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

$\omega_B=90rad/s$ $\omega_A=180rad/s$
The required angular velocity can be determined as follows: We know that $\vec{v_D}=\vec{v_P}+\vec{\omega_B}\times \vec{r_{D/P}}~$[eq(1)] The velocity of $D$ is given as $\vec{v_D}=\omega_{DE}\times \vec{r_{DE}}=18\hat k\times (0.5)\hat i=9\hat j$ and $\vec{r_{D/P}}=-0.1\hat j$ We plug in the known values in eq(1) to obtain: $9\hat j=0+(-\omega_B)\hat k\times (-0.1)\hat i=0.1\omega_B \hat j$ Comparing $j$ components on both sides, we obtain: $9=0.1\omega_B$ $\implies \omega_B=90 rad/s$ We also know that $\vec{v_P^{\prime}}=\vec{v_P}+\vec{\omega_B}\times \vec{r_{P^{\prime}/P}}~$[eq(2)] As $\vec{r_{P^{\prime}/P}}=-(0.1+0.3)\hat i=-0.4\hat i$ We plug in the known values in eq(2) to obtain: $\vec{v_{P^{\prime}}}=0+(-90\hat k)\times (-0.4\hat i)=36\hat j$ $\implies vec{v_{P^{\prime}}}=36m/s$ Now $\omega_A=\frac{v_{P^{\prime}}}{r_A}$ $\omega=\frac{36}{0.2}=180rad/s$