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A uniform circular disc of radius 50 cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of 2.0 rad $s^{-2}$. Its net acceleration in $ms^{-2}$ at the end of 2.0 s is approximately
A
7.0
B
6.0
C
3.0
D
8.0
Explanation
Combine centripetal and tangential components at the rim.
Detailed Solution
$\omega = \omega_0 + \alpha t = 0 + 2\times2 = 4$ rad/s
Centripetal acceleration $a_c = R\omega^2 = 0.5\times16 = 8\ m/s^2$
Tangential acceleration $a_t = R\alpha = 0.5\times2 = 1\ m/s^2$
$a_{net} = \sqrt{a_c^2 + a_t^2} = \sqrt{64 + 1} = \sqrt{65} \approx 8\ m/s^2$
Centripetal acceleration $a_c = R\omega^2 = 0.5\times16 = 8\ m/s^2$
Tangential acceleration $a_t = R\alpha = 0.5\times2 = 1\ m/s^2$
$a_{net} = \sqrt{a_c^2 + a_t^2} = \sqrt{64 + 1} = \sqrt{65} \approx 8\ m/s^2$