Physics - Rotation Question with Solution | TestHub

The moment of inertia of a circular disc of mass $m$ and radius $R$ about an axis passing through its centre and perpendicular to its plane is $I_z=\frac{1}{2}m{R}^2$.

By the theorem of perpendicular axes, the moment of inertia about its diameter is $I_d=\frac{I_z}{2}=\frac{1}{4}m{R}^2$.

The radius of gyration $k$ is given by $I_d=m{k}^2$.

Equating the two expressions for $I_d$:

$m{k}^2=\frac{1}{4}m{R}^2$

$k^2=\frac{R^2}{4}$

$k=\frac{R}{2}$

Answer: $R/2$