 Shall we take up the next one next situation is this you will learn a lot of integration before starting integration in mathematics itself Total mass is m for the rod Okay Mass is m of the rod and here is a point this point is at a perpendicular distance d from these from this Straight this thing mass straight wire mass Okay, what else is given is the angle theta 1 and this angle is theta 2 Okay, so basically We have been given the angles the two ends of this rod makes with the with this point and The perpendicular distance d Okay, you need to find gradation field So none of this will be as in school. Okay, don't worry about it. Just focus on whether you learn something new today find the gradation field at Point P due to the mass m Which is a rod should I solve it? ATM stick at the integral Yeah, bull wrap Barat ATM stick Okay, I am stuck Okay, so let me solve this now all of you focus it. It's not very easy Okay, so don't worry if you are not able to get it. It's all right focus on how we are doing it, okay? Again the usual process I will find out the small mass along the rod. Okay, are you able to see the screen? Let's say this is See the thing is the length of the rod is not given right just the values of theta 1 theta 2 is given So in terms of theta, let me try to find out the integral over here. Okay, so let's say this is 5 Okay, this The small angle Let's say this can be written as d phi Okay, this This if you if I say that this is equal to let's say x, okay This will be DX DX is the length of a small mass Which I am considering. Okay, this is a perpendicular sense D Okay, now field because of just this DM mass is what G into Dm divided by this length Let's say this length is R R is square. This is D E all of you clear. This is D E now There is no direct symmetry over here. So we'll not try to look for it unnecessarily. So we'll just integrate D E X This will give us E X Okay, and E Y Will be integral of D E Y This will give us field along Y direction. Okay, once you get these two field Total field will be root over E X square plus E Y square Okay, and tan of angle it makes with the x-axis will be tan theta equal to E Y by E X All right. So let's try to first get the integral E D E X D E X will be G into Dm by R square Into cos of phi Okay, now there are three variables Dm R and Phi so everything let's try to write in terms of Phi only Okay, so R is equal to D divided by D divided by cos of Phi isn't it? Is this incorrect? R is D by cos Phi Okay Or you can say that this is D sec of Phi All right, and Dm Dm is equal to Let's assume the total length is L so m by L Into DX. Okay, fine mass per unit length into DX Right now X is equal to what? X is equal to D tan of Phi D into tan of Phi so DX by D Phi is D into sec square Phi Right. So from here, I'll get DX is equal to sec square Phi into Into D into D Phi this D is making D is not a differential. Okay, so that is why let's take it as capital D This is capital D. Then it'll be easier to visualize This is capital D Anywhere else D is there This is capital D Okay So this is DX and this is This is the expression for DEX So DEX Will become equal to G times Dm is what m by L into DX, right? So m by total length into DX, which will be sec square Phi D into D Phi. This is Dm. All right divided by R square. R is D tan theta. So it will become D square sec square Phi All right, and this into cos of Phi So DEX so EX will be integral of this Integral of that Okay, it has become a little involved Okay, but second square theta goes away. So this will become equal to One D is also gone This will be GM by DL Integral of cos of Phi D Phi Fine and the Phi value goes from minus theta 1 to plus theta 2 Okay, so EX Integral of cos Phi sine Phi right so sine of theta 2 minus of sine of minus theta 1. So you will get GM by DL or LD Sine of theta 2 plus sine of theta 1 All of you understood how EX has come All of you understood how EX has come L is the length of the rod. Let's say length is given If it is not even you can always find also in terms of D and theta, but let's say L is All of you clear Okay, can you get the value of EY now get the value of EY How much is EY? Since EY rather than this cos Phi sine Phi will come. That's the only difference. Yes or no That is the only difference So integral of sine Phi D Phi will come limits will be exactly the same Okay, so EY will come out to be equal to Yes, one GM by DL cos of Theta 1 minus cos of theta 2 Okay, this is what you get EY. Looks like I'm teaching mathematics today. I Can feel that okay, so Tell me if the rod has infinite length If rod has infinite length, what would the value of theta 1 and theta 2? Theta 1 is pi by 2 theta 2 is also pi by 2 right the angle theta 1 and theta 2 are increasing Yes, as you increase a length theta will become more and more and slowly and slowly this line will become vertical Okay, so this line will become perfectly vertical if the Line this rod is infinite length. Okay, so for infinite length wire or infinite length rod EY is 0 and that has to be because then it becomes symmetrical about x axis Okay, so Theta 1 equal to theta 2 equal to 90 degree so this EY becomes 0 and EX is Theta 2 plus theta 1 become 2 to pi by 2 pi by 2 if you substitute so we'll get 2 GM by DL Okay, so like that you can get the value of field because of these because of the small length of the small mass length Okay Understood all of you You have a disc Okay, mass is m and radius is r Okay This distance is z Okay, along the axis Distance is z Okay now Again the screen is gone So find out the field at this point at this point What is the gravitational field because of the disc