Answer
$T=1.111s$
Work Step by Step
The natural period of vibration can be determined as follows:
$I_{\circ}=\frac{1}{2}mr^2$
$\implies I_{\circ}=\frac{1}{2}(10)(0.15)^2=0.1125Kg\cdot m^2$
Now $\omega_n=\sqrt{\frac{K_{eq}}{m_{eq}}}$
$\implies \omega_n=\sqrt{\frac{3.6}{0.1125}}=4\sqrt{2}rad/s$
The natural period of vibration is given as
$T=\frac{2\pi}{\omega_n}$
We plug in the known values to obtain:
$T=\frac{2\pi}{4\sqrt{2}}$
$\implies T=1.111s$
