Answer
(a) $\alpha = -96~rad/s^2$
(b) The engine makes 98 revolutions.
Work Step by Step
(a) $\Delta \omega = 1200~rpm - 3500~rpm = -2300~rpm$
$\Delta \omega = (-2300~rpm)(2\pi \frac{rad}{rev})(\frac{1~m}{60~s})$
$\Delta \omega = \frac{-230~\pi}{3}~rad/s$
We can find the angular acceleration $\alpha$ as:
$\alpha = \frac{\Delta \omega}{t} = \frac{\frac{-230~\pi}{3}~rad/s}{2.5~s}$
$\alpha = -96~rad/s^2$
(b) $\omega_0 = (3500~rpm)(2\pi \frac{rad}{rev})(\frac{1~m}{60~s}) = \frac{350\pi}{3}~rad/s$
$\theta = \omega_0 t + \frac{1}{2}\alpha t^2$
$\theta = (\frac{350\pi}{3}~rad/s)(2.5~s) + \frac{1}{2}(-96~rad/s^2)(2.5~s)^2$
$\theta = 616~radians$
We can then find the number of revolutions:
$rev = \frac{\theta}{2\pi} = \frac{616~rad}{2\pi} = 98~rev$
