Answer
$1.041\ rev/s^2$
Work Step by Step
The angular speed is given by $\omega_o =10\ rev/s$
Angular position $\Delta\theta = 60\ rev$
Final angular speed is $\omega = 15\ rev/s$
From rotational kinematics equations, we have the formula:
$\omega^2 =\omega_o^2+2\alpha \Delta\theta$
$\alpha =\frac{\omega^2-\omega_o^2}{2\Delta\theta}$
$\alpha =\frac{(15\ rev/s)^2-(10\ rev/s)^2}{2(60\ rev)}$
$\alpha =1.041\ rev/s^2$
