Answer
$\omega_2=1.01~ rad/s$
Work Step by Step
We can determine the required angular velocity as follows:
According to the conservation of angular momentum
$I_{A_1}\omega_1=I_{A_2}\omega_2~~~$[eq(1)]
As $I_{A_1}=\frac{1}{2}m_dr_d^2$
$\implies I_{A_1}=\frac{1}{2}(40)(0.3)^2=1.8Kg/m^2$
Similarly
$I_{A_2}=\frac{m_r l^2}{3}+\frac{1}{2}m_d r_d^2+m_d r^2$
$\implies I_{A_2}=\frac{12(2)^2}{3}+\frac{40}{2}(0.3)^2+40(2)^2$
This simplifies to:
$I_{A_2}=177.8Kg/m^2$
We plug in the known values in eq(1) to obtain:
$1.8(100)=177.8\omega_2$
This simplifies to:
$\omega_2=1.01~ rad/s$
