Answer
$\omega_2=3.47 rad/s$
Work Step by Step
We can determine the required angular velocity as follows:
We know that
$\omega_2=\frac{(v_B)_2}{r}$
$\implies \omega_2=\frac{(v_B)_2}{2}$ [eq(1)]
$e=\frac{(v_B)_2-(v_G)_2}{(v_G)_1-(v_B)_1}$
$\implies 0.8=\frac{(v_B)_2-(v_G)_2}{6-0}$
This simplifies to:
$(v_B)_2=(v_G)_2+4.8$ [eq(2)]
According to the conservation of angular momentum
$H_{z_1}=H_{z_2}$
$\implies (mr(v_G)_1)_{ball}=(mr(v_G)_2)_{ball}+I_{z_2}\omega_2$ [eq(3)]
We know that
$I_{z_2}=\frac{ml^2}{12}=(\frac{5}{32.2})(\frac{(4)^2}{12})=0.207Kg/m^2$
Now from eq(3), we obtain:
$\frac{3}{32.2}\times 2\times 6=(\frac{3}{32.2}\times \times (v_G)_2)+0.207(\frac{(v_{B})_2}{2})$
$\implies 1.118=0.18633(v_G)_2+0.1035(v_B)_2$ [eq(4)]
We plug in the known values from eq(2) to into eq(4) to obtain:
$1.118=0.18633(v_G)_2+0.1035(v_G)_2+4.8$
This simplifies to:
$(v_G)_2=2.1435ft/s$
Now from eq(2) $(v_B)_2=2.1435+4.8=6.9435ft/s$
We plug in this value in eq(1) to obtain:
$\omega_2=\frac{6.9435}{2}=3.47 rad/s$
