JEE Main 2018 — Rotational Motion Question with Solution

Moment of inertia of any disc form its center perpendicular to the plane of disc = $$\frac{1}{2}M{R}^2$$

Moment of inertia of any one of the outer disc about an axis passing through point O and perpendicular to the plane

$$I_1=\frac{1}{2}M{R}^2+M{\left(2R\right)}^2$$ = $$\frac{9}{2}M{R}^2$$

Moment of inertia of the entire system about point O

$$I_o=\frac{1}{2}M{R}^2+6I_1$$

= $$\frac{1}{2}M{R}^2+6\times \frac{9}{2}M{R}^2$$

= $$\frac{55}{2}M{R}^2$$

By symmetry, point $O$ is center of mass of the complete configuration. The point $\mathrm{P}$ is located at a distance $d_p=3R$ from the center of mass $\mathrm{O}$. The parallel axis theorem gives moment of inertia of the configuration about a normal axis through $\mathrm{P}$ as

$$I_P=I_O+7M{\left(3R\right)}^2$$

= $$\frac{55}{2}M{R}^2+63M{R}^2$$

= $$\frac{181}{2}M{R}^2$$