Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 16 - Planar Kinematics of a Rigid Body - Section 16.5 - Relative-Motion Analysis: Velocity - Problems - Page 357: 70


$v_C=28.98m/s \rightarrow$ $\omega_{BC}=10.6 rad/s \circlearrowleft$

Work Step by Step

We can determine the required velocity and angular velocity as follows: $v_B=\omega_{AB}\times r_{AB}$ $\implies v_B=(5\hat k)\times (-0.3cos 45\hat i-0.3sin45\hat j)$ $\implies v_B=(\frac{15\sqrt{2}}{2}\hat i-\frac{15{\sqrt{2}}}{2}\hat j)m/s$ We know that $v_C=v_B+\omega_{BC}\times r_{C/B}$ $\implies v_{C \hat i}=(\frac{15\sqrt{2}}{2}\hat i-\frac{15\sqrt{2}}{2}\hat j)+(\omega_{BC}\hat k)\times (2sin30\hat i-2cos30\hat j)$ $\implies v_{C \hat i}=(\frac{15\sqrt{2}}{2}+\omega_{BC} \sqrt{3})\hat i+(\omega_{BC}-\frac{15\sqrt{2}}{2})\hat j$ Comparing $j$ components on both sides, we obtain: $0=\omega_{BC}-\frac{15\sqrt{2}}{2}$ $\implies \omega_{BC}=\frac{15\sqrt{2}}{2}=10.6 rad/s \circlearrowleft$ and comparing $i$ components on both sides, we obtain: $v_C=\frac{15\sqrt{2}}{2}+\frac{15\sqrt{2}}{2}\sqrt{3}$ This simplifies to: $v_C=28.98m/s \rightarrow$
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