Answer:
[tex]\omega = 0.16 s^{-1}[/tex]
Explanation:
As we know that there is no torque on the system of merry go round
so the angular momentum will remain conserved
so here we have
[tex]I_1\omega_0 = (I_1 + I_2)\omega[/tex]
so we will have
[tex]I_1 = 500 kg m^2[/tex]
[tex]I_2 = 30(2.1^2)[/tex]
[tex]I_2 = 132.3 kg m^2[/tex]
[tex]\omega_0 = 0.20 s^{-1}[/tex]
now from above formula
[tex]500(0.20) = (500 + 132.3)\omega[/tex]
[tex]\omega = 0.16 s^{-1}[/tex]