## College Physics (4th Edition)

The frictional torque opposing the rotation of the gear is $~4.24~N \cdot m$
We can express the final angular speed in units of rad/s: $1.35~rev/s\times \frac{2\pi~rad}{1~rev} = 8.48~rad/s$ We can find the angular acceleration: $\omega_f = \omega_0+\alpha~t$ $\alpha = \frac{\omega_f - \omega_0}{t}$ $\alpha = \frac{8.48~rad/s-0}{1.70~s}$ $\alpha = 5.0~rad/s^2$ The frictional torque $\tau_f$ opposes the torque exerted by the chain $\tau_c$. We can find the frictional torque: $\sum \tau = I~\alpha$ $\tau_c-\tau_f = \frac{1}{2}MR^2~\alpha$ $\tau_f = \tau_c-\frac{1}{2}MR^2~\alpha$ $\tau_f = (R~F_T)-\frac{1}{2}MR^2~\alpha$ $\tau_f = (0.650~m)(72.5~N)-\frac{1}{2}(40.6~kg)(0.650~m)^2~(5.0~rad/s^2)$ $\tau_f = 4.24~N \cdot m$ The frictional torque opposing the rotation of the gear is $~4.24~N \cdot m$