## Engineering Mechanics: Statics & Dynamics (14th Edition)

$a_B=16.5m/s^2$
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: $a_B=16.5m/s^2$