Answer
Ans:
$$
\begin{aligned}
\alpha_D r_D=0.4 \mathrm{rad} / \mathrm{s}^2
\end{aligned}
$$
Work Step by Step
$$
\begin{aligned}
& \alpha_A d \theta_A=\omega_A d \omega_A \\
& \alpha_A d \theta_A=\left(\theta_A+1\right) d\left(\theta_A+1\right) \\
& \alpha_A d \theta_A=\left(\theta_A+1\right) d \theta_A \\
& \alpha_A=\left(\theta_A+1\right)
\end{aligned}
$$
At $\theta_A=3 \mathrm{rad}$,
$$
\begin{aligned}
& \alpha_A=3+1=4 \mathrm{rad} / \mathrm{s}^2 \\
& \alpha_B r_B=\alpha_A r_A \\
& \alpha_B=\left(\frac{r_A}{r_B}\right) \alpha_A=\left(\frac{15}{50}\right)(4)=1.20 \mathrm{rad} / \mathrm{s}^2
\end{aligned}
$$
Since gears $C$ and $B$ share the same shaft $\alpha_C=\alpha_B=1.20 \mathrm{rad} / \mathrm{s}^2$.
$$
\begin{aligned}
& \alpha_D r_D=\alpha_C r_C \\
& \alpha_D=\left(\frac{r_C}{r_D}\right) \alpha_C=\left(\frac{25}{75}\right)(1.20)=0.4 \mathrm{rad} / \mathrm{s}^2
\end{aligned}
$$
