Answer
$k_{\circ}=2.167m$
Work Step by Step
We can determine the required radius of gyration as follows:
$I_{\circ}=\Sigma(I_G+md^2)$
We plug in the known values to obtain:
$I_{\circ}=[\frac{1}{12}(2)(2)^2+2(1)^2]+[\frac{1}{12}(4)(0.5)^2+4(2.5)^2]$
$\implies I_{\circ}=28.17Kg\cdot m^2$
Now the radius of gyration is given as
$k_{\circ}=\sqrt{\frac{I_{\circ}}{m}}$
We plug in the known values to obtain:
$k_{\circ}=\sqrt{\frac{28.17}{4+2}}$
$\implies k_{\circ}=2.167m$
