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

$\omega=-4rad/s$ $v_P=4.838m/s \leftarrow$
We can determine the required velocity and angular velocity as follows: We know that $v_D=v_B+\omega\times r_{D/B}$ $\implies -v_Dcos30\hat i-v_Dsin30\hat j=(-2.4cos30\hat i+2.4sin 30\hat j)+(\omega\hat k)(0.6\hat i)$ $\implies -v_Dcos30\hat i-v_Dsin30\hat j=-2.4cos 30\hat i +(2.4sin30+0.6\omega)\hat j$ Comparing $i$ components on both sides, we obtain: $-v_Dcos30=-2.4cos30$ $\implies v_D=2.4m/s$ and for $j$componetnts $-v_Dsin30=2.4sin30+0.6\omega$ This simplifies to: $\omega=-4rad/s$ The relative velocity between $B$ and $P$ is given as $v_P=v_B+\omega\times r_{P/B}$ We plug in the known values to obtain: $v_P=-2.4cos30\hat i+2.4sin30\hat j+(-4\hat k)\times (0.3\hat i-2\times \frac{0.3}{cos30}\hat j)$ $\implies v_{Px}\hat i+v_{Py}\hat j=-2.4cos30\hat i+2.4sin30\hat j+(-4\hat k)\times (0.3\hat i-0.69\hat j)=(-2.4cos30-2.76)\hat i+(2.4sin30-1.2)\hat j$ Comparing $i$ components on both sides, we obtain: $v_{Px}=-4.838m/s$ and for $j$ components $v_{Py}=0$ $\implies v_P=4.838m/s \leftarrow$