Answer
His angular velocity when he left the diving board was $3.15~rad/s$
Work Step by Step
We can find the final angular velocity:
$\omega_f = \frac{(2.00)(2\pi~rad)}{1.33~s} = 9.45~rad/s$
We can use conservation of angular momentum to find the initial angular velocity:
$L_0 = L_f$
$I_0~\omega_0 = I_f~\omega_f$
$\omega_0 = \frac{I_f~\omega_f}{I_o}$
$\omega_0 = \frac{(\frac{I_0}{3.00})~(\omega_f)}{I_0}$
$\omega_0 = \frac{\omega_f}{3.00}$
$\omega_f = \frac{9.45~rad/s}{3.00}$
$\omega_f = 3.15~rad/s$
His angular velocity when he left the diving board was $3.15~rad/s$
