Answer
$k_x=57.7m$
Work Step by Step
We can determine the required radius of gyration as follows:
$dm=\rho \pi y^2 dx=\rho \pi (50x)dx$
$I_x=\int y^2 dm/2$
$\implies I_x=\frac{1}{2}\int_{0}^{200} 50x\pi \rho(50x)dx$
$\implies I_x=\rho \pi (\frac{(50)^2}{2})|_{0}^{200}$
$\implies I_x=\rho \pi [(50)^2\times (200)^3/6]$
Now $m=\int dm$
$\implies m=\int_{0}^{200}\pi \rho (50x)dx$
$\implies m=50\rho \pi (x^2/2)|_{0}^{200}$
$\implies m=\rho \pi (\frac{50}{2})(200)^2$
The radius of gyration is given as
$k_x=\sqrt{\frac{I_x}{m}}$
We plug in the known values to obtain:
$k_x=\sqrt{\frac{\rho \pi [(50)^2\times (200)^3/6]}{\rho \pi (\frac{50}{2})(200)^2}}$
This simplifies to:
$k_x=57.7m$
