Question

Solid sphere $A$ is rotating about an axis $PQ$. If the radius of the sphere is $5 cm$ then its radius of gyration about $PQ$ will be $\sqrt{x} cm$. The value of $x$ is __

Answer 1

$I _{ cm }=\frac{2}{5} MR ^2$
$I _{ PQ }= I _{ cm }+ md ^2$
$I _{ PQ }=\frac{2}{5} mR ^2+ m (10 cm )^2$
For radius of gyration
$I _{ PQ }= mk ^2$
$k ^2=\frac{2}{5} R ^2+(10 cm )^2$
$=\frac{2}{5}(5)^2+100$
$=10+100=110$
$k =\sqrt{110} cm$
$x =110$