Answer
$9000\ rev/min^2$
Work Step by Step
Given:
Initial angular acceleration speed $\omega_i=1200\ rev/s$
Final angular acceleration speed $\omega_f=3000\ rev/s$
Time interval $\Delta t =12\ s$
From the rotational kinetic equations, we have:
$\omega_f = \omega_i+\alpha \Delta t$
$3000\ rev/min = 1200\ rev/min +\alpha (12\ s) (\frac{1\ min}{60\ s}) $
$1800\ rev/min =\alpha (12\ s) (\frac{1\ min}{60\ s}) $
$\alpha =9000\ rev/min^2$
