Answer
$\omega_3=2.7309 rad/s$
Work Step by Step
We can determine the required angular velocity as follows:
We know that
$mv_1r_1+I\omega_1=mv_2r_2+I\omega_2$ [eq(1)]
As $v_1=\omega_1r=0.15(5)=0.75m/s$
and $v_2=\omega_2 r=0.15\omega_2$
Similarly $I=mk_G^2=(50)(0.125)^2=0.78125Kg/m^2$
We plug in these values in eq(1) to obtain:
$50(0.75)(0.15-0.025)+0.78125(5)=50(0.15\omega_2)(0.15)+0.78125\omega_2$
This simplifies to:
$\omega_2=4.50819 rad/s$
According to the conservation of energy
$T_2+V_2=T_3+V_3$ [eq(2)]
Now $T_2=\frac{1}{2}mv_2^2+\frac{1}{2}I\omega_2^2$
$\implies T_2=\frac{1}{2}(0.15\omega_2)^2+(\frac{0.78125}{2})\omega_2^2$
$\implies T_2=(0.953125)(4.5098)^2=19.3711J$
and $V_2=mgh_2=50(9.81)(0)=0$
$T_3=\frac{1}{2}mv_3^2+\frac{1}{2}I\omega_3^2$
$\implies T_3=\frac{50}{2}(0.15\omega_3)^2+(\frac{0.78125}{2})(\omega_3^2)$
$\implies T_3=0.953125\omega_3^2$
$V_3=mgh_3=(50)(9.81)(0.025)=12.2625J$
We plug in the known values in eq(2) to obtain:
$19.3711+0=0.953125\omega_3^2+12.2625$
This simplifies to:
$\omega_3=2.7309 rad/s$
