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Rotational kinematics – centripetal and tangential acceleration
Concepts tested here
- net-acceleration-rotating-disc
All Questions
2016 1 question
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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
Combine centripetal and tangential components at the rim.
$\omega = \omega_0 + \alpha t = 0 + 2\times2 = 4$ rad/s
Centripetal acceleration $a_c = R\omega^2 = 0.5\times16 = 8\ m/s^2$
Tangential acceleration $a_t = R\alpha = 0.5\times2 = 1\ m/s^2$
$a_{net} = \sqrt{a_c^2 + a_t^2} = \sqrt{64 + 1} = \sqrt{65} \approx 8\ m/s^2$