Answer
-4.48$\times10^{-3} \frac{rad}{sec^{2}}$
Work Step by Step
$\theta = (2π\times40)$ rad = $80π$ rad
Initial angular velocity, $\omega_{0}$ = 1.5 rad/sec
Final angular velocity, $\omega$ = 0
We know,
$\omega^{2}-\omega_{0}^{2}$ = 2α$\theta$ (where α is the angular acceleration)
Substituting values,
α = $\frac{0^{2}-(1.5rad/sec)^{2}}{2\times80π.rad}$ = $-4.48\times10^{-3} \frac{rad}{sec^{2}}$
