Answer:
3.14946 rad/s
Explanation:
[tex]I_i[/tex] = Intial moment of inertia
[tex]I_f[/tex] = Final moment of inertia
[tex]\omega_i[/tex] = Initial angular velocity
[tex]\omega_f[/tex] = Final angular velocity = [tex]\dfrac{2}{1.33}\times 2\pi\ rad/s[/tex]
[tex]\dfrac{I_f}{I_i}=\dfrac{1}{3}[/tex]
In this system the angular momentum is conserved
[tex]L_i=L_f\\\Rightarrow I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_i=\dfrac{I_f\omega_f}{I_i}\\\Rightarrow \omega_i=\dfrac{1\times \dfrac{2}{1.33}\times 2\pi}{3}\\\Rightarrow \omega_i=3.14946\ rad/s[/tex]
The angular velocity when the diver left the board is 3.14946 rad/s