Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2024 / Paper 1 / Section B / 8 / b / Solution

The contact radius is $r_1$, so rolling without slipping gives $v=r_1\omega$. With $I=M{r}_2^2/2$, the energy at the minimum of the stated parabolic motion model isConsequentlyThe corresponding spin angular frequency is ${\omega }_{\max }=v_{\max }/r_1=14.8\mathrm{rad}{\mathrm{s}}^{-1}$. It is the rod radius, not the disc radius, that relates translation and rotation; the discs determine the larger rotational inertia.