Visual Understanding of Centripetal Acceleration Formula Visual understanding of how centripetal acceleration relates to velocity and radius
Visual Understanding of Centripetal Acceleration Formula
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- 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.
- So this right over here is going to be v one, velocity vector one,
- and 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 the magnitude
- of the velocity vectors is constant or another way to think about this
- is that the speed is constant.
- So I'll just say, lower case V without the arrow on
- top so this is going to be a scaler 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
- We're going to assume that it is traveling in a path, a circle with radius r.
- What I'm going to do is I'm going to draw a position vector at each point
- Let's call r one, over there; that's position vector one, r one.
- That is position vector r two.
- So the position is clearly changing, that's position vector r two,
- and that is position vector r three
- but the magnitude of our position vectors are clearly the same,
- and I'm going to call the magnitude of out position vectors, r.
- That's just the radius of the circle; it's this distance right over here.
- So R is equal to the magnitude of R one, which is equal to the magnitude of R two,
- which is equal to the magnitude of R three.
- Now what I what to do in this video is to 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 an 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 re-plot 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 V one, so copy and paste so that is V
- actually and want to do it from the center, so that is V one,
- the same thing for V two. So, let me copy and paste it. That is V two
- and then let me do it also for V three.
- So, V three I'll just get the vector part of the label. So, copy and paste it.
- That right over there is vector, V three and let me clean this up a little bit.
- just do that we don't, so that's clearly Vtwo, I don't think we have to label anymore
- we know that Vtwo is in orange, we know that Vtwo 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 scaler 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 vectors, what's giving us 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 will call this A1, we will call this A2
- And I'll call this A3
- and I want to make sure you get the analogy that's going on here.
- As we go around the circle the position vector, first they point out to the left,
- then the upper, up, kind of the 11:00 position or I guess the top, the top left, then to the top
- So it's pointing to these different positions like a hand in a clock.
- And what's moving it along there, is
- the change in the 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,
- they are tangential to the radius, or sorry, they are tangential to
- the path which is a circle, they are perpendicular to the radius.
- you learn that in geometry,
- that a line that is tangent to a circle is perpendicular to a radius
- and it is 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 will be going just like that, it will be going towards the center.
- A2, once again is going towards the center.
- A3, if you translate that, that is going towards the center.
- So all of these are actually center seeking vectors, you see that right over here.
- So all of these are actually centripetal acceleration vectors right over here.
- Here we're talking about just the magnitude of it.
- And we are going to assume that these all have the same
- magnitude, so we are going to assume that our centripetal,
- they all have a magnitude of what we call A sub C. So that's the magnitude,
- 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 to get to
- this point on this circle to the point on that circle right over there.
- So, the way to think about it is, what is 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 2pi(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 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 your actual, 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 scaler.
- So this is going to be the time, the time to travel along that path.
- Now, the time to travel along this path is going to be
- exact the 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 we can set these two things equally to 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 2pi, get rid of that.
- Let me re-write 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 multiplying, cross multiplying over here
- V times V you get Vsquared is equal to AC times R.
- Cross-multiplying remember is just the same thing
- as multiplying both sides by both 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 AC these V'a cancel
- out, these AC's cancel out, you get V times V is Vsquared 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 I guess we've earned the drumroll now
- We're left with the magnitude of our centripetal acceleration.
- It's equal to the constant magnitude of our velocity, i.e. our speed
- divided by the radius of the circle and we're done
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