Answer
$\omega_2=57 rad/s$
$U_F=366.66J$
Work Step by Step
The required angular velocity and the work done can be determined as follows:
According to the principle of work and energy
$T_1+\Sigma U_{1\rightarrow 2}=T_2$ [eq(1)]
We know that
$T_1=\frac{1}{2}I_{z_1}\omega_1^2$
$\implies I_{z_1}=2(\frac{ml^2}{12}+mr^2)$
$\implies I_{z_1}=2(\frac{10(0.6)^2}{12}+10(0.3+0.1)^2)$
$\implies I_{z_1}=3.8Kg/m^2$
Now $T_1=\frac{1}{2}(3.8)(3)^2=17.1J$
Similarly $\Sigma U_{1\rightarrow 2}=U_W+U_F$
$\implies U_W=2(-mgh)=2[-10(9.81)(0.3)]=-58.86J$
$\implies \Sigma U_{1\rightarrow 2}=-58.86+U_F$
We also know that
$T_2=\frac{1}{2}I_{z_2}\omega_2^2$
$\implies I_{z_2}=2(\frac{ml^2}{12}+mr^2)$
$\implies I_{z_2}=2[\frac{10(0)^2}{12}+10(0.1)^2]$
$\implies I_{z_2}=0.2Kg/m^2$
and $T_2=\frac{1}{2}(0.2)(\omega_2)^2$
$\implies T_2=0.1\omega_2^2$ [eq(2)]
According to the conservation of angular momentum
$I_{z_1}\omega_1=I_{z_2}\omega_2$
We plug in the known values to obtain:
$3.8(3)=0.2\omega_2$
This simplifies to:
$\omega_2=57 rad/s$
We plug in this value in eq(2) to obtian:
$T_2=0.1(57)^2=324.9J$
Now from eq(1), we obtain:
$17.1-58.86+U_F=324.9$
This simplifies to:
$U_F=366.66J$
