Welcome back. In the last couple of videos,

actually the very first video on centripetal acceleration, I

told you that the necessary centripetal acceleration, and

we’re just talking about the magnitude, and actually the c

tells you the direction, it’s centripetal so its inward

acceleration, it equals the velocity squared over

the radius. I told you that the real

rigorous proof has to be done with calculus. But I looked on Wikipedia and

there’s actually a pretty neat proof, although when you read

Wikipedia, it’s not so obvious on what they’re saying. So I thought I would do a video

on it, because this is cool and you don’t need calculus

to understand it. So let’s just do something. Let’s just plot the distance

vector, the velocity vectors and the acceleration vectors

as something goes around a circle. Let me draw two circles. So let’s say this is

my first circle. Let me draw another circle,

another color just for fun. OK. This is my other circle. This is the center

of the circle. And this is the center

of the circle. And so what’s the position

vector in any point in time? Well, the position vector,

you could just draw it as a radius. So the position vector at

any given time looks something like this. So let’s say initially this

is the position vector. That’s the initial

position vector. Its magnitude is the radius

of the circle. And this direction is right

here in the positive x-direction. And at that point, what is

the velocity vector? Well, the velocity– let’s

assume we’re going counterclockwise. I don’t know why

I assumed that. It could go the other way. Let’s say that this is the

velocity vector at that point. The velocity vector is going to

look something like that. That is the velocity vector. That’s v. It’s going tangent

to the circle. Let’s plot the velocity vector

as a function of time on this circle. So if that’s the velocity vector

of that time, I can draw the velocity vector here. This the the same vector. Remember, I’m just saying, at

a particular time, what does the distance vector look like

and what does the velocity vector look like? So at that time, the velocity

vector looks like this. I’m trying to make it the

same size to show you it’s the same vector. And I’m doing it in the same

color to show you it’s the same vector. This is the exact same vector. This is the actual circle. Like if you were to draw it,

this is the path of the dot or whatever is moving around

the circle. And this, you can just view it

as I’m plotting the velocity over time, the velocity

vector over time. So let’s say a few seconds

later, or a few moments later, what does the radius

vector look like? Well, then the radius vector

looks like this. I’m trying to write as

neatly as possible. And what does the velocity

vector look like? Go back to the purple. The velocity vector once again

is tangent to the circle. And it’ll look something

like that. It’s going to have the

same magnitude, just different direction. Actually, let me do it in a

different color, just to show you what I’m doing. So let me do it in brown. So the velocity vector

is going to look something like this. It’s brown, supposed

to be a brown. That’s the velocity vector. So after a few seconds, or a

few moments, where is the velocity vector here? Well, it’s going to

look like this. Remember, this vector

I’m just plotting here after a few seconds. Let me use that line tool. So it’s the same magnitude and

now the velocity is just at a different angle. It should be the exact same

angle as what I just drew in the other circle. So that’s this velocity

vector. So when we start here, the

velocity is going to change at the circle. After a few moments, the

object’s rotating around the circle, so now the velocity

is the same magnitude. It’s just switched directions. And so what’s happening here? When we plot the velocity vector

over time, it has the same magnitude, so it’ll draw

out a circle, but its direction changes. Just as in this case, what takes

us from this point to this point? This was the velocity vector

and the velocity vector’s always changing. But, in general, this is

the change in position. And what causes the change

in position? Well, the velocity, or at least

the speed, because the vector’s always changing. So in this case, what’s

changing the velocity? Well, just like velocity changes

radius, acceleration changes velocity. Let me draw an acceleration

vector, and acceleration is going to be in this direction. Because if the velocity is

changing from here to here, the acceleration is going along

the direction of the change in velocity. So the acceleration

vector might look something like that. It’s going to be tangent

to this velocity path. And that’s interesting, too,

because if this is the acceleration vector when the

velocity is here, so when the velocity is here, the

acceleration vector’s going directly to the left, so that

means that acceleration vector’s going directly to left,

and so this is also the acceleration vector, which

coincides with what we learned about centripetal

acceleration. The acceleration has to

be going inwards. And we see that when

we actually plot the velocity vector. And if we plot acceleration

vector here, once again it’s going to be going tangent

to this plot of the velocity vector. Let me do it in a

different color. I’ve already used that color. I’ll do it in yellow. So then the acceleration

vector is going to look something like this. Remember, acceleration

is nothing but change in velocity. So at this point, when this is

the velocity vector, the acceleration vector

will just be this. Once again, it’s just the

opposite direction of the position vector. So why am I doing all this? Well, I’m doing all of

