Q. The moment of inertia of a solid sphere about an axis passing through centre of gravity is $\frac {2}{5}MR^2,$ then its radius of gyration about a parallel axis at a distance 2R from first axis is
Solution:
$I=MK^2= \boldsymbol{\Sigma}MR^2$
$\Rightarrow K=\sqrt {\left(\frac {I}{M}\right)} $
where M is the total mass of the body.
According to theorem of parallel axis
$I=I_{CG}+M(2R)^2$
where, $I_{CG} $ is moment of inertia about an axis through centre of gravity.
$\therefore I=\frac {2}{5}MR^2+4MR^2= \frac {22}{5}MR^2 $
or $MK^2= \frac {22}{5}MR^2$
$\therefore K=\sqrt {\frac {22}{5}}R$
