 Okay, so this is the solid cylinders moment of inertia, just right rod, okay and you need to find out moment of inertia of this rod about the axis which passes through the end and perpendicular to the rod, the axis is this axis, so how do you rotate like this or it should rotate like that, this is also that axis about which it rotates, find out moment of inertia, the magnet exactly same the way you have derived the center of mass for the rod earlier, so if you get answer which is more than mL square or equal to mL square then that is wrong, doing nonsense, in the case of this the mass was distributed on the surface, so that is why in the case of this you have taken mass divided by area because it was distributed in area, now it is distributed in the length, that is why mass per unit length, okay, so when you substitute it x square into m by its constant comes out of integral, what is integral x square? x cube by 3, 0 to l it will be l cube by 3 or 1 l from denominator gets cancelled away, so mL square by from one of its ends, so that is at this moment of inertia is important, now find out moment of inertia about n perpendicular to rod, same mass m total length is l, x square m by l the limits, what is the distance, yes or no, you decide and then m by 2 is m by 2 to plus l by 2, to cover the, what is the answer mL square by 2 is the answer, minus l cube by 8, so you are going to get mL square by, so what do you have to remember it when you solve problem, okay, so it passes through the rod, the rod is 0, moment of inertia about 0 from the axis, passing through the center exist which are perpendicular to the rod, how many, is the answer mL square by integration, okay, so z axis this is your x, suppose there is a small mass dm, this is dm mass, where the mass y and z, x comma y comma, is the center of the sphere, okay, now let us say I want to find out moment of inertia about z axis, I z, okay, I z for this small mass should be what perpendicular distance of the small mass, square times dm, yes or no, now tell me what is the perpendicular distance of this mass from the z axis, what is the perpendicular distance of this point having x, y, z coordinate from the z axis, suppose this is and y, okay, here is a point whose coordinates are x distance from the floor, this is z axis, I want a perpendicular distance from this line, so perpendicular root over moment of inertia of the dm mass, let us call it as di because just for the small mass, so di z will be equal to root over x comma y square of the whole square, so x square plus y square dm, right, any doubts here, now there will be particles, right, so you have to add up all that to find out total moment of inertia, so I z is summation of xi square plus yi square times dm, this is the moment of inertia of the entire sphere about the z axis, any doubt, now tell me what should be ix, ix will be what summation of yi square plus zi square times dm, i, y will be x square plus z square times dm, you can integrate here also, summation integration is same, okay, listen here, I want to tell you how to fit without integration also, now is i z, ix and iy all are equal, it is a sphere, any line passing through the center is same in all aspects, right, so ix is equal to iy, so if you add it up, i z is equal to what two times of summation, please do it your own, x square plus y square plus z square times dm, do you get this, distance of the surface from the center what it is, right, so this entire bracket constant for whichever x, y, z you take, getting it, so you will get three times of i z r square dm, so r square is constant, so comes out three times, i z is two r square summation of dm, what is summation of dm, total mass, so i z is, is r square, this is a distance of a point from the center, why did you do it, for these three i's, see if it is a sphere, moment of ratio about any line passing through the center should be same, because this axis is no different from that axis, it is a sphere, it is three dimensional symmetrical figure, then i z, i x and i y all three equal, due to symmetry all of this comes, okay, now you can choose to integrate also, the way of this location of a hemisphere, similar way you can follow and get the answer if i am integrating, that is the homework, please write it down, the homework is to determine the moment of inertia using integration for the hollow sphere, integration is easier or this is easier, but this is tricky actually, integration is straightforward, when you learn it, it becomes simpler, okay, so the homework is to find out moment of inertia of the hollow sphere using integration, how you integrate, you assume a ring, here you assume a ring like this, the way we have done it for hemisphere, okay, and you take angle theta like this and this is the theta, you remember, we got the radius of this ring as r, okay, so you can do mass per unit area, try to do it your own, okay, next is solid sphere, this is the last one, there is a moment of inertia, they will not ask you to derive it, integrating and all that, okay, but these methods you must understand how to do it, because J e likes to test you on the basics, they will, the mass changes with length and find out moment of inertia of the rod now, so if you know how to exactly solve it, you will have a question, solid sphere, but actually this is r and the width is gr, can you find out integral of d, so every which we did to find the central mass of the solid hemisphere, same, how you find dm is same, and it is same not only for this chapter, it will be same for some