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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

A box of mass M is positioned at the edge of a frictionless turntable of radius R that is rotating at a constant angular speed with the turntable. An ideal string passing over a massless pulley and through a hole in the middle of the table connects the box to a hollow ball of mass m and radius r with the string wrapped many times around the edge of the ball. The ball is allowed to fall and unwind. Determine the angular speed of the turntable.

Learn how to calculate the angular speed of a rotating frictionless turntable connected to a falling mass using principles of rotational dynamics and energy conservation. The problem involves calculating rotational kinetic energy and translating it to angular speed.

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