Proof of centripetal acceleration formula ($a_c = \frac{v^2}{r}$) for non-uniform circular motion
The formula for centripetal (radial) acceleration is well known, and there exist many proofs for it: $$||a_c|| = \frac{||v||^2}{r}$$
However, all the proofs I've seen rely on the fact that it is uniform circular motion and the magnitude of the tangential velocity vector does not change. For instance, take the classic proof using similar triangles — the similarity can only be established if the final tangential velocity vector and the initial one are of the same length.
In addition, take this calculus based proof from Khan Academy, outlined as follows:
For this proof to work, $\frac{\text{d$\theta $}}{\text{dt}}$ must be considered a constant, $\omega$, that does not depend on time. In the case of non-uniform circular motion, however, this is not always true as since there exists a tangential acceleration along with a radial one, $\omega$ must depend on time and is not necessarily a constant value.
Intuitively I understand that the centripetal/radial acceleration depends only on the difference in orientation between two tangential velocity vectors, and that their magnitudes do not matter -- hence the formula intuitively holds true in the non-uniform case. 1002
However, how would you go about modifying either of the proofs presented so that they are still valid in this case? Or alternatively, is there another proof that holds valid even when there exists tangential acceleration?
As per the suggestion of one of the answers, I let $\omega$ vary with time and took its derivative as $\alpha(t)$. This is my work so far. Unfortunately, I am stuck after the last step.
$$\overset{\rightharpoonup }{p}(t)=r \cos (\theta (t))\cdot\hat{i}+r \sin (\theta (t))\cdot\hat{j}$$ $$\overset{\rightharpoonup }{v}(t)=-r \sin (\theta (t))\cdot\omega(t)\cdot\hat{i}+r \cos (\theta (t))\cdot\omega(t)\cdot\hat{j}$$ $$\overset{\rightharpoonup }{a}(t)=(-rcos(\theta(t))\cdot\omega(t)^2 - rsin(\theta(t))\cdot\alpha(t))\hat{i} + (-rsin(\theta(t))\cdot\omega(t)^2 + rcos(\theta(t))\cdot\alpha(t))\hat{j}$$
From here on, $\theta(t)$ is represented as just $\theta$ for brevity and clarity
$$\overset{\rightharpoonup }{a}(t)= -\omega(t)^2(rcos\theta\cdot\hat{i} + rsin\theta\cdot\hat{j}) -\alpha(t)(rsin\theta\cdot\hat{i} - rcos\theta\cdot\hat{j})$$ 521