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Q. The moment of inertia of a solid sphere of mass $M$ and radius $R$ about a tangent to the sphere is

System of Particles and Rotational Motion

Solution:

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Moment of inertia of the solid sphere of mass $M$ and radius $R$ about any diameter is
$I_{\text{diameter}} = \frac{2}{5}MR^2$
According to theorem of parallel axes
$I_{\text{tangent}} = I_{\text{diameter}} + MR^2$
$ = \frac{2}{5} MR^2 + MR^2 $
$= \frac{7}{5}MR^2$