Physics - Rotation Question with Solution | TestHub

The moment of inertia of a thin hollow cylinder of mass $M$ and radius $R$ about its geometrical axis is $I_{cm}=M{R}^2$.

The tangential axis is parallel to the geometrical axis and is at a perpendicular distance $d=R$ from it.

Using the parallel axis theorem, the moment of inertia about the tangential axis is:

$I=I_{cm}+M{d}^2$

$I=M{R}^2+M{R}^2=2M{R}^2$

The radius of gyration $K$ is given by $I=M{K}^2$.

$M{K}^2=2M{R}^2$

$K=\sqrt{2}R$

Answer: $\sqrt{2}R$