Physics - Rotation Question with Solution | TestHub

Moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is $I_C=\frac{1}{2}M{R}^2$.

By the theorem of parallel axes, the moment of inertia about a parallel axis tangential to its rim is:
$I_1=I_C+M{R}^2=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$

The moment of inertia about a parallel axis passing through a point midway between the centre and the rim (at a distance of $\frac{R}{2}$ from the centre) is:
$I_2=I_C+M{\left(\frac{R}{2}\right)}^2=\frac{1}{2}M{R}^2+\frac{1}{4}M{R}^2=\frac{3}{4}M{R}^2$

The required ratio is:
$\frac{I_1}{I_2}=\frac{\frac{3}{2}M{R}^2}{\frac{3}{4}M{R}^2}=\frac{3}{2}\times \frac{4}{3}=\frac{2}{1}$

Thus, the ratio is $2:1$.

Answer: 2: 1