## Physics: Principles with Applications (7th Edition)

$22 N \cdot m$.
Calculate the torque using $\tau = I \alpha$. The rotational inertia of the rod around an end is $\frac{1}{3}ML^2$. The angular acceleration is: $$\alpha=\frac{\Delta \omega}{\Delta t}=\frac{(2 \pi )(2.6) rad/s-0}{0.2s}$$ $$\tau = I \alpha=\frac{1}{3}(0.90kg)(0.95 m)^2 \alpha=22 N \cdot m$$