Regarding circular motion...

The way that I know of how to derive the centripetal acceleration is based on the geometrical representation of two instantaneous linear velocities of equal magnitudes on a circle, and comparing the triangles to obtain the relationship $a = \frac{v^2}{r}.$

However, I've seen it stated in the textbook that this formula still holds even when there is angular acceleration and hence the magnitude of both angular and linear velocity are not constant.

My question is, wouldn't the change in magnitude of the linear velocity in subsequent instances render the formula for centripetal acceleration inaccurate under these circumstances, as the derivation above relies on the magnitudes being the same? Or do the shape of the triangles still retain similar relationships, thus $a = \frac{v^2}{r}$ still hold?


It looks like you are missing one major point in derivation of centripetal acceleration: the points in which you consider velocities have to be infinitely close to each other, lest you get the wrong direction. And for infinitely close points you can disregard change of magnitude.

| cite | improve this answer | |
  • $\begingroup$ Sorry for such a late reply, but according to what you said about disregarding magnitude changes, wouldn't the same apply to the direction? At very close points, there are very small changes in magnitude (if there is any) but also very small changes in direction. When differentiating functions of velocity to find acceleration, magnitude changes are still important even though they barely change over infinitely small amount of time or displacement right? $\endgroup$ – user147526 Dec 5 '17 at 11:33
  • $\begingroup$ Sure, but when you go to the limit of zero displacement, magnitude changes will vanish. $\endgroup$ – Tajimura Dec 5 '17 at 12:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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