Answer
$\alpha=11rad/s^2$
Work Step by Step
The moment of inertia of a hoop is equal to $$I=mr^2$$ Substituting known values of $m=0.75kg$ and $r=35cm=0.35m$ yields a moment of inertia of $$I=(0.75kg)(0.35m)^2=0.092 kg \times m^2$$ Since $\tau=I\alpha$, angular acceleration equals to $$\alpha=\frac{\tau}{I}$$ Substituting known values of $I=0.092kg \times m^2$ and $\tau=0.97 N\times m$ yields an angular acceleration of $$\alpha=\frac{0.97N \times m}{0.092kg \times m^2}=11rad/s^2$$