of? Mass m and radius are Okay, the hint is that you can consider the entire disc as if it is made up of concentric circles of different Radio the point is above the plane. Yes along the axis. I can't draw 3d on the Screen so that is why it may not be very clear, but it is around the axis Yes, yes You should consider the entire disc as if it is made up of Multiple concentric rings So for a ring you already know the expression right use that expression over here For a ring we have derived the expression as G Z divided by This is for the ring Yes Okay, now, let me solve this small field Small amount of field because of a ring of radius are having width dr is At this point will be equal to G times Z Divided by now. Tell me what will be capital R? What I should write instead of this Small r right because that is a ring. So small r square ratio power 3 by 2 M is dm Okay, dm is the mass of the ring of width dr Okay, so total mass is capital M that is distributed in an area of Of a circle of radius capital R. So dm will be what? mass per unit area that is m divided by pi r square Into the area of the ring which is 2 pi r Into dr all of you clear right in case of any doubts immediately ping me. Okay, don't wait to m by R square Into r dr. This is dm. Okay, so small amount of field will be G times Z To m by r square All right into r dr. Not able to see the screen. I think we didn't have break today, right? so Continuously three hours Maybe YouTube doesn't support. Okay. I think today was very heavy a Mathematically very heavy class a lot of concept we could discuss Anyways, so only 10 minutes Well, it's already 120. Okay, we'll end at 130 alright so de will be equal to 2 g m Z by R square Rdr divided by So I've taken all the constant terms outside of The integral let me take two inside of this so the value of R The value of R small R will go from where to where zero to capital R, right? You need to cover the entire Desk so the rings will have radius starting from zero. It'll go all the way up to R capital R, right? So let's say R square is T Okay, so to our dr by dt is DT. No, it is one Good, this is one So to our DR Will be equal to DT Okay, let me put Z square plus R square itself as T Z is a constant so when you differentiate it will come out to be like this So overall integral E will become equal to g m Z divided by R square To our DR is DT Okay, and Z square plus R square is T. So this T ratio power 3 by 2 Okay, but the limits will be changed now when R is 0 T is Z square This will be Z square and when R is capital R T is capital R square plus Z square Understood till now. Okay, so this will come out integral of T ratio power minus 3 by 2 DT How much is this? This integral is T ratio power minus 3 by 2 Plus 1 divided by minus 3 by 2 plus 1 Okay, so this will come out to be this will come out to be 1 divided by root T and Denominator will have This is minus 2 by 2 T right minus 2 by root T. So this is g m Z okay, don't g m Z by R square Integral is minus 2 by root T Is there any cellular that we have done? Just quickly check I think not To our DR is DT now This is g m Z by R square 1 by instead of T put Z square should become Z Minus 2 will come out. Let me take to outside Z square plus R square Okay So if I take this Z inside the bracket, I'll get 2 g m By R square 1 minus Z divided by root over Z square plus R square Fine, so this is the field This is the field At a distance Z Along the axis of the disc Okay, so if you place a point mass at that location where field is e the force will be m into e Okay Isn't that cos theta Z divided by Z square minus R square theta kind theta It depends how you define theta She put it like this and that Z divided by this distance. How if you say that this if this angle is given That is very nice. If this is theta, okay theta is given Z is not given Then sorry Z should be given What should not be given Z is Anyways, if theta is given you can write in terms of theta also Yeah, Z need not be given if theta is given Z need not be given then e will be equal to just 2 g m By R square into 1 minus cos theta Okay, and 1 minus cos theta is 2 cos square theta by 2. This will be 4 g m by R square Cos square theta by 2 But it theta should be given Okay, usually this formula is used All of you got it All right, so similarly we need to Also define say this is field, okay So another one hour class is remaining where we define something called gravitational potential gravitational potential We'll do it next class So just like field is force per unit mass Gravitation potential is Potential energy per unit mass Okay, so we'll be defining potential similarly for ring for desk for wire for sphere for sector All that we'll be doing it in the next class Niranjan, how many how much integration practice you need? I think we have done enough, right? We have like in bits and pieces. We have done a lot of integration practice already We have already done one. I think two hour class on integration sometime back and for You were there in the bridge course where we did integration 1 minus cos theta is 2 sin square theta by 2 Small correction lies This is sin All right guys, so that's it for today. Thanks for coming. We did a lot of Advanced level stuff today. I hope it did not scare you But at the same time don't take it lightly Make sure you spend one or two hours reading what our concept we have discussed today because Nobody gets it 100% while sitting in just in a class, right? You need to struggle at home solve few questions. Then only you'll master the concept. Okay Okay. Bye. Bye. See you and like the YouTube link. Okay in the channel Hit the like button and subscribe