Answer
$48\ rev$
Work Step by Step
Given:
initial angular speed $\omega_o = 0\ rev/s$
angular speed is given by $ω=10 rev/s$
angular acceleration $\alpha =1.041\ rev/s^2$
From rotational kinematics equations, we have
$\omega^2 = \omega_o^2+2\alpha (\Delta\theta)$
$\Delta\theta = \frac{\omega^2 - \omega_o^2}{2\alpha}$
$\Delta\theta = \frac{(10\ rev/s)^2 - 0}{2(1.041\ rev/s^2}$
$\Delta\theta = 48\ rev$
