Question #42436

This question tests the non-intuitive coupling between rotational dynamics and the normal force for a non-uniform body.

The primary trap is the assumption that the normal force $N$ equals $mg\cos\theta$.

Because the center of mass is offset and the cylinder undergoes angular acceleration, the center of mass has a component of acceleration perpendicular to the incline ($\vec{a}_{Gy} = -\alpha R/2$).

This dynamic effect reduces the required normal force.

Students who blindly use $N = mg\cos\theta$ will incorrectly calculate $\mu_s = 1/4$.

Furthermore, the problem elegantly allows the use of $\tau_O = I_O \alpha$ despite $O$ not being the center of mass, because the velocity of $O$ is instantaneously zero at release, eliminating the $\vec{v}_O \times \vec{p}$ pseudo-torque term.