Consider the system consisting of the child, the disk, and the stone. Since there are no external torques acting on the system about the axis of rotation, the total angular momentum is conserved.

The initial angular momentum of the system is zero, as everything is at rest.

$$L_i = 0$$

After the stone is thrown, the final angular momentum $L_f$ is the sum of the angular momentum of the stone ($L_{stone}$) and the angular momentum of the child-disk system ($L_{child+disk}$).

The angular momentum of the stone, which moves tangentially at radius R with velocity v, is:

$$L_{stone} = m v R$$

The child and the disk rotate together with a final angular velocity $\omega$. The moment of inertia of the child is $MR^2$. The total moment of inertia of the rotating system is $I_{total} = I + MR^2$. Their combined angular momentum is:

$$L_{child+disk} = (I + MR^2)\omega$$

By conservation of angular momentum, $L_i = L_f$:

$$0 = L_{stone} + L_{child+disk}$$ $$0 = m v R + (I + MR^2)\omega$$

Solving for the angular velocity $\omega$:

$$\omega = - \frac{m v R}{I + MR^2}$$

The negative sign indicates that the child and disk rotate in the direction opposite to the stone's velocity.