Answer
$12 rad/s$
$0.6 rad/s^2$
Work Step by Step
We can determine the required angular velocity and angular acceleration as follows:
$\omega_A r_A=\omega_B r_B$
$\implies \omega_B=\frac{r_A}{r_B}\omega_A$
$\implies \omega_B=\frac{40}{100}(60)=28rad/s$
and $\alpha_B=\frac{r_A}{r_B}\alpha_A$
$\implies \alpha_B=\frac{40}{100}(3)=1.2 rad/s^2$
As gear C is linked to gear B; therefore $\omega_B=\omega_C$ and $\alpha_B=\alpha_C$
Similarly, $\omega_Cr_C=\omega_D r_D$
This can be rearranged as:
$\omega_D=\frac{r_C}{r_D}\omega_C$
We plug in the known values to obtain:
$\omega_D=\frac{50}{100}(24)=12 rad/s$
and $\alpha_D=\frac{r_C}{r_D}\alpha_C$
We plug in the known values to obtain:
$\alpha_D=\frac{50}{100}(1.2)=0.6 rad/s^2$
