Answer
We choose $\pm$ directions such that the initial angular velocity is $\omega_{0}=-317 \mathrm{rad} / \mathrm{s}$ and the values for $\alpha, \tau,$ and $F$ are positive.
Combining Eq. 10-12 with Eq. 10-45 and Table $10-2(\mathrm{f})$
(and using the fact that $\omega=0$) we arrive at the expression
$$
\tau=\left[\frac{2}{5} M R^{2}\left]\left[-\frac{\omega_{0}}{t} \right]=-\frac{2}{5} \frac{M R^{2} \omega_{0}}{t}\right.\right.
$$
With $t=15.5 \mathrm{s}, R=0.226 \mathrm{m},$ and $M=1.65 \mathrm{kg},$ we obtain $$\tau=0.689 \mathrm{N} \cdot \mathrm{m}$$
Work Step by Step
We choose $\pm$ directions such that the initial angular velocity is $\omega_{0}=-317 \mathrm{rad} / \mathrm{s}$ and the values for $\alpha, \tau,$ and $F$ are positive.
Combining Eq. 10-12 with Eq. 10-45 and Table $10-2(\mathrm{f})$
(and using the fact that $\omega=0$) we arrive at the expression
$$
\tau=\left[\frac{2}{5} M R^{2}\left]\left[-\frac{\omega_{0}}{t} \right]=-\frac{2}{5} \frac{M R^{2} \omega_{0}}{t}\right.\right.
$$
With $t=15.5 \mathrm{s}, R=0.226 \mathrm{m},$ and $M=1.65 \mathrm{kg},$ we obtain $$\tau=0.689 \mathrm{N} \cdot \mathrm{m}$$
