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 356: 63


$\omega_{BC}=2.45 rad/s$ $v_C=2.20ft/s$

Work Step by Step

The required velocity and angular velocity can be determined as follows: We know that $\vec{v_C}=\vec{v_B}+\vec{\omega_{BC}}\times \vec{r_{C/B}}~$[eq(1)] Similarly $\vec{v_C}=-\vec{v_C}\hat i$ and $\vec{r_{C/B}}=3cos45\hat i+3sin45\hat j=2.1213\hat i+2.1213\hat j$ The velocity of $B$ is given as $\vec{\omega_B}=\vec{\omega_{AB}}\times \vec{r_{AB}}$ $\vec{\omega_B}=3\hat k\times (-2cos 30\hat i-2sin 30\hat j)=-5.1961\hat j +3\hat i$ We plug in the known values in eq(1) to obtain: $\vec{-v_C}\hat i=(-5.1961\hat j+3\hat i)+(\omega_{BC})\hat k\times (2.1213\hat i+2.1213 \hat j)$ $\implies \vec{-v_C}\hat i=(3-2.1213\omega_{BC})\hat i+(2.1213\omega_{BC}-5.1961)\hat j$ Comparing $i$ components, we obtain: $-v_C=3-2.1213\omega_{BC}$ $\implies v_C=2.1213\omega_{BC}-3~$[eq(2)] Now comparing $j$ component, we obtain: $0=2.1213\omega_{BC}-5.1961$ This simplifies to: $\omega_{BC}=2.45 rad/s$ We plug in this value in eq(2) to obtain: $v_C=2.1213(2.449)-3$ $v_C=2.20ft/s$
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