Physics - Rotation Question with Solution | TestHub

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

Using the parallel axis theorem, the moment of inertia of the disc about an axis passing through a point on its rim and perpendicular to its plane is:

$I_{disc}=I_{cm}+M{R}^2=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$

The four particles of mass $m$ are placed symmetrically on the rim. Let the axis pass through one of these particles. The distances of the four particles from this axis are $0$, $\sqrt{2}R$, $2R$, and $\sqrt{2}R$.

The moment of inertia of the four particles about this axis is:

$I_{particles}=m(0{)}^2+m(\sqrt{2}R{)}^2+m(2R{)}^2+m(\sqrt{2}R{)}^2$

$I_{particles}=0+2m{R}^2+4m{R}^2+2m{R}^2=8m{R}^2$

The total moment of inertia of the system is:

$I_{total}=I_{disc}+I_{particles}=\frac{3}{2}M{R}^2+8m{R}^2=\frac{(3M+16m)R^2}{2}$

Answer: $(3\mathrm{M}+16\mathrm{m})\frac{{\mathrm{R}}^2}{2}$