JEE Main 2023 — Rotational Motion Question with Solution

Solid Sphere :

$$\begin{array}{rl}& I_{\text{tangent }}=I_{\mathrm{cm}}+m{R}^2\\ & \\ & =\frac{2}{5}m{R}^2+m{R}^2=\frac{7}{5}m{R}^2\\ & \\ & =7R^2(m=5\mathrm{kg})\end{array}$$

CIRCULAR DISC :

$$\begin{array}{rl}{I}_{\text{disc }}& =I_{\mathrm{cm}}+m{R}^2\\ & \\ & =\frac{m{R}^2}{4}+m{R}^2\\ & \\ & =\frac{5}{4}m{R}^2\\ & \\ & =5R^2\end{array}$$

$\frac{l_{\text{disc }}}{l_{\text{tangent }}}=\frac{5}{7}$

Concept :

1. For Solid Sphere :

Position of the axis of rotation : About its diametric axis which passes through its centre of mass


Moment of Inertia (I) = $$\frac{2}{5}M{R}^2$$

Radius of gyration (K) = $$\sqrt{\frac{2}{5}}R$$

Position of the axis of rotation : About a tangent to the sphere


Moment of Inertia (I) = $$\frac{7}{5}M{R}^2$$

Radius of gyration (K) = $$\sqrt{\frac{7}{5}}R$$

2. For CIRCULAR DISC :

Position of the axis of rotation : About an axis perpendicular to the plane and passes through the centre


Moment of Inertia (I) = $$\frac{1}{2}M{R}^2$$

Radius of gyration (K) = $$\frac{R}{\sqrt{2}}$$

Position of the axis of rotation : About the diametric axis


Moment of Inertia (I) = $$\frac{1}{4}M{R}^2$$

Radius of gyration (K) = $$\frac{R}{2}$$