Chapter 11 - Rotational Dynamics and Static Equilibrium - Problems and Conceptual Exercises - Page 367: 22

$0.290N.m$

We can find the magnitude of the frictional torque as follows: $\Sigma \tau=-r(m_1g)+r(m_2g)=0$ $\implies \tau_{fr}=rg(m_1-m_2)$ We plug in the known values to obtain: $\tau=(0.0940)(9.81)(0.635-0.321)$ $\tau=0.290N.m$