Answer
The magnitude of the average torque due to frictional forces is $~0.0905~N \cdot m$
Work Step by Step
We can express the initial angular speed in units of rad/s:
$4.00~rev/s\times \frac{2\pi~rad}{1~rev} = (8.00~\pi)~rad/s$
We can find the angular acceleration:
$\omega_f = \omega_0+\alpha~t$
$\alpha = \frac{\omega_f - \omega_0}{t}$
$\alpha = \frac{0-8.00~\pi~rad/s}{50~s}$
$\alpha = -0.503~rad/s^2$
The torque due to frictional forces brings the wheel to a stop. We can use the magnitude of the angular acceleration to find the torque:
$\tau = I~\alpha$
$\tau = MR^2~\alpha$
$\tau = (2~kg)(0.30~m)^2~(0.503~rad/s^2)$
$\tau = 0.0905~N \cdot m$
The magnitude of the average torque due to frictional forces is $~0.0905~N \cdot m$.
