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  4. If the moment of inertia of a solid sphere of mass M and radius R about an axis passing through its centre of mass is (2/5) MR2 , then its radius of gyration about an axis which is parallel to given axis and at a distance 2R from the center will be

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Q. If the moment of inertia of a solid sphere of mass M and radius R about an axis passing through its centre of mass is $\frac {2}{5} MR^2$ , then its radius of gyration about an axis which is parallel to given axis and at a distance 2R from the center will be

Solution:

$I = \frac {2}{5}MR^2+M(2R)^2$
$\Rightarrow \, Mk^2\,=\,\frac {22MR^2}{5}\Rightarrow \,k\,=\,R \sqrt{\frac{22}{5}}$