1. Given data

• Mass of disc: $M = 1.5\,\text{kg}$
• Radius of disc: $R = 0.5\,\text{m}$
• Each force: $F = 0.5\,\text{N}$
• Time for which forces act: $t = 1\,\text{s}$

The disc is on a frictionless horizontal surface, so only the torque about the center matters for angular acceleration.

2. Geometry of the forces

The three forces act along the three sides of an equilateral triangle whose vertices lie on the circumference of the disc.

For an equilateral triangle inscribed in a circle of radius $R$, the perpendicular distance from the center of the circle to any side is the inradius of the triangle:

$d = R\cos 60^\circ = \frac{R}{2}$

Since $R = 0.5\,\text{m}$,

$d = \frac{0.5}{2} = 0.25\,\text{m}$

So each force has moment arm $0.25\,\text{m}$ about the center.

3. Torque due to one force

Torque magnitude due to one force:

$\tau_1 = Fd = 0.5 \times 0.25 = 0.125\,\text{N m}$

All three forces act along the sides in the same rotational sense, so torques add:

$\tau_{\text{net}} = 3\tau_1 = 3\times 0.125 = 0.375\,\text{N m}$

4. Moment of inertia of the disc

For a uniform disc about its central axis:

$I = \frac{1}{2}MR^2$

Substitute values:

$I = \frac{1}{2}(1.5)(0.5)^2 = \frac{1}{2}(1.5)(0.25) = 0.1875\,\text{kg m}^2$

5. Angular acceleration

Using

$\tau_{\text{net}} = I\alpha$

we get

$\alpha = \frac{\tau_{\text{net}}}{I} = \frac{0.375}{0.1875} = 2\,\text{rad s}^{-2}$

6. Angular speed after 1 s

Initially the disc is at rest, so $\omega_0 = 0$.

Using

$\omega = \omega_0 + \alpha t$

$\omega = 0 + 2\times 1 = 2\,\text{rad s}^{-1}$

7. Final answer

The angular speed after 1 second is

$\boxed{2\,\text{rad s}^{-1}}$

So the required integer answer is:

$\boxed{2}$

8. Comparison with stored answer

Stored correct answer = $2$

My derived answer is also $2$, so they agree.