Chapter 11 - Rotational Dynamics and Static Equilibrium - Problems and Conceptual Exercises - Page 369: 41

$18.7cm$

We know that $\Sigma \tau=0$ $\Sigma \tau=(\frac{1}{2L})m_{cereal}+rm_{juice}g-(\frac{1}{2}L)m_{milk}g=0$ This simplifies to: $r=\frac{\frac{1}{2}L(m_{cereal}+m_{milk})}{m_{juice}}$ We plug in the known values to obtain: $r=\frac{\frac{1}{2}(0.620)(0.722+1.81)}{1.80}$ $r=0.187m=18.7cm$