Answer
$19.0rad/s$
Work Step by Step
The required angular velocity can be calculated as follows:
$K=\frac{W}{\delta}$
$K=\frac{150}{1in\times \frac{1}{12}ft/in}=1800lb/ft$
As $\omega_n=\sqrt{\frac{K}{m}}$
$\implies \omega_n=\sqrt{\frac{Kg}{W}}$
$\implies \omega_n=\sqrt{\frac{1800\times 32.2}{150}}$
$\implies \omega_n=19.7rad/s$
$F_{\circ}=mr\omega_{\circ}^2=\frac{W}{g}r\omega_{\circ}^2$
$F_{\circ}=\frac{0.25}{32.2}\times \frac{10in}{12in/ft}$
$\implies F_{\circ}=2.588lb$
The amplitude is given as
$x=|\frac{F_{\circ}/K}{1-(\frac{\omega_{\circ}}{\omega_{n}})^2}|$
We plug in the known values to obtain:
$\frac{0.25in}{12in/ft}=|\frac{2.588/1800}{1-(\frac{\omega_{\circ}}{19.7})^2}|$
This simplifies to:
$\omega_{\circ}=18.9rad/s\approx 19.0rad/s$
