to mass it is dm it is mr omega square it is mass into radii of rotation into it is angular velocity square but dm is equal to that is the mass of the element is equal to. So we are going to have volume into density so in case of volume again it is area of cross section into the width into density again we are going to have the cross section area as a it is now in case of this the cross section area here in this view it will not be seen so it is now this area is only seen as a line here. So when we are going to have the rotation of this area that is this one about the center we are going to get the volume generated with the help of this area. So now the area has to be moved so this area which is seen as a line it is to be moved from this particular point to this particular point and during that one it is covering this small arc it is. Now this arc it is having radius equal to it is r and it is subtending an angle equal to it is d theta. So in case of that one we are going to have the width of this one it is now it will be equal to r into it is d theta into the density we are having it is equal to rho. So this one is now is equal to it is df is equal to so it is area into the width here we are observing width of this particular element so it is r into it is d theta then it is rho into the remaining part we will have it is r and then it is omega square. So here we can observe that here r it is appearing twice so this one will be equal to so this df is equal to a into it is r square then we will take omega square here and then we are going to have the density and then it is d theta. Now here in case of this centrifugal force the components of this centrifugal force we can take which is one is horizontal and the second one it is vertical so it is making this df it is making an angle equal to it is d theta with the horizontal so the front side of this one that is the vertical component it is going to be df it is sin theta. So vertical component of df is equal to a r square omega square omega d theta into it is sin theta. Now total vertical force is equal to so now we have to go for the integration of this element now we are going to have this a r square this omega square rho into the integration sin it is sin theta and it is d theta. Now we have to put the limits to the integration so we have to think of so what should be the lower limit of the integration and on the upper side what we have to put here we can think of 0 then it is 0 to pi by 2 0 to pi or 0 to 2 pi or it is say even not starting with 0 it is starting from pi etc. So we are interested in the total vertical force where it is going to have only vertical upward sin it is. See here from here to here if we are moving for this upper portion above x we are going to get the vertical upward direction here if the element it comes into this particular portion the vertical downward force will be there so we are going to put the values as 0 to it is pi and this total vertical force this will be equal to it is a r square it is omega square rho into we are going to have it is minus cos theta then it is 0 to pi. So if we are solving this one we will get this one as 2 a r square omega square it is rho. Now the resisting force is equal to so here this resisting force is equal to so resistance will be provided by the material and here at 2 positions we are going to have so the tensile stress or hoop stress will be there and at 2 positions so it is 2 into the p it is there so this will be nothing but it is equal to 2 into that sigma which is say tensile stress or the hoop stress into the cross section area. So this one it is number 2 and this one it is equation number 1 so equating 1 and 2 we are getting this one as 2 a r square omega square rho is equal to this 2 sigma into it is a 2 2 will get cancelled and this will get cancelled so we are going to have this one as a r square omega square is equal to sigma but r omega we are knowing that it is linear velocity it is v square is equal to sigma upon it has been cancelled here and it is rho and so it is rho into it is r square omega square v square is equal to sigma upon rho v is equal to it is under root sigma upon it is rho so v is equal to also it is pi dn upon it is 60 from this one the diameter can be determined so diameter we are able to determine next these are the references which are used thank you.