A uniform circular disc of mass 1.5 kg and radius 0.5 m is initially at rest on a horizontal frictionless surface. Three forces of equal magnitude $F = 0.5 N$ are applied simultaneously along the three sides of an equilateral triangle XYZ with its vertices on the perimeter of the disc (see figure). One second after applying the forces, the angular speed of the disc in rad/s is ____.

Step 1: Determine the perpendicular distance of the forces from the center.

The vertices of the equilateral triangle $XYZ$ lie on the perimeter of the disc of radius $R$, and the forces act along its sides. The perpendicular distance $d$ from the center of the disc to any side of the inscribed equilateral triangle is the inradius.

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

Step 2: Calculate the net torque acting on the disc.

Since all three forces act in the same rotational direction, their torques add up.

$$\tau = 3 \left( F \cdot \frac{R}{2} \right)$$ $$\tau = \frac{3FR}{2}$$ $$\tau = \frac{3 \times 0.5 \times 0.5}{2}$$ $$\tau = 0.375\,\text{N}\,\text{m}$$

Step 3: Calculate the moment of inertia of the disc.

The moment of inertia $I$ of a uniform circular disc about its central axis is given by the standard formula.

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

Step 4: Calculate the angular acceleration.

Use the rotational analogue of Newton's second law.

$$\alpha = \frac{\tau}{I}$$ $$\alpha = \frac{3F}{mR}$$ $$\alpha = \frac{3 \times 0.5}{1.5 \times 0.5}$$ $$\alpha = 2\,\text{rad}\,\text{s}^{-2}$$

Step 5: Calculate the angular speed after one second.

Using the first equation of rotational motion with the disc starting from rest:

$$\omega = \alpha t$$ $$\omega = 2 \times 1$$ $$\omega = 2\,\text{rad}\,\text{s}^{-1}$$

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