This question already has an answer here:
In the case of a particle on a line, if I prepare an (although non-normalizable) eigenstate of the momentum so that the uncertainty in the measurement of the momentum vanishes, the uncertainty principle is (at least in a physicist's sense) still satisfied as the position measurement would yield a value from $-\infty$ to $\infty$ with an equal probability and the uncertainty in the outcome of the position measurement thus diverges.
But, in the case of a particle on a ring, if I prepare an eigenstate of the momentum (which will even be normalizable) so that the uncertainty in the measurement of the momentum vanishes, there seem to be some subtleties around how to think of the uncertainty principle. Naively, the eigenvalues of the position operator range only from $0$ to $2\pi$ and thus, although they are all equiprobable outcomes and the uncertainty in the outcome of the position measurement is maximal, this uncertainty (the variance) can only be finite as the range of the eigenvalues is of finite span. Thus, in this naive look, it seems that the product of the uncertainties in the measurements of the position and the momentum vanishes as the uncertainty in the measurement of the momentum vanishes and that in the measurement of the position is yet finite.
In the derivation of the uncertainty principle, we essentially only take into account the commutator of the operators which remains the same for the position and the momentum operators whether we consider a particle on a ring or a line. Moreover, we tacitly assume normalizable states which would suggest that the case of a particle on a line with a momentum eigenstate should be more subtle than the case of a perfectly normalizable momentum eigenstate of a particle on a ring. But, in fact, the roles seem reversed.
I can only think of one place from which the subtlety is arising. The uncertainty principle somehow takes all the identified positions such as $0,2\pi,4\pi,...$ as distinct and computes a divergent uncertainty as if it were simply a particle on a line. But, I don't know why it should be the case or how we could prevent it from doing it (via deriving the uncertainty relations more carefully) if it is indeed doing so.