Answer
$I=0.984kg \times m^2$
Work Step by Step
To find the angular acceleration, use the definition of angular acceleration $$\alpha=\frac{\Delta \omega}{\Delta t}$$ Substituting known values of $\Delta \omega=-2.75rad/s$ and $\Delta t=22.5s$ yields an angular acceleration of $$\alpha=\frac{-2.75rad/s}{22.5s}=-0.122rad/s^2$$ Since $\tau=I\alpha$, moment of inertia $I$ is equal to $$I=\frac{\tau}{\alpha}$$ Substituting known values of $\tau=-0.120N\times m$ and $\alpha=-0.122rad/s^2$ yields a moment of inertia of $$I=\frac{-0.120N \times m}{-0.122rad/s^2}=0.984kg \times m^2$$
