Answer
$3.14\;rad/s$
Work Step by Step
The maximum kinetic energy of a torsion pendulum is given by
$E^{max}_k=\frac{1}{2}I\omega^2_m$
And the maximum potential energy of a torsion pendulum is given by
$E^{max}_p=\frac{1}{2}\kappa\theta^2_m$
The value of $E^{max}_k$ should be equal to $E^{max}_p$
Therefore,
$\frac{1}{2}I\omega^2_m=\frac{1}{2}\kappa\theta^2_m$
or, $\omega_m=\theta_m\sqrt {\frac{\kappa}{I}}$
From the given graph, $\kappa=\frac{4\times10^{-3}}{0.2}\;N.m/rad=0.02\;N.m/rad$ and $\theta_m=0.2\;rad$. We have already obtained that $I=8.106\times10^{-5}\;kg.m^2$
Putting the given values, we obtain
$\omega_m=(0.2)\sqrt {\frac{0.02}{8.106\times10^{-5}}}\;rad/s$
or, $\omega_m=3.14\;rad/s$
Therefore, the maximum angular speed of the disk is $3.14\;rad/s$
