Answer
The torque applied to the fan by the motor is $~0.437~N \cdot m$
Work Step by Step
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 $~0.437~N \cdot m$