this to set up the analogy to show you. As this object completes one

entire path, what’s happening on this circle? Well, this circle, the velocity

is completing one entire path, right? However long it takes to go

around this circle is the same amount of time it takes to go

from this velocity back to this velocity. The magnitude is the same the

whole time, but the direction is changing. So if this takes 10 seconds to

go around the circle in real position, then it takes 10

seconds for the acceleration to change the direction of this

velocity enough that it goes back to the original

velocity direction. So why am I doing all of that? Well, how long does it take

for the object to do one rotation around this path? Well, it’s just the distance

divided by the velocity. So the time to do one revolution

around this path is the distance– well,

that’s just the circumference of the circle. Well, that’s 2 pi r divided

by your speed, which is v. And how much time does it take

to go around this path? Well, we know it’s going be the

same time, and now we’re going to say the time’s

going to be the change in velocity, right? In this circle, this is the

change in distance, but it’s the change in velocity. So here we had this much

change in velocity. I know this might be a

little non-intuitive. Then we have a little bit more

change in velocity, a little bit more change in velocity,

a little bit more change in velocity. So the total change in velocity

is just going to be the circumference

of this circle. And what’s the circumference

of this circle? The radius of the circle is the

magnitude of the velocity. So it’s 2 pi times the velocity

of the circle. And so, what is the amount

of time it takes a do one revolution? Well, it’s the total

change in velocity. And that’s 2 pi times the

magnitude of the velocity divided by acceleration. And if this doesn’t make

complete sense, just remember, acceleration is change in

velocity over change in time. I’m glossing over it a little

bit, because you might say, oh, well, the net change

in velocity is zero. But we’re actually more

concerned about what’s the total change in the velocity? So it changes a lot, then it

goes back to its original, so that the net is still zero. But this should hopefully give

you a little intuition. So think about what I said. But the bottom line is we know

that this expression and this expression have to be equal,

because in the same amount of time it takes the object to do

one complete revolution on this circle, its velocity,

direction, has also done one complete revolution. So we can set these two

equal to each other. So we could say 2 pi r over v

is equal to 2 pi v over a. Let me switch colors. We can cross the 2 pi

out on both sides. And then we could multiply

both sides times v. You get r is equal to

v squared over a. Multiply both sides times a,

I’m doing this little bit circuitously, and you get ar

is equal to v squared. Divide both sides by r, and

you get a is equal to v squared over r. So hopefully, that gives you a

little bit of intuition about why this works. And I got this from Wikipedia,

so I want to give them proper credit. And hopefully, my explanation

helps clarify what the people on Wikipedia are talking

about a little bit. Anyway, I’ll also do the

calculus proof, because that’s more just straight up math. And I’ll see you in

the next video.

omg we are learnin this in physics class…thx cause this has helped a lot

This is a very good video. Thank you.

boobs

great visual presentation, helps understand the idea of derivatives of position and speed.

This is a brilliant proof without using calculus! I never thought it was possible to prove ac = v^2 / r without using derivatives.

rofl this intuitive proof screamed tangents and derivatives at me the whole time, was cool

Thank you! thank you so much for making this video!!!

I suffered for this formula for a while…

and finally I can use it by knowing where did it came from…

the coefficient of static friction between teflon and scrambled eggs is about 0.04, what is the smallest angle from the horizontal that will cause the eggs to slide across the buttom of the teflon-coated skillet the answer in said its 2degrees but i don get how they get that answer could u please help

1:11: BOOBIES!

@Kylethornton Einstein was 12 years old when he learnt Calculus

you shud put something that says thumbs up for this video or smth cuz sometimes i just move on to another video and when i realize it i feel bad for not giving a thumbs up cuz that is the least i can do right now!!:)thanks sal for getting me through calculus and physics!!!:)

2 people are teachers who can't explain this as good as khan 😉

incridible

Hey,

Please explain me this:

1. Why should the magnitude of the velocity vector be constant? what reasons us to think so? Please give Mathematical proof as well.

2. Acceleration Vector should be in the direction of the velocity. Why have you take it to be perpendicular in the second circle? If the acceleration vector and velocity vector are perpendicular- do they hold any connection at all? remembering that both are vectors.

Pls verify or point out my misconception.

Thanks

Could we go on one step further and relate 2 * pi * V /a to 2 * pi * a / (first derivative of acceleration)

And, What is first derivative of acceleration called?

I thank you , really, because in my class I didn't get it clearly .

thanx thanx thanx

me ha tomado dos años entender esto gracias a udsted de una forma muy clara

thankz a lot…..good job…keep it up!!! 😀

that such genius imo :O

Thanks ^^

omg i love u, such a genius

Thank you so much!!

full of win

yu have expanded so far bro very brilliant

Wonderful.

Consider this alternate proof.

Subtract the two velocity vectors by appending the inverse of the first to the tip of second in order to determine change in velocity. In doing so you draw two sides, each = v, of a triangle The third side is v times the angle; clearly, the angle is omega·t = vt/r; so the third side is (v^2t)/r. This is the module of the difference of velocities. Dividing by time, you get a_c = (v^2)/r.

