in a previous video i had mentioned that in two-dimensional motion it's entirely possible to have the velocity and the acceleration of an object point in pretty much any which way that you would like um the piece of the acceleration that is either parallel or anti-parallel to the velocity is responsible associated with the change in the speed of the object and the bit that's perpendicular to the velocity is responsible for the change in direction it's that latter part that i want to treat right now which we so we're going to go ahead and figure out what we call usually called the centripetal acceleration the the piece of the acceleration that's pointing perpendicular to the motion or towards the center of a circle so i'm imagining i've got an object that's going around a circle like so um at speed v and i'm having go clockwise it doesn't matter if it's clockwise or counterclockwise um and i want to know the magnitude and the direction of the acceleration right at this very top point here but in order for me to figure that out i'm going to have to look at a point a brief period of time before and an equal period of time afterwards and the game will be will make the time intervals ever shorter and shorter and shorter eventually churning our finite differences into differentials all right so first i need to find the velocity at the midpoint at the midpoint here and so to do that what we'll do is we'll go ahead and find the average velocity for this interval here and then we'll repeat that over here once i so to get the average velocity for this interval first thing i need to do is get the displacement so let's go ahead and do that oops so my final so if i look at the interval from one to two my final is r2 like so and my initial is r1 but i'll be adding negative r1 to that and so i will be i get that by the here i use the tailed head rule that the change in the velocity over this first time interval so i'll call that delta r1 points like so if i parallel transport that back here no big surprise that's just taking it from here to here because hey that's the displacement okay and similar we can do exactly the same thing to get the displacement for this interval here that's delta r 2. all right now the deal here is that my delta r1 points up and to the right my delta r2 points down to the left so i can draw corresponding vectors v1 and v2 that are parallel to my delta r1 and delta r2 because i'm looking at equal time intervals and remember the velocity is the average velocity is the displacement over the time but i can use this to find the change in the direction um so you can use this to find the change in the velocity and from that i can get the direction of the acceleration so we'll go ahead and play the same game here we will go ahead and draw a v2 like so i'll add to that negative v1 so i can find my final velocity minus my initial velocity so the velocity before and after the time i'm interested in so there's negative v1 and so delta v um the change in velocity points straight down it turns out cool now with a little bit of fiddle faddling um or right actually probably in one second here and since delta v points down since the acceleration is the change in velocity over the change in time so multiplying by 1 over the change in time is a positive scalar that means the acceleration points the same way so we get at this point right here that the acceleration points in to the center of the circle and there's nothing particularly special about this exact point that i chose um it could have been i could have repeated this for any point on the circle and i would have gotten the same result that the acceleration points into the center of the circle now if we do a little bit of geometry here these two angles are congruent by construction because i said these are equal time intervals and we're going at a constant speed around the circle so we cover the same arc length in the same amount of time so by definition of an angle being arc length over radius these angles are congruent you should also be able to convince yourself that this angle is the same as that angle and because my velocity vector my delta r's here um are sorry um and you should also be able to convince yourself ultimately that this angle is also congruent to all of those angles and this is just because um instantaneous as we go to smaller and smaller time intervals what we'll find is that our velocity becomes tangent to the curve in fact it's looking at a point right here so we we split this split that that's that angle there my velocity vectors are perpendicular so that carries the angle through there now what that lets me do is it lets me say that this triangle this vector triangle here and this vector triangle here are similar triangles so i can set up a proportionality i can say that delta v is to either of these v's because v1 and v2 are both the same so i can just say delta v over v i can say that that is the same difference as delta r over r that that those have to be that those proportions have to be the same by similar triangles okay now we know from the um from our definition of acceleration it's the time rate of change in the velocity and yes this is technically an average acceleration but don't worry at the end we'll take a limit as delta t goes to zero and everything will become derivative so it's all good um so i can rearrange this to get that delta v is equal to a delta t so i can go ahead and stick that in there so i get that a delta