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

$v_B=10.2m/s$ $a_{Bt}=8m/s^2$ $a_{Bn}=258.66m/s^2$
The required velocity and acceleration can be determined as follows: As $\omega^2=\omega_{\circ}^2+2\alpha (\theta-\theta_{\circ})$ $\implies \omega^2=(12)^2+2(20)[2(2\pi)-0]$ $\implies \omega=25.43rad/s$ We know that $v_B=\omega r_B$ We plug in the known values to obtain: $v_B=(25.43)(0.4)$ $\implies v_B=10.2m/s$ Now the tangential and normal components of acceleration are given as $a_{Bt}=\alpha r_B$ We plug in the known values to obtain $a_{Bt}=(20)(0.4)=8m/s^2$ Similarly, $a_{Bn}=\omega^2 r_B$ We plug in the known values to obtain: $a_{Bn}=(25.43)^2(0.4)$ $\implies a_{Bn}=258.66m/s^2$