Answer
$\omega_B=312rad/s$
$\alpha_B=176rd/s^2$
Work Step by Step
The required angular velocity and angular acceleration can be determined as follows:
As $\omega_A r_A=\omega_B r_B$
This can be rearranged as:
$\omega_B=\frac{r_A}{r_B}\times \omega_A$
$\implies \omega_B=\frac{0.2}{0.125}\times \omega_A$
$\implies \omega_B=1.6\omega_A~~~$[eq(1)]
We know that
$\theta_A=5t^3+10t^2$
Taking derivative with respect to $t$ on both sides, we obtain:
$\implies \omega_A=15t^2+20t$
At $t=3s$
$\omega_A=15(3)^2+20(3)=195rad/s$
We plug in this value in eq(1) to obtain:
$\omega_B=1.6\times 195=312 rad/s$
Now we find the angular acceleration
$\alpha_B=\frac{r_A}{r_B}\alpha_A$
$\implies \alpha_B=\frac{0.2}{0.125}\alpha_A=1.6\alpha_A~~~$[eq(2)]
We know that
$\omega_A=15t^2+20t$
Taking the derivative with respect to $t$ on both sides, we obtain:
$\alpha_A=30t+20$
At $t=3s$
$\alpha_A=30(3)+20=110 rad/s^2$
We plug in this value in eq(2)to obtain:
$\alpha_B=1.6\times 110=176rd/s^2$
