## Fundamentals of Physics Extended (10th Edition)

$9000\ rev/min^2$
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: $\omega_f = \omega_i+\alpha \Delta t$ $3000\ rev/min = 1200\ rev/min +\alpha (12\ s) (\frac{1\ min}{60\ s})$ $1800\ rev/min =\alpha (12\ s) (\frac{1\ min}{60\ s})$ $\alpha =9000\ rev/min^2$