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 16.3 - Rotation about a Fixed Axis - Problems - Page 336: 32


$a_{Pt}=6~m/s^2$ $a_{Pn}=2592~m/s^2$

Work Step by Step

We can determine the required components of acceleration as follows: We know that $\omega_A=(\omega_A)_{\circ}+\alpha_A t$ $\implies \omega_A=0+30t=30t$ At $t=3s$ $\omega_A=30(3)=90 rad/s$ As $\omega_B=\frac{\omega_A r_A}{r_B}$ $\omega_B=\frac{90(0.2)}{0.125}=144 rad/s$ and $\alpha_B=\frac{\alpha_A r_A}{r_B}=\frac{30(0.2)}{0.125}=48rad/s^2$ Now $a_{Pt}=\alpha_B r_P$ $\implies a_{Pt}=48\times 0.125=6m/s^2$ and $a_{Pn}=\omega_B^2 r$ $\implies a_{Pn}=(144)^2\times 0.125= 2592~m/s^2$
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