## Essential University Physics: Volume 1 (3rd Edition)

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: $\tau=\alpha I = 1055.9 \times .001166 = \fbox{1.23 Nm}$