Visual understanding of centripetal acceleration formula | Physics | Khan Academy

[@coverup9248]

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

[@SahibKhan-un4ek]

Oh no!

[@Three60DrSports]

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!

[@zarrafsharif3099]

Thanks man

[@ImAshod]

Than you a ton! This helped me a lot❤

[@pablofernandezesteberena7456]

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" 😊.

[@jonahfox1503]

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

[@stavshmueli6932]

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.

[@zarrafsharif3099]

Thanks man i appreciate it

[@sazob]

I just don't get why he's saying that t=d/a on the second one

[@saminm1]

I still don't understand :(

[@saminm1]

@@user-tn2zw7qk9k I think I kinda get it now but not too sure lol

[@noor-cm3eg]

​@@user-tn2zw7qk9klol😂😂😂

[@niharika3280]

Mood

[@tlexus0525]

12 years later do you still not understand?

[@rakinrahman890]

@@tlexus0525 lmao

[@ptyptypty3]

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

[@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🤣

[@DaughterofPersephone0630]

Yeah, sometimes I get really frustrated over "it is what it is, just use it" kind of education mathematics and physics offer. But then again, not everyone want or need to understand the formulas fundamentally, right? I think that's reserved for the majors.

[@spaceengineering6529]

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

[@nikhila6536]

Acceleration is in same direction of force

[@goliathstone3110]

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.

[@TheAlphaRalph]

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

[@biochem86]

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

[@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?

[@robinkovacic8145]

demystified the whole concept. thank you

[@Coolgiy67]

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

[@mithileshloveskaitru]

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

[@khizarhayat1387]

Yes then the formula will be a=underoot a tengential accleration aquare +centripetal accelartion square

[@pellurupadmaja675]

Sir plz do a video on angular acceleration 🙏🙏

[@ralphy1054]

Why can you simply consider the change in velocity to be equal to the geometric distance in the circle representation? Why exactly is it correct to use a circle representation where the radius is equal to the velocity's magnitude? Yes, the direction of vector v is constantly changing, but why is it quantifiable by an abstract geometric representation based off the vector's magnitude?

[@vortyx090]

AND WE'RE DONE!!!

[@riimzo]

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

[@Wackaflaka89]

If this is what you're asking, Because both equal T and so you can set them equal to eachother, and rearrange/solve for whatever variable. And all circumferences aren't all (1/4)(2)(pi)(r), it cancels out so it doesn't matter.

[@peterfred445]

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

[@skylar011]

Thankyou for this.

[@mayaraysin3284]

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?

[@Three60DrSports]

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!

[@frankivinstutoring3775]

Best explanation of a=v^2/r I've seen!

[@WHY70122]

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 :D

[@soulcutters375]

WHY70122 this is so hard, could you explain why the centripetal acceleration pointed toward the center? Or isn’t it the reason an object can moving around in circle? Or is there a reason to this?

[@MMuhammad-e7j]

What is the direction of the linear acceleration in circular motion and is it always the same?

[@ligma...]

Sir you are great !!!!!!

[@evelinelsa7806]

THANK YOU ...THIS HELPED ME A LOT

[@hvvcvgggh7595]

sir u taught that velocity is constant in this motion but thats not true. speed is constant here. speed is magnitude of velocity only when the direction is constant throughout the motion.

[@ddcll9538]

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

[@alissom987]

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

[@glennreece1999]

THANKS A MILLION! REALLY LOVE HOW YOU TEACH!

[@hafizagungmaulana8936]

thank u for ur amazing video!

[@h7opolo]

6:23 you should define a variable for arc-length for greater understanding

[@nhmllr725]

@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.

[@leticiazarate9553]

Thank you!

[@MurderBirdExpress]

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

[@june23]

Spawn I think it’s because the amount of time it took for velocity to move the position is equal to the amount of time for acceleration to accelerate the velocity

[@davidaubrey2037]

I don't think this has anything to do with velocity or time..what he did i think is parallel translation of the velocity vectors.Since the vectors are on the same plane you can translate it to another position without changing its magnitude.Its a property of vectors.Anyone here with an alternative explanation?

[@yw5617]

The reason time is equal is because by the time the direction of the velocity has changed by 90 degrees the moving object has moved across one quarter of the circle, thus the direction shifted by one quarter

[@bryanchristian5849]

Understandable, have a good day

[@soulcutters375]

Is centripetal acceleration the reason an object can moving around in circle?

