I wanted to derive centripetal acceleration from scratch and tried using differential equations. But no matter what I did I hit a snag as follows:

$\alpha=$ centripetal acceleration

$\omega=$ angular velocity

$v =$ linear velocity

$s =$ distance traveled (i.e. arc length)

$r =$ radius

$t =$ time

As $\omega = \frac{\mathrm d\theta}{\mathrm dt}$, we have $\alpha =\frac{\mathrm d^2\theta}{\mathrm dt^2}$.

From $\mathrm d\theta = \frac{\mathrm ds}{r}$ we get $\mathrm d^2\theta=\frac{\mathrm d^2s}{r}$.

Also, from $\mathrm dt=\frac{\mathrm ds}{v}$ we get $\mathrm dt^2=\frac{\mathrm ds}{v}\mathrm dt$.

Putting this together, $$ \begin{align} \alpha&=\frac{\mathrm d^2s/r}{\mathrm ds\mathrm dt/v}=\frac{\mathrm d^2s v}{r\mathrm ds\mathrm dt}=\\ &=\frac{\mathrm ds v}{rs\mathrm dt}=\frac{v\mathrm dv}{rs\mathrm dt}=\\ &=\frac{v^2}{rs} \end{align} $$

So you can see the "snag" in the final result - I have a random $1/s$ getting in the way of the actual formula $\alpha=v^2/r$.

I have had a look at the calculus proof (using vectors) of this formula and am aware that there are many others. Obviously something has gone wrong here, and it may be as simple as "you just can't do that". But this has my head tied in knots so I was wondering whether anyone could actually explain why this doesn't work and why $1/s$ pops up in the answer?

  • $\begingroup$ Why is $\frac{d^2s}{ds} = \frac{ds}{s}$? $\endgroup$ Feb 17, 2016 at 10:52
  • $\begingroup$ Why is $$d^{2} t = \frac{ds}{v} dt$$? Why not $$d^{2} t = \left(\frac{d^{2}s}{v^{2}}\right)$$? $\endgroup$ Feb 17, 2016 at 13:46

2 Answers 2


First of all:

[...] the actual formula $α=v^2/r$.

This is not the "actual" formula for angular acceleration $\alpha$. You might be thinking about the cirular motion's radial or centripetal acceleration $a_{rad}=v^2/r$, which is another quantity.

But if that's the case, then you start off with a wrong definition:

As $ω=dθ/dt$, we have $α=d^2θ/dt^2$.

If $\alpha$ is centripetal (or radial) acceleration, then $α=d^2θ/dt^2$ is a wrong formula. This is the formula for angular acceleration, which is usually what we write with the symbol $\alpha$, while centripetal acceleration is usually just written $a_{rad}$ or $a_c$.

If you are still not sure of this, then consider the units:

  • $α=d^2θ/dt^2$ will give $\alpha$ units of $\mathrm{[s^{-2}]}$ (or $\mathrm{[radians/s^2]}$, but the "radians" are always omitted).
  • $a_c=v^2/r$ will give $a_c$ units of $\mathrm{[(m/s)^2/m]=[m/s^2]}$.

To your derivation:

There seems to be a mistake in the line:

$$α = \frac{d^2sv}{rdsdt} = \frac{dsv}{rsdt}$$

You seem to cancel one $ds$ out with one $d$. Remember that the $d$ is just a symbol indicating a change in $s$. It is not a variable by itself and cannot simply be cancelled out - only the $ds$ as a whole can be considered, not $d$ by itself.

  • $\begingroup$ I don't think it's a mistake seeing as the d^2s in question does not represent (dsds) but rather (dds). This comes from dds=d^2s=dds/r=d^2s/r . So more explicitly the canceling out looks like this: α = (dds)v/r(d*s)dt = dsv/rsdt . However, I may be wrong so please do explain if I am. $\endgroup$
    – H.Turki
    Feb 17, 2016 at 9:22
  • $\begingroup$ @H.Turki Aha, I understand, but how can d then cancel out, when d is not a variable? It is just a symbol meaning change in the ds variable. $\endgroup$
    – Steeven
    Feb 17, 2016 at 9:26
  • $\begingroup$ I was under the impression that d can cancel out as long as it pertains to the same variable (in this case s) - also in my edited comment I mistakenly wrote dds=d^2s=... when I meant ddθ=d^2θ=... my apologies $\endgroup$
    – H.Turki
    Feb 17, 2016 at 9:29
  • $\begingroup$ Remember that ds is a small change. That is the name we have given to a small change in s. If a change in distance s equals the change in some other distsnce x, then we can say ds=dx. But that doesn't mean that s=x. If this was the case, then v=ds/dt would equal v=s/t, which is not at all the case (the later is rather the average speed over the entire distance s and not the instantaneous speed, which the former is). d's don't cancel out since they are not variables $\endgroup$
    – Steeven
    Feb 17, 2016 at 9:34
  • $\begingroup$ Ok I see your point :) But now instead of canceling if we replace ds with x (representing a small change in s) we end up with α = dxv/rxdt , and then isn't dx/dt approximately v? or is this a different v as ds/dt is not approximately equal to d(ds)/dt? $\endgroup$
    – H.Turki
    Feb 17, 2016 at 9:58

The goof is in "... we have $\alpha =\frac{\mathrm d^2\theta}{\mathrm dt^2}$." That is the angular acceleration, for example speeding up while traveling along a circle.

Centripetal acceleration is toward the center of the circle, perpendicular to the motion. It quantifies the change in direction of velocity while travelling at a steady speed along the circle.

I assume you are thinking of an object moving steadily along a circle - such as a ball on a rope, a motorcycle stunt rider in a "Ball of Death" - in which case angular acceleration is zero (after being started up) and centripetal is according to the standard formula.


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