Answer
$a_B=16.5m/s^2$
Work Step by Step
We can determine the required acceleration as follows:
$\omega_A^2=(\omega_A)^2_{\circ}+2\alpha_A(\theta_A-(\theta_A)_{\circ})$
$\implies \omega_A^2=(5)^2+2(6)[2(2\pi)-0]$
This simplifies to:
$\omega_A=13.2588rad/s$
We know that
$\omega_Cr_C=\omega_A-r_A$
$\omega_C\times 40=13.2588\times 50$
$\implies \omega_C=16.5735rad/s$
Similarly, $\alpha_C r_C=\alpha_Ar_A$
$\implies \alpha_C\times 40=6\times 50$
$\implies \alpha_C=7.5rad/s^2$
The tangential component of acceleration at B is given as
$a_{Bt}=\alpha_C r_B$
$\implies a_{Bt}=7.5\times 0.06=0.45m/s^2$
and the normal component of acceleration at point B is given as
$a_{Bn}=\omega_C^2 r_B$
$\implies a_{Bn}=(16.5735)^2(0.06)=16.4809m/s^2$
Now, $a_B=\sqrt{(a_{Bt})^2+(a_{Bn})^2}$
We plug in the known values to obtain:
$a_B=\sqrt{(0.45)^2+(16.4809)^2}$
This simplifies to:
$a_B=16.5m/s^2$
