The moment of inertia is\(I\).

The angular speed is \(\omega \).

In this problem, there is no external torque exerted on the nonrotating cylindrical disk, which means that the angular momentum will be conserved for this system.

The relation of angular momentum can be written as:

\(\begin{aligned}{c}L &= L'\\I\omega + {I_{\rm{r}}}{\omega _{\rm{r}}} &= 2I{\omega _{\rm{f}}}\end{aligned}\)

Here, \(L\) and \(L'\) are the initial and final angular momentum, \({\omega _{\rm{r}}}\) is the angular velocity of the disk whose value is zero, \({\omega _{\rm{f}}}\) is the final combined angular velocity, and \(I\) and \({I_{\rm{r}}}\) are the cylinder and disk moment of inertia, respectively.

On plugging the values in the above relation, you get:

\(\begin{aligned}{c}I\omega + 0 &= 2I{\omega _{\rm{f}}}\\{\omega _{\rm{f}}} &= \frac{\omega }{2}\end{aligned}\)

Thus, \({\omega _{\rm{f}}} = \frac{{{\omega _{\rm{c}}}}}{2}\) is the final angular speed of the two disks.