as I promised in the previous class I'm going to solve this problem before going ahead for the next topic the steel ball a as you can see here of diameter D slides freely on the horizontal Rod which leads to the p face of the electromagnet going from this way to this way right the force of attraction obeys an inverse Square law and the resulting acceleration of the ball is this here we go k is a measure of strength of the magnetic field that if you go to electromagnetics we name it as reluctance if the ball is released from rest at x0 here we go this is x0 it means that initial velocity is how much zero determine the velocity with which it strikes the pole face velocity when hitting here okay so uh ising right right uh so this is the problem right so it says acceleration is k/ l - x² and it says velocity is zero because it's starting at rest you want to see the velocity when it strikes the pole face solution is very easy because I told you about the a DX this formula is equal to v d v right and as you know integration integration from V not to V so what's the upper and lower Bound for the X you tell me so obviously starting from here look at here guys here is this x0 it's starting from here the lower bond is x0 right what's the upper bound when Bal which is here is L right everyone obviously no what's what's upper bound is not L because X is measured from what Center of that okay so it's Center going to be here going L minus L minus D / 2 that's right that's the point because it sets the coordinate system x0 at the center of the ball here so when ites here we are measuring everything with respect to center of the ball that's right right everyone so here going to be L minus radius of the wall d/2 so you have to be very careful of so from0 to L minus d/2 here we go so we just have to replace here L minus B2 K DX I'm just replacing acceleration inside here l - x² and you know for this is zero so you're going to be 1 / 2 v² V strike Square the other side is very clear if you remember from high school for refreshing your mind integration of DX over x + a s power of n right going to be 1 / 1 - n and here is x + a power of um uh 1 - that's right right so your n is here two your n is two right so you get from here this integration not Dynamics so you're going to get here 1/ 1 - 2 and here is l - x to the power of 1 - 2 right from where from 0 to L d/ 2 right so this is equal to 1 / 2 V s² so this means this is one and goes down l - x 0 L - d/ 2 = to s² I'm going to replace there so we're going to put minus upper bound 1 / by L minus l - P / 2us 1 - L here we go this equals to 1 / 2 V s² so this L with this L goes away so you get here 2/ d - 1 / L = to 1 / 2 strike plus d s this going to be we can put like this plus by putting the side this not be l d it going to be - 2 L + D it's going to be like that 1/ 2 s² right everybody H so you're going to say we s² going to be uh 2 in D minus 2 L / by l d or you can WR like this you can write as 4 d/ 2 - L over LD right every so when you are done so your re your V going to be 2 of d/ 2 - L by L fing D I did by you oh I see I miss see that's what I'm telling you and here is K here is you have a k here you have a k here so you have a k here you have a k here and also you have obviously K here okay so now I see there's a sign I my mistake guys this is plus this minus minus and plus and this is negative always double check your answers because I'm with this negative the negative comes here so going to be plus minus going to be this right see this is plus everything is right now actually um let me double check so this going to be negative inside here yeah yeah uh it's going to be Pa right yeah yeah um where did like that where did these two come from huh I'm stupid these to so not come from oh these two comes from there these two good it's going to be four right yeah everything's good except that okay there we go okay let's continue our topics the next topic is uh plan curve linear motion before going ahead so now suppose this is exactly what you are doing every day by coming to University as I promised you about the normal tangential coordinate system and the polar coordinate system so this is o and suppose you on the curve your car on the ramp of coming from I8 West to College Avenue and see you car as a particle huh compared to the San Diego or San Diego State going to the another part of the curve or ramp right and you see there there center of the rotation on the grass of the ramp right you can connect to here if it's the radius of the rotation Center of rotation it going to be r + Delta R and these are vectors obviously so you know the difference is going to be if you connect this to this point and it's going to be Delta R right and here this R going to be Delta s real distance that your car travels right here on on the on the Curve there is Delta s there we go okay as as we Define velocity velocity is v um average velocity is delt R over delta T of your car say but the speed as we defined previously is V of U speed its average value is Delta s over delta T is R that's right if time scale that you're measuring is very small value this means very very smaller than one so you can say