Thank you brilliant

Please visit Khan Academy's website.

Incredible man!

If these two equalities T=2π*r/v and T=2π*v/a are not obvious for everyone we could use simple geometry to get the a=v^2/r.

These two triangles are similar (both are isoscel triangles and the top angle is the same). In the left triangle the base is change in position (Δp) and in the right triangle the base is change in velocity (Δv).

So, using the fact that the two triangles are similar we get Δv/Δp=v/r. If we multiply both sides with Δp/Δt we get Δv/Δt=(v/r)*(Δp/Δt) <=> a=(v/r)*v=v^2/r (qed)

dude goood job

that was kewl

Thank you so much, I preferred it over the similar triangles proof, it seemed more "real" to me, and easier to relate to.

I'll spread my legs for you this is awesome!

Thanks. I have a test tomorrow and it helps.

anyone could also check this video on derivation of centripetal acceleration expression.

thank you so much it helped me a looooooootttttttttt

Thank you!

please mad things easy and explain it a sooner that made it interestin

this proof videos are the best!!! thank you sal!

Shouldn't the change in velocity for the second circle be =0, because displacement is 0 and velocity = displacement / time ?

Calculus method is better

Calculus method is better

Shouldn't the acceleration vector on the red circle diagram be placed so that its tail is on the head of its corresponding velocity vector?

The only step i don't understand is at 7:49 , the "total change in velocity" is the "circumference of this circle [with radius of |v|]. Does it mean change in the velocity's magnitudes is 2pir, or the change in the velocity's directions is 2pir, or the change in the vector values of the velocities is 2pir? I'm not sure how going from 5 m/s north to 5m/s west to 5m/s south to 5m/s east and back to 5m/s north can mean that my velocity changed by (2*3.14*5) m/s, for example. I do understand that this doesn't involve the net change in the velocity, because that is obviously 0.

why acceleration is tangent to velocity vector?

Wait so why did you just do a random circle for velocity… is that the "path" of the velocity?

how can plot the velocity vectors into a different circle?

"so let me draw this in brown"

* draws it in orange *

Love your videos but can you stop over-using the word intuition please, you use it for everything and sometimes incorrectly.

I love you Khan ! I want to work with you someday

I have a small question! What if the velocity is constant? will there be an acceleration when the body is rotating with a constant veloocity ?

oh my god this really helps!!!

Explanation in this video is simple and much easier to understand: https://www.youtube.com/watch?v=XgkzeUxChKU

Just amazing!

10/10 for Khan academy

0/10 for jason's quiz tho lol

That helped me alot. Thank you

Awesome. Calculus method and this combined gave me the most complete understanding of centripetal acceleration.

Idiot how are making me understand

teach your lessons in a black board with chalk..

it is irritating..

Now calculate this for Earth and you come to 6mm/s*2 … vector away from the sun. Yet this force is unobservable. Read 1Chr16:30 Psa93:1 Psa96:10 Isa40:22 Psa19:1 Gen1 Gen2 Jhn1:1-5 Rev19:13 Heb11:2 Jhn3:16 Jhn8:32 Jhn14:15 Exo20 KJV Bible 🙂 God bless!

Bit of confused.

amazing

It was very nice prospective

The stopping distance at that acceleration were it opposite the velocity would be half the radius. So the maximum radius bend a car can drive around (assuming its turn stability and grip is the same as its braking grip) is twice its braking distance.

Thanks a lot! Despite the video being10 years old, the explanation was clear and it helped me understand how to use it in a problem.

I still wonder why T= 2nv/a

Thank you very much ?

Brilliant……

Thank you so much, I was having trouble with the similar triangle proof but this made so much sense.

Still I'm confused that why is time=(2π×velocity)/a

???????????????

How do you have equal and opposite with a real centripetal and a fake centrifugal? That's stupid to me. Newton's trends are not causal.

I think there is a more intuitive proof than what was shown. Firstly, see how changing the radius changes the acceleration. If velocity stays constant, but the radius doubles, the acceleration has twice as much time to redirect the velocity, so it is halved. If the radius is halved, the acceleration has half the time to redirect the velocity, and has to double to keep velocity constant. Therefore, there is an inverse relationship between acceleration and radius. Now let's keep radius constant and vary velocity. If the velocity is doubled, the acceleration has double the speed to redirect and half the time to do it, because the object will take half the time to go all the way around the circle if it is moving twice as fast. Therefore, acceleration must quadruple because acceleration equals velocity divided by time, and 2/.5=4. Therefore, there is a square relationship between a and v. And behold, a=v^2/r. You're welcome.

Damm it's so niceeeee I'll never forget this <3