\(L = mv \times r \)\( = \left( {10 \times {{10}^{ - 3}}} \right) \times 500 \times \frac{1}{2} = 2.5\)
\(I = \frac{{M{L^2}}}{3} = \frac{{12 \times {{1.0}^2}}}{3} = 4kg{m^2}\)
\(L = I\omega \)
\(\therefore \omega = \frac{L}{I} = \frac{{2.5}}{4} = 0.625\,rad/\sec \)