cities, whatever you have done, what is dm? 4 pi r square d, when you multiply it the volume of dm you have to take, which is surface area, pi in pi goes dm by rq r square dm, this is my dm, so here you can substitute by 5, so you will get 2 square, 2 by 5 of mass okay, so we will shut it out, so till now we have done some basic integration to find out the moment of inertia of the some regular shape objects, but what is the issue, the issue is the moment of the issue is that moment of inertia of different axis is different for the same object, okay, so what different axis is and find the moment of inertia, so there has to be a way to find out going moment of inertia of one of the axis, so there should be a way, find for that we are going to learn two theorems which will relate moment of inertia of one axis with respect to the moment of inertia of other axis, very straight forward simple theorems, please write down I am going to done this topic, otherwise just 5 minutes topic I should not leave you to the next class, okay, let me finish this off, please write down parallel axis theorem, okay parallel axis theorem, so draw a rigid body like this, this is a need not be a regular shape, fine, this is a rigid body, this is an axis that passes through the center of mass, this is an axis that passes through the center of mass and moment of inertia above the center of mass is ICM let us say, okay, there is another axis, by the way the axis need not pass through the object, so this about this axis all the axis are parallel to each other, if I term ICM and I are parallel to each other and the distance between them is D, parallel to theorem I will be equal to ICM plus total mass of the rigid body into please write and what is it ICM is the least, so moment of inertia or inertia center of mass axis, fine, but one thing is sure if you are finding the least moment of inertia of all the axis possible it must pass through the center of mass, okay, there is a small proof of this, let us assume this is your z axis x and let us say this is y, whose coordinates are x, y, z, let us say mass is small m i, x i, y and z i, its perpendicular distance from the z is what, root of x square plus y square, so moment of inertia z axis which is also ICM is summation of right now x i, x i square plus y i square, yes or no perpendicular distance is root over x square plus y i square, okay, this is the parallel axis to the z axis, this axis z axis, so moment of inertia about that axis what do you think this will be summation of m i into what will be its perpendicular distance from this axis, the axis is displaced what d, perpendicular distance its x coordinate will be reduced by d, yes or no, so its x coordinate, if d is exactly equal to x the x coordinate for this line will become 0, so it will be summation of x i minus d whole square plus y i square, it is displaced only along the x axis and you can create such scenario, you can rotate your coordinate axis in such a way that it got displaced only along the x axis, fine, now expand this summation of m i x i square plus y i square x square plus y i square plus d square summation of m i minus 2 d summation of m i x i, did you like this, take the summation inside, this constant comes out of summation here and there both, okay, what is this is equal to ICM plus what is this d square into summation of all the masses is total mass summation of x summation of m i, this is what x center of mass, you remember that and what is x center of mass is x, so this is equal to 0 goes, this is the proof of parallel which is next theorem the last topic for today, just two steps done, you should talk about while I am telling, okay, what is the restriction in parallel axis theorem, one axis must pass through the center of mass, are you getting it, center of mass, the last topic for today because perpendicular axis theorem, this is valid only for planar objects, what are the planar objects tell me, planar object at this square planar objects, okay, so according this can also be irregular shape y z, moment of inertia about z axis will be the sum of the moment of inertia about x y, there is no restriction that it should pass through center of mass and please write down z axis, z axis must be perpendicular to the plane of the object, i z is perpendicular to the plane, i x and i y can be in the plane, but i z must be perpendicular to the plane and there is a reason why we have obtained i z, the reason is actually perpendicular to each other, getting it, okay, a quick proof of that for mass m i, what is it, 0 goes, we have been using it till now, it doesn't matter whether it is a plane or solid, okay, so i z is, we write down summation of m i, okay, as soon as this topic gets over all of you will die, what is happening, focus, you should help me write this, today this is stop talking, what is this, tell me what is x, the point from which x is y icon, so this is your i y and that is your i z, okay, so this is the proof of perpendicularity theorem, please write down these guys, this theorem is used quite often if i x is equal to i y due to symmetry, for example if it is this moment of inertia about any diameter should be same, so i x is equal to i y for a disk and i z is m r square by 2 anyways you know it, so i x plus i y becomes 2 times i x is m r square by 2, so about the diameter moment of inertia is m r square by 4 because this is m r square by 2 and these are 2 equal, so 2 times of this is equal to m r square by 2, this is equal to m r square by 4, we can do one problem, keep my time, what is the time,