velocity the alals to speed to speed average average because they are overlap if the time is very small look at here guys at here going to be here they are the same right by having that you know that instantaneous instant tus velocity from here is again the limit of Delta r or delta T when delta T goes to zero small value and you know that this is Dr R over DT from high school time derivative or r dot and instantaneous speed going to be DS as we did over DT is s do so can I go ahead so an acceleration obviously is instantaneous it's Delta V or delta T and delta T goes to Z going to be V Dot or rou dot change of radius of your car now we start the coordinate system I classifi for you cartisian coordinate system noral tangential cord po than the C and uh let's go with rectangular rectang rectangular coordinate system or XY 2D and this is very very clear because we have seen this in physics of high school so this is x y and O and suppose you the curve see your car and this uh coordinate system can be you can sit on the Sano state right and this is the curve towards Sano State and if you this is your car as Point p and it connect this to point p h here is R but you can de compose this on X and Y obviously based on the coordinate system so here going to be it's X component in I Direction and it's y component in J direction right but always you know velocity is tangential to the to trajectory tangential here is tangential to this so it's going to be V velocity but you can DEC compose based on X and Y so this going to be this Vector going to be VX if this angular Theta and this going to be v y so we say position Vector of Point P respect to point O this is your car Point P equals to x i plus y j obviously see this Vector plus this Vector y you have a time derivative you get the velocity so velocity of your car in coordinate system X do I + y do J you know that X is VX X going to be VX i+ v y j now from geometry you know that v² from this triangle is V x² plus v y 2 so absolute value of the V is root of 3x² + v y and you know that from the triangle is tangent of theta is y over VX or you can calculate the tangent Theta is tangent inverse of Vy or VX if you have time from this you get the acceleration of your car so acceleration going to be xou dot I + y dot j or ax+ a y the same Val is for that so absolute value of acceleration is ax s + a y now before going to normal tangential I'm going to teach you the most ever seen mov in Hollywood projectile motion when you throw up a ball or basketball player throughout the ball to basket or volleyball player or I don't know uh uh what's the name of that that you're shooting American sport I forgot the name uh baseball so you want to see when we shoot freely and the only force is gravity how we launch it what's the speed of that when what are we going to hit on the ground what's the angle of that so the name of that is projectile motion so and then I'm going to solve problems from book so uh projectile motion and this you have seen this million times in life so so suppose that this is your o because now you know the rectangular coordinate system o XY so we are shooting a ball this can be basketball this can be I don't know mile whatever okay so when you shoot now you learn that velocity is normal to trajectory this is velocity of here we go initial velocity basketball play when you shoot thing gives initial velocity to the ball uh moving toward basket right and I'm going to Sol a problem from book exactly based on basketball okay and and basketball player gives an angle to that Theta respect to X and Y right so as I told you you have to decompose it you have V not initial veloc in X direction as you can see here is V not cosine Theta respect to horizontal line and O sorry and come on there we go initial velocity in y direction from this triangle is V not sin Theta from high school here we go and then this ball is being is being going going going and reach is the maximum point again the velocity is the tangent to the trajectory we and you have get the maximum height right maximum height maximum y for projectile motion we have to make assumptions before going ahead assumptions assumptions okay one no aerodynamics drag forces drag forces means there's no huge wind or tornado coming and blocking the way of the projectile right two we assume no rotation of Earth what does it mean means we going to stop Earth no As you move do you feel the rotation of Earth no projec is the same as you different speed okay as I said one projecti is here the only acceleration it has is what I told you gravity okay as it is when you throw the ball only gravity effect on it so gravity here we go g so do you see any horizontal acceleration acting on it no only gravity right so ax is zero and the a y based on the coordinate system that is plus plus and minus and minus is minus G because looking down here's plus minus G you got it everyone there we go these two now let's stick to this we learned that ax is D VX over DT Lo in that or from here you get D VX equals to ax DT from initial velocity of shooting in X direction to final VX from time 0 to T this is zero problem says so integration of zero is a constant number