## College Physics (4th Edition)

The torque applied to the fan by the motor is $~0.437~N \cdot m$
We can express the final angular speed in units of rad/s: $1.8~rev/s\times \frac{2\pi~rad}{1~rev} = 11.3~rad/s$ We can find the angular acceleration: $\omega_f = \omega_0+\alpha~t$ $\alpha = \frac{\omega_f - \omega_0}{t}$ $\alpha = \frac{11.3~rad/s-0}{4.35~s}$ $\alpha = 2.6~rad/s^2$ We can find the torque: $\tau = I~\alpha$ $\tau = (4)(\frac{1}{3}ML^2)~(\alpha)$ $\tau = (4)(\frac{1}{3})(0.35~kg)(0.60~m)^2~(2.6~rad/s^2)$ $\tau = 0.437~N \cdot m$ The torque applied to the fan by the motor is $~0.437~N \cdot m$