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!

1st?

such a BOSS!

I still don't understand 🙁

I get it all, just wondering why the Time to travel 1/4 of both circumferences is the same?

@riimzo There is only one velocity vector on the circle for every position vector. If you know it's position, you know it's velocity. So if the position changes at a rate, the velocity vectors much change at the same rate.

@nhmllr725 Ohh.. Alright thanks 🙂

@riimzo No problem. ;p

@khanacademy what program do you use to draw in these videos?

@PmQable1 It's useful to describe the change in direction of bodies. Especially when they have paths like circles and ellipses. No?

That's great I miss getting your VHS everyday, Salmon, are you okay?

I don't understand a shit about what you are painting or talking about.. but it's fun to look at ur videos 🙂

I got my degree at Khan Academy!

Thanks Sal!

Brilliant!

Your awesome, Sal

This is awesome. Now I understand where this equation comes from. For those of you guys who will take Dynamics in the future, you better learn this stuff clearly cuz your gonna bump into it again in Curvilinear motion in Dynamics. I didn't understand anything about that kind of motion at first cuz my professor skipped the Centripetal Acceleration section when we were in Phys 1… I think i got it now 😀

you're awesome

beautiful intuition and explanation! Thank you!

i like sal he not only explains thing good but being able to have a clear and concise illustration of what your talking about as you mention it is important in my opinion

Your accent is kind of distracting

OH MY FRIGGEN GLOB THANK YOU SO MUCH. This was beautiful.

wait how did the velocity get to be the velocity?

and to find the time to go the second half why not just use the first formula

I get it now thanks!. Also, during class, I always have to stop listening to my professor to think about what she is saying. Ex. "The direction of velocity is moving in clockwise rotation." But when this guy says "like hands on a clock," less thinking needs to be done on my part. Many thanks.

Why is the radius of the second circle is the magnitude of the velocity?

i wonder why there is acceleration ??

like if you actually did a drum roll 🙂

this was such an awesome video!!

Thank You! I've been looking for while to find this

Just someone from 2015 passing through

great explanation

AM I CORRECT?

if we assume that angle is small so that arc lenght is flat hence equal to Δv so that velocity x θ = Δv and then dividing both sides with t gives Δv / t = velocity x θ/ t as θ / t = w hence Δv/t = a = velocity x w

Why would first T and second T be the same?

THANKS A MILLION! REALLY LOVE HOW YOU TEACH!

So if there is a change in direction there is an acceleration??

How come there is an acceleration, if the magnitude is constant? aren't you gonna get zero or there is an exception in circular motion?

Hi Sal, can you explain me why does the velocity vector and radius vector takes equal time period to travel 1/4circle ??

torsion

the black screen makes harder to see

AND WE'RE DONE!!!

Sorry, this didn't help at all, too many questions during the video, it might be my lack of knowledge on the subject

this was extremely intuitive and interesting. Thank You Khan Academy!

just started physics 12, thank you so much for making my life easier

Really great and innovative derivation without using calculus.

Why do you repeat yourself constantly?

What does the second circle represent? In what case do velocities come from the origin? Or does the second circle just represent the first circle with the forces redrawn?

I feel so much I understand it now

If there is a nobel prize for education, this guy deserves it 🙂

so interesting……

It was not nice for the study of physics

thxx buddy it really helped mee…..😜

This helped so much! Thank you!

My new PHYSICS SOLVING APP.More then 150+ formulas,Solves for any variable you want,Covers up all physics.download now.https://play.google.com/store/apps/details?id=com.physics.lenovo.myapplication

why did he made radius velocity?

I don't know what to say.

Wow thank you so much, I was staring at my notebook trying to visualize why would that make sense, and I couldn't get anything at all, this is perfectly what I tried to look up

I dont get it

Why Khan!!?????

😂 😂 😬❓❓❓

Why draw an acceleration vector tangent to a velocity vector?

Thank you so much for this video

I have never been so confused in my life

Why is the time same in both cases

Sir Ur a legend.thank you so much.

physics is crazy bro. I understand this video but it also got me thinking why does nature work this way? like think about it… you have an object moving at a constant speed in a circle and that object has a force pushing it towards the center of the circle. The only reason why it doesn't come crashing towards the center is because the force tangent to the circle is greater. Like bruh if that isn't crazy then idk what is.

5:42 why are the centripetal acceleration vectors perpendicular to velocity?

OMG such a complicated geometric derivation

xmtutor does a much better job

please sir let me know why the velocity is supposed to be constant

THANK YOU …THIS HELPED ME A LOT

Sal is STILL THE BEST!!.. thank you!!… perfect VIDEO from beginning to end!! they don't get any better than this video!

Are you God?

wait a second i thought acceleration was a vector quantity?? 2:03

Is 4π²r/T² another formula for centripetal acceleration?

All the centripetal acceleration is is the vertical component of a parabolic motion. To understand, let me use satellites as an example. The curvature of earth drops about 16 feet every 5 miles. Every single object which falls on earth, (excluding air resistance), falls almost exactly 16 feet after the first second of free fall. If an object is launched at 5 mi/s just above the surface of earth, it will cover 5 miles horizontally after one second and 16 feet vertically after one second. Any other circular motion is just the same. The earth is so big and the horizontal speed is so small relative to its size that the vertical componet/ centripetal acceleration is noticeable. Zooming in on any circular motion object, and slowing down time, the vertical component is noticeable.

Stil I am not able to understand how he got this equation

Samjh me nahi aaya

Woo!

Pfffff look at this guy thinking we remember geometry…..

Mind blowing sal. Thank you

Who says videos from 2011 doesn't help?

Why is the time (T) in the left circle is the same as the one in the right? Like, how can we compare the change in the velocity vector direction and the acceleration vector direction?

r, v and a are considered by only magnitude (they are constant) and the direction change that respect to time is neglected. In this sense, the change of r , v are just position changes, in the track of motion, which is arc length. It is very good interpretation for the uniformly centripetal acceleration formula. It is indeed a miss for physics lectures that are supposed to cover. Thank you for teaching me something new that is an important formula derive in physics.

I didnt get the transfer of acceleration vector to first circle

thanks