The moment of inertia of a solid sphere about an axis passing through centre of gravity is 5 2 M R 2 where M is the mass and R is the radius. The radius of gyration about a parallel axis at a distan

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The moment of inertia of a solid sphere about an axis passing through centre of gravity is 5 2 M R 2 ​ where M is the mass and R is the radius. The radius of gyration about a parallel axis at a distance 2 R from first axis is

TestHub · Accepted Answer

Using the parallel axis theorem, the moment of inertia about the new axis is given by $I = I_{c m} + M d^{2}$ .

Given $I_{c m} = \frac{2}{5} M R^{2}$ and $d = 2 R$ .

Substituting the values, we get $I = \frac{2}{5} M R^{2} + M (2 R)^{2} = \frac{2}{5} M R^{2} + 4 M R^{2} = \frac{22}{5} M R^{2}$ .

The radius of gyration $K$ is given by $I = M K^{2}$ .

Equating the two expressions for $I$ , we get $M K^{2} = \frac{22}{5} M R^{2}$ .

Solving for $K$ , we get $K = \frac{22}{5} R$ .

Answer: $\frac{22}{5} R$

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