Answer
$\omega_1=7.173 rad/s$
Work Step by Step
The required angular velocity can be determined as follows:
First, we apply the equation of motion in normal direction
$\Sigma F_n=ma_n$ [eq(1)]
As $a_n=\omega_2^2 r=0.2\omega_2^2$
and $Wcos\theta=20(0.2)\omega_2^2$
$\implies cos\theta=\frac{0.2-0.03}{0.2}=0.85$
Now from eq(1), we obtain:
$20(9.81)(0.85)=4\omega_2^2$
This simplifies to:
$\omega_2=6.456 rad/s$
Now according to conservation of angular momentum
$mv_1r_1+I\omega_1=mv_2r_2+I\omega_2$ [eq(2)]
We know that
$v=0.2\omega$
and $I=\frac{1}{2}mr^2=\frac{1}{2}(20)(0.2)^2=0.4Kg/m^2$
Now we plug in the known values in eq(2) to obtain:
$20(0.2\omega_1)(0.17)+0.4\omega_1=20(0.2\omega_2)(0.2)+0.4\omega_2$
This simplifies to:
$\omega_1=1.2\omega_2$
$\implies \omega_1=1.111\times 6.456=7.173 rad/s$
