Answer
$a_{Pt}=6~m/s^2$
$a_{Pn}=2592~m/s^2$
Work Step by Step
We can determine the required components of acceleration as follows:
We know that
$\omega_A=(\omega_A)_{\circ}+\alpha_A t$
$\implies \omega_A=0+30t=30t$
At $t=3s$
$\omega_A=30(3)=90 rad/s$
As $\omega_B=\frac{\omega_A r_A}{r_B}$
$\omega_B=\frac{90(0.2)}{0.125}=144 rad/s$
and $\alpha_B=\frac{\alpha_A r_A}{r_B}=\frac{30(0.2)}{0.125}=48rad/s^2$
Now $a_{Pt}=\alpha_B r_P$
$\implies a_{Pt}=48\times 0.125=6m/s^2$
and $a_{Pn}=\omega_B^2 r$
$\implies a_{Pn}=(144)^2\times 0.125= 2592~m/s^2$
