This problem involves both rotational and translational dynamics. The turntable rotates with a box of mass $M$ at its edge, and the motion is coupled to a falling ball of mass $m$, which is connected to the box via a string. We are tasked with determining the angular speed of the turntable when the ball is allowed to fall.

Step-by-step solution:

1. Energy Considerations: Since the system is frictionless, energy conservation is a good approach. The potential energy of the ball is converted into both rotational kinetic energy (of the turntable) and the translational kinetic energy of the falling ball.

   • The potential energy lost by the falling ball is $mgh$, where $h$ is the height it falls.
   • The rotational kinetic energy of the turntable is given by $\frac{1}{2}I_{\text{turntable}}\omega^2$, where $I_{\text{turntable}} = MR^2$ is the moment of inertia of the box (assuming the turntable itself is massless) and $\omega$ is the angular speed we want to find.
   • The translational kinetic energy of the falling ball is $\frac{1}{2}mv^2$, where $v$ is the linear velocity of the falling ball.

2. Relation Between the Ball’s Linear Velocity and Angular Speed: As the string unwinds, the velocity of the falling ball is related to the angular speed of the turntable. The linear velocity of the mass $M$ on the edge of the turntable is $v = \omega R$, and this must match the linear speed of the ball, so: $v = \omega R.$

3. Energy Conservation Equation: Using conservation of energy: $mgh = \frac{1}{2}I_{\text{turntable}}\omega^2 + \frac{1}{2}mv^2.$ Substituting $v = \omega R$ and $I_{\text{turntable}} = MR^2$, we get: $mgh = \frac{1}{2}MR^2\omega^2 + \frac{1}{2}m(\omega R)^2.$ Simplifying the right-hand side: $mgh = \frac{1}{2}MR^2\omega^2 + \frac{1}{2}mR^2\omega^2.$ Factoring out $R^2\omega^2$ gives: $mgh = \frac{1}{2}R^2\omega^2 (M + m).$

4. Solving for $\omega$: Solving for $\omega$, we get: $\omega = \sqrt{\frac{2mgh}{R^2(M + m)}}.$ Thus, the angular speed of the turntable is: [ \boxed{\omega = \frac{\sqrt{2mgh