Answer
We find values for } $t_{2}$ when the angular displacement (relative to its orientation at t=0) is $\theta_{2}=-10.5$ rad. Using $\mathrm{Eq} .10-13$ and the quadratic formula, we have
$$
\theta_{2}=\omega_{0} t_{2}+\frac{1}{2} \alpha t_{2}^{2} \Rightarrow t_{2}=\frac{-\omega_{0} \pm \sqrt{\omega_{0}^{2}+2 \theta_{2} \alpha}}{\alpha}
$$
which yields the two roots $-2.1 \mathrm{s}$ and 40 $\mathrm{s}$ .
Thus, at $t=-2.1 \mathrm{s}$ the reference line will be at
$$\theta_{2}=-10.5 \mathrm{rad} $$
Work Step by Step
We find values for } $t_{2}$ when the angular displacement (relative to its orientation at t=0) is $\theta_{2}=-10.5$ rad. Using $\mathrm{Eq} .10-13$ and the quadratic formula, we have
$$
\theta_{2}=\omega_{0} t_{2}+\frac{1}{2} \alpha t_{2}^{2} \Rightarrow t_{2}=\frac{-\omega_{0} \pm \sqrt{\omega_{0}^{2}+2 \theta_{2} \alpha}}{\alpha}
$$
which yields the two roots $-2.1 \mathrm{s}$ and 40 $\mathrm{s}$ .
Thus, at $t=-2.1 \mathrm{s}$ the reference line will be at
$$\theta_{2}=-10.5 \mathrm{rad} $$
