Angular Speed of Circular Disc Under Applied Forces

To find the angular speed of the circular disc after $t = 1\text{ s}$, we analyze the net force and net torque acting on the disc.

1. Geometry of the Setup

For an equilateral triangle $XYZ$ inscribed in a circular disc of radius $R$, the perpendicular distance $r_\perp$ from the center $O$ of the disc to any side of the triangle is given by: $r_\perp = R \cos 60^\circ = \frac{R}{2}$

2. Net Force and Net Torque

Three forces of equal magnitude $F = 0.5\text{ N}$ act along the three sides of the equilateral triangle in a cyclic order.

Since the forces are of equal magnitude and their directions are inclined at $120^\circ$ with respect to one another, the vector sum of these forces is zero: $\vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 = 0$

Therefore, the center of mass of the disc does not accelerate and remains at rest. The system undergoes pure rotation about its center of mass.

Each force exerts a torque in the same rotational sense about the center $O$. The total torque $\tau_{\text{net}}$ about the center of the disc is: $\tau_{\text{net}} = 3 \times (F \cdot r_\perp) = 3 \times \left(F \cdot \frac{R}{2}\right) = \frac{3}{2}FR$

3. Angular Acceleration

The moment of inertia $I$ of a uniform circular disc of mass $m$ and radius $R$ about its central axis perpendicular to its plane is: $I = \frac{1}{2} m R^2$

Using the rotational analogue of Newton's second law ($\tau_{\text{net}} = I \alpha$), the angular acceleration $\alpha$ is: $\frac{3}{2} FR = \frac{1}{2} m R^2 \alpha \implies \alpha = \frac{3F}{m R}$

Substituting the given values ($F = 0.5\text{ N}$, $m = 1.5\text{ kg}$, $R = 0.5\text{ m}$): $\alpha = \frac{3 \times 0.5}{1.5 \times 0.5} = 2\text{ rad s}^{-2}$

4. Angular Speed

Starting from rest ($\omega_0 = 0\text{ rad s}^{-1}$), the angular speed $\omega$ after $t = 1\text{ s}$ is: $\omega = \omega_0 + \alpha t = 0 + 2 \times 1 = 2\text{ rad s}^{-1}$