t over v is equal to delta r over r now here's where i need to cheat just a little bit this length right here from here to here that is the difference in arc length and i am going to say that my a delta t over v is approximately equal to my arc my arc length delta s for that interval over the radius now at first that looks really bad um but it's actually okay the reason it's okay is remember at the end of the day here we're going to be taking our time intervals down to approaching 0. so if i look at a sequence of chords that i strike off against my circle here here the arc length and the rays don't look anything like each other here they look more like each other here they look more like each other and in the limit of smaller and smaller and periods of time the smaller smaller angles the arc length and the radius approach each other so it's all good given where we're going with it so i can go ahead and rearrange this to get the acceleration done it's going to be equal to v over r times the time rate of change of the arc length and here i can go ahead now and take the limit as delta t approaches zero and in this limit this is going to become a derivative but the point of that is for me to realize that this is also the speed i'm going so we say that the magnitude of the acceleration which we call the centripetal acceleration is equal to v squared over r so we say this is the centripetal acceleration and centripetals just if you break the roots of the word apart just means center seeking so this is the piece associated with associated with changing direction it's not associated with changing speed at all um now we could have also said that my um that we were going around with angular velocity omega so if you remember from the last video we had that our tangential velocity um was equal to r omega which is just v so if i put that in i get i can get another expression for the centripetal acceleration v squared will be r squared omega squared over r cancel cancel and i get that an equally useful expression as r omega squared and again here just to look at units um the for an acceleration we're expecting meters per second squared um radius is meters this is radians per second is our omega but we're squaring it so we have to square the units again remember that radians squared that's going to be the same thing as 1 squared which is the same thing as 1. so then we do indeed have meters per second squared and adult checks remember radians are not a unit they are a symbol that reminds us that we're dealing with an angle but it doesn't have an actual physical dimension okay so just to maybe make this a little more concrete let's say that um we've got a mass going around on a string um like so let's say that we are going at um well let's say two meters per second and let's say our string has a radius of 50 centimeters or half a meter the acceleration if we're going to constant speed it will accelerate because as we go around the circle the direction of the velocity will keep changing so the acceleration is pointing to the center of the circle and the magnitude of the acceleration here will be the centripetal acceleration because we said we're going to constant speed so it's v squared over r so this will be two meters per second squared over half a meter we're gonna get eight meters per second squared so even though an object is going around moving at constant speed if it's changing direction it is accelerating because velocity is both um and the cell array is sorry velocity has both a speed and a direction okay however it is possible that we could be picking up speed and so to that end we want to think about how rates of rotation could be changing so we'll go ahead and define angular acceleration and just like we had said before with translational acceleration that it was the time rate of change or the derivative time derivative of velocity this will be the time rate change or the time derivative of angular velocity so the symbol is a lowercase greek alpha so this will be the limit as our time interval approaches zero of the change in the angular velocity with respect to time or we can write that as d omega dt and our radian units will be radians per second squared and remember radians don't have an actual physical dimension same difference as writing the number one and this symbol here is a lowercase greek alpha so just um to go ahead and draw that like that you do something like that um basically you're just drawing yourself a fish but you resist with all of your might to finish off the fin and the face and the little bubbles come out sorry um but anyway that's how you draw one okay so how do we interpret this um the way i like to think about it is if the sign of the angular acceleration the rate at which um my angular velocity is changing if it's positive what that i'll say is that omega is becoming more positive and if it's negative then we'll say that the angular velocity is becoming more negative now what do i mean by that let's go and take a look at the two possibilities we could have for angular acceleration being positive one possibility we can have is that the thing is rotating counterclockwise which means that the angular angular velocity is positive and it's speeding up however the other possibility would be that it's rotating clockwise which would make the angular velocity negative but if it's speeding down slowing down then that means that the um angular velocity is becoming more positive you start out with