[@williamwolseley-charles2045]

I believe centripetal acceleration is the calculation we use in which we can know where an object will be at any given point in time around an elliptical circle, this video though assumes that the velocity of the object is constant, unfortunately, planets in our solar system are not constant and have changing velocity vectors so it becomes harder to predict. The sun's gravitational pull on planets ensures planets don't leave the solar system but planets also a relative "forward" motion and therefore don't get sucked in by it and instead orbit.

[@soulcutters375]

@@williamwolseley-charles2045 wow this is much easier to understand. Thank u

[@sonic66646]

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

[@mizelcluett2105]

And we're done!

[@FibroniaShenoudaMahrous]

it was so amazing explain

[@moumusicharmoniumtutorial1835]

Can understand easily than my textbook

[@disturbedlol]

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

[@raydredX]

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

[@shreoshiroy]

Really great and innovative derivation without using calculus.

[@jojogirl2008]

beautiful intuition and explanation! Thank you!

[@BoZhaoengineering]

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.

[@plasticelephant1969]

Why schools refuse to explain this and demand us to memorize centripetal acceleration = v^2/r

[@arshisid47]

this was such an awesome video!!

[@arifkarim768]

Mind blowing sal. Thank you

[@azaezalbelthood9188]

Sir Ur a legend.thank you so much.

[@saminm1]

Brilliant! Your awesome, Sal

[@kalebmendez439]

I lost it when he says R2 😂😂😂😂

[@virupannamedikinal]

I have never understood centripetal acceleration this much better,thank you so much Sir

[@PKMNFan4664]

This helped so much! Thank you!

[@kavishshah8282]

so interesting......

[@TheSteinmetzen]

Thanks Sal!

[@angiem7725]

I have never been so confused in my life

[@naveennandhanan547]

Angie M I hv no idea why you feel that way...

[@silverfoxidm]

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

[@Hino_55]

Great visual derivation sir Sal!

[@iPhone3GSCase]

such a BOSS!

[@ProlificPianist]

Looks kind of like if you could take the derivative of a circle. If the velocity directions is your "slope" of sorts (shown in the first, left-most circle example), and you take the derivative of that circle, the center would be the velocity fxn. For every position of velocity in this circle, it would have its own directional velocity (as if you're finding the velocity of a velocity position), which as we know would be acceleration, shown in the 2nd circle. Probably not the CORRECT way of looking at it but I thought it was a cool perspective.

[@SwellBread]

I got my degree at Khan Academy!

[@MrABDURREHMAN1]

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

[@brandontea3815]

I’m sorry, may I know where you get the “w”?

[@lucasgerosa4177]

🤯

[@rakeshkulkarni4323]

appealing derivations compared to NCERT TEXTBOOK

[@anantmanglani5241]

Awesome explanation 👍

[@trevorwills3654]

Thank you so much for this video

[@muhammedansar3558]

I didnt get the transfer of acceleration vector to first circle

[@vecrleker7407]

Same!! I feel sad that no one has replied to you, guess I will have to keep scrolling

[@shubhamgund9616]

thxx buddy it really helped mee.....😜

[@sivanathanvithurshan2076]

thanks

[@ryandward]

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?

[@orchoose]

great explanation

[@akulakala9330]

N

[@mohammadjj]

you're awesome

[@Hanin_Alharbi]

The best thing that is have a Arabic caption💞❤️

[@MahmoodSaeedB]

Did you get the video?

[@strahinjanikolic4559]

Why would first T and second T be the same?

[@carultch]

Think of a hand traveling along a clock face. At the 12 O'clock position, the position is straight up, and the velocity is pointing to the right. At the 3 O'clock position, the position is rightward, and the velocity is downward. At the 6 O'clock position, the position is downward, and the velocity is leftward. At the 9 O'clock position, the position is leftward and the velocity is upward. As you can see, the velocity vector is foreshadowing what the position will be in the next quarter cycle. The position chases the velocity. The velocity chases the acceleration. The two cycles have the same period and same frequency, and it is the phase that is different.

[@harshdeepchhabra2193]

Idan Gelbard yeah but the distance changes as the first circle has radius R and second circle has radius V

[@sccm100]

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.

