Question 31:
easy
A uniform circular disc of radius $50\text{ cm}$ at rest is free to turn about an axis which is perpendicular to its plane and passes through its center. It is subjected to a torque which produces a constant angular acceleration of $2.0\text{ rad s}^{-2}$. Its net acceleration in $\text{ms}^{-2}$ at the end of $2.0\text{ s}$ is approximately:
(2016-I)
Concept: Combination of tangential and centripetal accelerations. Formula: $a = \sqrt{a_c^2 + a_t^2}$. Solution: $a_t = r\alpha = 1.0$, $a_c = \omega^2 r = 8.0$, giving $a = \sqrt{8^2 + 1^2} \approx 8.0\text{ ms}^{-2}$.