a bigger negative number you end up with a smaller negative number that is a positive change to the angular velocity so either of these cases are possible if the angular acceleration is pi is positive the angular acceleration is negative then our possibilities are that we could be rotating clockwise which would give me an angular velocity is negative and if we're speeding up we're just heading to ever bigger negative numbers or that's rotating counterclockwise which would mean that omega is positive and it's slowing down so then you're starting the big positive number you're ending with a smaller positive number that's a negative change to the angular veloc angular velocity and so thus the angular acceleration is negative where that helps us to tie in is if you remember i said that's also possible that we could have a component of accelerations either parallel or anti-parallel to the motion so let's say and so this will give rise to what we call the tangential acceleration strictly speaking it's a tangential component of acceleration but we're lazy in how we say that so let's say i have a mass uh string let's say we're going around at some angular velocity omega and importantly we're picking up speed so since so um just do a quick pause and ponder on this for a second and ask yourself um we can assume that this is the horizontal plane going around a circle we're picking up speed which way does the acceleration not the angular acceleration the acceleration point pause the video and get back with me okay so it turns out that it's going to point inward and forward and exactly how it gets divvied up will depend on how fast we're going how big the radius is etc but we will have a centripetal component vector here um this is the bit that's responsible for changing the direction so we're going around circle then since we're changing speed we also have to have a tangential piece pointing in the same direction as the tangential velocity and so here i'll go ahead and add them up by the parallelogram method the actual acceleration of my mass on string is actually pointing forward and inward now how big is this tangential piece since it's responsible for changing the speed the tangential piece will be equal to db dt the time rate of change of the tangential velocity so this will be the time rate of change of the quantity r times omega where omega is the angular velocity since the radius doesn't change the time rate of change of the product of these will be the radius times the time rate of change of just the angular velocity so this would be r times d omega dt so we get that the tangential acceleration is equal to r alpha so it turns out that the the rate at which the speed of something is changing is directly proportional both to the radius of the circle you're going around in and the rate at which the angular velocity is changing so to make this maybe just a little more concrete let's imagine that we have a car that's going around in a circle so and we'll say it's a pretty fast car so i have to go around a quarter of a circle like so we'll say that a grand total of five seconds elapsed in the process and we will say that we started at rest and we ended up going at um 10 sorry at 50 meters per second and we'll say that the radius of curvature of our road is 100 meters all right so let's go ahead and find the acceleration right here at this point we'll assume it's still continuing to pick up speed at the same rate so following a clue from what we had before we know that it's going to be inward and forward because we're picking up speed so we can just say that the tangential acceleration since the rate of change um so dv without a vector dt so this is rate of change of the speed since it's changing uniformly i can say that this is the same difference as the average tangential acceleration the average rate of change in the speed so this would be 50 meters per second over 5 seconds is 10 meters per second squared and to get the centripetal piece we'll go ahead and say that's v squared over r so my final velocity here is 50 meters per second quantity squared over 100 meters that works out to be 25 meters per second squared so it's going to look something like acceleration is going to look something like that and i can break it into a tangential piece of 10 meters per second and a centripetal piece of 25 meters per second squared that's supposed to be directed towards the center of curvature sorry i made a hash of the artwork there let's try that one more time to make it look a little less awful yeah something like that and we will make my center of the road look a little better too there we go much better so if we wanted to um if we were using a conventional up down right left coordinate system um so plus x plus y like that we could ultimately then say that my acceleration was equal to negative 10 meters per second squared i hat plus negative 25 meters per second squared j hat or if i wanted to get the magnitude of it um i could set the magnitude but the pythagorean theorem is the square root of the tangential p squared plus the centripetal piece squared since they're at right angles to each other this would be the square root of 10 meters per second squared squared plus 25 meters per second squared squared gives me 26.9 meters per second squared awesome so in the next video we'll take a look at situations where the um angular acceleration is a constant so we'll be doing rotational kinematics we'll catch you over there