Answer
$\omega_D=4rad/s$
$\alpha_D=0.4rad/s^2$
Work Step by Step
We can determine the required angular velocity and angular acceleration as follows:
As $\omega_Ar_A=\omega_B r_B$
This can be rearranged as
$\omega_B=\frac{r_A}{r_B}\omega_A$
$\implies \omega_B=\frac{40}{100}(20)=8rad/s$
and $\alpha_B=\frac{r_A}{r_B}\alpha_A$
$\implies \alpha_B=\frac{40}{100}(2)=0.8rad/s^2$
As the gear C is linked to gear B, therefore $\omega_B=\omega_C$ and $\alpha_B=\alpha_C$
Similarly, $\omega_C r_C=\omega_D r_D$
This can be rearranged as
$\omega_D=\frac{r_C}{r_D}\omega_C$
We plug in the known values to obtain:
$\omega_D=\frac{50}{100}(8)=4rad/s$
and $\alpha_D=\frac{r_C}{r_D}\alpha_C$
We plug in the known values to obtain:
$\alpha_D=\frac{50}{100}(0.8)$
$\implies \alpha_D=0.4rad/s^2$
