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Current time:0:00Total duration:9:48

let's say I have some object that's travelling in a circular path just like this and what I've drawn here is its velocity vector at different points along that path and so this right over here is going to be v1 velocity vector one this is going to be velocity vector two and this right over here is going to be velocity vector three and we're going to assume in this video is that the magnitude of these velocity vectors is constant or another way to think about the speed is constant so I'll just say lowercase V without the arrow on top so this is going to be a scalar quantity I'll call this the speed or you could call this the magnitude of these vectors and this is going to be constant so this is going to be equal to the magnitude of vector one which is equal to the magnitude of vector two the direction is clearly changing but the magnitude is going to be the same which is equal to the magnitude the magnitude of vector three and we're going to assume that is traveling in a path in a circle with radius R what I'm going to do is I'm going to draw a position vector at each point so let's call let's call R 1 actually I'll just do it in pink let's call R 1 that right over there that's position vector 1 R 1 that is position vector R 2 so the position is clearly changing that's position vector R 2 and that is position vector r 3 but the magnitude of our position vectors are clearly the same and I'm going to call the magnitude of our position vectors R and that's just the radius of the circle is this distance right over here so R is equal to the magnitude of R 1 which is equal to the magnitude of r 2 which is equal to the magnitude magnitude of r 3 now what I want to do in this video is prove to you visually that given this radius and given this speed that the magnitude of the centripetal acceleration and I'll just write that as a sub C I don't have an arrow on top so this is a scalar quantity so the magnitude of the centripetal acceleration is going to be equal to our speed squared our constant speed squared divided by divided by the radius of the circle this is what I want you I want you to feel good that this is indeed the case by the end of this video and to understand that what I want to do is I want to reap lot these these velocity vectors on another circle and just think about how the vectors themselves are changing so let's copy and paste this so let me copy and paste v1 so copy and paste so that is V actually I want to do it from the center so that is v1 then we do the same thing for v2 so let me copy and paste it that is v2 and then let me do it also for v3 so v3 I'll just get the vector part I don't have to get the label so copy and paste it and that right over there is vector v3 and let me clean this up a little bit just so that we don't so that's clearly v2 I don't think we have to label anymore we know that v2 is in orange we know that v2 is in orange and what is the radius of this circle going to be right over here well the radius of this circle is going to be the magnitude of the velocity vectors and we already know the magnitude of the velocity vectors is this quantity V the scalar quantity so the radius of this circle is V the radius of this circle we already already know is equal to R and just as the velocity vector is what's giving us the change in position over time the change in position vector over time what's the vector that's going to give us a change in our velocity vector over time well that's going to be our acceleration vectors so you will have some acceleration we'll call this a one we'll call this a two a two and I'll call this a three a three and I want to make sure that you get the analogy that's going on here as we go around in this circle just put the position vectors first to point out to the left then the upper up kind of in the maybe eleven o'clock position or I guess at the top the the top left then to the top so it's pointing in these different directions like a hand in a clock and what's moving it along there is the change in position vector over time which is are these velocity vectors over here the velocity vectors are moving around like the hands of a clock and what it what is doing the moving around is are these acceleration vectors and over here the velocity vectors they are tangential to the radius or I started they're tangential to the path which is a circle they're perpendicular to a radius and you learn that in geometry that a that a line that is tangent to a circle is perpendicular to a radius and it's also going to be the same thing right over here and just going back to what we learned when we learn about the intuition of centripetal acceleration if you look at a one right over here and you translate this vector it'll be going just like that it is going towards the center a - once again is going towards the center a three once again if you translate that that is going towards the center so all of these are actually Center seeking vectors and you see that right over here these are all these are actually centripetal acceleration vectors right over here here we're talking about just the magnitude of it and we're going to assume that all of these have the same magnitude so we're going to assume that our centripetal they all have a magnitude that we'll call a sub C so that's the magnitude it's equal to the magnitude of a one that vector it's equal to the magnitude of a two and it's equal to the magnitude of a three now what I want to think about is how long is it going to take for this thing to get from this point on this circle to that point on that circle right over there so the way to think about it is what's the length of the arc that it traveled the length of this arc that it traveled right over there that's one fourth around the circle it's going to be 1/4 of the circumference the circumference is 2 pi R it is going to be 1/4 of that so that is the length of the arc that is the length of the arc and then how long will it take it to go that well you divide the length of your / the actual speed the actual thing that's nudging it along that path so you want to divide that by your actual the magnitude of your velocity or your speed this is magnitude of velocity not velocity this is not a vector right over here this is a scalar so this is going to be the time the time to travel along that path now that the time to travel along this path is going to be the exact same amount of time it takes to travel along this path for the velocity vector so this is for the position vector to travel like that this is for the velocity vector to travel like that so it's going to be the exact same T and what is the length of this path and now think of it in the purely geometrical sense we look we're looking at a circle here the radius of the circle is V so the length of this path right over here is going to be one fourth is going to be I'll do it in that same color so you see the analogy it's equal to 1/4 times the circumference of this circle the circumference of this circle is 2 pi times the radius of the circle which is V now what is nudging it along this circle what is the nudging along this path what is the analogy for speed right over here speed is what's nudging along the path over here it is the magnitude of the velocity vector so it's not geing it along this arc right over here is the magnitude of the acceleration vector so it is going to be it is going to be a sub C and these times are going to be the exact same thing the amount of time it takes to go for this vector to go like that for the position vector is the same amount of time it takes except the velocity vector to go like that so we can set these two things equal each other so we get on this side we get 1/4 2 pi R over V is equal to 1/4 2 pi V 2 pi V over over the magnitude of our acceleration vector and now we can simplify a little bit we can divide both sides by 1/4 get rid of that we can divide both sides by 2 pi get rid of that let me rewrite it so that we get R over V is equal to V over the centripetal acceleration and now you could cross multiply and so you get V times V so I'm just multiplying I'm cross multiplying right over here V times V you get V squared is equal to AC times R and cross multiplying remember is really just the same thing as multiplying both sides by both the denominators by multiplying both sides times V and AC times V and AC so it's not some magical thing if you multiply both sides times V and a see these V's cancel out these a C's cancel out you get V times V is V squared is equal to a sub C times R is equal to a sub C times R and now to solve for the magnitude of our centripetal acceleration you just divide both sides by r you divide both sides by r and you are left with and i guess we we've earned a drumroll now you are left with the magnitude of our centripetal acceleration is equal to the magnitude our constant magnitude of our velocity so this right here is our speed divided by the radius of the circle and we're done