A horizontal circular platform of radius 0.5 m and mass 0.45 kg is free to rotate about its axis. Two massless spring toy-guns, each carrying a steel ball of mass 0.05 kg are attached to the platform at a distance 0.25 m from the centre on its either sides along its diameter (see figure). Each gun simultaneously fires the balls horizontally and perpendicular to the diameter in opposite directions. After leaving the platform, the balls have horizontal speed of $9 m/s$ with respect to the ground. The rotational speed of the platform in rad/s after the balls leave the platform is ____

Step 1: Identify the given parameters.

The parameters for the platform and the steel balls are:

$$R = 0.5\,\text{m}$$ $$M = 0.45\,\text{kg}$$ $$m = 0.05\,\text{kg}$$ $$r = 0.25\,\text{m}$$ $$v = 9\,\text{m/s}$$

Step 2: Apply conservation of angular momentum.

Since there are no external torques acting on the system about the vertical axis of rotation, the total angular momentum is conserved.

$$L_{\text{initial}} = 0$$ $$L_{\text{final}} = L_{\text{balls}} + L_{\text{platform}} = 0$$

Step 3: Calculate the angular momentum of the balls.

Because the two balls are fired in opposite directions from opposite sides of the diameter, their angular momenta about the central axis add up in the same direction.

$$L_{\text{balls}} = 2mrv$$

Step 4: Calculate the moment of inertia of the platform.

The platform is a uniform circular disk rotating about its central axis.

$$I = \frac{1}{2}MR^2$$ $$I = \frac{1}{2}(0.45)(0.5)^2$$ $$I = 0.05625\,\text{kg}\cdot\text{m}^2$$

Step 5: Determine the angular speed of the platform.

Equate the magnitude of the angular momentum of the platform to that of the balls.

$$I\omega = 2mrv$$ $$\omega = \frac{2mrv}{I}$$ $$\omega = \frac{2(0.05)(0.25)(9)}{0.05625}$$ $$\omega = 4\,\text{rad/s}$$

Answer: 4