c so from here here you get the other side integration of derivative function so vxus v x and let us assume for generality C is0 zero so you get VX is v x mean duration of the flight always velocity next direction is V not angle of shooting here we go it never changes so velocity next Direction only depends on the initial velocity and the angle of shooting that's it you got it can I go to the next page so now wex will learn from that is DX or DT or DX integration from X notot initial shooting to final x from 0 to T VX DT but we just derived that the VX only depends on this we not angle of shooting d from here you get the second equation x is we not cosine Theta t + x not you got it so now let's go to the other aspect of problem we saw that the only acceleration is G here we go so here he have a y based on definition is derivative of velocity in y direction over DT again dvy from V not y to v y is a y DT from 0 to integration of derivative is function itself so going to be y- v y 0 to T minus G DT this is simplest integration of Life minus GT so Bingo v y is minus GT plus v not sin Theta yeah we already established we not why at here you see other relation that we going to use in problems for for uh project once you have v y we know this uh W is equal to Dy / DT integration from y not to Y Dy is equal to from 0 to T minus GT plus v not sin Theta T this is super simple this side going I be y equals to I'm taking the other side integration of this is 1 - 2 G t² and this is constant we not sin Theta in t another ration for project in y direction you got for X Direction here for y direction and for y direction velocity clear everybody now compare this with whatever I Tau you can I go to next page I previously Tau you this remember you have seen this if acceleration is constant we have this remember not for projectile we drive this I just replaced acceleration by real value minus G and initial value of shooting this t plus the initial value on you are the same also I told you about this relation a DS to V DV but only acceleration is in the y direction so we're going to have this and this going to be y so integration from y not again y a y p y from V not y initial velocity in y direction VY VY D VY as we know acceleration ping y not Y is minus g d and here going to be same value it's going to be super simple - g y - y 1 / 2 v y 2us v y 2 or the other relation is we y² you have seen this in other format we know the sore is - 2G y - y here we go compare this with whatever I told you previously time independent like that like v² - V not s was 2 a x - x have your notes only I implemented your Aus G but in y direction because this Pro and the last relation almost we going through problems is this we learned that the V not X for projectile is we not cine we not y based on high school is we not sign and also we drive X's projectile is we not cine Theta in t we just D plus X and we just derived Y is = to - 1 / 2 G t² plus we not sin Theta in t also you have a t here and this is plus plus one for Simplicity assume that we set of our coordinates system at the origin of shooting at the hand of the basketball player so I say x not is z huh can I do that so by doing that I'm going to find time from here time is X or we not cosine I'm going to implement time here and here being said y becomes 1 - 2 G t² x² / v² cine s of theta plus v not sin Theta in t t is this X over V cosine Theta + y this with this goes to Chicago s over cosine is how much from kindergarten tangent of theta so we got the most important relation in projectile motion Y is equal to 1 / 2 G x² we not s cosine s of theta plus X so you got here guys x² X tangent of theta + 1 this is very important why I want that you look at the board and tell me why it is important why we are miss what we are missing in this equation what we eliminated time this is time independent formula and you're going to you in the book problems that I'm going to solve hopefully these are next class how we overwrite time so this is time independent time independent formula so if I want to summarize onto this point I'm not done yet so we got this for project WX always is constant is V not cosine the function of angle of shooting initial loss v y isus GT plus v sin Theta sorry plus X we derived as V cine Theta in t plus X Y we got as - 1 / 2 G t² plus we not sin Theta + y and you have t here also and we got time independent relation 1 / 2 G x² over we not sare cine sare theta plus X tangent of theta plus y but the basket forall play is say Michael Jordan or LeBron James when they are shooting the ball uh when you throw a ball to friends when you're playing like this you have a pick Point pick pick point or maximum height that we want to calculate based on the angle of shoting and the initial velocity why it is important in the battlefield when they shoot some ball to the enemy with some even the missile in the US Navy F right it goes based on gravity they exactly do this calculation to hit the target where is the maximum it goes when it lands where it lands okay so guys let's take you back to the first year of your high school still look at this graph forget about Dynamics let's go back to algebra what