Answer
$t\approx 34s$
Work Step by Step
If the flywheel has constant angular acceleration, then its angle $\theta (t) $ will be:
$\theta \left( t\right) =\theta _{0}+w_{0}t+\dfrac {\alpha t^{2}}{2}.......\left( 1\right) $
where $\theta_{0}=0$ , $w_{0}=4.7 \frac{rad}{s}$ and $\alpha =-0.25\dfrac {rad}{s^{2}}$
Solving equation (1), we get:
$\dfrac {α}{2}t^{2}+w_{0}t+\left( \theta _{0}-\theta \left( t\right) \right) =0\Rightarrow t_{1,2}=\dfrac {-w_{0}\pm \sqrt {\left( w_{0}^{2}-4\times \dfrac {α}{2}\times \left( \theta _{0}-\theta \right) \right) }}{2\times \dfrac {α}{2}}....(2)$
If we simplify (2), we get:
$t_{1,2}=\dfrac {-w_{0}\pm \sqrt {\left( w_{0}^{2}-2α\times \left( \theta _{0}-\theta \right) \right) }}{α}............(3)$
Given the numbers $\theta _{0}=0;w_{0}=4.7\dfrac {rad}{s};\alpha =-0.25\dfrac {rad}{s^{2}}$, using (3) we get:
$t_{1,2}=\dfrac {-4.7\pm \sqrt {\left( 4.7^{2}+0.5\left( 0-\theta _{0}\right) \right) }}{-0.25}=\dfrac {\pm \sqrt {\left( 4.7^{2}-0.5\theta _{0}\right) }-4.7}{0.25}\Rightarrow 0.5\theta _{0}\leq 4.7^{2}\Rightarrow \theta _{0}\leq \dfrac {4.7^{2}}{0.5}=44.18rad\left( 4\right) $
$\theta _{\max }=44.18rad$
Now lets calculate the time needed for flywheel to turn using $\theta \left( t\right) =\dfrac {\theta _{\max }}{2}$,
Using equation (3).we get:
$t_{1,2}=\dfrac {-w\pm \sqrt {\left( w^{2}-2a\times \left( \theta _{0}-\dfrac {\theta _{\max }}{2}\right) \right) }}{a}=\dfrac {-4.7\pm \sqrt {\left( 4.7^{2}-2\times \left( -0.25\right) \left( 0-\dfrac {44.18}{2}\right) \right) }}{-0.25}\Rightarrow t_{1,2}\approx 34s,3.6s$
So the second time the flywheel turns with the given angle will be at $t\approx 34s$.
