The angular speed of a fly wheel moving with uniform angular...

For uniform angular acceleration we can write,

$\omega = {\omega}_{0} + \alphat ⇒ \alpha = \dfrac{\omega - {\omega}_{0}}{t}$

Given, $\omega = 3120rpm = \dfrac{3120 \times 2\pi}{60} = 104\pi rad{s}^{- 1}$

and ${\omega}_{0} = 1200rpm = \dfrac{1200 \times 2\pi}{60} = 40\pi rad{s}^{- 1}$

Therefore, $\alpha = \dfrac{104\pi rad{s}^{- 1} - 40\pi rad{s}^{- 1}}{16} = 4\pi rad{s}^{- 2}$