Question

The moment of inertia of a solid sphere about an axis passing through the center of gravity is, $\frac{2}{5}M{R}^2$ then its radius of gyration about a parallel axis at a distance $2R$ from the first axis is

• A. $5R$
• B. $\sqrt{\frac{22}{5}}R$
• C. $\frac{5}{2}R$
• D. $\sqrt{\frac{12}{5}}R$

Answer 1

The correct option is B

$\sqrt{\frac{22}{5}}R$

Given data:

Moment of inertia of a solid sphere about an axis passing through the center of gravity $=\frac{2}{5}M{R}^2$

By parallel axis theorem, the moment of inertia about a parallel axis at a distance $2R$ from the first axis is

$I=\frac{2}{5}M{R}^2+M{\left(2R\right)}^2$

$\Rightarrow I=\frac{22M{R}^2}{5}$

If $K$ is he radius of gyration then the moment of inertia of the sphere about the said axis is

$M{K}^2=\frac{22M{R}^2}{5}$

Therefore $K=\sqrt{\frac{22}{5}}R$

Hence option B is correct.