Answer
$\omega_1=3.9778 rad/s$
Work Step by Step
The required angular velocity can be determined as follows:
According to the conservation of angular momentum equation
$mv_1 r_1+I\omega_1=mv_2r_2+I\omega_2$ [eq(1)]
We know that
$v_1=\omega_1r$
and $v_2=\omega_2 r=0.15\omega_2$
Similarly $I=mk_G^2=50(0.125)^2=0.78125Kg/m^2$
Now from eq(1), we have
$50(0.15\omega_1)(0.15-0.025)+0.78125\omega_1=50(0.15\omega_2)(0.15)+0.78125\omega_2$
This simplifies to:
$\omega_1=1.109\omega_2$ [eq(2)]
Now, we apply the conservation of energy equaiton
$T_2+V_2=T_3+V_3$ [eq(3)]
As $T_2=\frac{1}{2}mv_2^2+\frac{1}{2}I\omega_2^2$
$\implies T_2=\frac{50}{2}(0.15\omega_2)^2+\frac{0.78125}{2}\omega_2^2=0.953125\omega_2^2$
Similarly $V_2=0J$
$T_3=\frac{1}{2}mv_3^2+\frac{1}{2}I\omega_3^2=\frac{50}{2(0.15\times 0)^2}+(\frac{0.78125}{2})(0)^2=0J$
and $V_3=mgh_3=50(9.81)(0.025)=12.2625J$
We plug in the known values in eq(3) to obtain:
$0.953125\omega_2^2+0=0+12.2625$
This simplifies to:
$\omega_2=3.5868rad/s$
Now from eq(2), we obtain:
$\omega_1=1.109\times 3.5868=3.9778 rad/s$
