ok, just to be sure I got this right: the opposite torque you mention should correspond to a force acting upwards on the particle. This force does negative work so the particle is losing mechanical energy (the rod also exerts a centripetal force on the particle but this one does no work.) Right?

You are right, I made a mistake in my drawing: the rod should be attached to the outer edge of the disc, not the central axis. I believe the answer provided by @gandalf61 still applies, though. Thanks for pointing out.

well, since I am studying this just by interest, I don't mind my error propagation being ignored. But it would really helpful if could point me to some source about that mathematical approach to sig figs propagation convention

... more logically consistent approach to handle uncertainty propagation, by questioning this kind of rules. I don't see any logical need to keep the same number of digits in $R$ as in $I$ (they don't even have the same dimension so it is strange to compare the number of digits.) Finally, you also force the relative uncertainty to keep only 1 fig. but again, this is just convention, I believe there is no logical need to this, right?

Thanks for providing a different insight. First, after some thought I believe there is no "true" uncertainty but different ways to get an estimation of it, e.g. in this case you propose the sum of variances rule to characterize the uncertainty using variance, but I think this rule involves, in general, a linearisation of the indirect measured value as a function of the direct measured values. Obviously some estimations are better than other. Secondly, I see that in your derivation you apply "the rules to prpagate sig figs" which is precisely what my question is about. I am looking for a ...

thanks for the answer and the article, it looks interesting. This is not homework, and I although I am not a mathematician, I have some maths background, including some probability and statistics and its application to estimate uncertainties. I am reading this book only for interest, so I hope to have a deeper insight after reading the article you provided

so to be sure, your opinion is that the number of digits in -the estimation of- the uncertainty should be chosen according to simplicity, and fewer digits make a simpler value. And it has nothing to do with a comparison between the original uncertainties and the final one, as the author proposes. Did I get your point?

Although I can see that $\sum F_y^{ext} \neq 0$ so that $p_y$ is not conserved, it really seems from the picture that $\vec{v}_1$ is tangential to the circumference, and this holds for the instants before and after the collision. I believe that stating that $p_{y1} \neq 0$ after the collision is against the hypothesis that the collision is instantaneous.

I understand your point. But then, if you claim that $p_y$ is not conserved during the instantaneous collision, then you are implying that $p_y$ is not zero just before and after the collision, as it looks in the picture. Since $p_y = p_{y1} + p_{y2}$, would you say that $p_{y1}$ or $p_{y2}$ (or perhaps both) are nonzero immediately before or after the collision?

it seems to me that if we assume an instantaneous collision, then the velocity of $m_1$ is horizontal, both immediately before and immediately after the collision, whereas the velocity of $m_2$ is zero before the collision and horizontal immediately after. Since all non-zero velocities are horizontal immediately before/after the collision, I would say that momentum is conserved in both axes.