Answer
The frictional torque opposing the rotation of the gear is $~4.24~N \cdot m$
Work Step by Step
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 $~4.24~N \cdot m$
