$\omega ^2 =\omega ^2_0+ 2\alpha \theta_n $
where $\omega $ is final angular velocity, $\omega_0$ is initial angular velocity, $\alpha$ is angular retardation and $\theta_n$ is the angle turned in number of rotations.
Here initially $\omega =\frac{\omega_0}{2},\theta =36 \times 2 \pi$
hence $\frac{\omega^2_0}{4},\theta =36 \times 2 \pi$
or $\frac{3\omega ^2_0}{4}=144 \pi \alpha $ or $\alpha =-\frac{3 \omega ^2_3}{4\times 144 \pi}$
Negative sign signifies angular retardation. Again, for the IInd case
$0=\frac{\omega^2_0}{4}-2 \times \frac{3 \omega^2_0}{4 \times 144 \pi}\times \theta'$
or$\theta =\frac{\omega^2_0 \times 4 \times 144 \pi}{4 \times 2\times 3\omega^2_0}$
$ 24 \pi$
Hence, number of rotations it makes before coming to rest
$=\frac{24 \pi}{2 \pi}=12$