Answer
The frictional torque acting on the flywheel is $26.67~N \cdot m$
Work Step by Step
We can find the angular acceleration:
$\omega_f = \omega_0+\alpha~t$
$\alpha = \frac{\omega_f - \omega_0}{t}$
$\alpha = \frac{0 - 20.0~rad/s}{300.0~s}$
$\alpha = -0.0667~rad/s^2$
We can use the magnitude of the angular acceleration to find the torque:
$\tau = I~\alpha$
$\tau = (400.0~kg~m^2)(0.0667~rad/s^2)$
$\tau = 26.67~N \cdot m$
The frictional torque acting on the flywheel is $26.67~N \cdot m$
