Answer
$v=5.96~ ft/s$
Work Step by Step
We can determine the required velocity as follows:
We know that the angular momentum is conserved about D:
$H_{D_1}=H_{D_2}$
$\implies mv_1r_1=I_D\omega_2$ [eq(1)]
As $I_D=\frac{m(I^2+\omega^2)}{12}+mr_{D/G}^2$
$\implies I_D=\frac{10}{32.2}(\frac{(1)^2+(1)^2}{12})+\frac{10}{32.2}(\sqrt{(0.5)^2+(0.5)^2})^2=0.207 slug/ft^2$
We plug in the known values in eq(1) to obtain:
$\frac{10}{32.2}v(0.5)=0.207\omega_2$
$\implies 0.155279v=0.207\omega_2$ [eq(2)]
According to the conservation of energy
$T_2+V_2=T_3+V_3$ [eq (3)]
We know that
$T_2=\frac{1}{2}I_D\omega_2^2=(0.1035)\omega_2$
$V_2=mgh_2=10(0.5)=5lb\cdot ft$
Similarly $V_2=mgh_2=(10)(0.5)=5lb\cdot ft$
$T_3=\frac{1}{2}I_D\omega_3^2=\frac{1}{2}(0.207)(0)^2=0lb\cdot ft$
$V_3=mgh_3$
$h_3=\sqrt{(0.5)^2+(0.5)^2}=0.707ft$
$\implies V_3=10(0.707)=7.07\space lb\cdot ft$
We plug in theses values in eq(3) to obtain:
$0.1035\omega_2^2+5=0+7.07$
This simplifies to:
$\omega_2=4.4721 rad/s$
Now, we plug in the known values in eq(2) to obtain:
$0.155279v=0.207(4.4721)$
This simplifies to:
$v=5.96~ ft/s$
