Answer
a) .001166
b) 1.23 Nm
Work Step by Step
We add the rotational inertia of the mass spread out in the rim and the mass spread out over the whole disk to obtain:
a) $I = mr^2+\frac{1}{2}mr^2 = \frac{3}{2}mr^2=\frac{3}{2}(.054)(.12)^2=\fbox{.001166 }$
b) We first must find the value of the angular acceleration:
$\alpha= \frac{\omega^2}{2\theta}=\frac{3456}{\pi}=1055.9$
Thus, we find torque:
$\tau=\alpha I = 1055.9 \times .001166 = \fbox{1.23 Nm}$
