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# Visual understanding of centripetal acceleration formula

Visual understanding of how centripetal acceleration relates to velocity and radius. Created by Sal Khan.

## Want to join the conversation?

• If the velocity is constant, then wouldn't the acceleration be 0? So how can Ac = v^2/r when the velocity is constant?
(28 votes)
• The MAGNITUDE of the velocity is constant although velocity has speed AND direction so the direction of the velocity is changing thus the velocity is changing, thus the object is accelerating, even when the magnitude of the velocity / the speed is the same.
(109 votes)
• Can somebody explain in details as how v can be the radius of the second circle?
(45 votes)
• If you take an arbitrary vector that always has the same magnitude, but varying directions, you can create a circle showing all possible variations of the vector. The radius of the circle will be equal to the magnitude of the vector.

Sal took the velocity vector, which has a fixed magnitude, but changes direction constantly, and made a circle showing all "versions" of it. Therefore the radius of the circle must be equal to v.
(60 votes)
• At Sal translates those acceleration vectors but they seem to be in the incorrect position. I do not see how they become center seeking.
(18 votes)
• The acceleration vectors he translates are the ones that correspond to each colored vector. For example, the acceleration vector that corresponds to the green vector in the circle on the right becomes translated to the green vector in the circle on the left. Both of the acceleration vectors are the same for that green vector, Sal just moved it to a different place.
(12 votes)
• at why does he put the velocity vectors in the middle? and how is the radius of the same circle equal to v wouldn't that imply that v=r?
(9 votes)
• I think the second diagram shows the forces acting on the object in the object's viewpoint or reference frame (I think), with the center representing the object itself. If you were somehow on the object, you would observe a velocity vector that points upwards initially but changes direction in a clockwise manner because of an acceleration acting perpendicularly to the velocity vector.

In contrast, the first diagram shows everything in a different reference point (i.e. we are observing the object from a fixed position away from the object).
(17 votes)
• I have noticed that all tangents are 90 degrees to the radius... Does that apply for all tangents? I understood everything else.
(9 votes)
• what is a position vector?
(7 votes)
• Hi Sal, can you explain me why does the velocity vector and radius vector takes equal time period to travel 1/4circle ??
(8 votes)
• I'm not 100% certain about this answer but if you imagine a race car that took 60 seconds to go around a track, you can think of the length of the track in terms of time. You can plug in values of time and receive the position of the car at that time. You can also plug in values for time and receive the velocity of the car at that time. Both the velocity and position paths are 60 seconds long which means that the length of each of their paths should be the same.
(2 votes)
• why is the position vector measured from centre of the circle? It could also be measured from an external point.
(4 votes)
• Yes of course, it is your choice of coordinate systems. And the results should be the same.
But the calculations simplify a lot if you observe the symmetry of the problem (something going around in a circle) and chose a coordinate systems that reflects this symmetry.
Here you can specify the position of the object by just one number (the angle variable) because the radius stays the same. If you take another coordinate system (eg x-y axes) you need two numbers to specify any position: the x-coordinate and the y-coordinate.
(3 votes)
• at , how did he write T=(1/4*pi*v)a
(4 votes)
• V is the new radius of the smaller circle representing with the magnitudes of the acceleration. For this circle time is the change in velocity over acceleration. Watch it again when he is making the smaller circle. Hope that helps.
(3 votes)
• why is it assumed that angle between r1 and r3 is 90 degrees? shouldn't the formula be T = angle/360 X 2pir / V
(3 votes)
• He is just using this as an example, and since he drew the diagram he can assume what he wants. However, I agree that he should have started with the most general form of the this formula, as you wrote.
(3 votes)

## Video transcript

Let's say I have some object that's traveling 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 1. This is going to be velocity vector 2. And this right over here is going to be velocity vector 3. And what we're going to assume, in this video, is that the magnitude of these velocity vectors is constant. Or another way to think about it is that 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 1, which is equal to the magnitude of vector 2. The direction is clearly changing, but the magnitude is going to be the same, which is equal to the magnitude of vector 3. And we're going to assume that it's traveling in a path, in a circle with radius r. And what I'm going to do is, I'm going to draw a position vector at each point. So let's call r1-- actually I'll just do it in pink-- let's call r1 that right over there. That's position vector r1. That is position vector r2. So the position is clearly changing. That's position vector r2. And that is position vector r3. 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. It's this distance right over here. So r is equal to the magnitude of r1, which is equal to the magnitude of r2, which is equal to the magnitude of r3. 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 the radius of the circle. 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 re-plot 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 let me 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. 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. So that's clearly v2. I don't think we have to label anymore. 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, this scalar quantity. So the radius of this circle is v. The radius of this circle, we 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 the 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 a1. We'll call this a2. And I'll call this a3. And I want to make sure that you get the analogy that's going on here. As we go around in this circle, the position vectors first they point out to the left, then the upper, kind of in a maybe 11 o'clock position, or I guess the top left, and 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 are these velocity vectors. Over here, the velocity vectors are moving around like the hands of a clock. And what is doing the moving around are these acceleration vectors. And over here, the velocity vectors are tangential to the path, which is a circle. They're perpendicular to a radius. And you learned that in geometry-- 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 learned about the intuition of centripetal acceleration, if you look at a1 right over here, and you translate this vector, it'll be going just like that. It is going towards the center. a2, once again, is going towards the center. a3, 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 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 they all have a magnitude of what we'll call a sub c. So that's the magnitude. And it's equal to the magnitude of a1. That vector, it's equal to the magnitude of a2. And it's equal to the magnitude of a3. Now what I want to think about is how long is it going to take for this thing to get from this point on the 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 1/4 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. And then how long will it take it to go that? Well, you would divide the length of your path divided by the actual speed, the actual thing that's nudging it along that path. So you want to divide that by the magnitude of your velocity, or your speed. This is the 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 to travel along that path. Now 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'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 1/4. It 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 the 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 nudging it along this path? What is the analogy for speed right over here? Speed is what's nudging it along the path over here. It is the magnitude of the velocity vector. So what's nudging it along this arc right over here is the magnitude of the acceleration vector. So 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 for this vector to go like that, for the position vector, is the same amount of time it takes the velocity vector to go like that. So we can set these 2 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 over the magnitude of our acceleration vector. And now we can simplify it 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 then we get r/v is equal to v over the centripetal acceleration. And now you can cross multiply. And so you get v times v. So I'm just cross multiplying right over here. v times v, you get v squared, is equal to a c times r. And cross multiplying, remember, is really just the same thing as multiplying both sides by both denominators, by multiplying both sides times v and ac. So it's not some magical thing. If you multiply both sides times v and ac, these v's cancel out. These ac's cancel out. You get v times v is v squared, 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. And you are left with-- and I guess we've earned a drum roll now-- 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!