is zero here at this point I don't want I don't want to I need technical Lan for it what is this guys in this point slope that's right so slope here is zero because the slope is like this look at here this is the slope say Delta y over Delta X there's no y change here that's right so dy over d d t is zero that's right slope here is zero soorry DX is z right or this is stupid sorry or dy over DT which is the slope is zero right as your friend said if you have time to from this you get to here right velocity so slope is here is zero so I want that so if you do that is minus g t of Maximum maximum time ret ret here is plus v not sin Theta is zero so TMax for going to the pick is velocity initial velocity shooting of that over gravity can find the time that you your projector goes to the maximum height other okay so when you find that just Implement inside here Bingo Bingo time to get the maximum height or pick height let's do that together the Y was 1 - 2 G t² we not Square sin squ of theta look at here I'm just over G sare plus we not s Theta in t v sin Theta over G + y you got it everyone so this G with one one of this goes to name it La so you get 1 / 2 we not sare sin s of theta / G plus we not Square let me fix this guys s s of this G + one so this gives you maximum height or maximum pick or pick of uh s the 1 - half going to be we not Square sin s of theta over 2 G + y we know the last relation right problem 279 of the book two 79 there we go we got it this on the slope a projectile is launched with an initial speed of 200 m/s at the angle of 60 with respect to horizontal this total guys is 60 compute the range this R and slope as measured up the incline okay let's do that in line and we are shooting from here and it goes here and this angle is 60° and the slope angle by itself problem says is 20° and this is 200 m/s right so we not is 200 m/ second so what I'm going to do I'm going to set up the coordinate system at here X and Y okay let me show the problem make sure I'm right yeah I'm going to set up I mean this here I'm going to put X and okay I'm going to fix over there why I'm doing that because if I select on the point of shooting guys on the point of the shooting what I make zero I make X not Zer and Y not zero this is y's motion I make X not zero and Y not is zero now this is the range that hitting point B all of you look at here guys look at here we want to calculate this R here R we want to calculate this problems range but you see that you have a y here you have a x here based on the coordinate system so from uh high school you know tangent of 20° equals to y/x y/x you get one relation from here y of the projecti is X tangent of 20° okay now we have the relations for a projectile we develop together you set X is V not cosine of shooting not this 20 shooting 60° that's right in time t plus x x is zero we just put it there Y is equal to weop 1 / 2 G t² + we 200 S of 60° in t plus y not we did we did y not zero we just by proper selecting the coordinate system on the point of shooting but not to Z so what I'm going to do simple like that I'm going to put X here I'm going to put y here so we going to be 1/ 2 G t² plus we not sin Theta in t equals to X we not cine Theta in t in tangent of 20° now I'm going to move to the other side 1/ 2 9.81 gravity t² plus look at here guys all of you we not I'm factoring out this right so you're going to be 200 I'm factoring out s of theta right s of 60° minus cine of 60° mtip by tangent of 20 tangent of 20° here we go now they share a Time T is zero this is very simple equation factor out T so you can say t equals to I say a ided by uh 9.81 / by two I have the number so T of impact T of impact going to be 2789 seconds just put in the calculator you can get you got it everybody simple like that once you got this time bingo just um uh just put inside here inside y or X H I'm just putting there get here I'm saying y = to 1 - 2 9.81 t² 2789 squ just see here here plus 200 S 60 * by 2789 I have the number going to be 10 15. 59 M almost 1 kilom now let's back to high school I'm going to show you from triangle s of 20 is y / R we got y so R is y/ s 20° I have the number is 29 67 M animal 3 km from the ground you got it everyone I'm going to solve two more problem next class because it's going to be a little challenging the next problem I'm going to solve I have to go to the math lab and I teach you how to solve it uh equation in math lab okay and the problem uh two of to let me show you one thing see I just told you look at here guys basketball player exact this projector problem so what the LeBron James does is to practice as much as he can to shoot on the basket right or the bombing say by the airplane is just projectile motion or uh tennis player projectile motion whatever you see is is the just shooting a shooting bomb huh [Music] um I'm going to solve this it's going to be a little complicated okay that's are two parts or uh horseshoe player here we go okay this is interesting problem or picture you love it this right it's project uh problem right okay I'm going to solve two problems next class and we continue from there to the normal tangential coordinate system just bear