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 16.6 - Instantaneous Center of Zero Velocity - Problems - Page 369: 83

Answer

$\omega_{BC}=6.79rad/s$

Work Step by Step

The required angular velocity can be determined as follows: According to the law of sines $\frac{0.125}{sin30}=\frac{r_{IC/B}}{sin45}=\frac{r_{IC/C}}{sin 105}$ This simplifies to: $r_{IC/B}=sin45\times 0.125\times \frac{1}{sin30}=0.1768m$ and $r_{IC/C}=sin105\times 0.125\times \frac{1}{sin30}=0.2415m$ We know that $v_B=\omega_{AB}r_{AB}$ $\implies v_B=4(0.3)=1.2m/s$ and $v_B=\omega_{BC}(r_{IC/B})$ We plug in the known values to obtain: $1.2=0.1768\omega_{BC}$ This simplifies to: $\omega_{BC}=6.79rad/s$
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