[@gughanmali4632]

Is 4π²r/T² another formula for centripetal acceleration?

[@johndoesson]

yes

[@gingerdrinkr]

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

[@benjaminneufeld6137]

wait how did the velocity get to be the velocity?

[@june23]

Benjamin Neufeld I lol’d at this, I wonder what the answer to your question is too

[@Celicaw8]

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

[@mistakenmeme]

Start at the beginning!!!!

[@khaledaltowairgi9528]

Me two😭

[@talal1458]

I dont get it Why Khan!!????? 😂 😂 😬❓❓❓

[@harshdeepchhabra2193]

Why is the time same in both cases

[@aminsaraj4786]

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

[@muhammedansar3558]

Amin Saraj because its mentioning about uniform motion along circular path.

[@carultch]

To keep it simple, he first considers the special case of speed being constant. Not the velocity, but the speed...i.e. the magnitude of velocity. The equation ac = v^2/r is still valid even if speed isn't constant, it just isn't 100% of the picture of acceleration. It is just the centripetal component of acceleration (perpendicular to velocity), in the event that speed isn't constant. Tangential acceleration, that is aligned or opposite velocity, is equal in magnitude to the rate of change in the speed.

[@surindersinghrawal8415]

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

[@mr.yuriinspector347]

Woo!

[@bimbarstudios]

Just someone from 2015 passing through

[@tamas972]

2018

[@cullenwatson8296]

2019

[@DeepaSingh-ey7jw]

2020

[@serenity7637]

2024

[@dq9021]

I don't know what to say.

[@codosacho5924]

i wonder why there is acceleration ??

[@mufasaamen3181]

as long as there's force, there's acceleration. the force is coming from the centripetal force

[@gonzaloayalaibarre]

Codo Sacho Remember that change in velocity = Acceleration. When he drew the vectors representing velocity he made a line between them, the change in velocity, this answer is no longer for you, it seems xD

[@systempatcher]

+Mufasa Amen That is only true for circular motion.

[@orchoose]

+systempatcher its true whenever you change speed... you experience force

[@carultch]

It is true for all curved path motion. Circular motion is a special case, where the radius of curvature is uniform. And uniform circular motion is a special case where the speed is constant. In a general sense, curvature can change, direction of curvature can change, and speed can change. So you introduce a lot more varying terms, when you consider non-circular curved path motion. Still, at an instantaneous condition, general curved-path motion still can be looked at as circular motion. There will be a well-representative radius of curvature of the path at any given point, and you can draw a circle that matches the direction and curvature called an osculating circle. When you work with the physics of general curved path motion, on an instantaneous basis, it matches what it would be for travelling around that osculating circle, instead of the much more complicated path. With non-uniform circular motion, there can be a change in speed and direction. There are two components to the acceleration vector. Centripetal acceleration that causes the change in direction, keeping it on the curved path, and tangential acceleration, which is its rate of change in speed.

[@DarkSwanz]

1st?

[@YAHOOOOOO5]

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?

[@TheRunescapeian]

+YAHOO i think i know what you want to know, it is an acceleration because you are now accelerating in another direction, right? Take a car, going straight 30m/s north then take the NE exit, you are now accelerating in another direction, remember the acceleration is a vector. --- it's something like that i'm pretty sure

[@darknut305]

vectors are straight. if you want it to change there needs to be an outside force. it's called radial acceleration. sure you can maintain constant speed around a curve, but the direction is changing.

[@carultch]

Acceleration does not necessarily mean changing speed. As a physics concept, it means changing velocity. Velocity is a vector quantity. If either the direction or the speed is changing, then the velocity is changing. Therefore, there would be an acceleration to cause the change in velocity.

[@Nabeelahmed-vj1rc]

Yup there is magnitude is constant but continuesly chamge in direction which means change in velocity that is equal to acceleration

[@nhmllr725]

@riimzo No problem. ;p

[@dhruvakashyap9331]

Are you God?

[@qualquan]

OMG such a complicated geometric derivation xmtutor does a much better job

[@benjaminneufeld6137]

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

[@autotitslinger]

torsion

[@ivininasuchungath8012]

why did he made radius velocity?

[@raccooncitymassacreofficia9902]

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