Answer
Ans:
$$
t=6.98s\\
\theta_D=34.9 \mathrm{rev}\\
$$
Work Step by Step
$$
\begin{array}{l}
\alpha_B r_B=\alpha_A r_A \\
\alpha_B=\left(\frac{r_A}{r_B}\right) \alpha_A=\left(\frac{15}{50}\right)(90)=27 \mathrm{rad} / \mathrm{s}^2
\end{array}
$$
$$
\begin{array}{l}
\alpha_D r_D=\alpha_C r_C \\
\alpha_D=\left(\frac{r_C}{r_D}\right) \alpha_C=\left(\frac{25}{75}\right)(27)=9 \mathrm{rad} / \mathrm{s}^2
\end{array}
$$
The final angular velocity of gear $D$ is $\omega_D=\left(\frac{600 \mathrm{rev}}{\min }\right)\left(\frac{2 \pi \mathrm{rad}}{1 \mathrm{rev}}\right)\left(\frac{1 \mathrm{~min}}{60 \mathrm{~s}}\right)=$ $20 \pi \mathrm{rad} / \mathrm{s}$.
$$
\begin{array}{l}
\omega_D=\left(\omega_D\right)_0+\alpha_D t \\
20 \pi=0+9 t \\
t=6.98 \mathrm{~s}
\end{array}
$$
$$
\begin{array}{l}
\omega_D^2=\left(\omega_D\right)_0^2+2 \alpha_D\left[\theta_D-\left(\theta_D\right)_0\right] \\
(20 \pi)^2=0^2+2(9)\left(\theta_D-0\right) \\
\theta_D=(219.32 \mathrm{rad})\left(\frac{1 \mathrm{rev}}{2 \pi \mathrm{rad}}\right) \\
=34.9 \mathrm{rev}
\end{array}
$$
