## College Physics (4th Edition)

The frictional torque acting on the flywheel is $26.67~N \cdot m$
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 $26.67~N \cdot m$