Answer
$\omega_2=0.0906 rad/s$
Work Step by Step
We can determine the required angular velocity as follows:
According to the conservation of angular momentum
$H_{C_1}=H_{C_2}$
$\implies 2I_{d}\omega_1=I_r\omega_2+2I_d\omega_2+2m_dv_d r_d~~~$[eq(1)]
We know that
$I_d=\frac{1}{2}m_dr_d^2$
$\implies I_d=\frac{1}{2}(4)(0.15)^2=0.045Kg/m^2$
and $I_r=\frac{m_r l_r^2}{12}$
$\implies I_r=\frac{2(1.5)^2}{12}=0.375Kg/m^2$
$v_d=\omega_2 r$
$\implies v_d=(0.75)\omega_2$
We plug in the known values in eq(1) to obtain:
$2(0.045)(5)=(0.375)\omega_2+2(0.045)\omega_2+2(4)(0.75\omega)(0.75)$
This simplifies to:
$\omega_2=0.0906 rad/s$
