enter image description here

I was trying to make my own proof for centripetal acceleration (a = (v^2)/r).

I took two points along the path of motion A and B, which are a distance (along the circle) of half the circumference of the circle.

I made them half way, as that would mean they would have exactly opposite velocities, making calculations easier .

I calculated the time taken to travel from A to B by getting the time period and halving it (using the speed and total circumference).

I then simply used an equation of motion to get acceleration, although I seem to have this dangling (2/pi), which has left me stumped.

Why does my proof not work?

  • 1
    $\begingroup$ I think you've calculated the average acceleration over half a cycle. In order to get the instantaneous acceleration you have to do the equilvalent calculation for an angle $\theta$ and let $\theta \rightarrow 0$. Note that in your method there's no well defined place for which you are expressing the acceleration. $\endgroup$
    – garyp
    Sep 7, 2016 at 19:38
  • 7
    $\begingroup$ Acceleration is not constant! Only its magnitude is constant. $\endgroup$
    – garyp
    Sep 7, 2016 at 19:41
  • 1
    $\begingroup$ @DavidSchwartz Toss a baseball straight up in a uniform gravitational field; it will fall back down to you, even though it has constant, non-zero acceleration. $\endgroup$ Sep 8, 2016 at 0:55
  • 1
    $\begingroup$ "I made them half way ... making the calculations easier." If you had used a full revolution, the calculation would be easier yet (namely zero), and even more clearly too low. $\endgroup$ Sep 8, 2016 at 0:57
  • 1
    $\begingroup$ @AndreasBlass Right. I should have said "to a state it was previously in". $\endgroup$ Sep 8, 2016 at 17:22

2 Answers 2


Because you are calculating an average acceleration rather than the magnitude of the instantaneous acceleration. You are certainly not the only one to try it that way.

Near the top of your circle the acceleration points mostly down. Near the bottom it points mostly up. In the average that you have computed those parts cancel out, resulting in your result being low.

You can recover the correct value by taking the limit of the above and the angle traversed gets very small.

  • 1
    $\begingroup$ Couldn't have said it better myself. :) $\endgroup$
    – garyp
    Sep 7, 2016 at 19:40

I don't know if you know calculus, but let me show you a simple proof:

The trajectory of a particle in uniform circular motion can be written as:

$$\mathbf{r}(t)=(R\cos (\omega t), R\sin (\omega t))$$

Then its velocity is just the derivative

$$\mathbf{v}(t)=(-\omega R \sin(\omega t), \omega R \cos(\omega (t)),$$

and differentiating again we get the acceleration:

$$\mathbf{a}(t)=(-\omega^2 R\cos (\omega t), -\omega^2 R\sin (\omega t)).$$

If you take the magnitude of these vectors you find

$$|\mathbf{v}|^2=\omega^2R^2, \qquad |\mathbf{a}|^2=\omega^4R^2,$$

Thus combining these results you see that the magnitude of the acceleration is as expected:

$$|\mathbf{a}|^2 = \dfrac{|\mathbf{v}|^2}{R}.$$

Where your derivation breaks down? In supposing the acceleration is actually constant and that the mean acceleration is equal to the instantaneous acceleration.

If you follow the steps you see that although the magnitude of acceleration is constant and equal to $\omega^2 R$, the direction keeps changing with time.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.