Given: Radius of the disc, $r = 50 \text{ cm} = 0.5 \text{ m}$ Initial angular velocity, $\omega_0 = 0$ (since it is at rest) Angular acceleration, $\alpha = 2.0 \text{ rad/s}^2$ Time, $t = 2.0 \text{ s}$

Angular velocity at $t = 2.0 \text{ s}$ is given by the first equation of rotational kinematics: $\omega = \omega_0 + \alpha t$ $\omega = 0 + (2.0 \text{ rad/s}^2) \times (2.0 \text{ s}) = 4.0 \text{ rad/s}$

The net linear acceleration of a particle on the rim of the disc is the vector sum of its tangential acceleration ($a_t$) and centripetal acceleration ($a_c$).

Tangential acceleration: $a_t = r\alpha = 0.5 \text{ m} \times 2.0 \text{ rad/s}^2 = 1.0 \text{ m/s}^2$

Centripetal (radial) acceleration: $a_c = r\omega^2 = 0.5 \text{ m} \times (4.0 \text{ rad/s})^2 = 0.5 \times 16.0 = 8.0 \text{ m/s}^2$

Since $a_t$ and $a_c$ are perpendicular to each other, the magnitude of the net acceleration is: $a = \sqrt{a_t^2 + a_c^2}$ $a = \sqrt{(1.0)^2 + (8.0)^2} = \sqrt{1 + 64} = \sqrt{65} \approx 8.06 \text{ m/s}^2$

Therefore, the net acceleration is approximately $8 \text{ m/s}^2$.