Visual understanding of centripetal acceleration formula | Physics | Khan Academy

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

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[coverup]

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

[ThreeDrSports]

I posted a response to someone, and just want to copy and paste it for anyone else who may be struggling to grasp this concept.

The key here is to treat both circles as describing the motion of ONE SINGLE particle. The way I visualize this is to think of these two circles as existing simultaneously; almost as if they were layered within each other as one large “thing”. So when its Velocity vector is pointed straight upwards (the Green velocity vector of both circles), this is occurring at a singular paused moment in time. Similarly, when the Velocity of this particle changes and points straight to the right (the Blue Velocity vector), that ALSO occurs at a singular paused moment in time.

Let’s look at the first circle on the left. But only focus on the Velocity vectors. Let’s take the Green Velocity vector in the picture that is pointed straight upward. NOW, go to the second circle, and locate where that same exact Velocity vector would be (same color, copied and pasted from the center pointing straight upward).

Now concentrate on just the first circle, and notice where the Velocity vector points 90 degrees clockwise (the BLUE Velocity vector). Trace the path traveled in this circle when the Green vector eventually changed its direction to the Blue vector. NOW, return to the second circle, and notice the path of the same Green vector and how far along the path of the circle it had to go until that Velocity vector was pointed 90 degrees (the Blue vector), and you’ll see that it traces out the same portion of that circle (basically 90 degrees).  

Because this is one single particle moving through space, it’s velocity vectors for each circle are when it is at that location in a singular point in time (the same time; since this is all representing one objects’ motion).

I hope this helps!

[jonahfox]

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

[pablofernandezesteberena]

I have a shorter derivation, inspired by this great video! 😄If the position vector "r" moves in a circular motion with constant angular velocity, then so does the velocity vector "v", as it is tangential to the circle at all times. Then these are two vectors which rotate at the same angular velocity "omega". For the position omega=v/r and for the velocity omega=a/v (by definition of omega). Notice how in the expresion for the rotating velocity, the radius of rotation is its modulus "v", and the role of the derivative is taken by the acceleration "a". We simply isolate "a" in "v/r=a/v" and we get "a=v^2/r" 😊.

[riimzo]

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

[TheAlphaRalph]

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

[ptyptypty]

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

[spaceengineering]

Hi Sal,

Thank you for your amazing videos

at 5:10 when you talk about the intuition of centripetal acceleration and point the acceleration towards the center of the circle, why is it pointed towards the center? Can you please explain.

Thank you

[stavshmueli]

At 6:54 Sal says: "The time it takes to travel this path is the exact same time it takes to travel this path". Why is this statement true?

Ok, so for anyone in the future reading this comment, I've understood why: On the left circle, we calculated the time it takes to get from position vector number 1 (r1) to position vector number 3 (r3). We've found the time to be: T = (0.5πr)/V, and just for clarification, the time was found to be equal to this quantity, because recall that distance = speed x time, so we simply divided our distance (0.5πr), by our speed (V), and we found the time.
So we found the time. But why the time on the left circle is equal to the time on the right circle?
Because if you look on the left circle, the time it takes us to change from r1 to r3, will be the same time v1 changes to v3, thus, the time on the left circle will be the exact time for the right circle.

[atriagotler]

Wow this is so so so different than my text book. And that much better.

My textbook just gives the formula and then shows how to use it :) And this is suppoused to be "world class education" since I live in Finland🤣

[mayaraysin]

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?

[biochem]

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

[robinkovacic]

demystified the whole concept. thank you

[peterfred]

Darn right, a drum role should be given for this video!

[Coolgiy]

You should have also went over how to find velocity and revs/min vs rads/min :(

[alissom]

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

[FreyMayBeHere]

So basically in this case, the position vector (the radius I think) is perpendicular to the velocity vectors and the velocity vectors are perpendicular to the acceleration vectors?

[mithileshloveskaitru]

This is applicable only if the linear velocity remains constant right?, Otherwise the rate of change vector wont be perpendicular to the vector

[goliathstone]

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.

[glennreece]

THANKS A MILLION! REALLY LOVE HOW YOU TEACH!