## Engineering Mechanics: Statics & Dynamics (14th Edition)

$k=0.87m$
We can determine the required radius of gyration as follows: $I=\Sigma (I^{\prime}+md^2)$ $\implies I=\frac{4(0.2)^2}{2}+4(1.2)^2+\frac{2(0.1)^2}{2}+2(0.6)^2+\frac{4(1.5)^2}{12}+4(0.5)^2=7.57Kgm^2$ and $m_p=4+2+4=10Kg$ Now $k=\sqrt{\frac{I}{m_p}}$ We plug in the known values to obtain: $k=\sqrt{\frac{7.57}{10}}$ $\implies k=0.87m$