Chapter 11 - Rotational Dynamics and Static Equilibrium - Problems and Conceptual Exercises - Page 369: 54

$5.9\times 10^{-4}Kg.\frac{m^2}{s}$

We know that $L=I\omega$ $L=(\frac{1}{2}MR^2)\omega$ We plug in the known values to obtain: $L=\frac{1}{2}(0.015)(0.15)^2)(33\frac{1}{3}\frac{rev}{min})(\frac{2\pi rad}{rev})(\frac{1min}{60s})$ $L=5.9\times 10^{-4}Kg.\frac{m^2}{s}$