## College Physics (4th Edition)

The magnitude of the average torque due to frictional forces is $~0.0905~N \cdot m$
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 $~0.0905~N \cdot m$.