Question

The radius of gyration of a uniform solid sphere of radius R is $\sqrt{\frac{2}{5}}$ R for rotation about its diameter. Show that its radius of gyration for rotation about a tangential axis of rotation is $\sqrt{\frac{7}{5}}$ R.

Answer 1

By parallel axis theorem

$I_o=I_c+M{h}^2$

$M{K}_o^2=M{K}_c^2+M{h}^2$

$K_o^2={\left(\sqrt{2}5R\right)}^2+R^2$

$K_o^2=\frac{2}{5}{R}^2+R^2$

$K_o^2=\frac{2R^2+5R^2}{5}$

$K_o^2=\frac{7R^2}{5}$

$K_o=\sqrt{\frac{7}{5}}R$