Deriving the centripetal acceleration formula

A body that moves in a circular motion (of radius r) at constant speed (v) is always being accelerated.  The acceleration is at right angles to the direction of motion (towards the center of the circle) and of magnitude v2 / r.

The direction of acceleration is deduced by symmetry arguments.  If the acceleration pointed out of the plane of the circle, then the body would leave the plane of the circle; it doesn’t, so it isn’t.  If the acceleration pointed in any direction other than perpendicular (left or right) then the body would speed up or slow down.  It doesn’t.

Now for the magnitude.  Consider the distance traveled by the body over a small time increment Δt:

We can calculate the arc length s as both the distance traveled (distance = rate * time = v Δt) and using the definition of a radian (arc = radius * angle in radians = r Δθ:)

The angular velocity of the object is thus v / r (in radians per unit of time.)

The right half of the diagram is formed by putting the tails of the two v vectors together.  Note that Δθ is the same in both diagrams.

Note the passing from sin to cos is via l’Hôpital’s rule.


Comments (51)

  1. Jack says:

    But it doesn't give you direction, whereas acceleration is a vector quantity

  2. Jake says:

    Jack I believe the direction is given by v, because it is also a vector. This equation can be rewritten vector component wise.

  3. jaz says:

    plz make this more easier .i does not understand this .

  4. Sid says:

    Express in a easy way. With out using vectors notation.

  5. student says:

    nothing special

  6. Daniel says:

    where did you get sin(dtheta/2)?

    1. Mehrudin says:

      We need linear velocity and the radius for solving… And we know that sin angle is equal to opp side upon hypotenuse…

  7. @Daniel

    I got sin(Δθ/2) = (Δv/2)/v as follows.

    Look at the right-hand side of the figure diagram; it's a triangle with two red arrows and a green arrow. There's a black line from the place where the two red arrows meet, to the midpoint of the green arrow.

    There's a right triangle made up of the top red arrow, the top half of the green arrow, and the black line.

    The hypotenuse of this right triangle is v. The side made up by half of the green arrow is Δv/2. The angle opposite the green side is Δθ/2.

    sin = opposite/hypotenuse, so sin(Δθ/2) = (Δv/2)/v.

    1. Angel says:

      thnqu very Much…….Now I UNDERSTOOD this derivation

  8. Daniel says:

    Oh I see it now, thank you for the explanation [MSFT].  

  9. Noor says:

    Can't get it,s0 tense ab0ut this and just g0nna cry :-(

  10. nicesourav says:

    thats nice

  11. sourav says:

    mr msft can you please explain 5line of the second derivation

  12. sin (Δθ/2) and Δθ/2 both tend to 0 as Δθ tends to 0, so we have the indeterminate form 0/0.

    We can use l'Hôpital's rule; take the derivative of the top and the bottom with respect to Δθ.

    Top: (sin (Δθ/2))' = (cos (Δθ/2))((Δθ/2)') = (cos (Δθ/2))(1/2). This tends to 1/2.

    Bottom: (Δθ/2)' = 1/2. This tends to 1/2.

  13. sourav says:


  14. ritzy says:

    it could be made much more easier without using trigonometry in it….

  15. Jean says:

    from sin delta theta/2 =deltaV/2 hw did deltaV become 2V sindeltatheta/2

  16. Jean says:

    from sin delta theta/2 =deltaV/2 hw did deltaV become 2V sindeltatheta/2

  17. Jean says:

    from sin delta theta/2 =deltaV/2 hw did deltaV become 2V sindeltatheta/2

  18. glizel says:

    i gets the lesson of my teacher

  19. Nizam jokhio says:

    very good but i can't understand anything in the formula!

  20. Mohammed says:

    Nicely done thank you Maurits , this is the simplest way to derive the centripetal acceleration

  21. yo yo says:

    bakwaas hai

  22. aditi says:

    Oh finaly I got it .. thx a ton

  23. Jeremy says:

    There's a far simpler derivation than this. Draw the circle with the two velocity vectors separated by a tiny angle. Draw a vector diagram that shows that dv = vdq. Divide both sides by dt, then a = vw.=v2/r.

  24. Venkat says:

    It's a beautiful proof :)

  25. aijie says:

    complicated ….

  26. raiya says:

    i have not understand the third stage????

  27. sarthak says:

    Mr Maurits u have explained it super … ur w

    ay man its very nice no use of those shitty components simple……. thnx man

  28. sarthak sharma says:

    Mr Maurits u have explained it super love ur way simple and easy the best part u have done is that u have avoided the use of those  components that confuse..can I use this method all this I hopitals rule for an 11 the class exam?

  29. Gaurav gupta says:

    Make it a bit simpler,please!

  30. Callum says:

    For the limit as delta theta [D0] approaches O of v*sin(D0/2)/D0/2, can't the sine component be thought of as its infinite series too, with (x – x^3/3! + x^5/5! +…)/x = 1 – x^2/3! + x^4/5! +…, so as x approaches O, or D0 does so, the series approaches 1; making the initial equation = v*1 = v

    I came here trying to understand the underlying mathematics, it's just when someone uses someone else's theorems or whatever in their derivation, I then spend forever trying to understand the utilised theorems and end up on an infinite search for understanding. The infinite series method just helped me jump one of those hurdles — and I'm hoping it helps someone else like minded :P

  31. Yes, that is sensible.

  32. prince says:

    Nice style

  33. Ashish says:

    Easiest one

  34. many says:

    coud you please explain me what is i'hopital's rule?

  35. sru.. says:

    Its soo difficult to understand…..

  36. Ana says:

    Not understanding

  37. Ikram says:

    Two bodies of masses 10kg and 5kg movingvin concentric orbits of radii R and r such that their periods are same. Then the ratio between their centripetal accelerations is:

    (a) R/r




    plz help me give me the solution of this question

  38. Ali says:

    fabulous explanation

  39. User says:

    Thank you for posting a simple and easy to understand mathematical derivation of the equations. I had to google this and the first 4 links weren’t helpful. You were the 5th link. Also thanks for posting the note Re: L’Hopital’s rule. I did not get that step until I saw your note. PERFECT!

    1. Arshad Ali says:

      Hey bro is it easy??? Then u make it more easy for us

  40. Varshini says:

    Hard to understand. Very lengthy derivation. Please shorten and post.

  41. Varshini says:

    Hard to understand. Very lengthy derivation. Please shorten and post..

  42. sejal says:

    Its very difficult plzz try to explain it in a easiest way😦

  43. sejal says:

    Can’t undrstand plzz try to explain in easist way😦

  44. Arshad Ali says:

    It is too much lengthy and difficult to understand sir

  45. Ryu says:

    Good helpful, but please mention mathematically

  46. Kurt says:

    Learning about oscillators and I haven’t seen this proof for a few years. thanks for putting it up.

    1. Kurt says:

      People are saying it’s confusing. i agree there are some parts that are kinda glossed over.
      First, the two velocity vectors on the right side are the velocities at different times. You can imagine how on the left side you could have an entire circle with a v vector on every point of the circumference extending tangent to the surface. Using this idea that the tangent v we are thinking of would trace out the circle that the object follows over one period, we can see delta theta is the same because of symmetry / the geometry of the circles is the same.
      Second, there is the l’hopital’s rule part. Basically, provided some conditions are met, the limit of a ratio of functions at a point is the same as the limit of the ratio of their derivatives at that point. The four lines on the wikipedia page explain it. For the top you take out dtheta/2 and change sin to cos and on the bottom dtheta/2 becomes 1/2. He also canceled dtheta
      Also, the last line is the chain rule. If you don’t get stuff, just look it up!