1. Given data

• Radius of platform: $R = 0.5\,\text{m}$
• Mass of platform: $M = 0.45\,\text{kg}$
• Each ball mass: $m = 0.05\,\text{kg}$
• Distance of each gun from centre: $r = 0.25\,\text{m}$
• Speed of each ball after firing (w.r.t. ground): $v = 9\,\text{m s}^{-1}$

The two balls are fired simultaneously, perpendicular to the diameter, and in opposite directions.

2. Use conservation of angular momentum

Initially, the whole system is at rest, so initial angular momentum about the axis is

$L_i = 0.$

Hence after firing,

$L_{\text{platform}} + L_{\text{balls}} = 0.$

So we first calculate angular momentum carried by the two balls.

3. Angular momentum of each ball

For a particle moving in a straight line, angular momentum about the centre is

$L = mvr_\perp,$

where $r_\perp$ is the perpendicular distance from the centre to the line of motion.

Each ball is fired from a point at distance $0.25\,\text{m}$ from the centre, and its velocity is perpendicular to the radius/diameter line, so

$L_1 = mvr = 0.05 \times 9 \times 0.25 = 0.1125\,\text{kg m}^2\text{/s}.$

Now check the direction: although the balls move in opposite directions from opposite sides, their angular momenta about the centre are in the same rotational sense, so they add.

Therefore,

$L_{\text{balls}} = 2 \times 0.1125 = 0.225\,\text{kg m}^2\text{/s}.$

Thus platform must get equal and opposite angular momentum:

$L_{\text{platform}} = 0.225\,\text{kg m}^2\text{/s}.$

4. Moment of inertia of platform

The platform is a uniform disc, so

$I = \frac{1}{2}MR^2 = \frac{1}{2}(0.45)(0.5)^2.$

$I = \frac{1}{2} \times 0.45 \times 0.25 = 0.05625\,\text{kg m}^2.$

5. Find angular speed

Using

$L = I\omega,$

$\omega = \frac{L_{\text{platform}}}{I} = \frac{0.225}{0.05625} = 4\,\text{rad s}^{-1}.$

6. Final answer

$\boxed{4}$

The rotational speed of the platform is $4\,\text{rad s}^{-1}$.