I tried to make sure I understand uniform circular motion by deriving the formula for centripetal acceleration, $a_c = v^2/r$, rigorously from first principles. I think I did it correctly, but it seems quite complicated - can you show me better ways?

Given a particle moving uniformly in a circule of radius $r$ around $O$ with period $T$, I take it as given that it has tangential velocity at each point with the same constant magnitude. So let's say its velocity at the topmost point, broken down into Cartesian components, is $v_0=(v,0)$. Now some small amount of time $\Delta t$ passes, and the particle rotates by the angle $\alpha = \frac{2\pi}{T} \Delta t$. The magnitude of velocity remains $v$, and it's now pointed in the new tangential direction, so it's easy to see its components as $v_1=(v\cos\alpha, v\sin\alpha)$.

Now acceleration is the limit of $\frac{v_1-v_0}{\Delta t}$ when $\Delta t \to 0$. I just want the magnitude of acceleration which will be

$$\lim_{\Delta t \to 0}\frac{\sqrt{(v\cos\alpha - v)^2 + (v\sin\alpha)^2}}{\Delta t}$$

I can simplify this, by passing to $\alpha$ as the dependent variable ($\Delta t=\frac{T}{2\pi}\alpha$), to

$$v \frac{2\pi}{T} \lim_{\alpha \to 0}\frac{\sqrt{2-2\cos\alpha}}{\alpha} $$

To evaluate the limit, I bring the denominator under the square sign and break down $\cos\alpha$ by its Taylor series:

$$\lim_{\alpha \to 0}\sqrt{\frac{2-2\cos\alpha}{\alpha^2}} = \lim_{\alpha \to 0}\sqrt{\frac{2-2(1-\frac{\alpha^2}{2} + \frac{\alpha^4}{24} - \cdots}{\alpha^2}} = \lim_{\alpha \to 0}\sqrt{1 - \frac{\alpha^2}{12} + \cdots} = 1$$

So in the end the magnitude of acceleration is $v\frac{2\pi}{T}$. But since the particle traces the circumference of the circle, of length $2\pi r$, under constant speed $v$ in time $T$, we have $vT = 2\pi r$, so $\frac{2\pi}{T} = \frac{v}{r}$, and finally the acceleration is the familiar formula $$a_c=\frac{v^2}{r}$$

Did I do this right? And even if I did, was it necessary to do the Taylor series stuff to obtain the trigonometric limit? What are some ways to make this derivation easier, but still rigorous under the formal definition of acceleration in Cartesian coordinates?

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    $\begingroup$ This derivation is in every mechanics book. $\endgroup$ – David G. Stork Mar 21 at 22:33
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    $\begingroup$ The Taylor series is unnecessary: write $$ \sqrt{2-2 \cos \alpha} \ = \ \sqrt2 · \sqrt{1-[1 \ - \ 2 \sin^2 \left( \frac{\alpha}{2} \right) ] } \ \ . $$ Usually, this derivation would be carried out using vectors, but your description of the rotation of the velocity direction is equivalent. $\endgroup$ – boojum Mar 21 at 22:41
  • $\begingroup$ Thank you! After doing this, I would still use the small angle approximation, right? I guess viewed like this, my use of the Taylor series was in effect a proof of the small angle sine approximation? $\endgroup$ – AnatolyVorobey Mar 24 at 23:52
  • $\begingroup$ Your limit expression becomes $$ \lim_{\alpha \to 0}\sqrt{\frac{2-2\cos\alpha}{\alpha^2}} \ = \ \lim_{\alpha \to 0} \ \frac{\sqrt{2} \ · \ \sqrt{2} \ | \sin \left(\frac{\alpha}{2} \right) | }{\alpha} \ = \ \lim_{\alpha \to 0} \ \frac{2 \ | \sin \left(\frac{\alpha}{2} \right) | }{2 \ · \left( \frac{\alpha}{2} \right) } \ = \ 1 \ \ . $$ This last is the "limit law" $ \lim_{u \to 0} \ \frac{\sin u}{u} \ = \ 1 \ $ which justifies the "small-angle approximation". $\endgroup$ – boojum Mar 26 at 3:41

You are making this much more difficult than it has to be. Hints:

$1).\ $Write $\vec r(t)=r\cos \theta(t)\vec i+r\sin\theta(t)\vec j$, note that $r$ is constant, differentiate once to obtain $\vec v(t)$, again to get $\vec a(t).$ You will use the chain rule to do this.

$2).\ $Now, with $\frac{d\theta}{dt}:=\omega, $ the angular velocity, you will get $\vec a(t)=-\omega^2\vec r(t).$

$3).\ $To finish, take the magnitude of both sides, and the formula you want will drop out as soon as you observe that since the speed is constant (why?), one has $\omega=\|\vec v(t)\|/r=v/r.$

  • $\begingroup$ Thank you, appreciate your answer! I guess I was making things difficult by re-deriving, partly, trigonometric derivatives. Why is the speed constant? I want to say that $\omega$ is constant by the definition (?) of "uniform circular motion", but maybe that's wrong-headed and rather more immediately $v$ is constant, but why? $\endgroup$ – AnatolyVorobey Mar 24 at 23:56
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    $\begingroup$ Uniform circular motion means that the angular speed is constant or equivalently that the tangential speed is constant. The force that keeps the particle on this trajectory is directed toward the center of the circle ("radial force"), so it is always perpendicular to the velocity and produces no work on the particle. The net force would no longer be purely radial for varying speed on a circular path. $\endgroup$ – boojum Mar 26 at 3:50
  • $\begingroup$ @AnatolyVorobey it would be a good exercise to repeat the argument, letting $\theta$ be twice differentiable and non-constant. The you can see how boojum's comment relates to the uniformity of the motion. $\endgroup$ – Matematleta Mar 26 at 4:05

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